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 solve() {
  int n;
  cin >> n;
  vector<int> factors;
  for (int i = 2; 1LL * i * i <= (long long int)n; ++i) {
    if (n % i == 0) {
      factors.push_back(n / i);
      if (n / i != i) {
        factors.push_back(i);
      }
    }
  }
  if (factors.size() < 2) {
    cout << "NO\n";
  } else {
    sort(factors.begin(), factors.end());
    int poss = 0;
    for (int i = 0; i < factors.size() and !poss; ++i) {
      for (int j = i + 1; j < factors.size() and !poss; ++j) {
        int a = factors[i];
        int b = factors[j];
        int c = n / a;
        if (c % b != 0) {
          continue;
        }
        c /= b;
        if (c >= 2 and c != a and c != b) {
          poss = 1;
          cout << "YES\n";
          cout << a << " " << b << " " << c << '\n';
          break;
        }
      }
    }
    if (!poss) cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}