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() {
  int q;
  cin >> q;
  while (q--) {
    long long int n, x;
    cin >> n;
    x = n;
    vector<long long int> result;
    int i = 1;
    while (i * i <= x) {
      if (x % i == 0) {
        result.push_back(i);
        if (x / i != i) {
          result.push_back(x / i);
        }
      }
      i++;
    }
    if (result.size() >= 5) {
      long long int c = 0, p = 1;
      vector<long long int> v;
      sort(result.begin(), result.end());
      for (int i = 0; i < result.size(); i++) {
        if (result[i] != 1) {
          if (n % result[i] == 0) {
            n = n / result[i];
            v.push_back(result[i]);
            c++;
            p = p * result[i];
          }
          if (c == 2) {
            break;
          }
        }
      }
      v.push_back((x) / p);
      if ((v[0] != v[1]) && v[0] != v[2] && v[1] != v[2] &&
          (v[0] * v[1] * v[2] == x)) {
        cout << "YES" << endl;
        cout << v[0] << " " << v[1] << " " << v[2] << endl;
      } else {
        cout << "NO" << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}