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 long mod = 1e9 + 7;
void printDivisors(long long n, set<long long> &s, vector<long long> &v) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        s.insert(i);
        v.push_back(i);
      } else {
        s.insert(i);
        s.insert(n / i);
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    set<long long> s;
    bool flag = 0;
    vector<long long> v;
    printDivisors(n, s, v);
    for (int i = 0; i < v.size(); i++) {
      for (int j = i + 1; j < v.size(); j++) {
        if (n % (v[i] * v[j]) == 0 && s.find(n / (v[i] * v[j])) != s.end() &&
            v[i] != v[j] && v[i] != n / (v[i] * v[j]) &&
            v[j] != n / (v[i] * v[j])) {
          cout << "YES" << '\n';
          cout << v[i] << " " << v[j] << " " << n / (v[i] * v[j]) << '\n';
          flag = 1;
          break;
        }
      }
      if (flag == 1) break;
    }
    if (flag == 0) cout << "NO" << '\n';
  }
  return 0;
}