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;
using LL = long long;
using VI = vector<int>;
using VL = vector<long long>;
using VVI = vector<VI>;
using VVL = vector<VL>;
using MII = map<int, int>;
using MIVI = map<int, VI>;
using MLL = map<LL, LL>;
int READ_INT() {
  int temp;
  cin >> temp;
  return temp;
}
LL READ_LONG() {
  LL temp;
  cin >> temp;
  return temp;
}
string READ_STRING() {
  string temp;
  cin >> temp;
  return temp;
}
const int MOD = int(1e9) + 7;
const int INF = 1e9 + 5;
const double PI = acos(-1.0);
const double EPS = 1e-9;
int n;
LL a;
string s;
map<int, int> subarray_sum_counter;
int main() {
  ios::sync_with_stdio(false);
  cin >> a >> s;
  n = s.size();
  VI arr(n);
  for (int i = 0; i < n; ++i) arr[i] = s[i] - '0';
  for (int i = 0; i < n; ++i) {
    int s = 0;
    for (int j = i; j < n; ++j) {
      s += arr[j];
      subarray_sum_counter[s] += 1;
    }
  }
  LL ans = 0;
  if (a != 0) {
    for (auto p : subarray_sum_counter) {
      LL x = p.first, f1 = p.second;
      if (x != 0 && a % x == 0) {
        LL y = a / x;
        LL f2 = subarray_sum_counter[y];
        ans += (f1 * f2);
      }
    }
  } else {
    LL f = subarray_sum_counter[0];
    LL N = (n * (n + 1)) / 2;
    ans = f * N + N * f - f * f;
  }
  cout << ans << "\n";
  return 0;
}
