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# Long year

Long year
You are encouraged to solve this task according to the task description, using any language you may know.

Most years have 52 weeks, some have 53, according to ISO8601.

Write a function which determines if a given year is long (53 weeks) or not, and demonstrate it.

## 11l

Translation of: C++
`F is_long_year(year)   F p(year)      R (year + (year I/ 4) - (year I/ 100) + (year I/ 400)) % 7   R p(year) == 4 | p(year - 1) == 3 L(year) 2000..2100   I is_long_year(year)      print(year, end' ‘ ’)`
Output:
```2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
```

## Action!

`BYTE FUNC P(CARD y)RETURN ((y+(y/4)-(y/100)+(y/400)) MOD 7) BYTE FUNC IsLongYear(CARD y)  IF P(y)=4 OR P(y-1)=3 THEN    RETURN (1)  FIRETURN (0) PROC Main()  CARD y  BYTE LMARGIN=\$52,oldLMARGIN   oldLMARGIN=LMARGIN  LMARGIN=0 ;remove left margin on the screen  Put(125) PutE() ;clear the screen   FOR y=1900 TO 2400  DO    IF IsLongYear(y) THEN      PrintC(y) Put(32)    FI  OD   LMARGIN=oldLMARGIN ;restore left margin on the screenRETURN`
Output:
```1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987
1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167
2172 2178 2184 2189 2195 2201 2207 2212 2218 2224 2229 2235 2240 2246 2252 2257
2263 2268 2274 2280 2285 2291 2296 2303 2308 2314 2320 2325 2331 2336 2342 2348
2353 2359 2364 2370 2376 2381 2387 2392 2398
```

The Ada calendar package handles dates for years 1901 through 2399. This program outputs all the long years within that range.

`--------------------------------------------------------------- Calculate long years-- Reference: https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year-------------------------------------------------------------with Ada.Text_IO;             use Ada.Text_IO;with Ada.Calendar;            use Ada.Calendar;with Ada.Calendar.Formatting; use Ada.Calendar.Formatting; procedure Main is   First_Day : Time;   Last_Day  : Time;   package AC renames Ada.Calendar;   type Counter is mod 10;   Count : Counter := 0;begin   for Yr in Year_Number loop       First_Day := AC.Time_Of (Year => Yr, Month => 1, Day => 1);      Last_Day  := AC.Time_Of (Year => Yr, Month => 12, Day => 31);       -- If Jan 1 is Thursday or Dec 31 is Thursday then      -- the year is a long year       if Day_Of_Week (First_Day) = Thursday        or else Day_Of_Week (Last_Day) = Thursday      then         if Count = 0 then            New_Line;         end if;         Put (Yr'Image);         Count := Count + 1;      end if;   end loop;end Main; `
Output:
``` 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953
1959 1964 1970 1976 1981 1987 1992 1998 2004 2009
2015 2020 2026 2032 2037 2043 2048 2054 2060 2065
2071 2076 2082 2088 2093 2099 2105 2111 2116 2122
2128 2133 2139 2144 2150 2156 2161 2167 2172 2178
2184 2189 2195 2201 2207 2212 2218 2224 2229 2235
2240 2246 2252 2257 2263 2268 2274 2280 2285 2291
2296 2303 2308 2314 2320 2325 2331 2336 2342 2348
2353 2359 2364 2370 2376 2381 2387 2392 2398
```

## ALGOL 68

Translation of: ALGOL W
`BEGIN # find "long years" - years which have 53 weeks this is equivalent to #      # finding years where 1st Jan or 31st Dec are Thursdays               #    # returns the day of the week of the specified date (d/m/y), Sunday = 1 #    PROC day of week = ( INT d, m, y )INT:         BEGIN            INT mm := m;            INT yy := y;            IF mm <= 2 THEN                mm := mm + 12;                yy := yy - 1            FI;            INT j = yy OVER 100;            INT k = yy MOD  100;            (d + ( ( mm + 1 ) * 26 ) OVER 10 + k + k OVER 4 + j OVER 4 + 5 * j ) MOD 7         END # day of week # ;    # returns TRUE if year is a long year, FALSE otherwise                  #    PROC is long year = ( INT year )BOOL:        day of week( 1, 1, year ) = 5 OR day of week( 31, 12, year ) = 5;    # show long years from 2000-2099                                        #    print( ( "long years 2000-2099:" ) );    FOR year FROM 2000 TO 2099 DO        IF is long year( year ) THEN print( ( " ", whole( year, 0 ) ) ) FI    ODEND`
Output:
```long years 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
```

## ALGOL-M

`BEGIN COMMENT  FIND ISO CALENDAR YEARS HAVING 53 WEEKS. THE SIMPLEST  TEST IS THAT A GIVEN YEAR WILL BE "LONG" IF EITHER THE  FIRST OR LAST DAY IS A THURSDAY; % CALCULATE P MOD Q %INTEGER FUNCTION MOD(P, Q);INTEGER P, Q;BEGIN   MOD := P - Q * (P / Q);END; COMMENT  RETURN DAY OF WEEK (SUN=0, MON=1, ETC.) FOR A GIVEN  GREGORIAN CALENDAR DATE USING ZELLER'S CONGRUENCE;INTEGER FUNCTION DAYOFWEEK(MO, DA, YR);INTEGER MO, DA, YR;BEGIN  INTEGER Y, C, Z;  IF MO < 3 THEN    BEGIN      MO := MO + 10;      YR := YR - 1;    END  ELSE MO := MO - 2;  Y := MOD(YR, 100);  C := YR / 100;  Z := (26 * MO - 2) / 10;  Z := Z + DA + Y + (Y / 4) + (C /4) - 2 * C + 777;  DAYOFWEEK := MOD(Z, 7);END; % RETURN 1 IF YEAR IS LONG, OTHERWISE 0 %INTEGER FUNCTION ISLONGYEAR(YR);INTEGER YR;BEGIN  INTEGER THURSDAY;  THURSDAY := 4;  IF (DAYOFWEEK(1,1,YR) = THURSDAY) OR      (DAYOFWEEK(12,31,YR) = THURSDAY) THEN       ISLONGYEAR := 1  ELSE       ISLONGYEAR := 0;END; % MAIN PROGRAM STARTS HERE %INTEGER YEAR;WRITE("ISO YEARS THAT WILL BE LONG IN THIS CENTURY:");WRITE("");FOR YEAR := 2000 STEP 1 UNTIL 2099 DO  BEGIN    IF ISLONGYEAR(YEAR) = 1 THEN WRITEON(YEAR);  END; END `
Output:
```ISO YEARS THAT WILL BE LONG IN THIS CENTURY:
2004  2009  2015  2020  2026  2032  2037  2043  2048  2054  2060  2065  2071 2076
2082  2088  2093  2099```

## ALGOL W

Uses the Day_of_week procedure from the Day_of_the_week task.

`begin % find "long years" - years which have 53 weeks %      % this is equivalent to finding years where     %      % 1st Jan or 31st Dec are Thursdays             %    % finds the day of the week - Sunday = 1          %    integer procedure Day_of_week ( integer value d, m, y );        begin            integer j, k, mm, yy;            mm := m;            yy := y;            if mm <= 2 then begin                mm := mm + 12;                yy := yy - 1;            end if_m_le_2;            j := yy div 100;            k := yy rem 100;            (d + ( ( mm + 1 ) * 26 ) div 10 + k + k div 4 + j div 4 + 5 * j ) rem 7        end Day_of_week;    % returns true if year is a long year, false otherwise %    logical procedure isLongYear ( integer value year );        Day_of_week( 1, 1, year ) = 5 or Day_of_week( 31, 12, year ) = 5;    % show long years from 2000-2099 %    write( "long years 2000-2099:" );    for year := 2000 until 2099 do begin        if isLongYear( year ) then writeon( I_W := 5, S_W := 0, year )    end for_yearend.`
Output:
```long years 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
```

## APL

`dec31weekday ← {7|(⍵+⌊(⍵÷4)+⌊(⍵÷400)-⌊⍵÷100)}isolongyear ← {(4 = dec31weekday ⍵) ∨ (3 = dec31weekday (⍵ - 1))}`
Output:
```      (isolongyear y)/y←1800+⍳300
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885
1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043
2048 2054 2060 2065 2071 2076 2082 2088 2093 2099```

