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 double pi = 3.14159265358979323;
long long mod = 1000000000 + 7;
long long modu = 998244353;
const long double pii = acos(-1.0);
const long long INF = 1e18;
const long long inf = 1e9;
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
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 / gcd(a, b) * b); }
bool isPrime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long add(long long a, long long b) { return ((a % mod + b % mod) % mod); }
long long sub(long long a, long long b) {
  return ((a % mod - b % mod + mod) % mod);
}
long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod);
}
long long divi(long long a, long long b) {
  return (mul(a, powe(b, mod - 2)) % mod);
}
void fun() {}
const long long N = 1e6 + 7;
void pre() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  pre();
  long long tttt = 1;
  while (tttt--) {
    long long n, k;
    cin >> n >> k;
    long long c = 0;
    long long kk = k;
    set<long long> s;
    while (k % 2 == 0) {
      k /= 2;
      c++;
    }
    if (c) s.insert(2);
    for (long long i = 3; i * i <= k; i++) {
      c = 0;
      while (k % i == 0) {
        k /= i;
        c++;
      }
      if (c) s.insert(i);
    }
    if (k > 1) s.insert(k);
    map<long long, long long> m;
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      for (auto it : s) {
        c = 0;
        while (x % it == 0) {
          x /= it;
          c++;
        }
        m[it] = max(m[it], c);
      }
    }
    k = kk;
    for (auto it : m) {
      while (k % it.first == 0 and it.second > 0) {
        k /= it.first;
        it.second--;
      }
    }
    if (k == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}