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;
#pragma comment(linker, "/STACK:1024000000,1024000000")
const int N = 30005, M = 100005;
const int inf = 0x3f3f3f3f;
const long long INF = 1000000000000000000ll;
const long long mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-6;
long long vis[4000 * 9 + 5];
long long sum[4005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  long long a;
  cin >> a;
  string str;
  cin >> str;
  int siz = str.size();
  sum[1] = str[0] - '0';
  for (int i = 2; i <= siz; i++) {
    sum[i] = str[i - 1] - '0' + sum[i - 1];
  }
  long long all = 0;
  for (int len = 1; len <= siz; len++) {
    for (int i = 1, j; i + len - 1 <= siz; i++) {
      j = i + len - 1;
      long long tmp = sum[j] - sum[i - 1];
      vis[tmp]++;
      all++;
    }
  }
  if (a == 0) {
    cout << 2 * vis[0] * (all - vis[0]) + vis[0] * vis[0] << endl;
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i * i <= a; i++) {
    if (vis[i] && a % i == 0 && a / i <= 36000 && vis[a / i]) {
      ans += vis[i] * vis[a / i] * (i * i == a ? 1 : 2);
    }
  }
  cout << ans << endl;
}
