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;
long long pref[4 * 1005];
class Solution {
 public:
  void solve(std::istream& in, std::ostream& out) {
    long long a;
    string s;
    in >> a >> s;
    for (int i = 0; i <= (int)s.size() - 1; i++) pref[i] = (s[i] - '0');
    for (int i = 1; i <= (int)s.size() - 1; i++) pref[i] += pref[i - 1];
    map<long long, int> num;
    for (int i = 0; i <= (int)s.size() - 1; i++) {
      for (int windowsiz = 1;
           windowsiz <= (int)s.size() && i + windowsiz - 1 < (int)s.size();
           windowsiz++) {
        int r = i + windowsiz - 1;
        long long sum;
        if (i == 0)
          sum = pref[r];
        else
          sum = pref[r] - pref[i - 1];
        num[sum]++;
      }
    }
    long long cnt = 0;
    long long totcnt = 0;
    for (auto itr : num) {
      totcnt += itr.second;
    }
    for (int i = 0; i <= (int)s.size() - 1; i++) {
      for (int windowsiz = 1;
           windowsiz <= (int)s.size() && i + windowsiz - 1 < (int)s.size();
           windowsiz++) {
        int r = i + windowsiz - 1;
        long long sum;
        if (i == 0)
          sum = pref[r];
        else
          sum = pref[r] - pref[i - 1];
        if (sum == 0) {
          if (a == 0) cnt += totcnt;
        } else if (a % sum == 0) {
          long long secsum = a / sum;
          cnt += num[secsum];
        }
      }
    }
    out << cnt << '\n';
    return;
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}
