#include <bits/stdc++.h>
using namespace std;
long long ceill(long long x, long long y) {
  if (x % y == 0) {
    return x / y;
  }
  return (x / y) + 1;
}
long long prime(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n /= 2;
  }
  for (long long i = 3; i < sqrt(n); i += 1) {
    while (n % i == 0) {
      v.push_back(i);
      n /= i;
    }
  }
  if (n > 2) {
    v.push_back(n);
  }
  return v.size();
}
int min_sum(int n) {
  if (n == 0)
    return 0;
  else if (n == 1)
    return 2;
  return (n + 1 + min_sum(n / 2) + min_sum(n - n / 2 - 1));
}
void printDivisors(long long n, long long m) {
  vector<long long> v;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        v.push_back(i);
      } else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  long long cnt = 0;
  for (auto x : v) {
    if (x > m) {
      cnt++;
    }
  }
  if ((v.size() - (2 * cnt)) < 0)
    cout << 0 << "\n";
  else
    cout << v.size() - (2 * cnt) << "\n";
}
void body() {
  long long n, x;
  cin >> n >> x;
  long long d = n * n;
  printDivisors(x, n);
}
long long b_Search(long long *a, long long l, long long r, long long x) {
  if (l > r) {
    if (r == -1) {
      return -2;
    } else {
      return min(l + 1, r - 1);
    }
  }
  int m = ((l + r) / 2);
  if (a[m] <= x) {
    if (m == 0 || a[m + 1] != x) {
      return m + 1;
    }
    return b_Search(a, m + 1, r, x);
  }
  if (x > a[m]) {
    return b_Search(a, m + 1, r, x);
  }
  return b_Search(a, l, m - 1, x);
}
int main() {
  body();
  return 0;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
long long ceill(long long x, long long y) {
  if (x % y == 0) {
    return x / y;
  }
  return (x / y) + 1;
}
long long prime(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n /= 2;
  }
  for (long long i = 3; i < sqrt(n); i += 1) {
    while (n % i == 0) {
      v.push_back(i);
      n /= i;
    }
  }
  if (n > 2) {
    v.push_back(n);
  }
  return v.size();
}
int min_sum(int n) {
  if (n == 0)
    return 0;
  else if (n == 1)
    return 2;
  return (n + 1 + min_sum(n / 2) + min_sum(n - n / 2 - 1));
}
void printDivisors(long long n, long long m) {
  vector<long long> v;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        v.push_back(i);
      } else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  long long cnt = 0;
  for (auto x : v) {
    if (x > m) {
      cnt++;
    }
  }
  if ((v.size() - (2 * cnt)) < 0)
    cout << 0 << "\n";
  else
    cout << v.size() - (2 * cnt) << "\n";
}
void body() {
  long long n, x;
  cin >> n >> x;
  long long d = n * n;
  printDivisors(x, n);
}
long long b_Search(long long *a, long long l, long long r, long long x) {
  if (l > r) {
    if (r == -1) {
      return -2;
    } else {
      return min(l + 1, r - 1);
    }
  }
  int m = ((l + r) / 2);
  if (a[m] <= x) {
    if (m == 0 || a[m + 1] != x) {
      return m + 1;
    }
    return b_Search(a, m + 1, r, x);
  }
  if (x > a[m]) {
    return b_Search(a, m + 1, r, x);
  }
  return b_Search(a, l, m - 1, x);
}
int main() {
  body();
  return 0;
}

```