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;
vector<int> adj[200001];
long long int vist[200001], maxx;
void dfs(long long int node) {
  vist[node] = 1;
  maxx = max(maxx, node);
  for (long long int child : adj[node])
    if (!vist[child]) dfs(child);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int x = sqrt(n);
    bool ok = false;
    int i, j, k;
    for (i = 2; i <= x; i++) {
      int n1 = n;
      if (n1 % i == 0) {
        n1 /= i;
        for (j = i + 1; j <= x; j++) {
          if (n1 % j == 0 && j != i) {
            k = n1 /= j;
            if (k != i && k != j) {
              ok = true;
              break;
            }
          }
        }
        if (ok) break;
      }
    }
    if (ok)
      cout << "YES\n" << i << " " << j << " " << k << endl;
    else
      cout << "NO\n";
  }
  return 0;
}