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.Arrays;
import java.util.Scanner;
import java.util.Stack;

import javax.sound.sampled.ReverbType;

import java.util.ArrayList;
import java.util.List;
import java.lang.reflect.Array;
import java.math.*;

public class Sec4 {

	public static void main(String[] args) {

		Scanner read = new Scanner(System.in);
		int t = read.nextInt();
		for (int i = 0;i<t;i++)
		{
		long x =read.nextInt();
		long y =read.nextInt();
		long z =read.nextInt();
		
		
		long a =0,b=0,c=0;
		if (x==y&&x==z)
		{
			a=x;
			b=x;
			c=x;
		}else if (x==y&&y!=z&&x>z)
		{
			a=x;
			b=z;
			c=z;
		}else if (x==z &&x!=y&&x>y)
		{
			b=x;
			a=y;
			c=y;
		}else if (y==z && x!=y&&y>x)
		{
			b=y;
			a=x;
			c=x;
			
			
		}
		if (a!=0&&b!=0&&c!=0)
		{
			System.out.println("YES");
			System.out.println(a+" "+b+" "+c);
		}else
		{
			System.out.println("NO");
		}
		}
	}
	
	



	
}