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.io.*;
import java.util.*;

public class SixFiveSixA {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while(t-->0) {
			int a = in.nextInt();int b = in.nextInt(); int c = in.nextInt();
			if(a==b&&b==c&&a==c)
			{
				System.out.println("YES");
				System.out.println(a+" "+a+" "+a);
			}
			else {
				int max = (a>b)?(a>c)?a:c : (b>c)?b:c;
				int least = (a<b)?(a<c)?a:c : (b<c)?b:c;
				int count = 0; if(max == a) count++; if(max == b) count++; if(max == c) count++;
				if(count<2)
					System.out.println("NO");
				else {
					System.out.println("YES");System.out.println(max+" "+least+" "+least);
				}
			}
		}
	}

}