## AppleScript

`on isLongYear(y)    -- ISO8601 weeks begin on Mondays and belong to the year in which they have the most days.    -- A year which begins on a Thursday, or which begins on a Wednesday and is a leap year,    -- has majority stakes in the weeks it overlaps at *both* ends and so has 53 weeks instead of 52.    -- Leap years divisible by 400 begin on Saturdays and so don't so need to be considered in the leap year check.     tell (current date) to set {Jan1, its day, its month, its year} to {it, 1, January, y}    set startWeekday to Jan1's weekday     return ((startWeekday is Thursday) or ((startWeekday is Wednesday) and (y mod 4 is 0) and (y mod 100 > 0)))end isLongYear set longYears to {}repeat with y from 2001 to 2100    if (isLongYear(y)) then set end of longYears to yend repeat return longYears`
Output:
`{2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099}`

On the other hand, since the cycle repeats every 400 years, it's possible to cheat with a precalculated look-up list:

`on isLongYear(y)    return (y mod 400 is in {4, 9, 15, 20, 26, 32, 37, 43, 48, 54, 60, 65, 71, 76, 82, 88, 93, 99, 105, 111, 116, 122, 128, 133, 139, 144, 150, 156, 161, 167, 172, 178, 184, 189, 195, 201, 207, 212, 218, 224, 229, 235, 240, 246, 252, 257, 263, 268, 274, 280, 285, 291, 296, 303, 308, 314, 320, 325, 331, 336, 342, 348, 353, 359, 364, 370, 376, 381, 387, 392, 398})end isLongYear set longYears to {}repeat with y from 2001 to 2100    if (isLongYear(y)) then set end of longYears to yend repeat return longYears`

## Arturo

Translation of: Nim
`longYear?: function [year][    date: to :date .format: "dd/MM/yyyy" ~"01/01/|year|"     or? date\Day = "Thursday"         and? leap? year             date\Day = "Wednesday"] print "Years with 53 weeks between 2000 and 2100:"print select 2000..2100 => longYear?`
Output:
```Years with 53 weeks between 2000 and 2100:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 ```

## AutoHotkey

`Long_year(y) {	A := Mod(y + floor(y/4) - floor(y/100) + floor(y/400), 7)	y--, B := Mod(y + floor(y/4) - floor(y/100) + floor(y/400), 7)	return A=4 || B=3}`
Examples:
`loop, 100{	y := 1999+A_Index	res .= Long_year(y) ? Y " ": ""}MsgBox % "Long Years 2000-2100 : " resreturn`
Output:
`Long Years 2000-2100 : 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 `

## AWK

` # syntax: GAWK -f LONG_YEAR.AWKBEGIN {    for (cc=19; cc<=21; cc++) {      printf("%2d00-%2d99: ",cc,cc)      for (yy=0; yy<=99; yy++) {        ccyy = sprintf("%02d%02d",cc,yy)        if (is_long_year(ccyy)) {          printf("%4d ",ccyy)        }      }      printf("\n")    }#    printf("\n%4d-%4d: ",by=1970,ey=2037)    for (y=by; y<=ey; y++) {      if (strftime("%V",mktime(sprintf("%d 12 28 0 0 0",y))) == 53) {        printf("%4d ",y)      }    }    printf("\n")    exit(0)}function is_long_year(year,  i) {    for (i=0; i<=1; i++) {      year -= i      if ((year + int(year/4) - int(year/100) + int(year/400)) % 7 == 4-i) {        return(1)      }    }    return(0)} `
Output:
```1900-1999: 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998
2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
2100-2199: 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195

1970-2037: 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037
```

## BASIC

### Applesoft BASIC

Translation of: Commodore BASIC
` 10  DEF  FN M7(N) = N - 7 *  INT (N / 7) 20  DEF  FN WD(Y) =  FN M7(Y +  INT (Y / 4) -  INT (Y / 100) +  INT (Y / 400)) 30  DEF  FN LY(Y) = (4 =  FN WD(Y)) OR (3 =  FN WD(Y - 1)) 40  HOME : INVERSE : PRINT "****     LIST OF ISO LONG YEARS     ****": NORMAL  50  INPUT "START YEAR? ";S 60  INPUT "END YEAR? ";E 70  PRINT : FOR Y = S TO E 80  IF  FN LY(Y) THEN  PRINT S\$Y;:S\$ = " " 90  NEXT Y`

### ASIC

Translation of: Commodore BASIC
` REM Long yearCLSPRINT "****     List of ISO long years     ****"PRINT "Start year";INPUT SPRINT "End year";INPUT EPRINTFOR Y = S TO E  GOSUB CALCLY:  IF LY <> 0 THEN    PRINT Y;  ENDIFNEXT YPRINTEND CALCLY:  REM Nonzero if Y is long  LY = 0  AY = Y  GOSUB CALCWD:  IF WD = 4 THEN    LY = -1  ENDIF  AY = Y - 1  GOSUB CALCWD:  IF WD = 3 THEN    LY = -1  ENDIFRETURN CALCWD:  REM Weekday of AY-12-31, 0 = Sunday  WD = AY  TMP = AY / 4  WD = WD + TMP  TMP = AY / 100  WD = WD - TMP  TMP = AY / 400  WD = WD + TMP  WD = WD MOD 7RETURN `
Output:
```****     List of ISO long years     ****
Start year?1995
End year?2045

1998  2004  2009  2015  2020  2026  2032  2037  2043
```

### BASIC256

`function p(y)	return (y + int(y/4) - int(y/100) + int(y/400)) mod 7end function function isLongYear(y)	return (p(y) = 4) or (p(y - 1) = 3)end function for y = 2000 to 2100	if isLongYear(y) then print ynext yend`
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

### BBC BASIC

`      INSTALL @lib\$ + "DATELIB"       REM The function as per specification.      DEF FNLongYear(year%)=FN_dow(FN_mjd(1, 1, year%)) == 4 OR FN_dow(FN_mjd(31, 12, year%)) == 4       REM Demonstrating its use.      PROCPrintLongYearsInCentury(20)      PROCPrintLongYearsInCentury(21)      PROCPrintLongYearsInCentury(22)      END       DEF PROCPrintLongYearsInCentury(century%)      LOCAL year%, start%      start%=century% * 100 - 100      PRINT "The long years between ";start% " and ";start% + 100 " are ";      FOR year%=start% TO start% + 99        IF FNLongYear(year%) PRINT STR\$year% + " ";      NEXT      PRINT      ENDPROC`
Output:
```The long years between 1900 and 2000 are 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998
The long years between 2000 and 2100 are 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
The long years between 2100 and 2200 are 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195
```

### Commodore BASIC

`100 REM M7(N) = N MOD 7110 DEF FNM7(N) = N - 7*INT(N / 7)120 :130 REM WD(Y) = WEEKDAY OF Y-12-31, 0 = SUNDAY140 DEF FNWD(Y) = FNM7(Y + INT(Y / 4) - INT(Y / 100) + INT(Y / 400))150 :160 REM LY(Y) = NONZERO IF Y IS LONG170 DEF FNLY(Y) = (4 = FNWD(Y)) OR (3 = FNWD(Y-1))180 :190 PRINT CHR\$(147); CHR\$(18); "****     LIST OF ISO LONG YEARS     ****"200 INPUT "START YEAR"; S210 INPUT "END YEAR"; E220 PRINT230 :240 FOR Y = S TO E250 : IF FNLY(Y) THEN PRINT Y,260 NEXT Y270 PRINT`
Output:
```****     LIST OF ISO LONG YEARS     ****

START YEAR? 1995
END YEAR? 2045

1998      2004      2009      2015
2020      2026      2032      2037
2043

### FreeBASIC

`function p(y as unsigned integer) as unsigned integer    return ( y + int(y/4) - int(y/100) + int(y/400) ) mod 7end function function islongyear( y as uinteger ) as boolean    if p(y) = 4 then return true    if p(y-1) = 3 then return true    return falseend function print islongyear(1998)print islongyear(2020)print islongyear(2021)`
Output:
```true
true
false
```

