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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long copy = n;
    vector<pair<long long, long long>> v;
    long long count = 0;
    while (!(n % 2)) {
      n >>= 1;
      count++;
    }
    if (count) v.push_back({2, count});
    for (long long i = 3; i <= sqrt(n); i += 2) {
      count = 0;
      while (n % i == 0) {
        count++;
        n = n / i;
      }
      if (count) v.push_back({i, count});
    }
    if (n > 2) v.push_back({n, 1});
    if (v.size() < 3) {
      if (v.size() == 2) {
        long long f1 = v[0].first;
        long long f2 = v[1].first;
        long long f3 = copy / (v[0].first * v[1].first);
        if (f1 != f2 && f1 != f3 && f2 != f3 && f1 >= 2 && f2 >= 2 && f3 >= 2) {
          cout << "YES\n";
          cout << f1 << " " << f2 << " " << f3 << '\n';
        } else
          cout << "NO\n";
      } else {
        if (v[0].second <= 5)
          cout << "NO\n";
        else {
          cout << "YES\n";
          cout << v[0].first << " " << v[0].first * v[0].first << " "
               << copy / (v[0].first * v[0].first * v[0].first) << "\n";
        }
      }
    } else {
      cout << "YES\n";
      cout << v[0].first << " " << v[1].first << " ";
      cout << copy / (v[0].first * v[1].first) << '\n';
    }
  }
}