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;
struct node {
  int l, r;
};
vector<node> quan[100005];
int main() {
  int a, i, j, k, t, len;
  long long int s;
  double jie;
  char ch[4005];
  scanf("%d", &a);
  getchar();
  scanf("%s", ch);
  len = strlen(ch);
  long long int ans = 0, ss = 0, l = 0;
  if (a == 0) {
    for (i = 0; i < len; i++) {
      s = 0;
      for (j = i; j < len; j++) {
        s += (ch[j] - '0');
        if (s == 0) {
          ss++;
        } else
          l++;
      }
    }
    ans = ss * ss + ss * l * 2;
  } else {
    jie = sqrt(a);
    for (i = 0; i < len; i++) {
      s = 0;
      for (j = i; j < len; j++) {
        s += (ch[j] - '0');
        if (s != 0 && a % s == 0 && s <= jie) {
          node ss;
          ss.l = i;
          ss.r = j;
          quan[s].push_back(ss);
        }
        if (s > jie) break;
      }
    }
    for (i = 0; i < len; i++) {
      s = 0;
      for (j = i; j < len; j++) {
        s += (ch[j] - '0');
        if (s != 0 && a % s == 0 && s >= jie) {
          t = a / s;
          if (t * t == a) {
            ans += quan[t].size();
          } else {
            ans += 2 * quan[t].size();
          }
        }
        if (s > a) break;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
