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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 1e6 + 5;
int cnt[N];
vector<int> factors;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int temp = k;
  int p = 2;
  while (temp > 1) {
    int count = 0;
    while (temp % p == 0) {
      count++;
      temp /= p;
    }
    cnt[p] = count;
    p++;
  }
  for (int i = 0; i < N; i++)
    if (cnt[i] != 0) factors.push_back(i);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    for (int j = 0; j < factors.size(); j++) {
      int count = 0;
      while (x % factors[j] == 0) {
        x /= factors[j];
        count++;
      }
      if (count >= cnt[factors[j]]) cnt[factors[j]] = 0;
    }
  }
  int leftover = 0;
  for (int i = 0; i < factors.size(); i++) leftover += cnt[factors[i]];
  if (leftover == 0)
    puts("Yes");
  else
    puts("No");
  return 0;
}