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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n, x, i, y;
    cin >> n;
    x = n;
    vector<pair<long long int, int>> v;
    for (i = 2; i * i <= x; i++) {
      y = 0;
      if (x % i == 0) {
        while (x % i == 0) {
          x /= i;
          y++;
        }
        v.push_back({i, y});
      }
    }
    if (x > 1) v.push_back({x, 1});
    if (v.size() >= 3) {
      x = v[0].first;
      y = v[1].first;
      cout << "YES\n" << x << " " << y << " " << n / (x * y) << "\n";
    } else if (v.size() == 2) {
      x = v[0].first;
      y = v[1].first;
      long long int c0 = v[0].second, c1 = v[1].second;
      if (c0 >= 3) {
        cout << "YES\n" << x << " " << x * x << " " << n / (x * x * x) << "\n";
      } else if (c1 >= 3) {
        x = y;
        cout << "YES\n" << x << " " << x * x << " " << n / (x * x * x) << "\n";
      } else if (c0 >= 2 && c1 >= 2) {
        cout << "YES\n" << x << " " << y << " " << x * y << "\n";
      } else
        cout << "NO\n";
    } else if (v.size() == 1) {
      x = v[0].first;
      y = v[0].second;
      if (y >= 6) {
        cout << "YES\n" << x << " " << x * x << " " << n / (x * x * x) << "\n";
      } else
        cout << "NO\n";
    } else
      cout << "NO\n";
  }
}