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 double PI = 3.14159265358979323846;
long long a, z[4005], len, lr[4005 * 4005], e = 0, sum = 0;
string s;
vector<long long> del;
void ans_0() {
  long long X = 0, W = 0, cnt = 0;
  for (int i = 1; i <= len; ++i) {
    W += i;
    if (s[i - 1] == '0')
      ++cnt;
    else
      while (cnt) {
        X += cnt;
        --cnt;
      }
  }
  while (cnt) {
    X += cnt;
    --cnt;
  }
  cout << (2 * W - X) * X;
}
long long BS(long long x) {
  long long l = 0, r = e, mid;
  while (r - l > 1) {
    mid = l + (r - l) / 2;
    if (lr[mid] < x)
      l = mid;
    else
      r = mid;
  }
  long long cnt = 0, ii = l;
  if (lr[ii] != x && l < e - 1) ++ii;
  for (; lr[ii] == x && ii < e; ++ii, ++cnt)
    ;
  return cnt;
}
int main() {
  cin >> a >> s;
  len = s.size();
  if (a == 0) {
    ans_0();
    return 0;
  }
  for (long long i = 0; i < len; ++i) z[i] = s[i] - '0';
  for (long long i = 0; i < len; ++i)
    for (int j = i; j < len; ++j, ++e)
      if (j == i)
        lr[e] = z[j];
      else
        lr[e] = lr[e - 1] + z[j];
  sort(lr, lr + e);
  for (long long i = 1; i * i <= a; ++i)
    if (a % i == 0) {
      long long t = BS(i) * BS(a / i);
      if (i * i != a) t <<= 1;
      sum += t;
    }
  cout << sum;
}
