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 threePairwiseMax {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		for(int tt=0;tt<t;tt++)
		{
			int x = sc.nextInt();
			int y = sc.nextInt();
			int z = sc.nextInt();
			if((x==y && x<z) || (x==z && x<y) || (y==z && y<x))
				{
					System.out.println("NO");
				}
			else if(x>Math.max(y, z) || y>Math.max(x, z) || z > Math.max(x, y))
				System.out.println("NO");
			else {
				System.out.println("YES");
				
				if(x==y && y==z)
					System.out.println(x + " " + x + " " + x);
				
				else if(x==y)
					{
					System.out.println(x + " " + z + " " + z);
					}
				else if(x==z)
					System.out.println(y + " " + x + " " + y);
				else if(y==z)
					System.out.println(x + " " + x + " " + y);
			}
		}
	}

}