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;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n < 24) {
      cout << "NO" << endl;
      continue;
    }
    long long p = 0;
    for (long long i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
        p = 1;
        break;
      }
    }
    if (!p) {
      cout << "NO" << endl;
      continue;
    }
    long long a = 0, b = 0, c = 0;
    for (long long i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
        a = n / i;
        b = i;
        break;
      }
    }
    for (long long i = 2; i <= sqrt(a); i++) {
      if (a % i == 0) {
        if (a / i != i && i != b && a / i != b) {
          c = a / i;
          a = i;
          break;
        }
      }
    }
    if (a != b && b != c && a != c && a >= 2 && b >= 2 && c >= 2) {
      cout << "YES" << endl;
      cout << a << " " << b << " " << c << endl;
    } else
      cout << "NO" << endl;
  }
}