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 MAX_BUF_SIZE = 16384;
char BUFOR[MAX_BUF_SIZE];
int BUF_SIZE, BUF_POS;
char ZZZ;
int _MINUS;
const int MXN = 2000010;
const int C = 262144;
const int INF = 1000000001;
int A;
char str[MXN];
int a[MXN];
int sum[MXN];
int ile[MXN];
void test() {
  scanf("%d", &A);
  scanf("%s", str + 1);
  int n = strlen(str + 1);
  for (int i = (1); i <= (n); i++) {
    a[i] = str[i] - '0';
    sum[i] = sum[i - 1] + a[i];
  }
  long long int res = 0;
  for (int i = (1); i <= (n); i++)
    for (int j = (i); j <= (n); j++) ile[sum[j] - sum[i - 1]]++;
  for (int i = (1); i <= (n); i++)
    for (int j = (i); j <= (n); j++) {
      int k = sum[j] - sum[i - 1];
      if (k == 0 && A == 0)
        res += n * (n + 1) / 2;
      else if (k > 0 && A % k == 0 && A / k < MXN)
        res += ile[A / k];
    }
  printf("%I64d\n", res);
}
int main() {
  int te = 1;
  while (te--) test();
  return 0;
}
