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 comment(linker, "/stack:30000000")
using namespace std;
const int mx[4] = {1, -1, 0, 0};
const int my[4] = {0, 0, 1, -1};
const int Direct[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
const int Nmax = 200000;
long long Count[Nmax];
long long p[Nmax];
int main() {
  ios_base::sync_with_stdio(0);
  string s;
  long long a;
  cin >> a;
  getline(cin, s);
  getline(cin, s);
  s = '#' + s;
  for (int i = 1; i < s.size(); i++) p[i] = p[i - 1] + (s[i] - '0');
  for (int l = 1; l < s.size(); l++)
    for (int r = l; r < s.size(); r++) Count[p[r] - p[l - 1]]++;
  long long answer = 0;
  for (long long d = 1; d * d <= a; d++)
    if (a % d == 0) {
      long long Res;
      if (d >= Nmax || a / d >= Nmax)
        Res = 0;
      else
        Res = (long long)Count[d] * Count[a / d];
      answer += (long long)Res;
      if (d * d == a) continue;
      answer += Res;
    }
  if (a == 0) {
    int n = s.size() - 1;
    answer = (long long)Count[0] * (n * (long long)(n + 1) - Count[0]);
  }
  cout << answer << endl;
  return 0;
}
