Problem: Today Pari and Arya are playing a game called Remainders.

Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value <image>. There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya <image> if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value <image> for any positive integer x?

Note, that <image> means the remainder of x after dividing it by y.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).

Output

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.

Examples

Input

4 5
2 3 5 12


Output

Yes


Input

2 7
2 3


Output

No

Note

In the first sample, Arya can understand <image> because 5 is one of the ancient numbers.

In the second sample, Arya can't be sure what <image> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
void maximize(T& x, T y) {
  if (x < y) x = y;
}
template <class T>
void minimize(T& x, T y) {
  if (x > y) x = y;
}
template <typename t>
ostream& operator,(ostream& os, const t& x) {
  return os << ' ' << x;
}
template <class T>
ostream& operator<<(ostream& os, vector<T>& x) {
  for (int i = 0; i < x.size(); i++) os << x[i] << ' ';
  os << '\n';
  return os;
}
vector<int> divisor;
vector<int> a;
int n, k;
bool check[105];
void factorize() {
  for (long long i = 2; i * i <= k; i++) {
    if (k % i == 0) {
      int cur = 1;
      while (k % i == 0) {
        cur *= i;
        k /= i;
      }
      divisor.push_back(cur);
    }
  }
  if (k != 1) divisor.push_back(k);
}
void solve() {
  for (int i = 0; i < a.size(); i++) {
    for (int j = 0; j < divisor.size(); j++) {
      if (check[j] || a[i] % divisor[j] != 0) continue;
      check[j] = 1;
    }
  }
  for (int i = 0; i < divisor.size(); i++) {
    if (!check[i]) {
      cout << "No";
      exit(0);
    }
  }
  cout << "Yes";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  factorize();
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    a.push_back(t);
  }
  solve();
}