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 myCeil(long long num, long long divider) {
  long long result = num / divider;
  if ((result * divider) == num) return result;
  return result + 1;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    long long limit = sqrt(n) + 1;
    vector<long long> V;
    long long tempN = n;
    for (long long i = 2; i <= limit; i++) {
      if (tempN % i == 0) {
        V.push_back(i);
        tempN /= i;
        if (V.size() == 2) {
          V.push_back(tempN);
          break;
        }
      }
    }
    if (V.size() == 3) {
      if (V[1] < V[2]) {
        cout << "YES"
             << "\n";
        for (long long i = 0; i < 3; i++) {
          cout << V[i] << " ";
        }
        cout << "\n";
        continue;
      }
    }
    cout << "NO"
         << "\n";
  }
  return 0;
}