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>
const int MaxN = 500, MaxV = 1000010;
int n, k, tot;
int prime[MaxV], val[MaxV], fac[MaxV], num[MaxV], lcm[MaxV], C[MaxV];
char not_prime[MaxV];
void linear_sieve(int n) {
  fac[1] = -1;
  for (int i = 2; i <= n; ++i) {
    if (!not_prime[i]) {
      fac[i] = tot;
      val[tot] = -1;
      prime[tot++] = i;
    }
    for (int j = 0; j != tot; ++j) {
      long long t = 1ll * prime[j] * i;
      if (t > n) break;
      not_prime[t] = 1;
      fac[t] = j;
      if (i % prime[j] == 0) break;
    }
  }
}
int count(int x, int p) {
  int cnt = 0;
  while (x % p == 0) {
    x /= p;
    ++cnt;
  }
  return cnt;
}
bool check(int f, int num) {
  int p = prime[f], cnt = 0;
  for (int i = 0; i != n; ++i) {
    int z = count(C[i], p);
    cnt = std::max(z, cnt);
  }
  return cnt >= num;
}
int main() {
  linear_sieve(1000000);
  std::scanf("%d %d", &n, &k);
  for (int i = 0; i != n; ++i) std::scanf("%d", C + i);
  bool flag = true;
  while (k != 1) {
    int cnt = 0, f = fac[k];
    while (fac[k] == f) {
      ++cnt;
      k /= prime[f];
    }
    if (!check(f, cnt)) {
      flag = false;
      break;
    }
  }
  std::puts(flag ? "Yes" : "No");
  return 0;
}