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 code1{
  public static void main(String args[]){
         Scanner sc=new Scanner(System.in);
         int t=sc.nextInt();
         while(t-->0){
             List<Long> list=new ArrayList<>();
            int n=sc.nextInt();
            long k=sc.nextLong(); 
              for(long i=0;i<n;i++)
              {long a=sc.nextInt();
                list.add(a);}
                //System.out.println(list);
             long max=Collections.max(list);
             if(k%2==1){
             
                for(int i=0;i<n;i++)
                {System.out.print(max-list.get(i)+" ");}}
                else
                {
                    long min=Collections.min(list);
                   
                 for(int i=0;i<n;i++)
                {System.out.print(list.get(i)-min+" ");} 
                }
                System.out.println();
            }
    }
}