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 MAXN = 4e3 + 1;
int N;
long long targetSum, cumSum[MAXN];
string s;
map<long long, long long> haveCumSumDiff;
void findCumsum() {
  cumSum[0] = 0;
  for (int i = 1; i <= N; i++) cumSum[i] = cumSum[i - 1] + (s[i - 1] - '0');
}
long long solve() {
  findCumsum();
  long long totalCumSumDiff = 0;
  for (int i = 0; i <= N; i++)
    for (int j = i + 1; j <= N; j++) {
      haveCumSumDiff[cumSum[j] - cumSum[i]]++;
      totalCumSumDiff++;
    }
  long long ans = 0LL;
  for (int i = 0; i <= N; i++)
    for (int j = i + 1; j <= N; j++) {
      long long curCumSumDiff = cumSum[j] - cumSum[i];
      if (targetSum == 0) {
        if (curCumSumDiff == 0)
          ans += totalCumSumDiff;
        else
          ans += haveCumSumDiff[0];
      } else {
        if (curCumSumDiff == 0) continue;
        if (targetSum % curCumSumDiff) continue;
        if (not haveCumSumDiff.count(targetSum / curCumSumDiff)) continue;
        ans += haveCumSumDiff[targetSum / curCumSumDiff];
      }
    }
  return ans;
}
int main() {
  cin >> targetSum;
  cin >> s;
  N = s.size();
  cout << solve() << endl;
  return 0;
}
