Problem: You are given three positive (i.e. strictly greater than zero) integers x, y and z.

Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.

You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).

Output

For each test case, print the answer:

  * "NO" in the only line of the output if a solution doesn't exist; 
  * or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order. 

Example

Input


5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000


Output


YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000

Solution (in JAVA):

Solution:
import java.util.*;
public class ThreePairwiseMaximums {
	public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		for(int t1=1;t1<=t;t1++)
		{
			long x=sc.nextLong();
			long y=sc.nextLong();
			long z=sc.nextLong();
			long a,b,c;
			if(x==y)
			{
				a=x;
				b=z;
				c=Math.min(a,b);
				if(z>x)
					System.out.println("NO");
				else
				{
					System.out.println("YES");
					System.out.print(a+" "+b+" "+c);
					System.out.println();
				}
			}
			else
			{
				b=x;
				c=y;
				a=Math.min(b,c);
				if(z!=Math.max(b,c))
					System.out.println("NO");
				else
				{
					System.out.println("YES");
					System.out.print(a+" "+b+" "+c);
					System.out.println();
				}
			}
			
			
		}
		sc.close();
	}

}