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;
int T;
int main(int argc, char const *argv[]) {
  cin >> T;
  for (int kase = 1; kase <= T; kase++) {
    int n;
    cin >> n;
    map<int, int> mp;
    int tmp = n;
    int cnt = 3;
    while (cnt) {
      if (cnt == 1) {
        mp[tmp] = 1;
        break;
      }
      for (int i = 2; i <= sqrt(tmp); i++) {
        if (tmp % i == 0 && mp[i] == 0) {
          mp[i] = 1;
          tmp /= i;
          break;
        }
      }
      cnt--;
    }
    int ans[5];
    int tot = 0;
    for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
      if ((*it).second == 1 && (*it).first >= 2) {
        ans[++tot] = (*it).first;
      }
    }
    if (tot != 3) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      cout << ans[1] << " " << ans[2] << " " << ans[3] << endl;
    }
  }
  return 0;
}