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 Clock {
    
    public static void main (String[] args) {
        
        Scanner scan = new Scanner (System.in);
        int T = scan.nextInt();
        for (int l = 0; l < T; l++) {
            
            int n = scan.nextInt();
            long k = scan.nextLong();
            long[] arr = new long[n];
            for (int a = 0; a < n; a++) arr[a] = scan.nextLong();
            if (k % 2 == 0) show (solve(n, solve(n, arr)), n);
            else show (solve(n, arr), n);
        }
    }
    public static long[] solve (int n, long[] arr) {
        
        long max = Long.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            
            if (max < arr[i]) max = arr[i];
        }
        for (int i = 0; i < n; i++) {
            
            arr[i] = max - arr[i];
        }
        return arr;
    }
    public static void show (long[] arr, int n) {
        
        for (int i = 0; i < n - 1; i++) System.out.print(arr[i] + " ");
        System.out.println(arr[n - 1]);
    }
}