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 s[4040];
int main() {
  ios_base::sync_with_stdio(false);
  long long a;
  string str;
  cin >> a >> str;
  int n = str.size();
  for (int _n(n), i(0); i < _n; i++) s[i] = (str[i] - '0') + (i ? s[i - 1] : 0);
  map<int, int> cnt;
  for (int _n(n), i(0); i < _n; i++)
    for (int _n(i + 1), j(0); j < _n; j++) {
      int x = s[i] - (j ? s[j - 1] : 0);
      cnt[x]++;
    }
  long long ans = 0;
  if (a == 0) {
    int z = cnt[0];
    long long al = 0;
    for (int _n(n), i(0); i < _n; i++)
      for (int _n(i + 1), j(0); j < _n; j++) ++al;
    for (int _n(n), i(0); i < _n; i++)
      for (int _n(i + 1), j(0); j < _n; j++) {
        int x = s[i] - (j ? s[j - 1] : 0);
        if (x != 0)
          ans += z;
        else
          ans += al;
      }
  } else {
    for (int _n(n), i(0); i < _n; i++)
      for (int _n(i + 1), j(0); j < _n; j++) {
        int x = s[i] - (j ? s[j - 1] : 0);
        if (x == 0) continue;
        if (a % x != 0) continue;
        int need = a / x;
        if (cnt.count(need)) ans += cnt[need];
      }
  }
  cout << ans << endl;
  return 0;
}
