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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int ans = 0;
  long long int a, mat[4005] = {0}, count[100005] = {0};
  string s;
  cin >> a >> s;
  long long int n = s.length();
  for (int i = 0; i < int(n); ++i) mat[i + 1] = 1LL * (s[i] - '0');
  for (int i = 0; i < int(n); ++i) mat[i + 1] += mat[i];
  for (int i = 0; i < int(n); ++i)
    for (int j = i + 1; j <= n; j++) count[mat[j] - mat[i]]++;
  for (int i = 0; i < int(100000); ++i)
    if (a % (i + 1) == 0 && (a / (i + 1)) <= 100000)
      ans += count[i + 1] * count[a / (i + 1)];
  if (a == 0) ans *= 2LL, ans += count[0] * count[0];
  cout << ans;
  return 0;
}
