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;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.ArrayList;


public class main {
	
	public static long maxValue(long array[]){
		  List<Long> list = new ArrayList<Long>();
		  for (int i = 0; i < array.length; i++) {
		    list.add(array[i]);
		  }
		 return Collections.max(list);
		
	}
	
	
	public static long[] operateArray(long[] array, int n, long max) {
		long newArray[], sortedArray[];
		
		newArray = new long[n];
		
		
		//long max = maxValue(array);
		int i;
		
		for (i=0; i<n; i++) {
			newArray[i] = max - array[i];
			
		}
		
		return newArray;
	}
	
	
	
	public static void main(String[] args) {
		
		
		Scanner read = new Scanner(System.in);
		
		int t = read.nextInt(), i,j, n;
		long k, array[], newarray[], l;
		
		for (i=0; i<t; i++) {
			n = read.nextInt();
			k = read.nextLong();
			
			array = new long[n];
			newarray = new long[n];
			
			for(j=0; j< n; j++) {
				array[j] = read.nextLong();
				newarray[j] = array[j];
			}
			
			l = 0;
			
			/*while (l < k) {
				long max = maxValue(array);
				newarray = operateArray(array, n, max);
				array = newarray;
				l++;
			}*/
			long max = maxValue(array);
			newarray = operateArray(array, n, max);
			array = newarray;
			
			if (k% 2 == 0) {
				max = maxValue(array);
				newarray = operateArray(array, n, max);
				array = newarray;
			}
			
			
			
			for(j=0; j< n; j++) {
				System.out.printf("%d ", newarray[j]);
			}
			System.out.printf("\n");
		}
	
	}
}