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;
vector<int> printDivisors(int n) {
  vector<int> v;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        v.push_back(i);
      else {
        v.push_back(n / i);
        v.push_back(i);
      }
    }
  }
  return v;
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    bool p = false;
    vector<int> v = printDivisors(n);
    sort(v.begin(), v.end());
    for (int r = 0; r < v.size(); r++) {
      int j = v.at(r);
      int n1 = n / j;
      for (int k = 2; k <= sqrt(n1); k++) {
        if (n1 % k == 0) {
          if (n1 / k == k)
            continue;
          else {
            if (j != k && j != n1 / k) {
              cout << "YES" << endl;
              cout << j << " " << k << " " << n1 / k << endl;
              p = true;
              break;
            }
          }
        }
      }
      if (p) break;
    }
    if (!p) cout << "NO" << endl;
  }
  return 0;
}