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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
const int N = 1e5 + 7;
const int M = 1e6 + 7;
const int INF = 1e9 + 8;
int p[N];
int c[N];
int m;
void divide(int n) {
  m = 0;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      p[++m] = i, c[m] = 0;
      while (n % i == 0) {
        n /= i;
        c[m]++;
      }
    }
  }
  if (n > 1) {
    p[++m] = n;
    c[m] = 1;
  }
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    divide(n);
    int a = 1;
    int b = 1;
    int cc = 1;
    int tt = 1;
    for (int i = 1; i <= m; i++) {
      int g = c[i];
      for (int j = 1; j <= g; j++) {
        if (tt <= 1) {
          a *= p[i];
          tt++;
        } else {
          if (b == 1 || b == a)
            b *= p[i];
          else
            cc *= p[i];
        }
      }
    }
    if (cc == 1 || b == 1 || a == 1) {
      cout << "NO" << endl;
      continue;
    }
    if (cc != a && a != b && b != cc) {
      cout << "YES" << endl;
      cout << a << " " << b << " " << cc << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}