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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (1 << 29) + 5;
const long long int LLINF = (1ll << 59) + 5;
const int MOD = 1000 * 1000 * 1000 + 7;
const int MAX_N = 1000 * 1000 + 5;
int n, k;
int arr[MAX_N];
vector<pair<int, int> > primes;
int main() {
  scanf("%d %d", &n, &k);
  long long int mult = 1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
  }
  int cnt;
  for (int i = 2; i <= 1000000; ++i) {
    cnt = 0;
    while (k % i == 0) {
      k /= i;
      ++cnt;
    }
    if (cnt != 0) {
      primes.push_back(make_pair(i, cnt));
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < primes.size(); ++j) {
      cnt = 0;
      while (arr[i] % primes[j].first == 0) {
        arr[i] /= primes[j].first;
        ++cnt;
      }
      if (cnt >= primes[j].second) {
        primes[j].second = 0;
      }
    }
  }
  bool good = true;
  for (int i = 0; i < primes.size(); ++i) {
    good &= (primes[i].second == 0);
  }
  if (good) {
    printf("Yes");
  } else {
    printf("No");
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}