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 unsigned long long P = 239017, MaxN = 2100000, INF = 1000000000;
int a, arr[1000000], sum[1000000], have[1000000];
char s[1000000];
int main() {
  scanf("%d%s", &a, s);
  int len = strlen(s);
  for (int i = 0; i < len; ++i) arr[i] = s[i] - '0';
  sum[0] = 0;
  for (int i = 1; i <= len; ++i) sum[i] = sum[i - 1] + arr[i - 1];
  int S = 0;
  for (int i = 1; i <= len; ++i)
    for (int j = 1; j <= i; ++j) ++have[sum[i] - sum[j - 1]], ++S;
  long long ans = 0;
  for (int i = 1; i <= len; ++i)
    for (int j = 1; j <= i; ++j) {
      if ((sum[i] - sum[j - 1]) && a % (sum[i] - sum[j - 1]) == 0) {
        if ((a / (sum[i] - sum[j - 1])) < 1000000)
          ans += have[a / (sum[i] - sum[j - 1])];
      }
      if (a == 0 && (sum[i] - sum[j - 1]) == 0) ans += S;
    }
  printf("%I64d", ans);
  return 0;
}
