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.lang.Math;

//System.out.println();


public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int ks = sc.nextInt();
        for (int i =0;i<ks;i++) {
        	int n = sc.nextInt();
        	long k = sc.nextLong();
        	int arr[] = new int[n];
        	int minimum = 1000000000;
        	int max = -1000000000;

        	for (int j = 0;j <n;j++) {
        		arr[j] = sc.nextInt();
        		if (arr[j] > max) {
        			max = arr[j];
        		}

        		if (arr[j] < minimum) {
        			minimum = arr[j];
        		}
        	}

        	int newMax = -1000000000;
        	for (int j = 0;j<n;j++) {
        		arr[j] = max - arr[j];
        		if (arr[j] > newMax) {
        			newMax = arr[j];
        		}
        	}

        	long temp = k - 1;

        	if (temp % 2 == 0) {
        		System.out.print(arr[0]);
        		for (int j = 1;j<n;j++) {
        			System.out.print(" " + arr[j]);
        		}
        		System.out.println("");
        	} else {
				System.out.print((newMax - arr[0]));
        		for (int j = 1;j<n;j++) {
        			System.out.print(" " + (newMax - arr[j]));
        		}
        		System.out.println("");       		
        	}


        }
    }
}