You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.

Otherwise, print "YES" and any possible such representation.

Example

Input


5
64
32
97
2
12345


Output


YES
2 4 8 
NO
NO
NO
YES
3 5 823 
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[]) {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int t;
  int n;
  cin >> t;
  while (t--) {
    cin >> n;
    set<int> used;
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0 && !used.count(i)) {
        used.insert(i);
        n /= i;
        break;
      }
    }
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0 && !used.count(i)) {
        used.insert(i);
        n /= i;
        break;
      }
    }
    if (int(used.size()) < 2 || n == 1 || used.count(n))
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      set<int>::iterator iter;
      iter = used.begin();
      used.insert(n);
      for (; iter != used.end(); iter++) {
        cout << *iter << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
