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 omkarandinfinityclock {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        try {

            int t = sc.nextInt();
            while (t-- > 0) {
                long n = sc.nextInt();
                long k = sc.nextLong();
                long a[] = new long[(int) n];
                long max = Integer.MIN_VALUE;
                for (int i = 0; i < n; i++) {
                    a[i] = sc.nextInt();
                    if (a[i] > max)
                        max = a[i];
                }
                long max2 = Integer.MIN_VALUE;
                if (k % 2 == 0) {
                    for (int i = 0; i < 2; i++) {

                        for (int j = 0; j < n; j++) {
                            a[j] = max - a[j];
                            if (max2 < a[j]) {
                                max2 = a[j];
                            }

                        }

                        max = max2;
                    }
                    for (int i = 0; i < n; i++) {
                        System.out.print(a[i] + " ");
                    }
                } else {


                    for (int j = 0; j < n; j++) {
                        a[j] = max - a[j];
                        System.out.print(a[j] + " ");
                    }
                }
                System.out.println();
            }


        } catch (Exception e) {
            return;
        }
    }
}