#include <bits/stdc++.h>
using namespace std;
int main() {
  int N, x;
  cin >> N >> x;
  int s1[100010] = {0};
  if (x == 1)
    cout << "1" << endl;
  else {
    int sum = 0;
    for (int i = 2; i <= N; i++) {
      int mid = i;
      if (s1[mid] == 0) {
        while (mid <= N) {
          if (x / mid <= N && x % mid == 0 && s1[mid] == 0) {
            sum++;
            s1[mid] = 1;
            mid += i;
          } else {
            s1[mid] = 1;
            mid += i;
          }
        }
      }
    }
    if (N >= x) sum++;
    cout << sum << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int main() {
  int N, x;
  cin >> N >> x;
  int s1[100010] = {0};
  if (x == 1)
    cout << "1" << endl;
  else {
    int sum = 0;
    for (int i = 2; i <= N; i++) {
      int mid = i;
      if (s1[mid] == 0) {
        while (mid <= N) {
          if (x / mid <= N && x % mid == 0 && s1[mid] == 0) {
            sum++;
            s1[mid] = 1;
            mid += i;
          } else {
            s1[mid] = 1;
            mid += i;
          }
        }
      }
    }
    if (N >= x) sum++;
    cout << sum << endl;
  }
  return 0;
}

```