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 Three_Pairwise_Maximums {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            if(a == b && a >= c) {
                System.out.println("YES");
                System.out.println(a + " " + c + " " + c);
            }
            else if(a == c && a >= b) {
                System.out.println("YES");
                System.out.println(a + " " + b + " " + b);
            }
            else if(b == c && b >= a) {
                System.out.println("YES");
                System.out.println(b + " " + a + " " + a);
            }
            else {
                System.out.println("NO");
            }
        }
        sc.close();
    }    
}