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);
  long long t;
  cin >> t;
  map<long long, long long> cn;
  set<long long> fac;
  while (t--) {
    fac.clear();
    cn.clear();
    long long n;
    cin >> n;
    long long temp = n;
    for (long long i = 2; i * i <= n; i++) {
      while (n % i == 0) {
        cn[i]++;
        n /= i;
        fac.insert(i);
      }
    }
    if (n > 1) cn[n]++, fac.insert(n);
    n = temp;
    if (fac.size() >= 3) {
      cout << "YES\n";
      long long a = *fac.begin(), b = *(++fac.begin());
      cout << a << " " << b << " " << n / (a * b) << "\n";
      continue;
    } else if (fac.size() == 1 && cn[*(fac.begin())] >= 6) {
      cout << "YES\n";
      cout << *(fac.begin()) << " " << (*(fac.begin())) * (*(fac.begin()))
           << " "
           << n / ((*(fac.begin())) * (*(fac.begin())) * (*(fac.begin())))
           << "\n";
      continue;
    }
    if (fac.size() == 2) {
      long long a, ac, b, bc;
      a = *(fac.begin());
      ac = cn[*(fac.begin())];
      b = *(++fac.begin());
      bc = cn[*(++fac.begin())];
      if (ac + bc >= 4) {
        cout << "YES\n";
        cout << a << " " << b << " " << n / (a * b) << "\n";
        continue;
      }
    }
    cout << "NO\n";
  }
}