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 Main {	
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		
		int T = Integer.parseInt(br.readLine());
		
		for (int t = 0; t < T; t++) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			long k = Long.parseLong(st.nextToken());
			
//			ArrayList<Pair> m = new ArrayList<>();
			int[] arr = new int[n];
			st = new StringTokenizer(br.readLine());

			for (int i = 0; i < n; i++) {
				int val = Integer.parseInt(st.nextToken());
//				m.add(new Pair(val,i));
				arr[i] = val;
			}
			
//			Collections.sort(m);
			
			k %= 2;
			//if 1 then do one
			//if 2 then do two operations
			int max = arr[0];
			for (int i = 1; i < n; i++) {
				if (arr[i] > max) {
					max = arr[i];
				}
			}
			if (k == 1) {				
//				int max = m.get(n-1).a;				
//				for (int i = 0; i < n; i++) {
//					m.set(i, new Pair(m.get(i).index,max - m.get(i).a));
//				}
				for (int i = 0; i < n; i++) {
					arr[i] = max- arr[i];
				}
			}
			if (k== 0) {
//				int max = m.get(n-1).a;				
//				for (int i = 0; i < n; i++) {
//					m.set(i, new Pair(max - m.get(i).a,m.get(i).index));
//				}
//				max = m.get(0).a;
//				for (int i = 0; i < n; i++) {
//					m.set(i, new Pair(m.get(i).index,max - m.get(i).a));
//				}
				for (int i = 0; i < n; i++) {
					arr[i] = max- arr[i];
				}
				max = arr[0];
				for (int i = 1; i < n; i++) {
					if (arr[i] > max) {
						max = arr[i];
					}
				}
				
				for (int i = 0; i < n; i++) {
					arr[i] = max- arr[i];
				}
			}
			
			
			
//			for (Pair p: m) {
//				System.out.print(m.get(p.index).a + " ");
//			}
//			System.out.println(Arrays.toString(arr));
			for (int m: arr) {
				System.out.print(m + " ");
			}
			System.out.println();
		
		}
		
		
		
	}

}