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.*;
import java.io.*;
public class MyClass {
    static PrintWriter w;
    public static void main(String args[]) {
      
        Scanner sc = new Scanner(System.in);
         w=new PrintWriter(System.out);
        
        int t=Integer.parseInt(sc.nextLine());
        
       A: while(t-->0){
        int n=sc.nextInt();
        long k=sc.nextLong();
        int[] ar = new int[n];
       for(int i=0; i<n; i++){
           ar[i]=sc.nextInt();
       }
      
           int[] odd = util(ar, n);
            int[] even=util(odd,n);
       if(k%2==0)
       print(even,n);
       else
       print(odd,n);
       
           
       }
        w.flush();
        w.close();
    }
    static void print(int[] ar, int n){
        for(int i=0; i<n; i++)
        w.print(ar[i]+ " ");
        w.println();
        return;
    }
    static int[] util(int[] ar, int n){
        int max=ar[0];
        int[] temp = new int[n];
        for(int i=1; i<n; i++)
        max=Math.max(max,ar[i]);
        for(int i=0; i<n; i++){
            temp[i]=max-ar[i];
            
        }
       return temp;
    }
 
}