# recursive routine to calculate c^d%n
# by breaking it down using residues.
#
define x ( c,d,n) {
  #-1;  d;  -1;
  if ( d == 1 ) return (c^d%n);
  return x(c^2,d/2,n)%n;
}

r = 1;

# recursive routine to break d into powers of 2
# and then to collect multiply the results
# of the 2^n exponents.
#
define t(c,d,n) {
  auto i
  i = 0;
  
  # find biggest power of 2 left in exponentiation
  # 
  while(d >= (2^i)) { i = i+1; }

  #
  # subtract from exponent
  #
  d = (d - (2^(i-1)));
 
  
  if (d>0) { 
        r=(r*x(c,2^(i-1),n))%n;  # calculate this large exponentiation
        t(c,d,n);                # recurse with new d value
        return (2^(i-1));
  }
  if (d == 0) {
         r = (r*x(c,2^(i-1),n))%n; # last one
         return (2^(i-1));
  }
}