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 N = 100010;
string s;
int a, tmp;
vector<int> divisors, null;
int cnt_sum[N];
int main() {
  scanf("%d", &a);
  cin >> s;
  memset(cnt_sum, 0, sizeof(cnt_sum));
  for (int i = 1; i * i <= a; ++i) {
    if (a % i == 0) {
      divisors.push_back(i);
      if (a / i != i) divisors.push_back(a / i);
    }
  }
  for (int i = 0; i < int(s.size()); ++i) {
    int sum = 0;
    for (int j = i; j < int(s.size()); ++j) {
      sum += (s[j] - '0');
      cnt_sum[sum]++;
    }
  }
  long long ans = 0;
  for (int i = 0; i < int(divisors.size()); ++i) {
    ans += ((divisors[i] > N || a / divisors[i] > N)
                ? 0
                : 1ll * cnt_sum[divisors[i]] * cnt_sum[a / divisors[i]]);
  }
  if (a == 0) {
    ans = 0;
    int cnt = 0;
    int n = s.size();
    for (int i = 0; i < int(s.size()); ++i)
      if (s[i] == '0')
        cnt++;
      else {
        if (cnt != 0) null.push_back(cnt);
        cnt = 0;
      }
    null.push_back(cnt);
    for (int i = 0; i < null.size(); ++i) {
      ans += 2ll * n * (n + 1) * null[i] * (null[i] + 1) / 4;
      for (int j = 0; j < null.size(); ++j) {
        int tmp1 = null[i];
        int tmp2 = null[j];
        ans -= 1ll * tmp1 * tmp2 * (tmp1 + 1) * (tmp2 + 1) / 4;
      }
    }
  }
  printf("%I64d", ans);
  return 0;
}
