Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i* × *j*. The rows and columns are numbered starting from 1.

You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
The single line contains numbers *n* and *x* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>, 1 ≤ *x* ≤ 10<sup class="upper-index">9</sup>) — the size of the table and the number that we are looking for in the table.
#include <bits/stdc++.h>
#define int long long
using namespace std;
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, x;
    cin >> n >> x;
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        /* code */
        if (!(x % i) && n * i >= x)
            ans++;
    }
    cout << ans << endl;
    return 0;
}