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.Scanner;

public class pairwise_max {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		for(int i=0;i<t;i++) {
			int x=s.nextInt();
			int y=s.nextInt();
			int z=s.nextInt();
			
			if(x==y && y==z) {
				System.out.println("YES");
				System.out.println(x+" "+" "+y+" "+z );
			}
			else if(x!=y && y!=z && x!=z) {
				System.out.println("NO");
			}
			else {
				if(x==z) {
					if(y>x) {
						System.out.println("NO");
					}
					else {
						System.out.println("YES");
						System.out.println(y+" "+x+" "+y);
					}
				}
				else if(y==z) {
					if(y<x) {
						System.out.println("NO");
					}
					else {
						System.out.println("YES");
						System.out.println(x+" "+x+" "+y);
					}
				}
				else {
					if(z>y) {
						System.out.println("NO");
					}
					else {
					System.out.println("YES");
					System.out.println(z+" "+x+" "+z);
				}
				}
				}
			}
		}

	}