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 Solution
{
       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();
                    int arr[] = new int[n];
                    int max = Integer.MIN_VALUE;
                    for(int i = 0 ; i < n ; i++)
                    {
                        arr[i] = sc.nextInt();
                        max = Math.max(max , arr[i]);
                    }
                    
                    int ans1[] = new int[n];
                    int ans2[] = new int[n];
                int mx = Integer.MIN_VALUE;
                for(int i = 0 ; i < n ; i++)
                {
                    ans1[i] = max - arr[i];
                    mx = Math.max(mx , ans1[i]);
                }
                
                for(int i = 0 ; i < n ; i++)
                {
                    ans2[i] = mx - ans1[i];
                }
                StringBuffer str = new StringBuffer("");
                for(int i = 0 ; i < n ; i++)
                {
                    if(k%2 == 1L)
                    str.append(ans1[i]+" ");
                    
                    else
                    str.append(ans2[i] + " ");
                }
                System.out.println(str);
              }
       }
}