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:
import heapq as hq
from heapq import heappop,heappush
from collections import deque,defaultdict,Counter


def inp():
	return (int(input()))


def inlt():
	return (list(map(int,input().split())))


def insr():
	s=input()
	return (list(s[:len(s)]))


def invr():
	return (map(int,input().split()))


def subset_sum_count(arr,n,sum):
	dp=[[0 for _ in range(sum+1)] for _ in range(n+1)]
	for i in range(n+1):
		for j in range(sum+1):
			if j==0:
				dp[i][j]=1
			elif arr[i-1]<=j:
				dp[i][j]=dp[i-1][j-arr[i-1]]+dp[i-1][j]
			else:
				dp[i][j]=dp[i-1][j]
	return dp[n][sum]


def prefix(a):
	pre=[]
	pre.append(a[0])
	for i in range(1,len(a)):
		pre.append(pre[i-1]+a[i])
	return pre


def binary_search(func,lo,hi,abs_prec=1e-7):
	""" Locate the first value x s.t. func(x) = True within [lo, hi] """
	while abs(hi-lo)>abs_prec:
		mi=lo+(hi-lo)/2
		if func(mi):
			hi=mi
		else:
			lo=mi
	
	return (lo+hi)/2


def ternary_search(func,lo,hi,abs_prec=1e-7):
	""" Find maximum of unimodal function func() within [lo, hi] """
	while abs(hi-lo)>abs_prec:
		lo_third=lo+(hi-lo)/3
		hi_third=hi-(hi-lo)/3
		
		if func(lo_third)<func(hi_third):
			lo=lo_third
		else:
			hi=hi_third
	
	return (lo+hi)/2


def discrete_binary_search(func,lo,hi):
	""" Locate the first value x s.t. func(x) = True within [lo, hi] """
	while lo<hi:
		mi=lo+(hi-lo)//2
		if func(mi):
			hi=mi
		else:
			lo=mi+1
	
	return lo


def discrete_ternary_search(func,lo,hi):
	""" Find the first maximum of unimodal function func() within [lo, hi] """
	while lo<=hi:
		lo_third=lo+(hi-lo)//3
		hi_third=lo+(hi-lo)//3+(1 if 0<hi-lo<3 else (hi-lo)//3)
		
		if func(lo_third)<func(hi_third):
			lo=lo_third+1
		else:
			hi=hi_third-1
	
	return lo
	
	return -1


def right_rotate(a,s):
	return a[s:]+a[:s]


def dec_to_bin(x):
	return int(bin(x)[2:])


def str_to_integer_list(n):
	a=[]
	for i in range(len(n)):
		a.append(int(n[i]))
	return a


def list_to_str(l):
	s=""
	for i in l:
		s+=str(i)
	return s


def dijkstra(s,N,E):
	visited=set()
	dist={}
	for i in range(1,N+1):
		dist[i]=1<<29
	queue=[(dist[i],i) for i in range(1,N+1)]
	hq.heappush(queue,(0,s))
	dist[s]=0
	while queue:
		d,u=hq.heappop(queue)
		if u in visited:
			continue
		#Relax all the neighbours of u
		for t in E[u]:
			v,r=t
			if dist[v]>d+r:
				dist[v]=d+r
				hq.heappush(queue,(dist[v],v))
		#Node u has been processed
		visited.add(u)
	return dist


def prime_sieve(n):
	"""returns a sieve of primes >= 5 and < n"""
	flag=n%6==2
	sieve=bytearray((n//3+flag>>3)+1)
	for i in range(1,int(n**0.5)//3+1):
		if not (sieve[i>>3]>>(i&7))&1:
			k=(3*i+1)|1
			for j in range(k*k//3,n//3+flag,2*k):
				sieve[j>>3]|=1<<(j&7)
			for j in range(k*(k-2*(i&1)+4)//3,n//3+flag,2*k):
				sieve[j>>3]|=1<<(j&7)
	return sieve


def prime_list(n):
	"""returns a list of primes <= n"""
	res=[]
	if n>1:
		res.append(2)
	if n>2:
		res.append(3)
	if n>4:
		sieve=prime_sieve(n+1)
		res.extend(3*i+1|1 for i in range(1,(n+1)//3+(n%6==1)) if not (sieve[i>>3]>>(i&7))&1)
	return res


