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;
template <class T>
inline bool amax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
map<T, int> factorize(T n) {
  map<T, int> ret;
  for (T i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ++ret[i];
      n /= i;
    }
  }
  if (n != 1) ret[n] = 1;
  return ret;
}
void Main() {
  int N, K;
  scanf("%d%d", &N, &K);
  auto fk = factorize(K);
  vector<int> pk;
  for (auto &&pn : fk) {
    int x = 1;
    for (int i = int(0); i < int(pn.second); ++i) x *= pn.first;
    pk.push_back(x);
  }
  vector<int> C(N);
  for (int i = int(0); i < int(N); ++i) {
    scanf("%d", &C[i]);
  }
  sort((C).begin(), (C).end());
  C.erase(unique(C.begin(), C.end()), C.end());
  for (auto &&c : C) {
    for (auto &&j : pk)
      if (j != 1) {
        if (c % j == 0) j = 1;
      }
  }
  bool ok = 1;
  for (auto &&x : pk) {
    if (x != 1) ok = 0;
  }
  puts(ok ? "Yes" : "No");
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  Main();
  return 0;
}