character*4 eol
      integer*4 iol
      logical*1 nin(4)
      character  buf*20, infile*50, outfile*50
      equivalence (iol,eol)


C   ----------  note that SGI fortran works in 4-byte-words (isgi=4)
C           in which case this code will not work in the header length
C           is not a multiple of four bytes.
C           for these cases, use the C-code provided.
C
C               most other fortrans can access single bytes (isgi=1)
C           and don't have this limitation
C
C           Note that this code uses the shortest possible block size
C           and direct file access to make it compatible with the widest
C           possible range of files and compilers. This also makes it
C           very slow. Increasing the block size, or using 'flat' access
C           whenever possible will speed it, in many cases substantially.
C   ----------------------------------------------------------------------
*     isgi=4        ! use this value for SGI fortran
      isgi=1        ! use for almost any other fortran

      iol=10*(1+256**3)        ! define end-of-line character

      write(*,*) 'input file?'
      read(*,'(a50)') infile
      write(*,*) 'output file?'
      read(*,'(a50)') outfile

      open(1,file=infile,access='direct', recl=20/isgi)
      read(1,rec=1) buf
      i1=index(buf,'=')          !check for = in HEADERLENGTH
      i2=index(buf,eol(4:4))     !check for end of line after length


*     write(*,*) 'i1=',i1,'  i2=',i2,'  buf=',buf(i1+1:i2-1)
      read(buf(i1+1:i2-1),'(i5)') lhead

      irec0=lhead/isgi     ! note that this may fail if isgi=4
      write(*,*) 'header length is =',lhead, irec0

      if(mod(lhead,isgi).ne.0) stop 'header not multiple of word length'

      close(1)

      open(1,file=infile,access='direct', status='old',recl=1)
      open(2,file=outfile,access='direct', recl=1)


      if (isgi.eq.1) then

         do i=1,irec0
            read (1,rec=i) nin(1)
            write(2,rec=i) nin(1)
         enddo
         i=irec0+1

         read (1,rec=i,iostat=iend) nin(1)
         do while(iend.eq.0)
            do k=1,3
               read (1,rec=i+k,iostat=iend) nin(k+1)
               if (iend.ne.0)  then
                  close(1)
                  close(2)
                  stop 'last word was incomplete'
               endif
            enddo
            do k=1,4
               write(2,rec=i+4-k) nin(k)
            enddo
            i=i+4
            read (1,rec=i,iostat=iend) nin(1)
         enddo

      else

         do i=1,irec0
            read (1,rec=i) nin
            write(2,rec=i) nin
         enddo
         i=irec0+1

         read (1,rec=i,iostat=iend) nin
         do while(iend.eq.0)
            write(2,rec=i) (nin(k),k=4,1,-1)
            i=i+1
            read (1,rec=i,iostat=iend) nin
         enddo

      endif

      close(1)
      close(2)



      end