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.*;
import java.lang.*;
import java.io.*;
public class Three_pairwise_maximums {

	public static void main(String[] args)  throws java.lang.Exception
	{		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine());
		while(t-->0) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int x=Integer.parseInt(st.nextToken()),y=Integer.parseInt(st.nextToken()),z=Integer.parseInt(st.nextToken());
			if(x==y) {
				if(z<=x) System.out.println("YES\n"+x+" "+z+" "+z);
				else System.out.println("NO");
			}else if(x==z) {
				if(y<=x) System.out.println("YES\n"+y+" "+x+" "+y);
				else System.out.println("NO");
			}else if(y==z) {
				if(x<=y) System.out.println("YES\n"+x+" "+x+" "+y);
				else System.out.println("NO");
			}else System.out.println("NO");
		}

	}

}