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.io.*;
import java.util.*;

public class CF1392B extends PrintWriter {
	CF1392B() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1392B o = new CF1392B(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int k = sc.nextLong() % 2 == 1 ? 1 : 2;
			int[] aa = new int[n];
			for (int i = 0; i < n; i++)
				aa[i] = sc.nextInt();
			while (k-- > 0) {
				int d = (int) -2e9;
				for (int i = 0; i < n; i++)
					d = Math.max(d, aa[i]);
				for (int i = 0; i < n; i++)
					aa[i] = d - aa[i];
			}
			for (int i = 0; i < n; i++)
				print(aa[i] + " ");
			println();
		}
	}
}