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 PYTHON3):

Solution:
#*****DO NOT COPY*****
#F
#U       ___                     ___                         ___           ___           ___           ___          _____
#C      /  /\      ___          /  /\          ___          /__/\         /  /\         /  /\         /  /\        /  /::\
#K     /  /::\    /  /\        /  /:/_        /  /\         \  \:\       /  /::\       /  /::\       /  /:/_      /  /:/\:\
#     /  /:/\:\  /  /:/       /  /:/ /\      /  /::\         \  \:\     /  /:/\:\     /  /:/\:\     /  /:/ /\    /  /:/  \:\
#Y   /  /:/~/:/ /__/::\      /  /:/ /::\    /  /:/\:\    ___  \  \:\   /  /:/~/::\   /  /:/~/:/    /  /:/ /:/_  /__/:/ \__\:|
#O  /__/:/ /:/  \__\/\:\__  /__/:/ /:/\:\  /  /:/~/::\  /__/\  \__\:\ /__/:/ /:/\:\ /__/:/ /:/___ /__/:/ /:/ /\ \  \:\ /  /:/
#U  \  \:\/:/      \  \:\/\ \  \:\/:/~/:/ /__/:/ /:/\:\ \  \:\ /  /:/ \  \:\/:/__\/ \  \:\/:::::/ \  \:\/:/ /:/  \  \:\  /:/
#    \  \::/        \__\::/  \  \::/ /:/  \  \:\/:/__\/  \  \:\  /:/   \  \::/       \  \::/~~~~   \  \::/ /:/    \  \:\/:/
#A    \  \:\        /__/:/    \__\/ /:/    \  \::/        \  \:\/:/     \  \:\        \  \:\        \  \:\/:/      \  \::/
#L     \  \:\       \__\/       /__/:/      \__\/          \  \::/       \  \:\        \  \:\        \  \::/        \__\/
#L      \__\/                   \__\/                       \__\/         \__\/         \__\/         \__\/
#
#
#*****DO NOT COPY******
from sys import stdin

def arr_sum(n , k ,arr):

    temp1 = []
    maxx = max(arr)
    for i in range(n):
        temp1.append(maxx - arr[i])

    maxx2 = max(temp1)
    temp2 = []
    for i in range(n):
        temp2.append(maxx2 - temp1[i])

    temp1 = map(str , temp1)
    temp2 = map(str , temp2)
    if k%2 ==1:
        return " ".join(temp1)

    else:
        return " ".join(temp2)
        








for _ in range(int(input())):
    n , k = map(int , input().split())
    #n = int(input())
    #str = input()
    arr = list(map(int , stdin.readline().strip().split()))

    print(arr_sum(n , k , arr))