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:102400000,102400000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
const long long mod = 1e9 + 7;
const unsigned long long mx = 133333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
int c[1000005];
int prime[1000005];
int pr[1000005];
int tot;
void init() {
  memset(prime, 0, sizeof(prime));
  tot = 0;
  for (int i = 2; i <= 1000000; i++) {
    if (!prime[i]) {
      for (int j = 2 * i; j <= 1000000; j += i) prime[j] = 1;
      pr[tot++] = i;
    }
  }
}
int main() {
  int n, k;
  init();
  while (cin >> n >> k) {
    for (int i = 0; i < n; i++) scanf("%d", &c[i]);
    bool ans = true;
    for (int i = 0; i < tot && pr[i] * pr[i] <= k; i++) {
      if (k % pr[i] == 0) {
        int tmp = 1;
        while (k % pr[i] == 0) {
          k /= pr[i];
          tmp *= pr[i];
        }
        int flag = 0;
        for (int j = 0; j < n; j++) {
          if (c[j] % tmp == 0) {
            flag = 1;
            break;
          }
        }
        if (!flag) {
          ans = false;
          break;
        }
      }
    }
    if (k != 1) {
      int flag = 0;
      for (int j = 0; j < n; j++) {
        if (c[j] % k == 0) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        ans = false;
      }
    }
    if (ans)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}