def dijkstra(n,graph,start):
	""" Uses Dijkstra's algortihm to find the shortest path between in a graph. """
	dist,parents=[float("inf")]*n,[-1]*n
	dist[start]=0
	queue=[(0,start)]
	while queue:
		path_len,v=heappop(queue)
		if path_len==dist[v]:
			for w,edge_len in graph[v]:
				if edge_len+path_len<dist[w]:
					dist[w],parents[w]=edge_len+path_len,v
					heappush(queue,(edge_len+path_len,w))
	
	return dist,parents


def path(start,end,parent):
	path=[end]
	while path[-1]!=start:
		path.append(parent[path[-1]])
	path.reverse()
	return path


def bfs(graph,start,goal):
	"""
	finds a shortest path in undirected `graph` between `start` and `goal`.
	If no path is found, returns `None`
	"""
	if start==goal:
		return [start]
	visited={start}
	queue=deque([(start,[])])
	
	while queue:
		current,path=queue.popleft()
		visited.add(current)
		for neighbor in graph[current]:
			if neighbor==goal:
				return path+[current,neighbor]
			if neighbor in visited:
				continue
			queue.append((neighbor,path+[current]))
			visited.add(neighbor)
	return None  # no path found. not strictly needed


'''for _ in range(int(input())):
	n,m=list(map(int,input().split()))
	graph=defaultdict(list)
	distance=[-1]*(n+1)
	visited=[False]*(n+1)
	for _ in range(m):
		x,y=list(map(int,input().split()))
		graph[x].append(y)
		graph[y].append(x)'''
'''for _ in range(int(input())):
	n,m=list(map(int,input().split()))
	graph=[[]for _ in range(n+1)]
	distance=[-1]*(n+1)
	visited=[False]*(n+1)
	for _ in range(m):
		x,y,w=list(map(int,input().split()))
		graph[x].append([y,w])
		graph[y].append([x,w])'''

import os
import sys
from io import BytesIO,IOBase


def DecimalToBinary(n):
	def helper(n):
		if n==0:
			return ''  # <- empty string
		else:
			if n%2==1:
				binaryDigit='1'
			else:
				binaryDigit='0'
			return helper(n//2)+binaryDigit
	
	if n==0:
		result='0'  # <- "leading zero" only for `n == 0`
	else:
		result=helper(n)
	return result


def p(board):
	for i in range(8):
		s=""
		for j in range(8):
			s+=board[i][j]
		print(s)


def binaryToDecimal(n):
	return int(n,2)


def decimalToBinary(n):
	return bin(n).replace("0b","")


def solve(a,b):
	x=DecimalToBinary(a);
	y=DecimalToBinary(b)
	z=x+y
	l=y+x
	
	r=binaryToDecimal(z)-binaryToDecimal(l)
	return r


def mostFrequent(arr,n):
	# Sort the array
	arr.sort()
	
	# find the max frequency using
	# linear traversal
	max_count=1;
	res=arr[0];
	curr_count=1
	
	for i in range(1,n):
		if (arr[i]==arr[i-1]):
			curr_count+=1
		
		else:
			if (curr_count>max_count):
				max_count=curr_count
				res=arr[i-1]
			
			curr_count=1
	
	# If last element is most frequent
	if (curr_count>max_count):
		max_count=curr_count
		res=arr[n-1]
	
	return res


def sub_lists(list1):
	# store all the sublists
	sublist=[[]]
	
	# first loop
	for i in range(len(list1)+1):
		
		# second loop
		for j in range(i+1,len(list1)+1):
			# slice the subarray
			sub=list1[i:j]
			sublist.append(sub)
	return sublist


def check(s):
	n=len(s)
	if n==1:
		return True
	return s==s[::-1]


for _ in range(inp()):
	n,k=invr()
	a=inlt()
	d=max(a)
	if k%2==0:
		d=max(a)
		for i in range(n):
			a[i]=d-a[i]
		d=max(a)
		for i in range(n):
			a[i]=d-a[i]
		print(*a)
		
	else:
		d=max(a)
		for i in range(n):
			a[i]=d-a[i]
		print(*a)