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 maxN = 4e3 + 2, maxX = 9 * 4e3 + 1;
int n;
long long int a, accSum[maxN], memo[maxN][maxN], cnt[maxX];
char s[maxN];
int main() {
  scanf("%lld", &a);
  scanf(" %s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) accSum[i + 1] = (s[i] - '0') + accSum[i];
  if (0)
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        printf("%d%c", (s[i] - '0') * (s[j] - '0'), j < n - 1 ? ' ' : '\n');
  long long int zeros = 0;
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      memo[i][j] = accSum[j + 1] - accSum[i];
      cnt[accSum[j + 1] - accSum[i]]++;
    }
  long long int ans = 0;
  if (a) {
    for (int sum = 1; sum < maxX; sum++)
      if (sum != 0 && a % sum == 0 && a / sum < maxX)
        ans += cnt[sum] * cnt[a / sum];
  } else {
    ans += cnt[0] * cnt[0];
    for (int sum = 1; sum < maxX; sum++) ans += cnt[0] * cnt[sum] * 2LL;
  }
  printf("%lld\n", ans);
  return 0;
}
