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>
const long long int mod = 1e9 + 7;
const long long int INF = 10000000000000;
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t = 0;
  cin >> t;
  while (t--) {
    long long int n = 0;
    cin >> n;
    long long int tn = n;
    vector<long long int> fac;
    for (long long int i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        fac.push_back(i);
        n /= i;
      }
    }
    fac.push_back(n);
    long long int p = 1;
    if (fac.size() < 3)
      cout << "NO\n";
    else {
      for (long long int i = 3; i < fac.size(); ++i) {
        fac[2] *= fac[i];
      }
      if ((fac[0] == fac[1]) || (fac[1] == fac[2]) || fac[2] == fac[0])
        cout << "NO\n";
      else {
        cout << "YES\n";
        cout << fac[0] << " " << fac[1] << " " << fac[2] << "\n";
      }
    }
  }
  return 0;
}