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;
vector<size_t> primes(int n) {
  vector<size_t> res;
  vector<bool> mask(n, true);
  int j = 2;
  for (; j * j < n; ++j) {
    if (mask[j]) {
      res.push_back(j);
      for (int k = j * j; k < n; k += j) {
        mask[k] = false;
      }
    }
  }
  for (; j < n; ++j) {
    if (mask[j]) res.push_back(j);
  }
  return res;
}
list<pair<size_t, size_t>> factorization(const vector<size_t>& primes,
                                         size_t k) {
  list<pair<size_t, size_t>> res;
  for (auto p : primes) {
    size_t f = 1;
    while (k && k % p == 0) {
      f *= p;
      k /= p;
    }
    if (f > 1) {
      res.push_back(make_pair(p, f));
    }
  }
  if (k > 1) {
    res.push_back(make_pair(k, k));
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  auto p = primes(1000);
  size_t n, k;
  std::cin >> n >> k;
  auto fact = factorization(p, k);
  vector<size_t> old(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> old[i];
  }
  for (auto next : old) {
    if (fact.empty()) {
      std::cout << "Yes";
      return 0;
    }
    for (auto it = fact.begin(); it != fact.end();) {
      if (next % it->second == 0) {
        fact.erase(it++);
      } else {
        ++it;
      }
    }
  }
  std::cout << (fact.empty() ? "Yes" : "No");
  return 0;
}