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 Solution {
    public static void main(String[] args) {
       /*
       x = max(a, b)
       y = max(a, c)
       z = max(b, c)

       case 1 : a > b && a > c
       -> x == y

       case 2 : b > a && b > c
       -> x == z

       case 3: c > a && c > b
       -> y == z
       */
     Scanner s = new Scanner(System.in);
     int t = s.nextInt();
     for (int i = 0; i < t; ++i){
         int x = s.nextInt(),
         y = s.nextInt(),
         z = s.nextInt();
         if (x == y && x >= z){
             System.out.println("YES");
             System.out.println(x+" "+z+" "+z);
         }else if (x == z && x > y){
             System.out.println("YES");
             System.out.println(y+" "+x+" "+y);
         }else if (y == z && y > x){
             System.out.println("YES");
             System.out.println(x+" "+x+" "+y);
         }else{
             System.out.println("NO");
         }
     }

    }
}