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;
vector<long long> vec;
void primeFactors(long long n) {
  while (n % 2 == 0) {
    vec.push_back(2);
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      vec.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) vec.push_back(n);
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    primeFactors(n);
    if (vec.size() < 3) {
      cout << "NO" << endl;
    } else {
      set<long long> s;
      s.insert(vec[0]);
      long long x = 1;
      for (i = 1; i < vec.size(); i++) {
        x *= vec[i];
        if (x != vec[0]) break;
      }
      s.insert(x);
      x = (n / x) / vec[0];
      if (x != 1) s.insert(x);
      if (s.size() == 3) {
        cout << "YES" << endl;
        for (auto it = s.begin(); it != s.end(); it++) {
          cout << *it << " ";
        }
        cout << endl;
      } else {
        cout << "NO" << endl;
      }
    }
    vec.clear();
  }
  return 0;
}