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 long long M = 1e9 + 7;
const long long INF = 1e9;
inline long long pwr(long long base, long long n, long long m) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans = (ans * base) % m;
    base = (base * base) % m;
    n /= 2;
  }
  return ans;
}
unsigned long long a, t;
string s;
map<unsigned long long, unsigned long long> cnt;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> a;
  cin >> s;
  int l = s.length();
  unsigned long long ans = 0, z = 0;
  for (int i = 0; i < l; i++) {
    t = 0;
    for (int j = i; j < l; j++) {
      t += (unsigned long long)s[j] - '0';
      if (t > a) break;
      if (t != 0 && a % t == 0) cnt[t]++;
      if (t == 0 && a == 0) {
        z++;
        ans += ((l + 1) * l) / 2;
      }
    }
  }
  if (a == 0) {
    for (int i = 0; i < l; i++) {
      t = 0;
      for (int j = i; j < l; j++) {
        t += (unsigned long long)s[j] - '0';
        if (t != 0) ans += z;
      }
    }
    cout << ans;
    return 0;
  }
  for (map<unsigned long long, unsigned long long>::iterator it = cnt.begin();
       it != cnt.end(); it++) {
    ans += ((*it).second * cnt[a / (*it).first]);
  }
  cout << ans;
  return 0;
}
