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>
int n;
using namespace std;
vector<long long> c;
long long sum(long long l, long long r) {
  return c[r] - ((l - 1) < 0 ? 0 : c[l - 1]);
}
signed main() {
  long long a;
  cin >> a;
  string s;
  cin >> s;
  n = s.length();
  long long b[n];
  for (long long i = 0; i < n; i++) {
    b[i] = s[i] - '0';
  }
  c.resize(n);
  c[0] = b[0];
  for (long long i = 1; i < n; i++) {
    c[i] = c[i - 1] + b[i];
  }
  vector<long long> cnt(c[n - 1] + 1);
  for (long long i = 0; i <= c[n - 1]; i++) {
    cnt[i] = 0;
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      cnt[sum(i, j)]++;
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      long long now = sum(i, j);
      if (!now && a) {
        continue;
      }
      if (!now) {
        ans += n * (n + 1) / 2;
        continue;
      }
      if (a % now == 0) {
        long long need = a / now;
        if (need <= c[n - 1]) {
          ans += cnt[need];
        }
      }
    }
  }
  cout << ans << endl;
}
