#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long lxt() {
  long long x;
  cin >> x;
  return x;
}
double dxt() {
  double x;
  cin >> x;
  return x;
}
long double ldxt() {
  long double x;
  cin >> x;
  return x;
}
bool cmp(const pair<int, int> &p, const pair<int, int> &q) {
  if (p.first < q.first)
    return 1;
  else if (p.first == q.first)
    return (p.second < q.second);
  else
    return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, x;
  cin >> n >> x;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (x % i == 0 && x / i <= n) {
      cnt++;
    }
  }
  cout << cnt << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long lxt() {
  long long x;
  cin >> x;
  return x;
}
double dxt() {
  double x;
  cin >> x;
  return x;
}
long double ldxt() {
  long double x;
  cin >> x;
  return x;
}
bool cmp(const pair<int, int> &p, const pair<int, int> &q) {
  if (p.first < q.first)
    return 1;
  else if (p.first == q.first)
    return (p.second < q.second);
  else
    return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, x;
  cin >> n >> x;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (x % i == 0 && x / i <= n) {
      cnt++;
    }
  }
  cout << cnt << endl;
  return 0;
}

```