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;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, a = 0, b = 0, c = 0, n1;
    cin >> n;
    for (int i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
        n1 = n / i;
        a = i;
        break;
      }
    }
    for (int i = a + 1; i <= sqrt(n1); i++) {
      if (n1 % i == 0) {
        b = i;
        break;
      }
    }
    if (a > 0 && b > 0) c = n / (a * b);
    if (c > 1 && c != a && c != b)
      cout << "YES" << endl << a << " " << b << " " << c << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}