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 
 
import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
import java.util.Map.*;
public class codeforces {
      public static void main(String [] args) throws IOException, InterruptedException {
	    Scanner sc=new Scanner(System.in);
	    PrintWriter pw=new PrintWriter(System.out);
	   int t=sc.nextInt();
	   
	    while(t-->0) {
	    	long y=sc.nextLong();
	    	TreeSet<Long> set=printDivisors(y);
	        set.pollFirst(); set.pollLast();
	    //  pw.println(set);
	        if(set.size()<3)pw.println("NO");
	        else {
	        	StringBuilder sb=new StringBuilder();
	        	boolean f=true;
	        	long ans=set.first();
	        	sb.append(ans+" ");
	        	long k=y/ans;
	        	TreeSet<Long> set2=printDivisors(k);
	        	set2.pollFirst(); set2.pollLast();
	        	while(!set2.isEmpty() && f) {
	        		if(set2.first()!=set2.last() && set2.first()!=ans) {pw.println("YES"+"\n"+ans+" "+set2.first()+" "+set2.last()); f=false; }
	        		else if(set2.size()==1) break ;
	        		set2.pollFirst(); set2.pollLast();
	        	}
	        	if(f)pw.println("NO");
	        }
	    }
	    pw.flush();
		pw.close();
	}
  
      static TreeSet<Long> printDivisors(long n) 
      { 
    	  TreeSet<Long> set=new TreeSet<>();
          for (long i=1; i<=Math.sqrt(n); i++) 
          { 
              if (n%i==0) 
              { 
                  if (n/i == i) 
                     set.add(i);
         
                  else {set.add(i); set.add(n/i);} 
              } 
          } 
          return set;
      } 
  	}