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 n, t, a, b, c, var;
  cin >> t;
  vector<long long> fact1, fact2, fact3;
  while (t--) {
    cin >> n;
    var = n;
    a = b = c = 1;
    fact1.clear();
    fact2.clear();
    fact3.clear();
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        fact1.push_back(i);
        fact1.push_back(n / i);
      }
    }
    sort(fact1.begin(), fact1.end());
    if (fact1.size() == 0) {
      cout << "NO\n";
      continue;
    }
    n = n / fact1[0];
    a = fact1[0];
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        fact2.push_back(i);
        fact2.push_back(n / i);
      }
    }
    fact2.push_back(n);
    sort(fact2.begin(), fact2.end());
    for (int i = 0; i < fact2.size(); i++) {
      if (fact2[i] != a) {
        b = fact2[i];
        break;
      }
    }
    c = var / (a * b);
    if (a != b && b != c && c != a && c != 1) {
      cout << "YES\n";
      cout << a << ' ' << b << ' ' << c << '\n';
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}