subroutine jday(imon, iday, iyear, julday)
! This routine calculates the Julian day (julday) from
! the month (imon), day (iday), and year (iyear). The
! appropriate correction is made for leap year.

! Modifications:
! 4/20/21 (J. Dostalek) Started this Fortran 90 version
!-------------------------------------------------------------------------------

  implicit none

  integer, intent(in) :: imon
  integer, intent(in) :: iday
  integer, intent(in) :: iyear

  integer, intent(out) :: julday

  integer :: i

  integer, dimension(12) :: ndmon

! Specify the number of days in each month
  ndmon(1)  = 31
  ndmon(2)  = 28
  ndmon(3)  = 31
  ndmon(4)  = 30
  ndmon(5)  = 31
  ndmon(6)  = 30
  ndmon(7)  = 31
  ndmon(8)  = 31
  ndmon(9)  = 30
  ndmon(10) = 31
  ndmon(11) = 30
  ndmon(12) = 31

! Correct for leap year
  if (mod(iyear,   4) == 0) ndmon(2) = 29
  if (mod(iyear, 100) == 0) ndmon(2) = 28
  if (mod(iyear, 400) == 0) ndmon(2) = 29

! Check for illegal input
  if (imon < 1 .or. imon > 12) then
    julday = -1
    return
  end if

  if (iday < 1 .or. iday > ndmon(imon)) then
    julday = -1
    return
  end if

! Calculate the Julian day
  julday = iday
  if (imon > 1) then
    do i = 2, imon
      julday = julday + ndmon(i - 1)
    end do
  end if

  return
end subroutine jday