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 len(int a) {
  int r = 0;
  while (a) {
    r++;
    a /= 10;
  }
  return r;
}
long long powmod(long long x, int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return x;
  else if (n % 2 == 1)
    return x * powmod((x * x) % 1000000007, (n - 1) / 2) % 1000000007;
  else
    return powmod((x * x) % 1000000007, n / 2) % 1000000007;
}
int main() {
  int t = 1;
  while (t--) {
    int n;
    scanf("%d", &n);
    char second[5000];
    scanf("%s", second);
    long long ans = 0;
    int l = strlen(second);
    map<int, long long> a;
    for (int i = 0; i < l; i++) {
      int c = 0;
      for (int j = i; j < l; j++) {
        c += (second[j] - '0');
        a[c]++;
      }
    }
    map<int, long long>::iterator it;
    if (n == 0) {
      ans += a[0] * a[0];
      long long x = a[0];
      a.erase(0);
      long long second = 0;
      for (it = a.begin(); it != a.end(); it++) second += ((*it).second);
      ans += 2 * second * x;
      cout << (ans);
      return 0;
    }
    a.erase(0);
    for (it = a.begin(); it != a.end(); it++) {
      if (n % ((*it).first) == 0 && (a.count(n / ((*it).first))))
        ans += (a[n / ((*it).first)] * ((*it).second));
    }
    cout << ans;
  }
  return 0;
}
