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;
long long q, n;
long long ans1, ans2, ans3;
long long cnt, a[1000005];
signed main() {
  cin >> q;
  while (q--) {
    cnt = 0;
    cin >> n;
    for (long long i = 2; i * i <= n; i++)
      if (n % i == 0) {
        a[++cnt] = i;
        a[++cnt] = n / i;
      }
    if (cnt < 3) {
      cout << "NO\n";
      continue;
    }
    bool f = 0;
    for (long long i = 1; i <= cnt; i++)
      for (long long j = i + 1; j <= cnt; j++) {
        long long s = a[i] * a[j];
        if (n % s == 0 && n / s != a[i] && n / s != a[j] && s != n) {
          f = 1;
          ans1 = a[i];
          ans2 = a[j];
          ans3 = n / s;
        }
      }
    if (!f)
      cout << "NO\n";
    else
      cout << "YES\n" << ans1 << ' ' << ans2 << ' ' << ans3 << endl;
  }
  return 0;
}