#include <bits/stdc++.h>
using namespace std;
void stringsplitter(string s, string delim, string arr[]) {
  char *ptr = strtok((char *)s.c_str(), (char *)delim.c_str());
  long long i = 0;
  while (ptr != NULL) {
    arr[i] = ((string)ptr);
    ptr = strtok(NULL, (char *)delim.c_str());
    i++;
  }
}
long long string_to_int(string s) {
  long long ans = 0;
  long long p = 1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    ans += (s[i] - '0') * p;
    p *= 10;
  }
  return ans;
}
void t_c_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a, b;
  cin >> a >> b;
  long long cnt = 0;
  for (long long i = 1; i <= a; i++) {
    if (b % i == 0 and (b / i) <= a) {
      cnt++;
    }
  }
  cout << cnt << endl;
}
int32_t main() {
  t_c_c();
  return 0;
}

### Prompt

Create a solution in Cpp for 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 stringsplitter(string s, string delim, string arr[]) {
  char *ptr = strtok((char *)s.c_str(), (char *)delim.c_str());
  long long i = 0;
  while (ptr != NULL) {
    arr[i] = ((string)ptr);
    ptr = strtok(NULL, (char *)delim.c_str());
    i++;
  }
}
long long string_to_int(string s) {
  long long ans = 0;
  long long p = 1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    ans += (s[i] - '0') * p;
    p *= 10;
  }
  return ans;
}
void t_c_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a, b;
  cin >> a >> b;
  long long cnt = 0;
  for (long long i = 1; i <= a; i++) {
    if (b % i == 0 and (b / i) <= a) {
      cnt++;
    }
  }
  cout << cnt << endl;
}
int32_t main() {
  t_c_c();
  return 0;
}

```