### GW-BASIC

`10 INPUT "Enter a year: ", Y20 X = Y30 GOSUB 10040 IF P = 4 THEN L = 150 X = Y - 160 GOSUB 10070 IF P = 3 THEN L = 180 IF L = 1 THEN PRINT Y; " is a long year." ELSE PRINT Y;" is not a long year."90 END100 P = X + INT(X/4) - INT(X/100) + INT(X/400)110 P = P MOD 7120 RETURN`

### IS-BASIC

`100 PROGRAM "Longyear.bas"110 DEF RD(Y)=Y*365+INT(Y/4)-INT(Y/100)+INT(Y/400)120 DEF LONGYEAR(Y)=(4=MOD(RD(Y),7)) OR(4=MOD((RD(Y-1)+1),7))130 INPUT PROMPT "Start year: ":S140 INPUT PROMPT "End year:   ":E150 FOR Y=S TO E160   IF LONGYEAR(Y) THEN PRINT Y,170 NEXT180 PRINT`

### Nascom BASIC

Translation of: Commodore BASIC
Works with: Nascom ROM BASIC version 4.7
` 10 REM Long year20 REM FNM7(N)=MOD(N,7)30 DEF FNM7(N)=N-7*INT(N/7)40 REM FNWD(Y)=Weekday of Y-12-31, 0 Sunday50 DEF FND(Y)=Y+INT(Y/4)-INT(Y/100)+INT(Y/400)60 DEF FNWD(Y)=FNM7(FND(Y))70 REM FNLY(Y)=Nonzero if Y is long80 DEF FNLY(Y)=(4=FNWD(Y))OR(3=FNWD(Y-1))90 CLS100 PRINT "****     ";110 PRINT "List of ISO long years";120 PRINT "     ****"130 INPUT "Start year";S140 INPUT "End year";E150 PRINT160 FOR Y=S TO E170 IF FNLY(Y) THEN PRINT Y;180 NEXT Y190 PRINT200 END `
Output:
```****     List of ISO long years     ****
Start year? 1995
End year? 2045

1998  2004  2009  2015  2020  2026  2032  2037  2043
```

### PureBasic

Translation of: BASIC256
`Procedure.b p(y)  ProcedureReturn (y + Int(y/4) - Int(y/100) + Int(y/400)) % 7EndProcedure Procedure.b isLongYear(y)  ProcedureReturn Bool((p(y) = 4) Or (p(y - 1) = 3))EndProcedure If OpenConsole()  For y = 2000 To 2100    If isLongYear(y)      PrintN(Str(y))    EndIf  Next y   Print(""): Input()  CloseConsole()EndIf`
Output:
```Igual que la entrada de BASIC256.
```

### Quick BASIC

Works with: QB version 4.x
Works with: PDS version 7.x
Works with: QBasic version 1.x
Works with: VB-DOS version 1.0

Translated from Delphi

` DEFINT A-Z DECLARE FUNCTION p% (Yr AS INTEGER)DECLARE FUNCTION LongYear% (Yr AS INTEGER) DIM iYi, iYf, i CLSPRINT "This program calculates which are 53-week years in a range."PRINTINPUT "Initial year"; iYiINPUT "Final year (could be the same)"; iYfIF iYf >= iYi THEN  FOR i = iYi TO iYf    IF LongYear(i) THEN      PRINT i; " ";    END IF  NEXT iEND IFPRINTPRINTPRINT "End of program."END FUNCTION LongYear% (Yr AS INTEGER)  LongYear% = (p%(Yr) = 4) OR (p%(Yr - 1) = 3)END FUNCTION FUNCTION p% (Yr AS INTEGER)  p% = (Yr + INT(Yr / 4) - INT(Yr / 100) + INT(Yr / 400)) MOD 7END FUNCTION `
Output:
```This program calculates which are 53-week years in a range.

Initial year? 1900
Final year (can be the same)? 1999
1903   1908   1914   1920   1925   1931   1936   1942   1948   1953   1959
1964   1970   1976   1981   1987   1992   1998

End of program.
```

### S-Basic

` \$lines rem - compute p mod qfunction mod(p, q = integer) = integerend = p - q * (p/q) comment  return day of week (Sun = 0, Mon = 1, etc.) for a  given Gregorian calendar date using Zeller's congruenceendfunction dayofweek (mo, da, yr = integer) = integer  var y, c, z = integer  if mo < 3 then    begin      mo = mo + 10      yr = yr - 1    end  else mo = mo - 2  y = mod(yr,100)  c = int(yr / 100)  z = int((26 * mo - 2) / 10)  z = z + da + y + int(y/4) + int(c/4) - 2 * c + 777  z = mod(z,7)end = z comment  The simplest of several possible tests is that  any calendar year starting or ending on a  Thursday is "long", i.e., has 53 ISO weeksendfunction islongyear(yr = integer) = integer  var thursday, result = integer  thursday = 4  if (dayofweek(1,1,yr) = thursday) or \     (dayofweek(12,31,yr) = thursday) then       result = -1  rem "true"  else       result = 0   rem "false"end = result rem - main program begins here var year = integerprint "ISO years that will be long in this century:"for year = 2000 to 2099  if islongyear(year) then print year;next year end`
Output:
```ISO years that will be long in this century:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088
2093 2099
```

### Tiny BASIC

`    PRINT "What year would you like?"    INPUT Y    LET X = Y    GOSUB 100    IF P = 4 THEN LET L = 1    LET X = Y - 1    GOSUB 100    IF P = 3 THEN LET L = 1    IF L = 1 THEN PRINT Y," is a long year."    IF L = 0 THEN PRINT Y," is not a long year."    END100 LET P = X + X/4 - X/100 + X/400110 IF P < 7 THEN RETURN    LET P = P - 7    GOTO 110 `
Output:
```What year would you like?
2020
2020 is a long year.

What year would you like?
2021
2021 is not a long year.
```

### True BASIC

Translation of: BASIC256
`FUNCTION p(y) = REMAINDER((y + INT(y/4) - INT(y/100) + INT(y/400)), 7) FUNCTION isLongYear(y)    IF p(y) = 4 THEN       LET isLongYear = 1    ELSEIF p(y-1) = 3 THEN       LET isLongYear = 1    ELSE       LET isLongYear = 0    END IFEND FUNCTION FOR y = 2000 TO 2100    IF isLongYear(y) > 0 THEN PRINT yNEXT yEND`
Output:
```Igual que la entrada de BASIC256.
```

### Yabasic

Translation of: BASIC256
`sub p(y)	return mod((y + int(y/4) - int(y/100) + int(y/400)), 7)end sub sub isLongYear(y)	return (p(y) = 4) or (p(y - 1) = 3)end sub for y = 2000 to 2100	if isLongYear(y) then print y : finext yend`
Output:
```Igual que la entrada de BASIC256.
```

## BCPL

`get "libhdr" let p(y) = (y + y/4 - y/100 + y/400) rem 7let longyear(y) = p(y)=4 | p(y-1)=3 let start() be     for y = 2000 to 2100        if longyear(y) do writef("%N*N", y)`
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

## C#

`using static System.Console;using System.Collections.Generic;using System.Linq;using System.Globalization; public static class Program{    public static void Main()    {        WriteLine("Long years in the 21st century:");        WriteLine(string.Join(" ", 2000.To(2100).Where(y => ISOWeek.GetWeeksInYear(y) == 53)));    }     public static IEnumerable<int> To(this int start, int end) {        for (int i = start; i < end; i++) yield return i;    } }`
Output:
```Long years in the 21st century:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099```

## C

`#include <stdio.h>#include <math.h> // https://webspace.science.uu.nl/~gent0113/calendar/isocalendar.htm int p(int year) {	return (int)((double)year + floor(year/4) - floor(year/100) + floor(year/400)) % 7;} int is_long_year(int year) {	return p(year) == 4 || p(year - 1) == 3;} void print_long_years(int from, int to) {	for (int year = from; year <= to; ++year) {		if (is_long_year(year)) {			printf("%d ", year);		}	}} int main() { 	printf("Long (53 week) years between 1800 and 2100\n\n");	print_long_years(1800, 2100);	printf("\n");	return 0;}`
Output:
```Long (53 week) years between 1800 and 2100

