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;
using ll = long long;
using lli = long long int;
using ld = long double;
const int N = 2e5 + 5, inf = 2e9;
bool valid(int a, int b, int c) {
  if (a == b || b == c || a == c) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int x = 0, y = 0;
    vector<pair<int, int>> v;
    set<int> st;
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        v.push_back({i, n / i});
      }
    }
    int a = 0, b = 0, c = 0;
    int found = false;
    for (auto p : v) {
      for (int i = 2; i * i <= p.first; i++) {
        if (p.first % i == 0) {
          a = i;
          b = p.first / i;
          c = p.second;
          if (valid(a, b, c)) {
            found = true;
            break;
          }
        }
      }
      if (found) break;
      for (int i = 2; i * i <= p.second; i++) {
        if (p.second % i == 0) {
          a = i;
          b = p.second / i;
          c = p.first;
          if (valid(a, b, c)) {
            found = true;
            break;
          }
        }
      }
      if (found) break;
    }
    if (found) {
      cout << "YES"
           << "\n";
      cout << a << " " << b << " " << c << "\n";
    } else
      cout << "NO"
           << "\n";
  }
  return 0;
}