#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int inf = 99999999;
int main() {
  long long int n, k;
  while (~scanf("%lld%lld", &n, &k)) {
    long long int ans = 0;
    for (long long int i = 1; i <= n; ++i) {
      if (k > n * i) continue;
      if (k % i == 0) ans++;
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

Input

The single line contains numbers n and x (1 ≤ n ≤ 105, 1 ≤ x ≤ 109) — the size of the table and the number that we are looking for in the table.

Output

Print a single number: the number of times x occurs in the table.

Examples

Input

10 5


Output

2


Input

6 12


Output

4


Input

5 13


Output

0

Note

A table for the second sample test is given below. The occurrences of number 12 are marked bold. 

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int inf = 99999999;
int main() {
  long long int n, k;
  while (~scanf("%lld%lld", &n, &k)) {
    long long int ans = 0;
    for (long long int i = 1; i <= n; ++i) {
      if (k > n * i) continue;
      if (k % i == 0) ans++;
    }
    printf("%lld\n", ans);
  }
}

```