1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
```

## C++

`// Reference:// https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year #include <iostream> inline int p(int year) {    return (year + (year/4) - (year/100) + (year/400)) % 7;} bool is_long_year(int year) {    return p(year) == 4 || p(year - 1) == 3;} void print_long_years(int from, int to) {    for (int year = from, count = 0; year <= to; ++year) {        if (is_long_year(year)) {            if (count > 0)                std::cout << ((count % 10 == 0) ? '\n' : ' ');            std::cout << year;            ++count;        }    }} int main() {    std::cout << "Long years between 1800 and 2100:\n";    print_long_years(1800, 2100);    std::cout << '\n';    return 0;}`
Output:
```Long years between 1800 and 2100:
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852
1857 1863 1868 1874 1880 1885 1891 1896 1903 1908
1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020
2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099
```

## Clojure

`(defn long-year? [year]   (-> (java.time.LocalDate/of year 12 28)       (.get (.weekOfYear (java.time.temporal.WeekFields/ISO)))       (= 53))) (filter long-year? (range 2000 2100))`
Output:
```(2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099)
```

## CLU

`% We can't hide one procedure inside another, but% we can hide the helper `p' in a cluster longyear = cluster is test    rep = null     p = proc (n: int) returns (int)        return ((n + n/4 - n/100 + n/400) // 7)    end p     test = proc (y: int) returns (bool)        return (p(y)=4 | p(y-1)=3)    end testend longyear start_up = proc ()    po: stream := stream\$primary_output()     for i: int in int\$from_to(2000, 2100) do        if longyear\$test(i) then            stream\$putl(po, int\$unparse(i))        end    endend start_up`
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

## Common Lisp

`(defun december-31-weekday (year)    (mod (+ year (floor year 4) (- (floor year 100)) (floor year 400)) 7)) (defun iso-long-year-p (year)     (or (= 4 (december-31-weekday year)) (= 3 (december-31-weekday (1- year))))) (format t "Long years between 1800 and 2100:~&~a~%"    (loop for y from 1800 to 2100 if (iso-long-year-p y) collect y))`
Output:
```Long years between 1800 and 2100:
(1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880
1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048
2054 2060 2065 2071 2076 2082 2088 2093 2099)```

## Dc

Translation of: Tcl
Works with: GNU dc version 1.3.95
`[0q]s0[1q]s1[1r- r 1r- * 1r-]sO                             # O = logical OR##.............................................................................#     C: for( initcode ; condcode ; incrcode ) {body}# .[q]           [1]        [2]        [3]       [4]# #           [initcode] [condcode] [incrcode] [body] (for)[ [q]S. 4:.3:.2:.x [2;.x 0=. 4;.x 3;.x 0;.x]d0:.x    Os.L.o]sF                                             # F = for##.............................................................................#                           [1]         [0]# (.)       [cond_code] [then_code] [else_code] (if_CTE)[ []S. 0:. 1:. x [0=0 1]x ;. s.L. x]sI          # I = if##-----------------------------------------------------------------------------[S. l. l.4/+ l.100/- l.400/+ 7% s.L.]sp         # p##.............................................................................[S. [l.    lpx 4=1 0]x    [l. 1- lpx 3=1 0]x lOx    s.L.]si                                             # i = is_long_year##.............................................................................[    # f = from    # t = to    # y = year    # c = count    st sf                       # fetch args from stack    [lfsy 0sc]        [ly lt <0 1]            # cond        [ly 1+ sy]              # incr y    [        [ly lix]                # is_long_year(y)        [            [lc 0 <1 0]         # 0<c            [                [ lc 10% 0=1 0]         # (c % 10) == 0                [ AP ]                [ [ ]P ]                lIx             # if            ]            []            lIx         # if            ly n            lc 1+ sc        ]        []        lIx             # if    ] lFx       # for]sD                     # D = doit = print_long_years##............................................................................. [Long years between 1800 and 2100:]P AP1800 2100 lDxAP`
Output:
```Long years between 1800 and 2100:
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852
1857 1863 1868 1874 1880 1885 1891 1896 1903 1908
1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020
2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099
```

## Delphi

Note: The Library System.DateUtils implement a WeeksInYear,but not working, return 52 always.

Translation of: C++
` program Long_year; {\$APPTYPE CONSOLE} {\$R *.res} uses  System.SysUtils; function p(const Year: Integer): Integer;begin  Result := (Year + (Year div 4) - (Year div 100) + (Year div 400)) mod 7;end; function IsLongYear(const Year: Integer): Boolean;begin  Result := (p(Year) = 4) or (p(Year - 1) = 3);end; procedure PrintLongYears(const StartYear: Integer; const EndYear: Integer);var  Year, Count: Integer;begin  Count := 0;  for Year := 1800 to 2100 do    if IsLongYear(Year) then    begin      if Count mod 10 = 0 then        Writeln;      Write(Year, ' ');      inc(Count);    end;end; var  Year: Integer; begin  Writeln('Long years between 1800 and 2100:');  PrintLongYears(1800, 2100);  Readln;end. `

## Elixir

`defmodule ISO do  def long_year?(y) do    {:ok, jan1} = Date.new(y,1,1)    {:ok, dec31} = Date.new(y,12,31)    Date.day_of_week(jan1) == 4 or Date.day_of_week(dec31) == 4  endend IO.inspect(Enum.filter(1990..2050, &ISO.long_year?/1))`
Output:
`[1992, 1998, 2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048]`

## Factor

Works with: Factor version 0.99 2019-10-06
`USING: calendar formatting io kernel math.ranges sequences ; : long-year? ( n -- ? ) 12 28 <date> week-number 53 = ; "Year  Long?\n-----------" print 1990 2021 [a,b][ dup long-year? "yes" "no" ? "%d  %s\n" printf ] each`
Output:
```Year  Long?
-----------
1990  no
1991  no
1992  yes
1993  no
1994  no
1995  no
1996  no
1997  no
1998  yes
1999  no
2000  no
2001  no
2002  no
2003  no
2004  yes
2005  no
2006  no
2007  no
2008  no
2009  yes
2010  no
2011  no
2012  no
2013  no
2014  no
2015  yes
2016  no
2017  no
2018  no
2019  no
2020  yes
2021  no
```

## Forth

`: dec31wd ( year -- weekday ) dup dup 4 / swap dup 100 / swap 400 / swap - + + 7 mod ;: long? ( year -- flag ) dup dec31wd 4 = if drop 1 else 1 - dec31wd 3 = if 1 else 0 then then ;: demo ( startyear endyear -- ) cr swap do i long? if i . then loop cr ;`
Output:
```1995 2045 demo
1998 2004 2009 2015 2020 2026 2032 2037 2043
ok```

## Fortran

` program longyear    use iso_fortran_env, only: output_unit, input_unit    implicit none     integer             :: start, ende, i, counter    integer, parameter  :: line_break=10     write(output_unit,*) "Enter beginning of interval"    read(input_unit,*) start    write(output_unit,*) "Enter end of interval"    read(input_unit,*) ende     if (start>=ende) error stop "Last year must be after first year!"     counter = 0    do i = start, ende        if (is_long_year(i)) then            write(output_unit,'(I0,x)', advance="no") i            counter = counter + 1            if (modulo(counter,line_break) == 0) write(output_unit,*)        end if    end docontains    pure function p(year)        integer, intent(in) :: year        integer             :: p         p = modulo(year + year/4 - year/100 + year/400, 7)    end function p     pure function is_long_year(year)        integer, intent(in) :: year        logical             :: is_long_year         is_long_year = p(year) == 4 .or. p(year-1) == 3    end function is_long_yearend program longyear `
Output:
``` Enter beginning of interval
1800
Enter end of interval
2100
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852
1857 1863 1868 1874 1880 1885 1891 1896 1903 1908
1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020
2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099 %
```

## Go

`package main import (    "fmt"    "time") func main() {    centuries := []string{"20th", "21st", "22nd"}    starts := []int{1900, 2000, 2100}     for i := 0; i < len(centuries); i++ {        var longYears []int        fmt.Printf("\nLong years in the %s century:\n", centuries[i])        for j := starts[i]; j < starts[i] + 100; j++ {            t := time.Date(j, time.December, 28, 0, 0, 0, 0, time.UTC)            if _, week := t.ISOWeek(); week == 53 {                longYears = append(longYears, j)            }        }        fmt.Println(longYears)    }}`
Output:
```Long years in the 20th century:
[1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998]

