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>
#pragma GCC optimize("03")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
const double pi = acos(-1);
const long long MODPRIMA = (long long)1e9 + 7;
const long long MAXX = (long long)1e18;
const long long MINN = -(long long)1e18;
const double EPS = 1e-9;
int a;
string s;
long long ans;
map<int, int> frek;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> s;
  for (int i = 0; i < (int)s.size(); i++) {
    int sum = 0;
    for (int j = i; j < (int)s.size(); j++) {
      sum += (s[j] - '0');
      frek[sum]++;
    }
  }
  if (a > 0) {
    for (int i = 1; i <= (int)sqrt(a); i++) {
      if (a % i == 0) {
        int j = a / i;
        ans += 1LL * frek[i] * frek[j];
        if (j != i) ans += 1LL * frek[j] * frek[i];
      }
    }
    cout << ans << '\n';
  } else {
    ans = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      int sum = 0;
      for (int j = i; j < (int)s.size(); j++) {
        sum += (s[j] - '0');
        if (sum == 0) ans += 1LL * ((int)s.size() * ((int)s.size() + 1));
      }
    }
    ans -= 1LL * frek[0] * frek[0];
    cout << ans << '\n';
  }
  return 0;
}
