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 javax.lang.model.type.IntersectionType;

import java.io.*;

public class A {
//	static ArrayList<Integer>[] g;

	static int[] a;
	static long[] memo;
	static int n, m, d;

	static long dp(int idx) {
		if (idx >= n)
			return 0;
		if (memo[idx] != -1)
			return memo[idx];
		long ans = 0;
		if (a[idx] <= m) {
			ans = a[idx] + dp(idx + 1);
		} else {
			ans = dp(idx + 1);
			ans = Math.max(ans, a[idx] + dp(idx + d));
		}
		return memo[idx] = ans;
	}

	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {

			StringTokenizer st = new StringTokenizer(br.readLine());
			n = Integer.parseInt(st.nextToken());
			long k = Long.parseLong(st.nextToken());
			st = new StringTokenizer(br.readLine());
			int[] a = new int[n];
			int max=Integer.MIN_VALUE;
			for (int i = 0; i < a.length; i++) {
				a[i] = Integer.parseInt(st.nextToken());
				max=Math.max(max, a[i]);
			}
			int max2=Integer.MIN_VALUE;
			for (int i = 0; i < a.length; i++) {
				a[i]=max-a[i];
				max2=Math.max(max2, a[i]);
			}
			if (k%2==0)
				for (int i = 0; i < a.length; i++) {
					a[i]=max2-a[i];
				}
			for (int i = 0; i < a.length; i++) {
				pw.print(a[i]+" ");
			}
			pw.println();
		}

		pw.flush();
	}

	static ArrayList<Integer> intersect(ArrayList<Integer> a, ArrayList<Integer> b) {
		ArrayList<Integer> res = new ArrayList<>();
		if (b.size() != 0) {
			HashSet<Integer> hm = new HashSet<>();
			for (int x : a)
				hm.add(x);
			for (int x : b)
				if (hm.contains(x))
					res.add(x);
		}
		return res;
	}

	static int maxSize;
	static boolean[][] exist;
	static ArrayList<Integer>[] startAt, endAt;

	static void compress(pair[] arr) {
		TreeSet<Integer> tr = new TreeSet<>();
		for (int i = 0; i < arr.length; i++) {
			tr.add(arr[i].x);
			tr.add(arr[i].y);
		}
		tr.add(0);
		tr.add((int) 1e9);
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : tr)
			hm.put(x, hm.size());
		for (int i = 0; i < arr.length; i++) {
			arr[i].x = hm.get(arr[i].x);
			arr[i].y = hm.get(arr[i].y);
		}
		maxSize = hm.size();
	}

	static class pair implements Comparable<pair> {
		int x;
		int y;

		public pair(int d, int u) {
			x = d;
			y = u;
		}

		@Override
		public int compareTo(pair o) {
			// TODO Auto-generated method stub
			int x1 = y - x;
			int x2 = o.y - o.x;
			return x1 - x2;
		}

	}

}