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.*;
public class Template {
	
	public static void main (String[] args) throws java.lang.Exception
	{
	
	Scanner sc=new Scanner(System.in);
	int t=sc.nextInt();
	while(t-->0) {
		long n=sc.nextLong();
	long k=sc.nextLong();
		ArrayList<Long> arr=new ArrayList();
		
	for(long i=0;i<n;i++ ) {
			long c=sc.nextLong();
			arr.add(c);
		}
	
	long max=Collections.max(arr);
	long min=Collections.min(arr);
	if(k%2!=0){
	for(long p=0;p<n;p++) {
		arr.set((int) p,max-arr.get((int) p));
	}
	}
	else{
	 for(long p=0;p<n;p++) {
		arr.set((int) p,arr.get((int) p)-min);
	}   
	}
	
	
	for(long o=0;o<n;o++) {
		System.out.print(arr.get((int) o)+" ");
	}
	System.out.println();
		
		
}
	
	}
	

}