#include <bits/stdc++.h>
using namespace std;
const long long mod1 = static_cast<long long>(1073741824);
const long long mod2 = static_cast<long long>(1000000007);
const long long INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long double ldOne = 1.0;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long exp(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n & 1)
    return x * exp(x * x, (n - 1) / 2);
  else
    return exp(x * x, n / 2);
}
bool isPrime(long long n) {
  for (long long i = 0; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
vector<long long> SieveOfEratosthenes(long long n) {
  vector<bool> prime(n + 1, true);
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p]) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  vector<long long> v;
  for (long long i = 2; i <= n; i++) {
    if (prime[i]) v.push_back(i);
  }
  return v;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x;
  cin >> n >> x;
  long long cnt = 0;
  for (long long i = 1; i < n + 1; i++) {
    if (x % i == 0 && x / i <= n) cnt++;
  }
  cout << cnt;
  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;
const long long mod1 = static_cast<long long>(1073741824);
const long long mod2 = static_cast<long long>(1000000007);
const long long INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long double ldOne = 1.0;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long exp(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n & 1)
    return x * exp(x * x, (n - 1) / 2);
  else
    return exp(x * x, n / 2);
}
bool isPrime(long long n) {
  for (long long i = 0; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
vector<long long> SieveOfEratosthenes(long long n) {
  vector<bool> prime(n + 1, true);
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p]) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  vector<long long> v;
  for (long long i = 2; i <= n; i++) {
    if (prime[i]) v.push_back(i);
  }
  return v;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x;
  cin >> n >> x;
  long long cnt = 0;
  for (long long i = 1; i < n + 1; i++) {
    if (x % i == 0 && x / i <= n) cnt++;
  }
  cout << cnt;
  return 0;
}

```