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 Solution {
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int t = scan.nextInt();
		while(t-->0) {
			int n = scan.nextInt();
			long ops = scan.nextLong();
			if(n==1) {
				scan.nextInt();
				System.out.println("0");
				continue;
			}
			if(ops%2==0) ops=2;
			else ops = 1;
			int[] ar = new int[n];
			int max = Integer.MIN_VALUE;
			int min = Integer.MAX_VALUE;
			for(int i=0;i<n;i++) {
				ar[i]=scan.nextInt();
				if(max<ar[i]) max = ar[i];
			}
			
			while(ops-->0) {
				for(int i=0;i<n;i++) {
					if(min>ar[i]) min = ar[i];
					ar[i]=max-ar[i];
				}
				max = max-min;
				min = Integer.MAX_VALUE;
			}
			for(int i=0;i<n;i++) {
				System.out.print(ar[i]+" ");
			}
			System.out.println("");
		}
		scan.close();
	}
}