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;
const long long N = 100005, INF = 2000000000000000000;
const long double EPS = 0.000000000001;
long long power(long long a, long long b, long long p) {
  if (a == 0) return 0;
  long long res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
bool prime[N] = {true};
long long sum[N] = {0};
void pre() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
  for (long long i = 2; i < N; i++) {
    if (prime[i]) sum[i] += i;
    sum[i + 1] = sum[i];
  }
  return;
}
class yup {
 public:
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
};
long long minCoins(long long coins[], long long m, long long V) {
  if (V == 0) return 0;
  long long res = INT_MAX;
  for (long long i = 0; i < m; i++) {
    if (coins[i] <= V) {
      long long sub_res = minCoins(coins, m, V - coins[i]);
      if (sub_res != INT_MAX && sub_res + 1 < res) res = sub_res + 1;
    }
  }
  return res;
}
bool com(long long a, long long b) { return a > b; }
long long ans = 0;
class gfg {
 public:
  unsigned long long getFirstSetBitPos(long long n) { return log2(n & -n) + 1; }
};
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  yup y;
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long lc = a[0];
  for (long long i = 0; i < n; i++) {
    lc = y.gcd(k, y.lcm(lc, a[i]));
  }
  if (lc % k == 0) {
    cout << "YES"
         << "\n";
  } else {
    cout << "NO"
         << "\n";
  }
}