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.InputStream;
import java.util.Scanner;

public class A656 {

    static int countOfDatasets;

    public static void main(String[] args) {
        readInput(System.in);
    }
    private static void readInput(InputStream s) {
        Scanner in = new Scanner(s);
        countOfDatasets=in.nextInt();
        for (int i = 0; i < countOfDatasets; i++) {
            int first=in.nextInt();
            int second=in.nextInt();
            int third=in.nextInt();

            if((first==second && first<third) ||
                    (first==third && first<second) ||
                    (second==third && first>second) ||
                    (first!=second && second!=third && first!=third)){
                System.out.println("NO");
            }else {
                System.out.println("YES");
                if(first==second && second==third){
                    System.out.println(first+" "+first+" "+first);
                }else {
                    int max=Math.max(Math.max(first,second),third);
                    int min=Math.min(Math.min(first,second),third);
                    System.out.println(max+" "+min+" "+min);
                }
            }

        }
    }
}