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.io.*;
import java.util.*;
public class Main
{
	public static void main(String[] args)throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int t=Integer.parseInt(br.readLine());
	    while(t-->0)
	    {
	        String s1[]=(br.readLine()).split(" ");
	        int n=Integer.parseInt(s1[0]);
	        long k=Long.parseLong(s1[1]);
	        String s[]=(br.readLine()).split(" ");
	        long a[]=new long[n];
	        long min=Long.MAX_VALUE;
	        long max=min;
	        max-=-1;
	        for(int i=0;i<n;i++)
	        {
	            a[i]=Long.parseLong(s[i]);
	            if(a[i]>max)
	            {
	                max=a[i];
	            }
	            if(a[i]<min)
	            {
	                min=a[i];
	            }
	        }
	       // Arrays.sort(a);
	        
	       // long odd[]=new long[n];
	        for(int i=0;i<n;i++)
	        {
	            a[i]=max-a[i];
	        }
	        if(k%2l==1l)
	        {
	            for(int i=0;i<n;i++)
	            {
	                pw.print(a[i]+" ");
	            }
	            pw.println();
	            continue;
	        }
	        
	        max=max-min;
	        
	       // long even[]=new long[n];
	        for(int i=0;i<n;i++)
	        {
	            a[i]=max-a[i];
	        }
	        if(k%2l==0l)
	        {
	            for(int i=0;i<n;i++)
	            {
	                pw.print(a[i]+" ");
	            }
	            pw.println();
	            continue;
	        }
	        
	    }
	    pw.flush();
	}
}