#include <bits/stdc++.h>
int n, x;
int len;
int ans[111111];
void fun(int num) {
  len = 0;
  int i = 2;
  while (num >= 2) {
    if (num % i == 0) {
      ans[len++] = i;
      num /= i;
    } else
      i++;
  }
}
int main() {
  int i, j;
  int count;
  while (~scanf("%d%d", &n, &x)) {
    count = 0;
    for (i = 1; i <= n; i++) {
      if (x % i == 0 && x / i <= n) {
        count++;
      }
    }
    printf("%d\n", count);
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
int n, x;
int len;
int ans[111111];
void fun(int num) {
  len = 0;
  int i = 2;
  while (num >= 2) {
    if (num % i == 0) {
      ans[len++] = i;
      num /= i;
    } else
      i++;
  }
}
int main() {
  int i, j;
  int count;
  while (~scanf("%d%d", &n, &x)) {
    count = 0;
    for (i = 1; i <= n; i++) {
      if (x % i == 0 && x / i <= n) {
        count++;
      }
    }
    printf("%d\n", count);
  }
  return 0;
}

```