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.Scanner;

public class OmkarAndInfinityClock {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
      Scanner scn=new Scanner(System.in);
      int t=scn.nextInt();
     // int test=1;
      while(t>0)
      {
    	  int n=scn.nextInt();
    	  long k=scn.nextLong();
    	  long[] arr=new long[n];
    	  long max=Long.MIN_VALUE;
    	  long min=Long.MAX_VALUE;
    	  for(int i=0;i<n;i++)
    	  {
    		  arr[i]=scn.nextLong();
    		  if(i==0)
    		  {
    			 max=arr[i];
    			 min=arr[i];
    		  }
    		  else{
    			  if(arr[i]>max)
    				  max=arr[i];
        		  else if(arr[i]<min)
        			  min=arr[i];
    		  }	  
    	  }
    	  getAns(arr,k,min,max);
    	  
    	  for(long aa:arr)
    		  System.out.print(aa+" ");
    	  
    	  System.out.println();
    	  t--;
      }
	}
	private static void getAns(long[] arr, long k,long min,long max)
	{
		if(k==1)
		{
			for(int i=0;i<arr.length;i++)
				arr[i]=max-arr[i];
			return;
		}
		else if(k>1)
		{
			for(int i=0;i<arr.length;i++)
				arr[i]=max-arr[i];
			
			if(k%2!=0)
				return;
			else {
				max=max-min;
				//System.out.println(max);
				for(int i=0;i<arr.length;i++)
					arr[i]=max-arr[i];
				return;
			}
		}
		
		
	}
}