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 ThreePair{
	public static void main(String[]args){
		Scanner Sc=new Scanner(System.in);
		int T=Sc.nextInt();
		for(int i=0;i<T;i++){
			int x=Sc.nextInt();
			int y=Sc.nextInt();
			int z=Sc.nextInt();
			int count=0;
			if(x==y){
				count++;
			}
			if(x==z){
				count++;
			}
			if(y==z){
				count++;
			}
			int max=x;
			if(y>max){
				max=y;
			}
			if(z>max){
				max=z;
			}
			int notMax=0;
			
				
			
			if(count==0){
				System.out.println("NO");
			}else{
				if(count==3){
					System.out.println("YES");
					System.out.print(x+" "+x+" "+x+"\n");
				}else{
					int count1=0;
					if(x==max){
						count1++;
					}
					if(y==max){
						count1++;
					}
					if(z==max){
						count1++;
					}
					if(count1==1){
						System.out.println("NO");
					}else{
						if(x!=max){
							notMax=x;
						}
						if(y!=max){
							notMax=y;
						}
						if(z!=max){
							notMax=z;
						}
						System.out.println("YES");
						System.out.print(max+" "+(notMax)+" "+(notMax)+"\n");
					}
				}
			}
	
		}
	}
}