#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
long long a[N], b[N], c, d, dp, dis;
char o;
pair<int, int> p;
bool mark, h;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t, n, m, k = 0, f = 0, sum = 0, ans = 0;
  cin >> n >> m;
  for (int i = 1; i <= sqrt(m); i++) {
    if (m % i == 0 && i <= n && (m / i) <= n) ans += 2;
  }
  f = sqrt(m);
  if ((f * f) == m && f <= n) {
    ans--;
  }
  cout << ans << endl;
}

### 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 N = 2e5 + 5;
const int mod = 1e9 + 7;
long long a[N], b[N], c, d, dp, dis;
char o;
pair<int, int> p;
bool mark, h;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t, n, m, k = 0, f = 0, sum = 0, ans = 0;
  cin >> n >> m;
  for (int i = 1; i <= sqrt(m); i++) {
    if (m % i == 0 && i <= n && (m / i) <= n) ans += 2;
  }
  f = sqrt(m);
  if ((f * f) == m && f <= n) {
    ans--;
  }
  cout << ans << endl;
}

```