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.security.SecureRandom;
import java.sql.Array;
import java.util.*;

public class Main {
    static void printArray(long arr[], long n) {
        for (long i = 0; i < n; i++) {
            System.out.print(arr[(int)i] + " ");
        }
        System.out.println();
    }

    // Function to remove the minimum value of
    // the array from every element of the array
    static void removeMin(long arr[], long n) {
        long i, minVal = arr[0];

        // Get the minimum value from the array
        for (i = 1; i < n; i++)
            minVal = Math.min(minVal, arr[(int)i]);

        // Remove the minimum value from
        // every element of the array
        for (i = 0; i < n; i++)
            arr[(int)i] = arr[(int)i] - minVal;
    }

    // Function to remove every element of the
    // array from the maximum value of the array
    static void removeFromMax(long arr[], long n) {
        long i, maxVal = arr[0];

        // Get the maximum value from the array
        for (i = 1; i < n; i++)
            maxVal = Math.max(maxVal, arr[(int)i]);

        // Remove every element of the array from
        // the maximum value of the array
        for (i = 0; i < n; i++)
            arr[(int)i] = maxVal - arr[(int)i];
    }

    // Function to print the modified array
    // after k operations
    static void modifyArray(long arr[], long n, long k) {

        // If k is odd then remove the minimum element
        // of the array from every element of the array
        if (k % 2 == 0)
            removeMin(arr, n);

            // Else remove every element of the array from
            // the maximum value from the array
        else
            removeFromMax(arr, n);

        // Print the modified array
        printArray(arr, n);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long t = sc.nextLong();
        while (t-- >= 1) {
            long n = sc.nextLong();
            long k = sc.nextLong();
            long a[] = new long[(int) n];
            for (long i = 0; i < n; i++) {
                a[(int) i] = sc.nextLong();
            }
            modifyArray(a, n, k);

        }
    }
}