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 double pi = acos(-1.0);
const double eps = 1e-11;
int main() {
  std::ios_base::sync_with_stdio(0);
  long long a;
  string s;
  cin >> a >> s;
  long long sum[5111];
  sum[0] = (s[0] - '0');
  for (int i = 1; i < int((s).size()); ++i) {
    sum[i] = sum[i - 1] + (s[i] - '0');
  }
  map<long long, long long> cach;
  for (int i = 0; i < int((s).size()); ++i) {
    for (int j = i; j < int((s).size()); ++j) {
      if (i == 0) {
        cach[sum[j]]++;
      } else
        cach[sum[j] - sum[i - 1]]++;
    }
  }
  long long all = 0;
  for (__typeof((cach).begin()) it = (cach).begin(); it != (cach).end(); it++) {
    all += (it->second);
  }
  long long rs = 0;
  for (int i = 0; i < int((s).size()); ++i) {
    for (int j = i; j < int((s).size()); ++j) {
      long long sm;
      if (i == 0)
        sm = sum[j];
      else
        sm = sum[j] - sum[i - 1];
      if (sm == 0 && a == 0) {
        rs += (int((s).size()) * (int((s).size()) + 1)) / 2;
      } else if (sm > 0 && a % sm == 0) {
        if (cach.count(a / sm)) {
          rs += (cach[a / sm]);
        }
      }
    }
  }
  cout << rs << endl;
  return 0;
}
