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 int ppr = 257;
const long long INF = 1e18;
const int inf = 1e9;
const int mod = 1e9 + 9;
const int N = 1e6 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
long long a, x[N], p[N], m[N], sum;
string second;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> a >> second;
  for (long long i = 0; i < second.size(); i++) x[i + 1] = second[i] - '0';
  long long n = second.size();
  for (long long i = 1; i <= n; i++) p[i] = p[i - 1] + x[i];
  for (long long r = 1; r <= n; r++)
    for (long long l = 1; l <= r; l++) {
      long long y = p[r] - p[l - 1];
      m[y]++;
    }
  if (a == 0) {
    for (long long r = 1; r <= n; r++)
      for (long long l = 1; l <= r; l++)
        if (p[r] - p[l - 1] == 0)
          sum += (n * (n + 1)) / 2;
        else
          sum += m[0];
  } else {
    for (long long r = 1; r <= n; r++)
      for (long long l = 1; l <= r; l++)
        if (p[r] - p[l - 1] != 0 && a % (p[r] - p[l - 1]) == 0 &&
            a / (p[r] - p[l - 1]) < 1000000)
          sum += m[a / (p[r] - p[l - 1])];
  }
  cout << sum;
  return 0;
}