Long years in the 21st century:
[2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099]

Long years in the 22nd century:
[2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195]
```

`import Data.Time.Calendar (fromGregorian)import Data.Time.Calendar.WeekDate (toWeekDate) longYear :: Integer -> BoollongYear y =  let (_, w, _) = toWeekDate \$ fromGregorian y 12 28   in 52 < w main :: IO ()main = mapM_ print \$ filter longYear [2000 .. 2100]`
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

## J

Translation of: C
`   p   =:  1 4 _100 400&(7 | [: <. +/ @: %~)"1 0    ily =:  (4=p) +. 3[email protected]:<:   ply =:  (#~ ily)@:([ + 1[email protected]:-~)`
Output:
```   ply/ 1800 2100
1801 1807 1812 1818 1823 1829 1835 1840 1846 1852 1857 1863 1869 1874 1880 1885 1891 1897 1902 1908 1914 1919 1925 1930 1936 1942 1947 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2021 2026 2032 2038 2043 2049 2054 2060 2066 2071 2077 2083 2...```

## Java

` import java.time.LocalDate;import java.time.temporal.WeekFields; public class LongYear {     public static void main(String[] args) {        System.out.printf("Long years this century:%n");        for (int year = 2000 ; year < 2100 ; year++ ) {            if ( longYear(year) ) {                System.out.print(year + "  ");            }        }    }     private static boolean longYear(int year) {        return LocalDate.of(year, 12, 28).get(WeekFields.ISO.weekOfYear()) == 53;    } } `
Output:
```Long years this century:
2004  2009  2015  2020  2026  2032  2037  2043  2048  2054  2060  2065  2071  2076  2082  2088  2093  2099  ```

## JavaScript

Translation of: TypeScript
`const isLongYear = (year) => {  const jan1 = new Date(year, 0, 1);  const dec31 = new Date(year, 11, 31);  return (4 == jan1.getDay() || 4 == dec31.getDay())} for (let y = 1995; y <= 2045; y++) {  if (isLongYear(y)) {    console.log(y)  }}`
Output:
```1998
2004
2009
2015
2020
2026
2032
2037
2043```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

### Using Zeller's congruence

` # Use Zeller's Congruence to determine the day of the week, given# year, month and day as integers in the conventional way.# Emit 0 for Saturday, 1 for Sunday, etc.#def day_of_week(\$year; \$month; \$day):  if \$month == 1 or \$month == 2 then    [\$month + 12, \$year - 1]  else    [\$month, \$year]  end   | \$day + (13*(.[0] + 1)/5|floor)    +  (.[1]%100)       + ((.[1]%100)/4|floor)    +  (.[1]/400|floor) - 2*(.[1]/100|floor)   | . % 7 ; def has53weeks:  day_of_week(.; 1; 1) == 5 or day_of_week(.; 12; 31) == 5; # To display results neatly:def nwise(\$n):  def n: if length <= \$n then . else .[0:\$n] , (.[\$n:] | n) end;  n; "Long years from 1900 to 2100 inclusive:",([range(1900;2101) | select(has53weeks)] | nwise(10) | join(", "))`
Output:
```Long years from 1900 to 2100 inclusive:
1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953
1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998, 2004, 2009
2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065
2071, 2076, 2082, 2088, 2093, 2099
```

### Using mktime and gmtime

`# Use jq's mktime and gmtime to produce the day of week,# with 0 for Sunday, 1 for Monday, etc# \$year \$month \$day are conventionaldef day_of_week_per_gmtime(\$year; \$month; \$day):   [\$year, \$month - 1, \$day, 0, 0, 1, 0, 0] | mktime | gmtime | .[-2]; # 4 corresponds to Thursdaydef has53weeks:  day_of_week_per_gmtime(.; 1; 1) == 4 or day_of_week(.; 12; 31) == 4; def nwise(\$n):  def n: if length <= \$n then . else .[0:\$n] , (.[\$n:] | n) end;  n; "Long years from 1900 to 2100 inclusive:",([range(1900;2101) | select(has53weeks)] | nwise(10) | join(", "))`
Output:

As above.

## Julia

`using Dates has53weeks(year) = week(Date(year, 12, 28)) == 53 println(" Year  53 weeks?\n----------------")for year in 1990:2021    println(year, "   ", has53weeks(year) ? "Yes" : "No")end `
Output:
``` Year  53 weeks?
----------------
1990   No
1991   No
1992   Yes
1993   No
1994   No
1995   No
1996   No
1997   No
1998   Yes
1999   No
2000   No
2001   No
2002   No
2003   No
2004   Yes
2005   No
2006   No
2007   No
2008   No
2009   Yes
2010   No
2011   No
2012   No
2013   No
2014   No
2015   Yes
2016   No
2017   No
2018   No
2019   No
2020   Yes
2021   No
```

## Kotlin

` fun main() {    val has53Weeks = { year: Int -> LocalDate.of(year, 12, 28).get(WeekFields.ISO.weekOfYear()) == 53 }    println("Long years this century:")    (2000..2100).filter(has53Weeks)        .forEach { year -> print("\$year ")}} `
Output:
```Long years this century:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
```

## Logo

Works with: UCB Logo
`to div :x :y  output int quotient :x :yend to dec31_weekday :year   output remainder (sum :year div :year 4 div :year -100 div :year 400) 7end to iso_long_year? :year   output or 4 = dec31_weekday :year 3 = dec31_weekday difference :year 1end for [y 1995 2045 1] [if iso_long_year? :y [print :y]]`
Output:
```1998
2004
2009
2015
2020
2026
2032
2037
2043```

## Mathematica/Wolfram Language

`firstyear = 2000;lastyear = 2099;years = Range[firstyear, lastyear];firstday = Table[DayName[{years[[n]], 01, 01}], {n, Length[years]}];lastday = Table[DayName[{years[[n]], 12, 31}], {n, Length[years]}];Table[If[years[[n]] >= 1582,   If[firstday[[n]] == Thursday || lastday[[n]] == Thursday,    Style[years[[n]] " long year \n", Bold, Red] ,    years[[n]] " short \n"], "error \n"], {n, Length[years]}]`
Output:
```{2000 short
, 2001 short
, 2002 short
, 2003 short
, 2004 long year
, 2005 short
, 2006 short
, 2007 short
, 2008 short
, 2009 long year
, 2010 short
, 2011 short
, 2012 short
, 2013 short
, 2014 short
, 2015 long year
, 2016 short
, 2017 short
, 2018 short
, 2019 short
, 2020 long year
, 2021 short
, 2022 short
, 2023 short
, 2024 short
, 2025 short
, 2026 long year
, 2027 short
, 2028 short
, 2029 short
, 2030 short
, 2031 short
, 2032 long year
, 2033 short
, 2034 short
, 2035 short
, 2036 short
, 2037 long year
, 2038 short
, 2039 short
, 2040 short
, 2041 short
, 2042 short
, 2043 long year
, 2044 short
, 2045 short
, 2046 short
, 2047 short
, 2048 long year
, 2049 short
, 2050 short
, 2051 short
, 2052 short
, 2053 short
, 2054 long year
, 2055 short
, 2056 short
, 2057 short
, 2058 short
, 2059 short
, 2060 long year
, 2061 short
, 2062 short
, 2063 short
, 2064 short
, 2065 long year
, 2066 short
, 2067 short
, 2068 short
, 2069 short
, 2070 short
, 2071 long year
, 2072 short
, 2073 short
, 2074 short
, 2075 short
, 2076 long year
, 2077 short
, 2078 short
, 2079 short
, 2080 short
, 2081 short
, 2082 long year
, 2083 short
, 2084 short
, 2085 short
, 2086 short
, 2087 short
, 2088 long year
, 2089 short
, 2090 short
, 2091 short
, 2092 short
, 2093 long year
, 2094 short
, 2095 short
, 2096 short
, 2097 short
, 2098 short
, 2099 long year
}```

