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;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long int lcm(int a, int b) {
  int g = gcd(a, b);
  return (long long int)a / g * b;
}
namespace number_theory {
int primes[1000000 + 5];
int pfac[1000000 + 5];
bool pflag[1000000 + 5];
int p;
void prepare() {
  p = 0;
  for (int i = 2; i <= 1000000; ++i) {
    if (!pflag[i]) {
      primes[p++] = i;
      pfac[i] = i;
    }
    for (int j = 0; j < p && i * primes[j] <= 1000000; ++j) {
      pflag[i * primes[j]] = true;
      pfac[i * primes[j]] = primes[j];
      if (i % primes[j] == 0) break;
    }
  }
}
void factorize(map<int, int>& f, int num) {
  f.clear();
  while (num > 1) {
    ++f[pfac[num]];
    num /= pfac[num];
  }
}
};  // namespace number_theory
int c[1000005];
map<int, int> fac[1000005];
bool vis[105];
int n, k;
int main(int argc, char* argv[]) {
  number_theory::prepare();
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
    number_theory::factorize(fac[i], c[i]);
  }
  map<int, int> kfac;
  map<int, bool> solved;
  number_theory::factorize(kfac, k);
  for (const auto& p : kfac) solved[p.first] = false;
  for (int i = 0; i < n; ++i) {
    for (const auto& p : fac[i]) {
      if (kfac.count(p.first)) {
        if (p.second >= kfac[p.first]) solved[p.first] = true;
      }
    }
  }
  bool ans = true;
  for (const auto& p : solved) {
    if (!p.second) {
      ans = false;
      break;
    }
  }
  if (ans)
    puts("Yes");
  else
    puts("No");
  return 0;
}