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:
#input template
from sys import stdin, stdout
from collections import Counter, OrderedDict
import math
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))

def chars(): #function for taking string input as character array since string in python is immutable
    s = cin()
    return(list(s[:len(s) - 1]))

#print list	
def pl(a):	
	for val in a:
		cout(str(val) + ' ')
	cout('\n')
		
#main

t, = mp()
for _ in range(t):
	n, k = mp()
	
	l = [[] for _ in range(3)]
	l [0] = mp()
	#a = mp()
	
	#l = [[] for _ in range(n)]
	
	for i in range(1,3):
		maxi = max(l[i-1])
		for j in range(n):
			l[i] += [maxi - l[i-1][j]]
			
	if k % 2:
		pl(l[1])
	else:
		pl(l[2])