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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pw(long long a, long long b, long long p) {
  if (b == 0) return 1;
  long long t = pw(a, b / 2, p);
  if (b % 2)
    return (((a * t) % p) * t) % p;
  else
    return ((t * t) % p);
}
long long z[4004];
set<long long> fact;
map<long long, long long> m;
void solve() {
  long long n;
  string s;
  cin >> n >> s;
  for (long long i = 0; i < s.size(); ++i) {
    z[i + 1] = s[i] - '0';
    z[i + 1] += z[i];
  }
  if (n)
    for (long long i = 1; i * i <= n; ++i) {
      if (n % i == 0) {
        fact.insert(i);
        fact.insert(n / i);
      }
    }
  else
    fact.insert(0);
  for (long long i = 1; i <= s.size(); ++i) {
    for (long long j = i; j <= s.size(); ++j) {
      long long sum = z[j] - z[i - 1];
      if (fact.find(sum) != fact.end()) m[sum]++;
    }
  }
  long long ans = 0;
  if (n)
    for (auto z : m) {
      long long a = z.first, b = z.second;
      if (a * a == n) {
        ans += b * b;
      } else {
        ans += b * m[n / a];
      }
    }
  else {
    ans = (s.size() * (1 + s.size()) - m[0]) * m[0];
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  long long t = 1;
  while (t--) solve();
  return 0;
}
