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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> v;
    long long int x = n;
    while (n % 2 == 0) {
      v.push_back(2);
      n /= 2;
    }
    for (long long int i = 3; i * i <= n; i += 2) {
      while (n % i == 0) {
        v.push_back(i);
        n /= i;
      }
    }
    if (n > 2) v.push_back(n);
    map<long long int, long long int> mp;
    for (long long int i = 0; i < v.size(); i++) mp[v[i]]++;
    vector<long long int> v1;
    if (mp.size() > 2) {
      cout << "YES\n";
      map<long long int, long long int>::iterator it;
      long long int cnt = 0;
      for (it = mp.begin(); it != mp.end(); it++) {
        if (cnt == 2) break;
        cout << (long long int)pow(it->first, it->second) << " ";
        cnt++;
      }
      long long int x = 1;
      for (; it != mp.end(); it++) x = (x * pow(it->first, it->second));
      cout << x << '\n';
    } else if (mp.size() == 2) {
      map<long long int, long long int>::iterator it = mp.begin();
      long long int cnt = it->second;
      it++;
      cnt += it->second;
      if (cnt < 4)
        cout << "NO\n";
      else {
        cout << "YES\n";
        it = mp.begin();
        long long int x, y, z;
        x = it->first;
        y = pow(it->first, it->second - 1);
        it++;
        z = it->first;
        y *= pow(it->first, it->second - 1);
        cout << x << " " << y << " " << z << '\n';
      }
    } else {
      map<long long int, long long int>::iterator it = mp.begin();
      if (it->second >= 6) {
        cout << "YES\n";
        cout << it->first << " " << (long long int)pow(it->first, 2) << " "
             << (long long int)pow(it->first, it->second - 3) << '\n';
      } else
        cout << "NO\n";
    }
  }
}