#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O2")
const long long MOD = 1e9 + 7;
const double PI = acos(-1);
long long add(long long x, long long y) {
  long long res = x + y;
  return (res >= MOD ? res - MOD : res);
}
long long mul(long long x, long long y) {
  long long res = x * y;
  return (res >= MOD ? res % MOD : res);
}
long long sub(long long x, long long y) {
  long long res = x - y;
  return (res < 0 ? res + MOD : res);
}
long long power(long long a, long long b, long long m = MOD) {
  long long ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1ll * a * ans) % m;
    b >>= 1;
    a = (1ll * a * a) % m;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isprime(long long n) {
  if (n < 2) return 0;
  long long i = 2;
  while (i * i <= n) {
    if (n % i == 0) return 0;
    i++;
  }
  return 1;
}
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
double distform(long long x, long long y, long long z, long long w) {
  return sqrt(1. * pow(x - z, 2) + 1. * pow(y - w, 2));
}
long long dx[] = {-1, 1, 0, 0, -1, 1, 1, -1};
long long dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
const long long MAXN = 2e5 + 10;
const long long inf = 1e18;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long intlog(double base, double x) {
  return (long long)(log(x) / log(base));
}
bool isvalid(long long x, long long y, long long n, long long m) {
  return x >= 0 and x < n and y >= 0 and y < m;
}
template <typename T>
void read(T &x) {
  for (long long i = 0; i < ((long long)(x).size()); ++i) cin >> x[i];
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void solve() {
  long long n, x;
  cin >> n >> x;
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    if (x % i == 0 and x / i <= n) {
      ans++;
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

### 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;
#pragma GCC optimize("O2")
const long long MOD = 1e9 + 7;
const double PI = acos(-1);
long long add(long long x, long long y) {
  long long res = x + y;
  return (res >= MOD ? res - MOD : res);
}
long long mul(long long x, long long y) {
  long long res = x * y;
  return (res >= MOD ? res % MOD : res);
}
long long sub(long long x, long long y) {
  long long res = x - y;
  return (res < 0 ? res + MOD : res);
}
long long power(long long a, long long b, long long m = MOD) {
  long long ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1ll * a * ans) % m;
    b >>= 1;
    a = (1ll * a * a) % m;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isprime(long long n) {
  if (n < 2) return 0;
  long long i = 2;
  while (i * i <= n) {
    if (n % i == 0) return 0;
    i++;
  }
  return 1;
}
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
double distform(long long x, long long y, long long z, long long w) {
  return sqrt(1. * pow(x - z, 2) + 1. * pow(y - w, 2));
}
long long dx[] = {-1, 1, 0, 0, -1, 1, 1, -1};
long long dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
const long long MAXN = 2e5 + 10;
const long long inf = 1e18;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long intlog(double base, double x) {
  return (long long)(log(x) / log(base));
}
bool isvalid(long long x, long long y, long long n, long long m) {
  return x >= 0 and x < n and y >= 0 and y < m;
}
template <typename T>
void read(T &x) {
  for (long long i = 0; i < ((long long)(x).size()); ++i) cin >> x[i];
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void solve() {
  long long n, x;
  cin >> n >> x;
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    if (x % i == 0 and x / i <= n) {
      ans++;
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

```