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:
//package round_Global_10;

import java.util.Scanner;

public class B {
    public static void main(String[] args){
        Scanner s=new Scanner(System.in);
        int a=s.nextInt();
        s.nextLine();
        for (int p=0;p<a;p++){
            String[] l=s.nextLine().split(" ");
            long n = Long.parseLong(l[0]);
            long k=Long.parseLong(l[1]);
            String[] t=s.nextLine().split(" ");
            long[] list = new long[t.length];
            for (int i=0;i< list.length;i++){
                list[i]=Long.parseLong(t[i]);
            }
            long max=list[0];
            for (long i:list){
                if (i>max){
                    max=i;
                }
            }
            long[] second=new long[list.length];
            for (int i=0;i<list.length;i++){
                second[i]=max-list[i];
            }
            max=0;
            for (long i:second){
                if (i>max){
                    max=i;
                }
            }
            long[] third=new long[list.length];
            for (int i=0;i<list.length;i++){
                third[i]=max-second[i];
            }
            if (k%2==1){
                for (long j:second){
                    System.out.print(j+" ");
                }
            }
            else{
                for (long j:third){
                    System.out.print(j+" ");
                }

            }
            System.out.println();
        }
    }
}