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;
inline int setBit(int N, int pos) { return N = N | (1 << pos); }
inline int resetBit(int N, int pos) { return N = N & ~(1 << pos); }
inline bool checkBit(int N, int pos) { return (bool)(N & (1 << pos)); }
int notPrime[1000000 + 10];
vector<int> v, prime;
void sieve() {
  for (int i = 2; i <= 1000000; i++) {
    if (notPrime[i] == 0) {
      prime.push_back(i);
      for (int j = i + i; j <= 1000000; j += i) notPrime[j] = 1;
    }
  }
}
int nd[1000000 + 10];
int main() {
  long long i, j, cs, t, d;
  long long n, k, lcm = 1;
  sieve();
  scanf("%lld %lld", &n, &k);
  long long bla = k;
  for (i = 0; i < prime.size(); i++) {
    int cnt = 0;
    while (k % prime[i] == 0) {
      cnt++;
      k /= prime[i];
    }
    nd[i] = cnt;
    if (cnt) v.push_back(i);
  }
  int ok = 0;
  k = bla;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &t);
    for (j = 0; j < v.size(); j++) {
      int p = prime[v[j]];
      int cnt = 0;
      while (t % p == 0) {
        cnt++;
        t /= p;
      }
      if (nd[v[j]] <= cnt) ok++, nd[v[j]] = 1000000;
    }
  }
  if (ok == v.size())
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}