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:
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;
import java.lang.*;

public class GFG {
	public static void main (String[] args) {Scanner sc=new Scanner(System.in);
	         int t=sc.nextInt();
	         for(int i=0;i<t;i++)
	            {int n=sc.nextInt();
	            long k=sc.nextLong();
	             long a[]=new long[n];
	             for(int j=0;j<n;j++)
	                 a[j]=sc.nextInt();
	                 long b[]=new long [n];
	                 long c[]=new long[n];
	                 long max1=-1000000007;
	                 for(int j=0;j<n;j++)
	                      max1=Math.max(max1,a[j]);
	                      for(int j=0;j<n;j++)
	                         b[j]=max1-a[j];
	                         max1=-1000000007;
	                         for(int j=0;j<n;j++)
	                      max1=Math.max(max1,b[j]);
	                      for(int j=0;j<n;j++)
	                         c[j]=max1-b[j];
	                         if(k%2==1)
	                           {for(int j=0;j<n;j++)
	                          System.out.print(b[j]+" ");
	                           System.out.println();}
	                           else {for(int j=0;j<n;j++)
	                          System.out.print(c[j]+" ");
	                           System.out.println();}
	                          
		
	}
}}