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;
vector<long long> calc_primes(int k) {
  vector<long long> lp(k + 1, 0);
  vector<long long> primes;
  for (int i = 2; i < k; ++i) {
    if (lp[i] == 0) {
      primes.push_back(i);
      lp[i] = i;
    }
    for (long long p : primes) {
      if (i * p > k || lp[i] < p) {
        break;
      }
      lp[i * p] = p;
    }
  }
  return lp;
}
vector<pair<long long, int> > factorize(long long n,
                                        const vector<long long> &lp) {
  vector<pair<long long, int> > ret;
  while (n > 1) {
    long long p = lp[n];
    if (p == 0) {
      ret.emplace_back(n, 1);
      break;
    }
    ret.emplace_back(p, 0);
    while (lp[n] == p) {
      ret.back().second += 1;
      n /= p;
    }
  }
  return ret;
}
long long gcd(long long a, long long b) {
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long calc_lcm(long long a, long long b) {
  long long g = gcd(a, b);
  return (a / g) * b;
}
long long get_lcm(const vector<long long> &a, const long long k) {
  long long lcm = a[0];
  for (int i = 1; i < a.size(); ++i) {
    lcm = calc_lcm(lcm, a[i]);
    if (lcm > k) {
      break;
    }
  }
  return lcm;
}
void solve(int n) {
  long long k;
  {
    int kk;
    scanf("%d", &kk);
    k = kk;
  }
  vector<long long> coefs(n);
  for (int i = 0; i < n; ++i) {
    int cur;
    scanf("%d", &cur);
    coefs[i] = cur;
  }
  const vector<long long> lp = calc_primes(1e6 + 1);
  auto k_facts = factorize(k, lp);
  map<long long, int> k_set(k_facts.begin(), k_facts.end());
  for (int i = 0; i < n; ++i) {
    long long c = coefs[i];
    auto facts = factorize(c, lp);
    for (const pair<long long, int> &f : facts) {
      auto it = k_set.find(f.first);
      if (it != k_set.end() && it->second <= f.second) {
        k_set.erase(it);
      }
    }
  }
  long long mul = -1;
  if (k_set.empty()) {
    mul = k + 10;
  }
  if (mul >= k) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    solve(n);
  }
  return 0;
}