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 X, class Y>
void amax(X& x, const Y& y) {
  if (x < y) x = y;
}
template <class X, class Y>
void amin(X& x, const Y& y) {
  if (x > y) x = y;
}
const int INF = 1e9 + 10;
const long long INFL = 1e18 + 10;
const int MAX = 1e6 + 10;
const int MAXP = 1e6;
int n, k;
int c[MAX];
int prime[MAX];
int cnt_k[MAX], cnt_c[MAX];
void yes() {
  cout << "Yes";
  exit(0);
}
void no() {
  cout << "No";
  exit(0);
}
void prime_sieve() {
  for (int i = 2; i * i <= MAXP; i++)
    if (!prime[i])
      for (int j = i; j <= MAXP / i; j++) prime[i * j] = i;
  for (int i = 1; i <= MAXP; i++)
    if (prime[i] == 0) prime[i] = i;
}
vector<pair<int, int> > get_factor(int x) {
  vector<pair<int, int> > res;
  while (x > 1) {
    pair<int, int> t = make_pair(prime[x], 0);
    while (x % t.first == 0) {
      x /= t.first;
      t.second++;
    }
    res.push_back(t);
  }
  return res;
}
void process() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> c[i];
  prime_sieve();
  vector<pair<int, int> > fk = get_factor(k);
  for (auto p : fk) cnt_k[p.first] = p.second;
  for (int i = 1; i <= n; i++) {
    vector<pair<int, int> > ft = get_factor(c[i]);
    for (auto p : ft) amax(cnt_c[p.first], p.second);
  }
  for (int i = 2; i <= MAXP; i++)
    if (cnt_k[i] > cnt_c[i]) no();
  yes();
}
int main() {
  ios_base::sync_with_stdio(false);
  process();
}