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;
char str[4005];
long long a, n, repSum[40005], accuSum[4005];
int main() {
  scanf("%lld%s", &a, str);
  n = (long long)strlen(str);
  for (long long(i) = (1); i <= (long long)(n); (i)++)
    accuSum[i] = (str[i - 1] - '0') + accuSum[i - 1];
  for (long long(l) = (1); l <= (long long)(n); (l)++)
    for (long long(i) = (1); i <= (long long)(n - l + 1); (i)++)
      ++repSum[accuSum[i + l - 1] - accuSum[i - 1]];
  long long ans = 0;
  if (!a) {
    ans = repSum[0] * (n * (n + 1) - repSum[0]);
  } else {
    for (long long(i) = (1); i <= (long long)(40000); (i)++)
      if (a % i == 0 && a / i < 40000) ans += repSum[i] * repSum[a / i];
  }
  printf("%lld\n", ans);
  return 0;
}
