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;
long long spf(long long n) {
  if (n % 2 == 0) {
    return 2;
  }
  while (n % 2 == 0) {
    n /= 2;
  }
  for (long long i = 3; i * i <= n; i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return n;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long no = 1;
    for (long long i = 2; i * i <= n && no; i++) {
      if (n % i == 0 && (n != i * i)) {
        long long next = n / i;
        for (long long j = i + 1; j * j <= next && no; j++) {
          if (next % j == 0 && (next / j) != j && (next / j) != i) {
            cout << "YES"
                 << "\n";
            cout << i << " " << j << " " << next / j << "\n";
            no = 0;
          }
        }
      }
    }
    if (no) {
      cout << "NO"
           << "\n";
    }
  }
  return 0;
}