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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Maij
{
	public static void main(String[] args)
	{
		new Thread(null, new Runnable() {
			public void run() {
                solve();
            }
        }, "1", 1 << 26).start();
	}
	static void solve () {
		FastReader fr =new FastReader();	PrintWriter op =new PrintWriter(System.out);
 
 		int t =fr.nextInt() ,n ,fct[][] ,i ,j ,k ;
 		while (t-- > 0) {
 			n =fr.nextInt() ;	fct =new int[(int)1e5+5][2] ;	j =-1 ;
 			if (n%2 == 0) {
 				fct[++j][0] =2 ;	k =0 ;
 				while (n%2 ==0 ) {	n/=2 ;	++k ;	}
 				fct[j][1] =k ;
 			}
 			for (i =3 ; i*i<=n ; i+=2) {
 				if (n%i == 0) {
 					fct[++j][0] =i ;	k =0 ;
 					
 					while (n%i ==0 ) {	n/=i ;	++k ;	}
 					fct[j][1] =k ;
 				}
 			}
 			if (n>2) {
 				fct[++j][0] =n ;	fct[j][1] =1 ;
 			}
 			if (j>=2) {
 				op.println("YES") ;
 				op.print((int)Math.pow(fct[0][0],fct[0][1])+" ") ;
 				op.print((int)Math.pow(fct[1][0],fct[1][1])+" ") ;
 				i =1 ;
 				for (k =2 ; k<=j ; ++k)	i *= (int)Math.pow(fct[k][0],fct[k][1]) ;
 					op.print(i+" ") ;	op.println() ;
 			}
 			else {
 				if (j==1) {
 					if (fct[1][1]>fct[0][1]) {
 						fct[1][1] ^= fct[0][1] ;	fct[0][1] ^= fct[1][1] ;	fct[1][1] ^= fct[0][1] ;
 						fct[1][0] ^= fct[0][0] ;	fct[0][0] ^= fct[1][0] ;	fct[1][0] ^= fct[0][0] ;
 					}
 					if (fct[1][1]+fct[0][1] > 3) {
 						op.println("YES") ;
 						i =(int)Math.pow(fct[0][0],1) ;
 						j =(int)Math.pow(fct[1][0],1) ;
 						k =(int)Math.pow(fct[0][0],--fct[0][1]) * (int)Math.pow(fct[1][0] , --fct[1][1]) ;
 						op.println(i+" "+j+" "+k) ;
 					}
 					else 	op.println("NO") ;
 				}
 				else {
 					if (fct[0][1]>5) {
 						op.println("YES") ;
 						i =(int)Math.pow(fct[0][0] , 1) ;	--fct[0][1] ;
 						j =(int)Math.pow(fct[0][0] , 2) ;	fct[0][1] -= 2 ;
 						k =(int)Math.pow(fct[0][0] , fct[0][1]) ;
 						op.println(i+" "+j+" "+k) ;
 					}
 					else 	op.println("NO") ;
 				}
 			}
 		}
		op.flush();	op.close();
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br =new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st==null || (!st.hasMoreElements())) 
			{
				try
				{
					st =new StringTokenizer(br.readLine());
				}
				catch(IOException e)
				{
					e.printStackTrace();
				}
				
			}
			return st.nextToken();
		}

		String nextLine() {
			String str ="";

			try
			{
				str =br.readLine();
			}
			catch(IOException e)
			{
				e.printStackTrace();
			}

			return str;
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next()) ;
		}
	}
}