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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
int main() {
  int a;
  scanf("%d", &a);
  char s[4001];
  scanf("%s", s);
  int n = strlen(s);
  int x[n + 1];
  x[0] = 0;
  for (auto i = 0; i < n; i++) {
    x[i + 1] = x[i] + s[i] - '0';
  }
  long long int y[n * 9 + 1];
  memset(y, 0, sizeof(y));
  for (auto i = 1; i < n + 1; i++) {
    for (auto j = i; j < n + 1; j++) {
      y[x[j] - x[i - 1]]++;
    }
  }
  long long int res = 0;
  for (auto i = 1; i < n * 9 + 1; i++) {
    if (a % i == 0 && a / i <= n * 9) {
      res += y[i] * y[a / i];
    }
  }
  if (a == 0) {
    res = 2 * res + y[0] * y[0];
  }
  printf("%lld\n", res);
  return 0;
}
