Problem: Being stuck at home, Ray became extremely bored. To pass time, he asks Lord Omkar to use his time bending power: Infinity Clock! However, Lord Omkar will only listen to mortals who can solve the following problem:

You are given an array a of n integers. You are also given an integer k. Lord Omkar wants you to do k operations with this array.

Define one operation as the following: 

  1. Set d to be the maximum value of your array. 
  2. For every i from 1 to n, replace a_{i} with d-a_{i}. 



The goal is to predict the contents in the array after k operations. Please help Ray determine what the final sequence will look like!

Input

Each test contains multiple test cases. The first line contains the number of cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 10^{18}) – the length of your array and the number of operations to perform.

The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (-10^9 ≤ a_{i} ≤ 10^9) – the initial contents of your array.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each case, print the final version of array a after k operations described above.

Example

Input


3
2 1
-199 192
5 19
5 -1 4 2 0
1 2
69


Output


391 0
0 6 1 3 5
0

Note

In the first test case the array changes as follows:

  * Initially, the array is [-199, 192]. d = 192.

  * After the operation, the array becomes [d-(-199), d-192] = [391, 0].

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class  Solver1 {
	public static void main(String args[]) throws IOException{
		
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

		
		int t=Integer.parseInt(br.readLine());
		
		while(t>0){
		   String str[]=br.readLine().split(" ");
		   int n=Integer.parseInt(str[0]);
		   long k=Long.parseLong(str[1]);
			
			long arr[]=new long[n];
			long max=-1111111111;
			long min=1111111111;
			String brr[]=br.readLine().split(" ");
			for(int i=0;i<n;i++){
				arr[i]=Long.parseLong(brr[i]);
				if(arr[i]>max)
					max=arr[i];
				if(arr[i]<min)
					min=arr[i];
			}
			long temp=max-min;
			for(int i=0;i<n;i++){
				if(k%2==1 )
					System.out.print(max-arr[i]+" ");
				else
					System.out.print(temp-max+arr[i]+" ");
				
			}
			System.out.println("");
					
		t--;	
	    }
	}
}