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 ll = long long;
using namespace std;
const int N = 1e6 + 1, mod = 1e9 + 7;
int t;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> divisors;
    int d = n, s = sqrt(n);
    for (int i = 2; i <= s; i++) {
      if (d % i == 0) {
        divisors.push_back(i);
        d /= i;
      }
    }
    if (d > 1) {
      divisors.push_back(d);
    }
    if (divisors.size() == 1) {
      int a = divisors[0], cnt = 0;
      while (n % a == 0) {
        n /= a;
        cnt++;
      }
      if (cnt >= 6) {
        cout << "YES\n";
        cout << a << ' ' << a * a << ' ' << n / (a * a * a) << '\n';
      } else {
        cout << "NO\n";
      }
    } else if (divisors.size() >= 3) {
      int a = divisors[0], b = divisors[1];
      int c = n / divisors[0] / divisors[1];
      if (a == c || b == c) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        cout << a << ' ' << b << ' ' << c << '\n';
      }
    } else {
      cout << "NO\n";
    }
  }
}