## Modula-2

`MODULE LongYear;FROM InOut IMPORT WriteCard, WriteLn; VAR year: CARDINAL; PROCEDURE isLongYear(year: CARDINAL): BOOLEAN;    PROCEDURE p(year: CARDINAL): CARDINAL;    BEGIN        RETURN (year + year DIV 4 - year DIV 100 + year DIV 400) MOD 7;    END p;BEGIN    RETURN (p(year) = 4) OR (p(year-1) = 3);END isLongYear; BEGIN    FOR year := 2000 TO 2100 DO        IF isLongYear(year) THEN            WriteCard(year, 4);            WriteLn;        END;    END;END LongYear.`
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

## Nim

`import times proc has53weeks(year: Positive): bool =  let dt = initDateTime(monthday = 1, month = mJan, year = year, hour = 0, minute = 0, second= 0)  result = dt.weekday == dThu or year.isLeapYear and dt.weekday == dWed when isMainModule:  echo "Years with 53 weeks between 2000 and 2100:"  for year in 2000..2100:    if year.has53weeks:      echo year`
Output:
```Years with 53 weeks between 2000 and 2100:
2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

## МК-61/52

`П0	ИП0	4	/	[x]	+	ИП0	1	ВП	2/	[x]	-	ИП0	4	ВП	2	/	[x]	+^	^	7	/	[x]	7	*	-	П1	4-	x#0	40	ИП1	3	-	x#0	40	0	С/П1	С/П`
Output:
```Result for 2020-2030 years: 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0.
```

## Pascal

`program long_year(input);  var    y: integer;   function rd_dec31(year: integer): integer;  begin    { Rata Die of Dec 31, year }    rd_dec31 := year * 365 + year div 4 - year div 100 + year div 400  end;   function rd_jan1(year: integer): integer;  begin    rd_jan1 := rd_dec31(year - 1) + 1  end;   function weekday(rd: integer): integer;  begin    weekday := rd mod 7;  end;   function long_year(year: integer): boolean;  var    jan1: integer;    dec31: integer;  begin    jan1 := rd_jan1(year);    dec31 := rd_dec31(year);    long_year := (weekday(jan1) = 4) or (weekday(dec31) = 4)  end;   begin    for y := 1990 to 2050 do      if long_year(y) then        writeln(y)  end.`
Output:
```1993
1999
2004
2010
2016
2021
2027
2032
2038
2044
2049```

### Free Pascal

Using DateUtils and WeeksInYear to not reinvent this.

` program Long_year; uses  SysUtils,  DateUtils;   procedure PrintLongYears(StartYear, EndYear: Uint32);  var    Year, Count: Uint32;    DateSep: char;  begin    DateSep := FormatSettings.DateSeparator;    Writeln('Long years between ', StartYear, ' and ', EndYear);    Count := 0;    for Year := StartYear to EndYear do      if WeeksInYear(StrToDate('01' + DateSep + '01' + DateSep + IntToStr(Year))) = 53 then      begin        if Count mod 10 = 0 then          Writeln;        Write(Year, ' ');        Inc(Count);      end;    if Count mod 10 <> 0 then      Writeln;    writeln('Found ', Count, ' long years between ', StartYear, ' and ', EndYear);  end; begin  PrintLongYears(1800, 2100);  {\$IFDEF WINDOWS}  Readln;  {\$ENDIF}end.`
Output:
```Long years between 1800 and 2100
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852
1857 1863 1868 1874 1880 1885 1891 1896 1903 1908
1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020
2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099
Found 54 long years between 1800 and 2100```

## Perl

`use strict;use warnings;use DateTime; for my \$century (19 .. 21) {  for my \$year (\$century*100 .. ++\$century*100 - 1) {    print "\$year " if DateTime->new(year => \$year, month => 12, day => 28)->week_number > 52  }  print "\n";}`
Output:
```1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195```

## Phix

```with javascript_semantics
function week_number(integer y,m,d)
integer doy = day_of_year(y,m,d),
dow = day_of_week(y,m,d),
week = floor((doy-dow+10)/7)
return week
end function

for c=20 to 22 do
sequence long_years = {}
integer century = (c-1)*100
for year=century to century+99 do
if week_number(year,12,28)=53 then
long_years &= year
end if
end for
printf(1,"Long years in the %d%s century:%v\n", {c,ord(c),long_years})
end for
```
Output:
```Long years in the 20th century:{1903,1908,1914,1920,1925,1931,1936,1942,1948,1953,1959,1964,1970,1976,1981,1987,1992,1998}
Long years in the 21st century:{2004,2009,2015,2020,2026,2032,2037,2043,2048,2054,2060,2065,2071,2076,2082,2088,2093,2099}
Long years in the 22nd century:{2105,2111,2116,2122,2128,2133,2139,2144,2150,2156,2161,2167,2172,2178,2184,2189,2195}
```

## PHP

`function isLongYear(\$year) {  return (53 == strftime('%V', gmmktime(0,0,0,12,28,\$year)));} for (\$y=1995; \$y<=2045; ++\$y) {  if (isLongYear(\$y)) {    printf("%s\n", \$y);  }}`
Output:
```1998
2004
2009
2015
2020
2026
2032
2037
2043```

## PowerShell

`Function Is-Long-Year {  param([Int]\$year)  53 -eq (Get-Date -Year \$year -Month 12 -Day 28 -UFormat %V)} For (\$y=1995; \$y -le 2045; \$y++) {  If (Is-Long-Year \$y) {    Write-Host \$y  }}`
Output:
```1998
2004
2009
2015
2020
2026
2032
2037
2043```

## Prolog

Translation of: C++
Works with: SWI Prolog
`% See https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year p(Year, P):-    P is (Year + (Year//4) - (Year//100) + (Year//400)) mod 7. long_year(Year):-    p(Year, 4),    !.long_year(Year):-    Year_before is Year - 1,    p(Year_before, 3). print_long_years(From, To):-    writef("Long years between %w and %w:\n", [From, To]),    print_long_years(From, To, 0),    nl. print_long_years(From, To, _):-    From > To,    !.print_long_years(From, To, Count):-    long_year(From),    !,    (Count > 0 ->        (0 is Count mod 10 -> nl ; write(' '))        ;        true    ),    write(From),    Count1 is Count + 1,    Next is From + 1,    print_long_years(Next, To, Count1).print_long_years(From, To, Count):-    Next is From + 1,    print_long_years(Next, To, Count). main:-     print_long_years(1800, 2100).`
Output:
```Long years between 1800 and 2100:
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852
1857 1863 1868 1874 1880 1885 1891 1896 1903 1908
1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020
2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099
```

## Python

Works with: Python version 3.7
`'''Long Year ?''' from datetime import date  # longYear :: Year Int -> Booldef longYear(y):    '''True if the ISO year y has 53 weeks.'''    return 52 < date(y, 12, 28).isocalendar()[1]  # --------------------------TEST---------------------------# main :: IO ()def main():    '''Longer (53 week) years in the range 2000-2100'''    for year in [            x for x in range(2000, 1 + 2100)            if longYear(x)    ]:        print(year)  # MAIN ---if __name__ == '__main__':    main()`
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099```

## Quackery

`dayofweek` is defined at Day of the week#Quackery

`  [ dup dip      [ 1 1 rot dayofweek 4 = ]    31 12 rot dayofweek 4 = or ] is longyear ( n --> b )   say "Long Years in the 21st Century" cr  cr  100 times    [ 2000 i^ + longyear if       [ 2000 i^ + echo sp ] ]`
Output:
```Long Years in the 21st Century

2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 ```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2019.11

December 28 is always in the last week of the year. (By ISO8601)

`sub is-long (\$year) { Date.new("\$year-12-28").week[1] == 53 } # Testingsay   "Long years in the 20th century:\n", (1900..^2000).grep: &is-long;say "\nLong years in the 21st century:\n", (2000..^2100).grep: &is-long;say "\nLong years in the 22nd century:\n", (2100..^2200).grep: &is-long;`
Output:
```Long years in the 20th century:
(1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998)

Long years in the 21st century:
(2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099)

Long years in the 22nd century:
(2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195)```

