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 OmkarInfinityClock {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t!=0){
            int n=sc.nextInt();
         long k=sc.nextLong();
           long a[]=new long[n];
            for (int i = 0; i < n; i++) {
                a[i]=sc.nextInt();
            }
            if(k%2==0){
                for (int i = 1; i <= 2; i++) {
               long max=a[0];
                for (int j = 1; j < n; j++) {
                    if(max<a[j]){
                        max=a[j];
                    }
                }
                for (int j = 0; j < n; j++) {
                    a[j]=max-a[j];
                }
            }
                for (int i = 0; i < n; i++) {
                    System.out.print(a[i]+" ");
                }}
            else {

                long max=a[0];
                for (int i = 1; i < n; i++) {
                    if(max<a[i]){
                        max=a[i];
                    }
                }
                for (int i = 0; i < n; i++) {
                    System.out.print(max-a[i]+" ");
                }
            }
            t--;
            System.out.println();
        }
    }
}