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 ThreePairwiseMaximums {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		if(1 <= t && t <= 20000) {
			for(int i=1;i<=t;i++) {
			int x =sc.nextInt();
			int y =sc.nextInt();
			int z =sc.nextInt();
			if(x==y && x>=z) {
				System.out.println("Yes");
				System.out.println(z + " " + z + " " + x);
			}
			else if(x==z && x>=y) {
				System.out.println("Yes");
				System.out.println(y+ " " + y +" "+ x);
			}
			else if(y==z && y>=x) {
				System.out.println("Yes");
				System.out.println(x + " " + x + " " + y );
			}
			else {
				System.out.println("No");
			}
		}
		}
	}

}