## REXX

`/*REXX program determines if a (calendar) year is a SHORT or LONG year (52 or 53 weeks).*/parse arg LO HI .                                             /*obtain optional args.   */if LO=='' | LO=="," | LO=='*'  then LO= left( date('S'), 4)   /*Not given?  Use default.*/if HI=='' | HI==","            then HI= LO                    /* "    "      "     "    */if                    HI=='*'  then HI= left( date('S'), 4)   /*an asterisk ≡ current yr*/        do j=LO  to  HI                           /*process single yr  or range of years.*/       say '     year '  j  " is a "   right( word('short long', weeks(j)-51),5)   " year"       end   /*j*/exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/pWeek: parse arg #;  return  (#    +    # % 4    -    # % 100    +    # % 400)     //   7weeks: parse arg y;  if pWeek(y)==4  |  pWeek(y-1)==3  then return 53;           return 52`
output   when using the inputs of:     1990   2030

(Shown at three-quarter size.)

```     year  1990  is a  short  year
year  1991  is a  short  year
year  1992  is a   long  year
year  1993  is a  short  year
year  1994  is a  short  year
year  1995  is a  short  year
year  1996  is a  short  year
year  1997  is a  short  year
year  1998  is a   long  year
year  1999  is a  short  year
year  2000  is a  short  year
year  2001  is a  short  year
year  2002  is a  short  year
year  2003  is a  short  year
year  2004  is a   long  year
year  2005  is a  short  year
year  2006  is a  short  year
year  2007  is a  short  year
year  2008  is a  short  year
year  2009  is a   long  year
year  2010  is a  short  year
year  2011  is a  short  year
year  2012  is a  short  year
year  2013  is a  short  year
year  2014  is a  short  year
year  2015  is a   long  year
year  2016  is a  short  year
year  2017  is a  short  year
year  2018  is a  short  year
year  2019  is a  short  year
year  2020  is a   long  year
year  2021  is a  short  year
year  2022  is a  short  year
year  2023  is a  short  year
year  2024  is a  short  year
year  2025  is a  short  year
year  2026  is a   long  year
year  2027  is a  short  year
year  2028  is a  short  year
year  2029  is a  short  year
year  2030  is a  short  year
```

## Ring

` see "long years 2000-2099: "for year = 2000 to 2100    num1 = (year-1900)%7     num2 = floor((year-1904)/4)    num3 = (num1+num2+5)%7     if num3 = 0 or (num1 = 6 and num3 = 1)       see "" + year + " "    oknext `
Output:
```long years 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
```

## Ruby

` require 'date' def long_year?(year = Date.today.year)  Date.new(year, 12, 28).cweek == 53end (2020..2030).each{|year| puts "#{year} is long? #{ long_year?(year) }." }  `
Output:
```2020 is long? true.
2021 is long? false.
2022 is long? false.
2023 is long? false.
2024 is long? false.
2025 is long? false.
2026 is long? true.
2027 is long? false.
2028 is long? false.
2029 is long? false.
2030 is long? false.
```

## Rust

`extern crate time; // 0.2.16 use time::Date; fn main() {    (2000..=2099)        .filter(|&year| is_long_year(year))        .for_each(|year| println!("{}", year));} fn is_long_year(year: i32) -> bool {    Date::try_from_ymd(year, 12, 28).map_or(false, |date| date.week() == 53)} `
Output:
```2004
2009
2015
2020
2026
2032
2037
2043
2048
2054
2060
2065
2071
2076
2082
2088
2093
2099
```

## Scala

Output:
Best seen running in your browser by Scastie (remote JVM).
`import java.time.temporal.TemporalAdjusters.firstInMonthimport java.time.temporal.{ChronoField, IsoFields}import java.time.{DayOfWeek, LocalDate, Month} import scala.util.{Failure, Try} private object LongYear extends App {  private val (currentCentury, maxWeekNumber) = (LocalDate.now().getYear / 100, ChronoField.ALIGNED_WEEK_OF_YEAR.range().getMaximum)  private val centuries = currentCentury * 100 until (currentCentury + 1) * 100  private val results = List(    centuries.filter(isThursdayFirstOrLast),    centuries.filter(year => maxIsoWeeks(year) == maxWeekNumber),    centuries.filter(mostThursdaysInYear)  )   // Solution 1, the first or respectively last day of the year is a Thursday.  private def isThursdayFirstOrLast(_year: Int): Boolean = {     LocalDate.of(_year, Month.DECEMBER, 31).get(ChronoField.DAY_OF_WEEK) == DayOfWeek.THURSDAY.getValue ||    LocalDate.of(_year, Month.JANUARY, 1).get(ChronoField.DAY_OF_WEEK) == DayOfWeek.THURSDAY.getValue  }   // Solution 2, if last week that contains at least four days of the month of December.  private def maxIsoWeeks(_year: Int) = {    // The last week that contains at least four days of the month of December.    LocalDate.of(_year, Month.DECEMBER, 28).get(IsoFields.WEEK_OF_WEEK_BASED_YEAR)  }   // Solution 3, if there are 52 Thursdays in a year  private def mostThursdaysInYear(_year: Int) = {    val datum = LocalDate.of(_year, Month.JANUARY, 1).`with`(firstInMonth(DayOfWeek.THURSDAY))     datum.plusDays(52 * 7).getYear == _year  }   println(s"Years in this \${currentCentury + 1}st century having ISO week \$maxWeekNumber :")   Try { // Testing the solutions    assert(results.tail.forall(_ == results.head), "Discrepancies in results.")  } match {    case Failure(ex) => Console.err.println(ex.getMessage)    case _ =>  }   results.zipWithIndex.foreach(solution => println(s"Solution \${solution._2}: \${solution._1.mkString(" ")}")) }`

## Scheme

Works with: MIT Scheme
Works with: Guile
Works with: Racket

The demo code uses iota as defined in SRFI-1, so won't work in Schemes lacking that function. Racket requires the addition of (require srfi/1).

`(define (dec31wd year)     (remainder (apply + (map (lambda (d) (quotient year d)) '(1 4 -100 400))) 7)) (define (long? year) (or (= 4 (dec31wd year)) (= 3 (dec31wd (- year 1))))) (display "Long years between 1800 and 2100:") (newline)(display (filter long? (iota 300 1800)))`
Output:
```Long years between 1800 and 2100:
(1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099)```

## Sidef

`func is_long_year(year) {    Date.parse("#{year}-12-28", "%Y-%m-%d").week == 53} say (  "Long years in the 20th century:\n", (1900..^2000).grep(is_long_year))say ("\nLong years in the 21st century:\n", (2000..^2100).grep(is_long_year))say ("\nLong years in the 22nd century:\n", (2100..^2200).grep(is_long_year))`
Output:
```Long years in the 20th century:
[1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953, 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998]

Long years in the 21st century:
[2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099]

Long years in the 22nd century:
[2105, 2111, 2116, 2122, 2128, 2133, 2139, 2144, 2150, 2156, 2161, 2167, 2172, 2178, 2184, 2189, 2195]
```

## Snobol

`               DEFINE('DEC31WD(Year)')       :(END_DEC31WD)DEC31WD        DEC31WD = REMDR(Year + (Year / 4) - (Year / 100) + (Year / 400), 7) :(RETURN)END_DEC31WD                DEFINE('ISOLONG(Year)')       :(END_ISOLONG)ISOLONG        EQ(DEC31WD(Year), 4)          :S(RETURN)               EQ(DEC31WD(Year - 1), 3)      :S(RETURN)F(FRETURN)END_ISOLONG                DEFINE('ISODEMO(Start,End)')  :(END_ISODEMO)ISODEMO        OUTPUT = 'ISO long years between ' Start ' and ' End ':'               Year = StartLOOP           OUTPUT = ISOLONG(Year) Year               Year = Year + 1               LE(YEAR, 2045)                :S(LOOP) F(RETURN)END_ISODEMO                 ISODEMO(1995, 2045)END`
Output:
```ISO long years between 1995 and 2045:
1998
2004
2009
2015
2020
2026
2032
2037
2043```

