subroutine invtime(timestamp,yr,mon,day,hr,min)
c      
c This subroutine will convert a minutes timestamp to a year/month
c date. Based on the function convtime by Shawn Smith (COAPS).
c
c Written the spring of 1995, several iterations.
c James N. Stricherz (stricherz@coaps.fsu.edu)
c
c This code returns correct results for the range of 01 Jan 1980 0000GMT
c thru 31 Dec 1999 2359GMT. I know it does, because I tried each minute
c of that range.
c
      integer year(1980:2000), month(13), leap_mon(13), min_day,
     +        min_hr, yr, mon, day, hr, min,timestamp,itime,tmon,ttime,
     +        thour
     
      data year /0, 527040, 1052640, 1578240, 2103840, 2630880, 3156480,
     +             3682080, 4207680, 4734720, 5260320, 5785920, 6311520,
     +             6838560, 7364160, 7889760, 8415360, 8942400, 9468000,
     +             9993600,10519200/    
      data month /0,  44640,  84960, 129600, 172800, 217440, 260640,
     +       305280, 349920, 393120, 437760, 480960,525600/
      data leap_mon /0,  44640,  86400, 131040, 174240, 218880, 262080,
     +          306720, 351360, 394560, 439200, 482400,527040/
      data min_day, min_hr /1440, 60/
*
*     ok, lets inverse the effects of the years -- subtract off the
*     number of minutes per year until it goes negative. iyr 
*     then gives the year that the time (in minutes) occurs...
*
      if (timestamp.ge.year(2000)) then
         yr=-9999
         return
      endif
      iyr=1979
      itime=timestamp
 10   iyr=iyr+1
      ttime=itime-year(iyr)
      if (ttime.le.0) then
         if (iyr.eq.1980) iyr=iyr+1
         iyr=iyr-1
         itime=itime-year(iyr)
      else
         goto 10
      endif
*
*     assign the return variable
*
      yr=iyr
*
*     ok, the remaining time is less than one full year, so convert
*     by the same method as above into months:
*
      imon=0
*
*     that darn leap month!
*
      if (mod(iyr,4).ne.0) then
*
*     increment the month, and subtract off the minutes from the
*     remaining time for a non-leap year
*
 20      imon=imon+1
         tmon=itime-month(imon)
         if (tmon.gt.0) then
            goto 20
         else if (tmon.lt.0) then
            imon=imon-1
            itime=itime-month(imon)
         else
            if (imon.gt.12) then
               imon=imon-12
               yr=yr+1
            endif
            mon=imon
            day=1
            hr=0
            min=0
            return
         endif
      else
*     
*     same thing, same code, but for a leap year
*
 30      imon=imon+1
         tmon=itime-leap_mon(imon)
         if (tmon.gt.0) then
            goto 30
         elseif (tmon.lt.0) then
            imon=imon-1
            itime=itime-month(imon)
         else
            if (imon.gt.12) then
               imon=imon-12
               yr=yr+1
            endif
            mon=imon
            day=1
            hr=0
            min=0
            return
         endif         
      endif
*
*     assign the return variable
*
      mon=imon
*
*     any remaining minutes will belong to day/hour/minutes
*     ok, let's get those pesky days!
*
      iday=0
 40   iday=iday+1
      ttime=itime-min_day
      if (ttime.ge.0) then
         itime=ttime
         goto 40
      endif
*
*     assign the return variable
*
      if (mod(iyr,4).eq.0.and.mon.gt.2) then
         day=iday-1
      else 
         day=iday
      endif
*
*     pick off the hours of the days...remember, hours can be 0, so
*     we start at -1
*
      ihour=-1
 50   ihour=ihour+1
      thour=itime-min_hr
      if (thour.ge.0) then
         itime=thour
         goto 50
      endif
*
*     assign the return variables
*
      hr=ihour
*
*     the remainder at this point is the minutes, so return them directly!
*
      min=itime
      return
      end