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;
const long long N = 1e5 + 10;
vector<long long> ans;
pair<long long, long long> a[N];
long long n, T;
void f(long long n) {
  ans.clear();
  long long x = n;
  for (long long i = 2; i * i <= x; i++) {
    while (x % i == 0) {
      ans.push_back(i);
      x = x / i;
    }
  }
  ans.push_back(x);
  if (x == n) {
    cout << "NO" << endl;
  } else {
    if (ans.size() >= 3) {
      long long k = 2;
      long long x1 = ans[0];
      long long x2 = ans[1];
      if (x2 == x1) x2 = ans[1] * ans[2], k = 3;
      long long x3 = 1;
      for (long long i = k; i < ans.size(); i++) {
        x3 *= ans[i];
      }
      if (x2 != x3 && x1 != x3 && x3 != 1) {
        cout << "YES" << endl;
        cout << x1 << " " << x2 << " " << x3 << endl;
      } else {
        cout << "NO" << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
}
int main() {
  cin >> T;
  while (T--) {
    cin >> n;
    f(n);
  }
  return 0;
}