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 crib[1000009];
void criba() {
  for (long long i = 2; i < 1000009; i++) {
    if (crib[i] == 0) {
      for (long long j = i * i; j < 1000009; j += i) {
        crib[j] = i;
      }
    }
  }
}
void factorize(long long n, map<long long, long long> &pf) {
  if (n < 2) return;
  while (crib[n] != 0) {
    pf[crib[n]]++;
    n /= crib[n];
  }
  pf[n]++;
}
void lcm2(vector<long long> &nums, map<long long, long long> &L) {
  L.clear();
  for (int i = 0; i < nums.size(); i++) {
    map<long long, long long> pf;
    factorize(nums[i], pf);
    for (auto it = pf.begin(); it != pf.end(); ++it) {
      if (it->second > L[it->first]) L[it->first] = it->second;
    }
  }
}
int divide(map<long long, long long> &L, long long k) {
  map<long long, long long> pf;
  factorize(k, pf);
  for (const pair<long long, long long> &p : pf) {
    if (p.second > L[p.first]) return 0;
  }
  return 1;
}
void print_pair(pair<long long, long long> num) {
  cout << num.first << " " << num.second;
}
void print_vll(vector<pair<long long, long long> > &v) {
  for (pair<long long, long long> x : v) {
    print_pair(x);
    cout << endl;
  }
}
int main() {
  criba();
  long long n, k;
  cin >> n >> k;
  vector<long long> ar(n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    ar[i] = x;
  }
  map<long long, long long> L;
  lcm2(ar, L);
  if (divide(L, k)) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  return 0;
}