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;
int a;
const int maxn = 36001;
string s;
vector<int> find_divisors(int x) {
  vector<int> ans;
  int X = sqrt(x);
  for (int j = 1; j <= X; j++) {
    if (x % j == 0) {
      ans.push_back(j);
      if (j * j != x) ans.push_back(x / j);
    }
  }
  sort(ans.begin(), ans.end());
  return ans;
}
int cnt[maxn];
vector<int> divisors;
long long AnswerProblem;
;
int main() {
  cin >> a >> s;
  divisors = find_divisors(a);
  for (int i = 0; i < s.size(); i++) {
    int tsum = 0;
    for (int j = i; j < s.size(); j++) {
      tsum += s[j] - '0';
      cnt[tsum]++;
    }
  }
  if (a == 0) {
    for (int i = 1; i < maxn; i++) {
      AnswerProblem += cnt[i];
    }
    AnswerProblem *= cnt[0];
    AnswerProblem <<= 1;
    AnswerProblem += 1ll * cnt[0] * cnt[0];
    cout << AnswerProblem << endl;
    return 0;
  }
  for (int i = 0; i < int(divisors.size()); i++) {
    if (divisors[i] < maxn)
      if (a / divisors[i] < maxn) {
        AnswerProblem += 1ll * cnt[divisors[i]] * cnt[a / divisors[i]];
      }
  }
  cout << AnswerProblem << endl;
  return 0;
}
