#include <bits/stdc++.h>
using namespace std;
vector<long long int> seve(1000001, 0);
void create_seve() {
  seve[0] = 1;
  seve[1] = 1;
  for (long long int i = 2; i <= 1000000; i++) {
    for (long long int j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        seve[i] = 1;
        break;
      }
    }
  }
}
bool is_prime(long long int n) { return seve[n]; }
vector<pair<long long int, long long int>> prime_factorization(
    long long int n) {
  vector<pair<long long int, long long int>> pf;
  for (long long int i = 2; i * i <= n; i++) {
    long long int c = 0;
    while (n % i == 0) {
      c++;
      n /= i;
    }
    if (c != 0) pf.push_back(make_pair(i, c));
  }
  if (n > 1) pf.push_back(make_pair(n, 1));
  return pf;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int power(long long int a, long long int n) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n = n >> 1;
  }
  return ans;
}
long long int modulo_power(long long int a, long long int n, long long int m) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    n = n >> 1;
  }
  return ans % m;
}
long long int sumofdig(long long int ds) {
  long long int ans = 0;
  while (ds) {
    ans += ds % 10;
    ds = ds / 10;
  }
  return ans;
}
void solve2() {}
void solve3() {
  long long int n;
  cin >> n;
  string s, s1 = "abacaba";
  cin >> s;
  long long int p = 0;
  for (long long int i = 0; i < n - 6; i++) {
    string a = s.substr(i, 7);
    if (a == s1) p++;
  }
  if (p == 1) {
    cout << "YES" << endl;
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '?')
        cout << 'd';
      else
        cout << s[i];
    }
    cout << endl;
    return;
  }
  if (p > 1) {
    cout << "NO" << endl;
    return;
  }
  for (long long int i = 0; i < n - 6; i++) {
    string a = s.substr(i, 7);
    bool f = true;
    string ans = "";
    for (long long int j = 0; j < 7; j++) {
      if (a[j] == s1[j]) {
        ans = ans + s1[j];
      } else if (a[j] == '?') {
        ans = ans + s1[j];
      } else {
        f = false;
        break;
      }
    }
    if (f) {
      cout << "YES" << endl;
      for (long long int j = 0; j < i; j++) {
        if (s[j] == '?')
          cout << 'd';
        else
          cout << s[j];
      }
      cout << ans;
      for (long long int j = i + 7; j < n; j++) {
        if (s[j] == '?')
          cout << 'd';
        else
          cout << s[j];
      }
      cout << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
void solve1() {}
void solve() {
  long long int n, x, ans = 0;
  cin >> n >> x;
  for (long long int i = 1; i <= n; i++) {
    if (x % i == 0) {
      if (x / i <= n) {
        ans++;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<long long int> seve(1000001, 0);
void create_seve() {
  seve[0] = 1;
  seve[1] = 1;
  for (long long int i = 2; i <= 1000000; i++) {
    for (long long int j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        seve[i] = 1;
        break;
      }
    }
  }
}
bool is_prime(long long int n) { return seve[n]; }
vector<pair<long long int, long long int>> prime_factorization(
    long long int n) {
  vector<pair<long long int, long long int>> pf;
  for (long long int i = 2; i * i <= n; i++) {
    long long int c = 0;
    while (n % i == 0) {
      c++;
      n /= i;
    }
    if (c != 0) pf.push_back(make_pair(i, c));
  }
  if (n > 1) pf.push_back(make_pair(n, 1));
  return pf;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int power(long long int a, long long int n) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n = n >> 1;
  }
  return ans;
}
long long int modulo_power(long long int a, long long int n, long long int m) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    n = n >> 1;
  }
  return ans % m;
}
long long int sumofdig(long long int ds) {
  long long int ans = 0;
  while (ds) {
    ans += ds % 10;
    ds = ds / 10;
  }
  return ans;
}
void solve2() {}
void solve3() {
  long long int n;
  cin >> n;
  string s, s1 = "abacaba";
  cin >> s;
  long long int p = 0;
  for (long long int i = 0; i < n - 6; i++) {
    string a = s.substr(i, 7);
    if (a == s1) p++;
  }
  if (p == 1) {
    cout << "YES" << endl;
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '?')
        cout << 'd';
      else
        cout << s[i];
    }
    cout << endl;
    return;
  }
  if (p > 1) {
    cout << "NO" << endl;
    return;
  }
  for (long long int i = 0; i < n - 6; i++) {
    string a = s.substr(i, 7);
    bool f = true;
    string ans = "";
    for (long long int j = 0; j < 7; j++) {
      if (a[j] == s1[j]) {
        ans = ans + s1[j];
      } else if (a[j] == '?') {
        ans = ans + s1[j];
      } else {
        f = false;
        break;
      }
    }
    if (f) {
      cout << "YES" << endl;
      for (long long int j = 0; j < i; j++) {
        if (s[j] == '?')
          cout << 'd';
        else
          cout << s[j];
      }
      cout << ans;
      for (long long int j = i + 7; j < n; j++) {
        if (s[j] == '?')
          cout << 'd';
        else
          cout << s[j];
      }
      cout << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
void solve1() {}
void solve() {
  long long int n, x, ans = 0;
  cin >> n >> x;
  for (long long int i = 1; i <= n; i++) {
    if (x % i == 0) {
      if (x / i <= n) {
        ans++;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}

```