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 Omkarinfi {
    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while (t-- > 0) {
            int len = sc.nextInt();
            long k = sc.nextLong();
            long[] a = new long[len];
            for (int i = 0; i < len; i++) {
                a[i] = sc.nextLong();
            }
            print(a, k);
            System.out.println();
        }
    }

    static void print(long[] a, long k) {
        long[] p = a.clone();
        Arrays.sort(p);
        long max = p[p.length - 1];
        long min = p[0];
        long dminmax = max - min;
        if (k == 0) {
            for (int i = 0; i < a.length; i++) {
                System.out.print(a[i] + " ");
            }
        } else if (k == 1) {
            for (int i = 0; i < a.length; i++) {
                a[i] = max - (a[i]);
                System.out.print(a[i] + " ");
            }
        } else {
            for (int i = 0; i < a.length; i++) {
                a[i] = max - (a[i]);
            }
            k--;
            if (k % 2 != 0) {

                for (int i = 0; i < a.length; i++) {
                    a[i] = dminmax - (a[i]);
                    System.out.print(a[i] + " ");
                }

            } else {
                for (int i = 0; i < a.length; i++) {
                    System.out.print(a[i] + " ");
                }
            }
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}