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;
const long long N = 1000001;
vector<long long> isprime(N, true);
vector<long long> prime;
vector<long long> SPF(N);
int cnt[N] = {};
void manipulatedsieve() {
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < N; i++) {
    if (isprime[i]) {
      prime.push_back(i);
      SPF[i] = i;
    }
    for (long long int j = 0;
         j < prime.size() && i * prime[j] < N && prime[j] <= SPF[i]; j++) {
      isprime[i * prime[j]] = false;
      SPF[i * prime[j]] = prime[j];
    }
  }
}
int main() {
  manipulatedsieve();
  int n, k, p, flag = 1;
  scanf("%d %d", &n, &k);
  int c[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    while (c[i] > 1) {
      p = SPF[c[i]];
      int co = 0;
      while (c[i] % p == 0) {
        co++;
        c[i] /= p;
      }
      cnt[p] = max(cnt[p], co);
    }
  }
  while (k > 1) {
    p = SPF[k];
    int co = 0;
    while (k % p == 0) {
      co++;
      k /= p;
    }
    if (co > cnt[p]) flag = 0;
  }
  if (flag)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}