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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> vp;
    for (int i = 2; i <= sqrt(n); i++) {
      while (n % i == 0) {
        vp.push_back(i);
        n /= i;
      }
    }
    if (n > 1) {
      vp.push_back(n);
    }
    sort(vp.begin(), vp.end());
    int a = 1, b = 1, c = 1;
    int i = 0;
    while (a < 2 && i < vp.size()) {
      a *= vp[i++];
    }
    while (b < 2 && i < vp.size()) {
      b *= vp[i++];
    }
    if (b == a && i < vp.size()) {
      b *= vp[i++];
    }
    while (i < vp.size()) {
      c *= vp[i++];
    }
    if (c == b && i < vp.size()) {
      c *= vp[i++];
    }
    if (a > 1 && b > 1 && c > 1 && a != b && b != c && a != c) {
      cout << "YES\n" << a << " " << b << " " << c << "\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}