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 long double INF = 1e18 + 7;
long long N = 1e6 * 5;
long double EPS = 1 / 1e18;
long long a[1005];
map<long long, long long> cnt;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long k = 0, err = 0, n;
    cin >> n;
    for (long long j = 2; j <= round(sqrt(n)); j++) {
      if (!(n % j)) {
        a[k++] = j;
        cnt[j]++;
        a[k++] = n / j;
        cnt[n / j]++;
      }
    }
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        if (!(a[i] % a[j]) && cnt[a[i] / a[j]] != 0 && a[j] != n / a[i] &&
            a[j] != a[i] / a[j] && n / a[i] != a[i] / a[j]) {
          cout << "YES\n"
               << a[j] << ' ' << a[i] / a[j] << ' ' << n / a[i] << '\n';
          err = 1;
          break;
        }
      }
      if (err) break;
    }
    if (!err) cout << "NO\n";
  }
  return 0;
}