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.ArrayList;
import java.util.Scanner;

public class abca {

	public static void main(String[] args) 
	{
		Scanner cin = new Scanner(System.in);
		int t=cin.nextInt();
		for (int i = 0; i < t; i++)
		{
			int n=cin.nextInt();
			long k= cin.nextLong();
			ArrayList<Long> arr= new ArrayList<>();
			arr.add(cin.nextLong());
			long max=arr.get(0);
			long min=arr.get(0);
			for (int j =1; j <n; j++)
			{
				arr.add(cin.nextLong());
				max=Math.max(max,arr.get(j));
				min=Math.min(min,arr.get(j));
			}
			StringBuffer str= new StringBuffer();
			if(k%2==1)
			{
				for (int j = 0; j <n; j++)
				{
					arr.set(j,max-arr.get(j));
					str.append(arr.get(j));
					str.append(" ");
				}
			}
			else
			{
				for (int j = 0; j <n; j++)
				{
					arr.set(j,max-arr.get(j));
				}
				max=max-min;
				for (int j = 0; j < n; j++)
				{
					arr.set(j,max-arr.get(j));
					str.append(arr.get(j));
					str.append(" ");
				}
			}
			if(n==1)
			{
				System.out.println("0");
			}
			else
			{
			System.out.println(str);
			}
		}
	}

}