## Swift

`func isLongYear(_ year: Int) -> Bool {  let year1 = year - 1  let p = (year + (year / 4) - (year / 100) + (year / 400)) % 7  let p1 = (year1 + (year1 / 4) - (year1 / 100) + (year1 / 400)) % 7   return p == 4 || p1 == 3} for range in [1900...1999, 2000...2099, 2100...2199] {  print("\(range): \(range.filter(isLongYear))")}`
Output:
```1900...1999: [1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953, 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998]
2000...2099: [2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099]
2100...2199: [2105, 2111, 2116, 2122, 2128, 2133, 2139, 2144, 2150, 2156, 2161, 2167, 2172, 2178, 2184, 2189, 2195]```

## Tcl

Translation of: C++
` ## Reference: https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year proc p {year} {    return [expr {(\$year + (\$year/4) - (\$year/100) + (\$year/400)) % 7}]} proc is_long_year {year} {    return [expr {[p \$year] == 4 || [p [expr {\$year - 1}]] == 3}]} proc print_long_years {from to} {    for {set year \$from; set count 0} {\$year <= \$to} {incr year} {        if {[is_long_year \$year]} {            if {\$count > 0} {                puts -nonewline [expr {(\$count % 10 == 0) ? "\n" : " "}]            }            puts -nonewline \$year            incr count        }    }} puts "Long years between 1800 and 2100:"print_long_years 1800 2100puts ""`
Output:
```Long years between 1800 and 2100:
1801 1807 1812 1818 1824 1829 1835 1840 1846 1852
1857 1863 1868 1874 1880 1885 1891 1896 1903 1908
1914 1920 1925 1931 1936 1942 1948 1953 1959 1964
1970 1976 1981 1987 1992 1998 2004 2009 2015 2020
2026 2032 2037 2043 2048 2054 2060 2065 2071 2076
2082 2088 2093 2099
```

## Terraform

Works with: Terraform version 0.13+

Contents of main module:

`module "iso-long-years" {  source = "./iso-long-years"  start_year = 1995  end_year = 2045} output "long-years" {   value = module.iso-long-years.long-years}`

Contents of iso-long-years module:

`variable start_year {  type = number} variable end_year {  type = number} locals {  year_list = range(var.start_year, var.end_year+1)} module "iso-long-year" {  for_each = toset([for y in local.year_list: tostring(y)])  source = "../iso-long-year"  year = each.key} output "long-years" {  value = compact([for y in [for n in local.year_list: tostring(n)]:    module.iso-long-year[y].isLong ? y : ""])}`

Contents of iso-long-year module:

`variable year {  type = string  default = ""} locals {  ystr = var.year != "" ? var.year : split("-",timestamp())[0]  y = tonumber(local.ystr)  e = local.y - 1  dec31 = local.y * 365 + floor(local.y/4) - floor(local.y/100) + floor(local.y/400)  jan1 = local.e * 365 + floor(local.e/4) - floor(local.e/100) + floor(local.e/400) + 1} output isLong {  value = (local.dec31 % 7 == 4 || local.jan1 % 7 == 4)}`
Output:
```Apply complete! Resources: 0 added, 0 changed, 0 destroyed.

Outputs:

long-years = [
"1998",
"2004",
"2009",
"2015",
"2020",
"2026",
"2032",
"2037",
"2043",
]```

## TypeScript

`const isLongYear = (year: number): boolean => {  const jan1: Date = new Date(year, 0, 1);  const dec31: Date = new Date(year, 11, 31);  return (4 == jan1.getDay() || 4 == dec31.getDay())} for (let y: number = 1995; y <= 2045; y++) {  if (isLongYear(y)) {    console.log(y)  }}`
Output:
```1998
2004
2009
2015
2020
2026
2032
2037
2043```

## UNIX Shell

### Thursdays check using cal(1) and grep(1)

`long_year() {  cal 1 \$1 | grep -q ' 3 *\$' && return 0  cal 12 \$1 | grep -q ' 26 *\$'}`

### Straightforward check using GNU date(1)

`long_year() {  expr \$(date -d "\$1-12-28" +%V) = 53 >/dev/null}`

### Direct computation with built-in arithmetic in newer shells

Works with: Bourne Again SHell
Works with: Korn Shell
Works with: Zsh
`dec31wd() {  typeset -i y=\$1  echo \$(( (y + y / 4 - y / 100 + y / 400) % 7 ))} long_year() {   typeset -i y=\$1   (( 4 == \$(dec31wd \$y) || 3 == \$(dec31rd \$(( y - 1 ))) ))}`

Demo code for any of the above:

`for y in \$(seq 1995 2045); do  if long_year \$y; then    echo \$y  fidone | column `
Output:
`1998    2004    2009    2015    2020    2026    2032    2037    2043`

## Visual Basic

Works with: Visual Basic version 5
Works with: Visual Basic version 6
Works with: VBA version Access 97
Works with: VBA version 6.5
Works with: VBA version 7.1
`Option Explicit Function IsLongYear(ByVal Year As Integer) As Boolean  Select Case vbThursday  Case VBA.DatePart("w", VBA.DateSerial(Year, 1, 1)), _       VBA.DatePart("w", VBA.DateSerial(Year, 12, 31))    IsLongYear = True  End SelectEnd Function Sub Main()'testDim l As Long  For l = 1990 To 2021    Select Case l    Case 1992, 1998, 2004, 2009, 2015, 2020      Debug.Assert IsLongYear(l)    Case Else      Debug.Assert Not IsLongYear(l)    End Select  Next lEnd Sub `

### Visual Basic for DOS

Translated from Delohi

`OPTION EXPLICIT DECLARE FUNCTION p (Yr AS INTEGER) AS INTEGERDECLARE FUNCTION LongYear (Yr AS INTEGER) AS INTEGER DIM iYi AS INTEGER, iYf AS INTEGER, i AS INTEGER CLSPRINT "This program calculates which are 53-week years in a range."PRINTINPUT "Initial year"; iYiINPUT "Final year (could be the same)"; iYfIF iYf >= iYi THEN  FOR i = iYi TO iYf    IF LongYear(i) THEN      PRINT i; " ";    END IF  NEXT iEND IFPRINTPRINTPRINT "End of program."END FUNCTION p (Yr AS INTEGER) AS INTEGER  p = (Yr + INT(Yr / 4) - INT(Yr / 100) + INT(Yr / 400)) MOD 7END FUNCTION FUNCTION LongYear (Yr AS INTEGER) AS INTEGER  LongYear = (p(Yr) = 4) OR (p(Yr - 1) = 3)END FUNCTION `

## Wren

Translation of: Go
Library: Wren-date
`import "/date" for Date var centuries = ["20th", "21st", "22nd"]var starts = [1900, 2000, 2100]for (i in 0...centuries.count) {    var longYears = []    System.print("\nLong years in the %(centuries[i]) century:")    for (j in starts[i]...starts[i]+100) {        var t = Date.new(j, 12, 28)        if (t.weekOfYear[1] == 53) {            longYears.add(j)        }    }    System.print(longYears)}`
Output:
```Long years in the 20th century:
[1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953, 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998]

Long years in the 21st century:
[2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099]

Long years in the 22nd century:
[2105, 2111, 2116, 2122, 2128, 2133, 2139, 2144, 2150, 2156, 2161, 2167, 2172, 2178, 2184, 2189, 2195]
```

## XPL0

Translation of: Commodore BASIC
Works with: EXPL-32
` \Long yearcode Rem=2, CrLf=9, IntIn=10, IntOut=11, Text=12, Clear=40;integer S, E, Y;   function integer Weekday(Y);  \Weekday of Y-12-31, 0 Sunday  integer Y;  return Rem((Y + Y / 4 - Y / 100 + Y / 400) / 7);   function integer IsLongYear(Y);  integer Y;  return 4 = Weekday(Y) ! 3 = Weekday(Y - 1); beginClear;   Text(0, "****     List of ISO long years     ****");CrLf(0);Text(0, "Start year: "); S:= IntIn(0);Text(0, "End year: "); E:= IntIn(0);CrLf(0);for Y:= S, E do  if IsLongYear(Y) then [IntOut(0, Y); Text(0, " ")];CrLf(0);end `
Output:
```****     List of ISO long years     ****
Start year: 1995
End year: 2045

1998 2004 2009 2015 2020 2026 2032 2037 2043
```

## zkl

`fcn isLongYear(y){ Time.Date.weeksInYear(y)==53 }foreach nm,y in (T(T("20th",1900), T("21st",2000), T("22nd",2100))){   println("\nLong years in the %s century:\n%s".fmt(nm,      [y..y+99].filter(isLongYear).concat(" ")));}`
Output:
```Long years in the 20th century:
1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998

Long years in the 21st century:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099

Long years in the 22nd century:
2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195
```