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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
clock_t START;
inline long double TIME() {
  return (long double)(clock() - START) / CLOCKS_PER_SEC;
}
void SHOW() {
  cout << fixed << setprecision(10);
  cout << TIME() << " SECONDS FROM START\n";
}
long double TL = 2.0;
int calls = 0;
inline bool IS() {
  if (TL - TIME() < 0.1) return true;
  return false;
}
template <typename T1, typename T2>
inline bool amin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
inline bool amax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &u : v) is >> u;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  for (auto &u : v) os << u << ' ';
  return os;
}
int hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937},
    mod = hash_mod[rnd() % 4];
int hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];
int binpow(int first, int p) {
  int res = 1;
  while (p) {
    if (p % 2) res = (res * first) % mod;
    first = (first * first) % mod;
    p /= 2;
  }
  return res;
}
const int N = 1e6 + 7, NS = 2e3 + 7, lg = 20, sq = 550, inf = 2e9 + 7;
const long double eps = 1e-9, pi = 3.14159265359;
int mx[N], pr[N];
vector<int> v[N];
void solve() {
  for (int i = 2; i < N; i++) {
    if (pr[i]) continue;
    for (int j = i; j < N; j += i) v[j].push_back(i);
    if (i < N / i) {
      for (int j = i * i; j < N; j += i) {
        pr[j] = 1;
      }
    }
  }
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int c;
    cin >> c;
    for (auto u : v[c]) {
      int cur = 0, c1 = c;
      while (c1 % u == 0) {
        c1 /= u;
        cur++;
      }
      amax(mx[u], cur);
    }
  }
  bool ok = true;
  for (int d = 2; d * d <= k; d++) {
    if (k % d) continue;
    int cur = 0;
    while (k % d == 0) {
      cur++;
      k /= d;
    }
    if (cur > mx[d]) {
      ok = false;
      break;
    }
  }
  if (k != 1) {
    if (mx[k] < 1) ok = false;
  }
  if (ok)
    cout << "Yes";
  else
    cout << "No";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}