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;
const int INF = 1e8 + 1;
const long double PI = 3.141592653;
bool isp(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    if (isp(n)) {
      cout << "NO" << endl;
      continue;
    } else {
      vector<int> pot;
      for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
          if (n / i == i)
            pot.push_back(i);
          else {
            pot.push_back(i);
            pot.push_back(n / i);
          }
        }
      }
      sort(pot.begin(), pot.end());
      int o = pot[0], t = pot[int(pot.size()) - 1];
      vector<int> xd;
      for (int i = 2; i * i <= t; i++) {
        if (t % i == 0) {
          if (t / i == i)
            xd.push_back(i);
          else {
            xd.push_back(i);
            xd.push_back(t / i);
          }
        }
      }
      sort(xd.begin(), xd.end());
      int i = 0, j = int(xd.size()) - 1;
      bool fin = 0;
      int p, pp;
      while (i < j) {
        if (!(xd[i] != xd[j] && xd[i] != o && xd[j] != o)) {
          i++;
          j--;
        } else {
          fin = 1;
          p = i;
          pp = j;
          break;
        }
      }
      if (fin) {
        cout << "YES" << endl;
        cout << o << " " << xd[p] << " " << xd[pp] << endl;
      } else {
        cout << "NO" << endl;
      }
    }
  }
}