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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
    public static void main(String ags[])throws Exception{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        StringBuilder ans=new StringBuilder();
        int t=Integer.parseInt(st.nextToken());
        for(int f=0;f<t;f++){
            st=new StringTokenizer(br.readLine());
            int a[]=new int[3];
            a[0]=Integer.parseInt(st.nextToken());
            a[1]=Integer.parseInt(st.nextToken());
            a[2]=Integer.parseInt(st.nextToken());
            Arrays.sort(a);
            if(a[0]!=a[1] && a[1]!=a[2]){
                ans.append("NO\n");
            }
            else if(a[0]==a[1] && a[1]==a[2]){
                ans.append("YES\n");
                ans.append(a[0]+" "+a[0]+" "+a[0]+"\n");
            }
            else if(a[0]==a[1] && a[1]!=a[2]){
                ans.append("NO\n");
            }
            else
            {
                ans.append("YES\n");
                ans.append(a[0]+" "+a[2]+" "+1+"\n");
            }
        }
        System.out.print(ans);
    }
}