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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class P2 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int T = s.nextInt();
        while (T-- >0) {
            long n = s.nextLong();
            long k = s.nextLong();
            List<Long> arr = new ArrayList<>();
            for(long i=0; i<n; ++i) {
                arr.add(s.nextLong());
            }
            Long min = Long.MAX_VALUE;
            Long max = Long.MIN_VALUE;
            for(Long val: arr) {
                min = Math.min(min, val);
                max = Math.max(max, val);
            }        
            if(k%2==1) {
                for(Long val: arr) {
                    System.out.format("%d ", max-val);
                }
            } else {
                for(Long val: arr) {
                    System.out.format("%d ", val-min);
                }
            }
            System.out.println();
        }
    }
}