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;
void find(vector<long long int>& v, long long int n) {
  for (long long int i = 2; i <= sqrt(n); i++) {
    while (n > 0 && n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, a, b, c;
    cin >> n;
    vector<long long int> v;
    find(v, n);
    if (v.size() < 2) {
      cout << "NO" << endl;
      continue;
    }
    a = v[0];
    if (v.size() > 2 && v[1] == v[0]) {
      b = v[1] * v[2];
    } else {
      b = v[1];
    }
    if (n % (a * b) != 0) {
      cout << "NO" << endl;
    } else {
      c = n / (a * b);
      if (a > 1 && b > 1 && c > 1 && a != b && a != c && b != c) {
        cout << "YES" << endl;
        cout << a << " " << b << " " << c << endl;
      } else {
        cout << "NO" << endl;
      }
    }
  }
  return 0;
}