subroutine imtql1(n,d,e,ierr)
c
      integer i,j,l,m,n,ii,mml,ierr
      double precision d(n),e(n)
      double precision b,c,f,g,p,r,s,tst1,tst2,pythag
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     calls pythag for  dsqrt(a*a + b*b) .
c
c     questions and comments should be directed to burton s. garbow,
c     mathematics and computer science div, argonne national laboratory
c
c     this version dated august 1983.
c
c     ------------------------------------------------------------------
c
      ierr = 0
      if (n .eq. 1) go to 1001
c
      do 100 i = 2, n
  100 e(i-1) = e(i)
c
      e(n) = 0.0d0
c
      do 290 l = 1, n
         j = 0
c     .......... look for small sub-diagonal element ..........
  105    do 110 m = l, n
            if (m .eq. n) go to 120
            tst1 = dabs(d(m)) + dabs(d(m+1))
            tst2 = tst1 + dabs(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) / (2.0d0 * e(l))
         r = pythag(g,1.0d0)
         g = d(m) - p + e(l) / (g + dsign(r,g))
         s = 1.0d0
         c = 1.0d0
         p = 0.0d0
         mml = m - l
c     .......... for i=m-1 step -1 until l do -- ..........
         do 200 ii = 1, mml
            i = m - ii
            f = s * e(i)
            b = c * e(i)
            r = pythag(f,g)
            e(i+1) = r
            if (r .eq. 0.0d0) go to 210
            s = f / r
            c = g / r
            g = d(i+1) - p
            r = (d(i) - g) * s + 2.0d0 * 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) = 0.0d0
         go to 105
c     .......... recover from underflow ..........
  210    d(i+1) = d(i+1) - p
         e(m) = 0.0d0
         go to 105
c     .......... order eigenvalues ..........
  215    if (l .eq. 1) go to 250
c     .......... for i=l step -1 until 2 do -- ..........
         do 230 ii = 2, l
            i = l + 2 - ii
            if (p .ge. d(i-1)) go to 270
            d(i) = d(i-1)
  230    continue
c
  250    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