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 =
    3.14159265358979323846264338327950288419716939937510582097494459230;
long long powm(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long ceil(const long long &a, const long long &b) {
  if (a % b == 0) return a / b;
  return a / b + 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a, n, i, j, ans = 0, sum;
  cin >> a;
  string second;
  cin >> second;
  n = second.length();
  map<long long, long long> mp;
  vector<long long> v;
  for (i = 0; i < n; i++) {
    sum = 0;
    for (j = i; j < n; j++) {
      sum += (second[j] - '0');
      mp[sum]++;
    }
  }
  ans = 0;
  for (int i = 0; i < n; i++) {
    sum = 0;
    for (int j = i; j < n; j++) {
      sum += second[j] - '0';
      if (sum != 0 && a % sum == 0) {
        ans += mp[a / sum];
      } else if (sum == 0 && a == 0) {
        ans += n * (n + 1) / 2;
      }
    }
  }
  cout << ans;
  return 0;
}
