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;
double squrt(int64_t num) {
  double high = num, low = 0, mid;
  while (high - low > 0.00001) {
    mid = low + (high - low) / 2;
    if (mid * mid > num) {
      high = mid;
    } else {
      low = mid;
    }
  }
  return mid;
}
void solve() {
  int64_t num;
  cin >> num;
  for (int64_t i = 2; i * i < num; i++) {
    if (num % i != 0) continue;
    for (int64_t j = i + 1; j * j <= (num / i); j++) {
      int64_t x;
      if (num % (i * j) == 0) {
        int64_t x = num / (i * j);
        if (x != i && i != j && x != j) {
          cout << "YES"
               << "\n";
          cout << i << " " << j << " " << x << "\n";
          return;
        }
      }
    }
  }
  cout << "NO"
       << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cout.precision(10);
  cout << fixed;
  ;
  int64_t test;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}