You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.

A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.

Input

The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).

Output

Print a single integer — the answer to a problem.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

10
12345


Output

6


Input

16
439873893693495623498263984765


Output

40
#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m;
int a[4010];
char p[4010];
map<int, int> M;
long long an;
int main() {
  scanf("%d%s", &n, p + 1);
  m = strlen(p + 1);
  for (i = 1; i <= m; i++) a[i] = a[i - 1] + p[i] - 48;
  for (i = 1; i <= m; i++)
    for (j = 1; j <= i; j++) {
      int x = a[i] - a[j - 1];
      if ((!x && !n) || x && n % x == 0) {
        if (!M.count(x))
          M[x] = 1;
        else
          M[x]++;
      }
    }
  if (!n) {
    if (M.count(0)) an = M[0];
    long long s = m * (m + 1) / 2;
    printf("%I64d\n", an * s * 2 - an * an);
    return 0;
  }
  map<int, int>::iterator A = M.begin();
  for (; A != M.end(); A++) {
    if (A->first == 0) continue;
    int x = n / (A->first);
    if (M.count(x)) an += (long long)M[x] * A->second;
  }
  printf("%I64d\n", an);
  return 0;
}
