#include <bits/stdc++.h>
using namespace std;
void imprimirVector(vector<int> &v) {
  if (!v.empty()) {
    int p = v.size();
    cout << "[";
    for (int i = 0; i < (int)(p - 1); i++) cout << v[i] << ",";
    cout << v[p - 1] << "]" << endl;
  } else
    cout << "[]" << endl;
}
int main() {
  unsigned long long N, X;
  cin >> N >> X;
  unsigned long long count = 0;
  for (int i = (1); i < (int)(N + 1); i++) {
    if (X % i == 0) {
      if (i * N >= X) count++;
    }
  }
  cout << count << endl;
  return 0;
}

### Prompt

Generate 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;
void imprimirVector(vector<int> &v) {
  if (!v.empty()) {
    int p = v.size();
    cout << "[";
    for (int i = 0; i < (int)(p - 1); i++) cout << v[i] << ",";
    cout << v[p - 1] << "]" << endl;
  } else
    cout << "[]" << endl;
}
int main() {
  unsigned long long N, X;
  cin >> N >> X;
  unsigned long long count = 0;
  for (int i = (1); i < (int)(N + 1); i++) {
    if (X % i == 0) {
      if (i * N >= X) count++;
    }
  }
  cout << count << endl;
  return 0;
}

```