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:
'''question 15
def check_22(num_list):
    for i in range(len(num_list)-1):
        if num_list[i]==2:
            if num_list[i]==num_list[i+1]:
                return True
    return False
        
print(check_22([3,2,5,1,2,1,2]))'''

'''question 22
def diagonal_stars(number):
   for i in range(number):
       s=('.')*i+'*'
       print(s)

number=6    
diagonal_stars(number)'''
'''def find_target_positions(input_string, target_word):
    target_list=[]
    for i in range(len(input_string)):
        if input_string[i]==target_word:
            target_list.append(i)
    return target_list

def find_inverted_index(input_string):
    target_dict={}
    input_string=input_string.split()
    for i in range(len(input_string)):
        if input_string[i] not in target_dict:
            x=find_target_positions(input_string,input_string[i])
            target_dict[input_string[i]]=x


    
    return target_dict
    
    
input_string="time times time times time squared equals time times time times time times time"
result_dict=find_inverted_index(input_string)
print(result_dict)'''
for _ in range(int(input())):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    if k%2==1:
        d=max(arr)
        for i in range(n):
            arr[i]=d-arr[i]
    else:
        d=max(arr)
        for i in range(n):
            arr[i]=d-arr[i]
        d=max(arr)
        for i in range(n):
            arr[i]=d-arr[i]
    print(*arr)