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;
const int maxN = 100000;
long long sum[maxN], ans, a, is[maxN * 10], n, mx;
int main() {
  cin >> a;
  string s;
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    sum[i] = s[i] - '0';
    if (i) sum[i] += sum[i - 1];
    mx = max(mx, sum[i]);
  }
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      long long now = sum[j];
      if (i) now -= sum[i - 1];
      is[now]++;
    }
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      long long now = sum[j];
      if (i) now -= sum[i - 1];
      if (now)
        if (a % now == 0 && a / now <= mx) ans += is[a / now];
      if (!now)
        if (!a) ans += (n * (n + 1) / 2);
    }
  cout << ans << endl;
}
