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<pair<long long, long long>> arr;
vector<pair<long long, long long>> arr2;
vector<long long> dels;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    long long v = -1;
    for (long long i = 2; i * i < n; i++) {
      if (n % i == 0) {
        for (long long j = 2; j * j <= i; j++) {
          if (i % j == 0 && i / j != 1 && i / j != n / i && n / i != j &&
              j != i / j) {
            cout << "YES\n" << i / j << ' ' << n / i << ' ' << j << endl;
            goto r;
          }
        }
        for (long long j = 2; j * j <= n / i; j++) {
          if ((n / i) % j == 0 && (n / i) / j != 1 && (n / i) / j != i &&
              i != j && j != (n / i) / j) {
            cout << "YES\n" << (n / i) / j << ' ' << i << ' ' << j << endl;
            goto r;
          }
        }
      }
    }
    cout << "NO\n";
  r:;
  }
  return 0;
}