subroutine imtql1(n,d,e,ierr)
c
      integer i,j,l,m,n,ierr
      real d(n),e(n)
      real b,c,f,g,p,r,s,tst1,tst2
c
c     this subroutine is a translation of the algol procedure imtql1,
c     num. math. 12, 377-383(1968) by martin and wilkinson,
c     as modified in num. math. 15, 450(1970) by dubrulle.
c     handbook for auto. comp., vol.ii-linear algebra, 241-248(1971).
c
c     this subroutine finds the eigenvalues of a symmetric
c     tridiagonal matrix by the implicit ql method.
c
c     on input
c
c        n is the order of the matrix.
c
c        d contains the diagonal elements of the input matrix.
c
c        e contains the subdiagonal elements of the input matrix
c          in its last n-1 positions.  e(1) is arbitrary.
c
c      on output
c
c        d contains the eigenvalues in ascending order.  if an
c          error exit is made, the eigenvalues are correct and
c          ordered for indices 1,2,...ierr-1, but may not be
c          the smallest eigenvalues.
c
c        e has been destroyed.
c
c        ierr is set to
c          zero       for normal return,
c          j          if the j-th eigenvalue has not been
c                     determined after 30 iterations.
c
c     Questions and comments should be directed to Alan K. Cline,
c     Pleasant Valley Software, 8603 Altus Cove, Austin, TX 78759.
c     Electronic mail to cline@cs.utexas.edu.
c
c     this version dated january 1989. (for the IBM 3090vf)
c
c     ------------------------------------------------------------------
c
      call xuflow(0)
      ierr = 0
      if (n .eq. 1) go to 1001
c
      do 100 i = 2, n
  100 e(i-1) = e(i)
c
      e(n) = 0.0e0
c
      do 290 l = 1, n
         j = 0
c     .......... look for small sub-diagonal element ..........
  105    do 110 m = l, n-1
            tst1 = abs(d(m)) + abs(d(m+1))
            tst2 = tst1 + abs(e(m))
            if (tst2 .eq. tst1) go to 120
  110    continue
c
  120    p = d(l)
         if (m .eq. l) go to 215
         if (j .eq. 30) go to 1000
         j = j + 1
c     .......... form shift ..........
         g = (d(l+1) - p) / (2.0e0 * e(l))
cccccccccccccccccccccccccccccccccccccccccccccccccccc
c *      r = pythag(g,1.0d0)
cccccccccccccccccccccccccccccccccccccccccccccccccccc
         if (abs(g).le.1.0e0) then
            r = sqrt(1.0e0 + g*g)
         else
            r = g * sqrt(1.0e0 + (1.0e0/g)**2)
         endif
cccccccccccccccccccccccccccccccccccccccccccccccccccc
         g = d(m) - p + e(l) / (g + sign(r,g))
         s = 1.0e0
         c = 1.0e0
         p = 0.0e0
c     .......... for i=m-1 step -1 until l do -- ..........
         do 200 i = m-1, l, -1
            f = s * e(i)
            b = c * e(i)
cccccccccccccccccccccccccccccccccccccccccccccccccccc
c *         r = pythag(f,g)
cccccccccccccccccccccccccccccccccccccccccccccccccccc
            if (abs(f).ge.abs(g)) then
               r = abs(f) * sqrt(1.0e0 + (g/f)**2)
            else if (g .ne. 0.0e0) then
               r = abs(g) * sqrt((f/g)**2 + 1.0e0)
            else
               r = abs(f)
            endif
cccccccccccccccccccccccccccccccccccccccccccccccccccc
            e(i+1) = r
            if (r .eq. 0.0e0) then
c     .......... recover from underflow ..........
               d(i+1) = d(i+1) - p
               e(m) = 0.0e0
               go to 105
            endif
            s = f / r
            c = g / r
            g = d(i+1) - p
            r = (d(i) - g) * s + 2.0e0 * c * b
            p = s * r
            d(i+1) = g + p
            g = c * r - b
  200    continue
c
         d(l) = d(l) - p
         e(l) = g
         e(m) = 0.0e0
         go to 105
c     .......... order eigenvalues ..........
c     .......... for i=l step -1 until 2 do -- ..........
  215    do 230 i = l, 2, -1
            if (p .ge. d(i-1)) go to 270
            d(i) = d(i-1)
  230    continue
c
         i = 1
  270    d(i) = p
  290 continue
c
      go to 1001
c     .......... set error -- no convergence to an
c                eigenvalue after 30 iterations ..........
 1000 ierr = l
 1001 return
      end