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;
using lli = long long int;
using ulli = unsigned long long int;
using db = double;
using vint = vector<int>;
using vlli = vector<vint>;
using vvint = vector<vint>;
using vvlli = vector<vlli>;
using cd = complex<double>;
const int inf = 1e9 + 5;
const lli infl = 1e18 + 5;
void solve() {
  int n;
  cin >> n;
  int sqn = sqrt(n);
  for (int a = 2; a <= sqn; ++a) {
    if (n % a == 0) {
      int sqna = sqrt(n / a);
      for (int b = 2; b <= sqna; ++b) {
        if ((n / a) % b == 0 and (a != b and b != (n / a) / b and
                                  a != (n / a) / b and (n / a) / b >= 2)) {
          cout << "YES"
               << "\n";
          cout << a << " " << b << " " << (n / a) / b << "\n";
          return;
        }
      }
    }
  }
  cout << "NO"
       << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}