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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> divisors;
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        divisors.push_back(i);
        if (i * i < n) divisors.push_back(n / i);
      }
    }
    divisors.push_back(n);
    vector<int> ans;
    int val;
    for (auto it1 : divisors) {
      for (auto it2 : divisors) {
        if (it1 == it2) continue;
        val = n;
        val /= it1;
        val /= it2;
        if (val >= 2 && ((val * it1) * it2) == n && val != it1 && val != it2) {
          ans.push_back(it1);
          ans.push_back(it2);
          ans.push_back(val);
          break;
        }
      }
      if (!ans.empty()) break;
    }
    if (!ans.empty()) {
      cout << "YES\n";
      for (auto it : ans) cout << it << " ";
    } else
      cout << "NO";
    cout << "\n";
  }
}