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;
vector<int> primefactors(int n) {
  vector<int> ans;
  while (n % 2 == 0) {
    ans.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i * i <= n; i += 2) {
    while (n % i == 0) {
      ans.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) {
    ans.push_back(n);
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> ans = primefactors(n);
    int count = 1;
    vector<int> un;
    un.push_back(ans[0]);
    int temp = 0;
    for (int i = 1; i < ans.size(); i++) {
      if (temp == 0) {
        temp = ans[i];
      } else {
        temp *= ans[i];
      }
      if (temp != un[i - 1]) {
        un.push_back(temp);
        break;
      }
    }
    if (un.size() != 2) {
      cout << "NO" << endl;
    } else {
      int num1 = un[0], num2 = un[1];
      int num3 = n / (num1 * num2);
      if (num3 == 1 || num3 == num2 || num1 == num3) {
        cout << "NO" << endl;
      } else {
        cout << "YES" << endl;
        cout << num1 << " " << num2 << " " << (n / (num1 * num2)) << endl;
      }
    }
  }
}