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 OmkarandPassword {

	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		
		int t=  in.nextInt();
		int i;
		
		for(i=0;i<t;i++) {
			int n=in.nextInt();
			long k = in.nextLong();
			
			long[] a = new long[n];
			int j;
			for(j=0;j<n;j++)
				a[j]=in.nextLong();
			
			long h=k;
			if(h>3) {
				if(h%2==0) {
					h=2;
				}else {
					h=3;
				}
			}
			k=h;
			
			while(k!=0) {
				
				long d = a[0];
				int e=0;
				for(e=0;e<n;e++)
				{
					if(d<a[e]) d=a[e];
				}
				for(e=0;e<n;e++)
				{
					a[e]=d-a[e];
				}
				k--;
			}
			
			
			for(j=0;j<n;j++)
				System.out.print(a[j]+" ");
			
			System.out.println();
		}
//		System.out.println("HE:LO");
		
		
		
	}
	
}