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 math
from decimal import Decimal
import heapq
import copy
import heapq
from collections import deque
from collections import defaultdict
MOD = 1000000007
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b
 
	
def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c
 
		
def dv():
	n, m = map(int, input().split())
	return n,m
 
 
def da():
	n, m = map(int, input().split())
	a = list(map(int, input().split()))
	return n,m, a 
 
 
def dva():
	
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b
 
 
def eratosthenes(n): 
	sieve = list(range(n + 1))
	for i in sieve:
		if i > 1:
			for j in range(i + i, len(sieve), i):
				sieve[j] = 0
	return sorted(set(sieve))
 
 
def lol(lst,k):
	k=k%len(lst)
	ret=[0]*len(lst)
	for i in range(len(lst)):
		if i+k<len(lst) and i+k>=0:
			ret[i]=lst[i+k]
		if i+k>=len(lst):
			ret[i]=lst[i+k-len(lst)]
		if i+k<0:
			ret[i]=lst[i+k+len(lst)]
	return(ret)
def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c
 
 
def dvs():
	n = int(input())
	m = int(input())
	return n, m 
 
 
def fact(n):
	tc = []
	ans = {}
	d = 2
	while d * d <= n:
		if n % d == 0:
			tc.append(d)
			n //= d
		else:
			d += 1
	if n > 1:
		tc.append(n)
	for i in tc:
		ans[i] = ans.get(i, 0) + 1
	return ans


def to_zero(n, s):
	lst = s[0]
	ans = []
	for i in range(n):
		if lst == s[i]:
			continue
		else:
			if lst == '0':
				lst = '1'
				ans.append(i)
			else:
				lst = '0'
				ans.append(i)
	if lst == '1':
		ans.append(n)
	return ans


for _ in range(int(input())):
  n,k=map(int,input().split())
  a=list(map(int,input().split()))
  d=max(a)
  k-=1
  a=[d-i for i in a]
  if k==0:
    print(*a)
  else:
    if k%2:
      d=max(a)
      a=[d-i for i in a]
    print(*a)