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() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    vector<long long> v;
    map<long long, long long> mp;
    while (n % 2 == 0) {
      v.push_back(2);
      mp[2]++;
      n = n / 2;
    }
    for (i = 3; i <= sqrt(n); i = i + 2) {
      while (n % i == 0) {
        v.push_back(i);
        mp[i]++;
        n = n / i;
      }
    }
    if (n > 2) {
      v.push_back(n);
      mp[n]++;
    }
    if (v.size() < 6 && mp.size() == 1)
      cout << "NO\n";
    else {
      if (mp.size() == 3 && v.size() == 3) {
        cout << "YES\n" << v[0] << " " << v[1] << " " << v[2] << endl;
      } else {
        long long ans1 = v[1], ans2 = 1;
        i = 2;
        while (ans1 == v[0]) {
          ans1 = ans1 * v[i];
          i++;
        }
        while (i < v.size()) {
          ans2 = ans2 * v[i];
          i++;
        }
        if (ans1 == v[0] || ans2 == v[0] || ans1 == ans2 || v[0] == 1 ||
            ans1 == 1 || ans2 == 1)
          cout << "NO\n";
        else
          cout << "YES\n" << v[0] << " " << ans1 << " " << ans2 << endl;
      }
    }
  }
  return 0;
}