Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t = 0;
  cin >> t;
  while (t--) {
    long long int n = 0;
    cin >> n;
    if (n < 24) {
      cout << "NO\n";
      continue;
    } else {
      vector<long long int> f;
      for (long long int i = 2; i * i <= n; ++i) {
        if (n % i == 0) {
          f.push_back(i);
          n /= i;
        }
      }
      f.push_back(n);
      sort(f.begin(), f.end());
      if (f.size() > 3) {
        for (long long int i = 3; i < f.size(); ++i) {
          f[2] *= f[i];
        }
      }
      if (f.size() < 3)
        cout << "NO\n";
      else if (f[0] != f[1] && f[0] != f[2] && f[1] != f[2]) {
        cout << "YES\n";
        cout << f[0] << " " << f[1] << " " << f[2] << "\n";
      } else
        cout << "NO\n";
    }
  }
  return 0;
}