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 sys,math
try:sys.stdin,sys.stdout=open('input.txt','r'),open('out.txt','w')
except:pass
from time import clock
a=clock()
from sys import stdin,stdout;mod=int(1e9 + 7);from statistics import mode
from collections import *;from math import ceil,floor,inf,factorial,gcd,log2,sqrt,log
ii1=lambda:int(stdin.readline().strip())
is1=lambda:stdin.readline().strip()
iia=lambda:list(map(int,stdin.readline().strip().split()))
isa=lambda:stdin.readline().strip().split()
# print('{:.3f}'.format(1),round(1.123456789,4))
# sys.setrecursionlimit(500000)
def lcm(a,b): return (a*b)//gcd(a,b)
def setbits(n):return bin(n).count('1')
def resetbits(n):return bin(n).count('0')
def modinv(n,p):return pow(n,p-2,p)
def ncr(n,r):
    num,den=1,1;r=min(n,n-r)
    for i in range(r):num*=(n-i);den*=(i+1)
    return num//den
def ncr_p(n, r, p):
    num,den=1,1;r=min(r,n-r)
    for i in range(r):num = (num * (n - i)) % p ;den = (den * (i + 1)) % p
    return (num * modinv(den,p)) % p
def isPrime(num) :
    if num<=1:return False
    if num==2 or n==3:return True
    if (num % 2 == 0 or num % 3 == 0) :return False
    m = int(num**0.5)+1
    for i in range(5,m,6):
        if (num % i == 0 or num % (i + 2) == 0) :return False
    return True
def bin_search(arr, low, high, val):
    while low <= high:
        mid = low + (high - low) // 2;
        if arr[mid] == val:return mid
        elif arr[mid] < val:low = mid + 1
        else:high = mid - 1
    return -1
def sumofdigit(num):
    count=0;
    while num : count+=num % 10;num //= 10;
    return count;
def inputmatrix():
    r,c=iia();mat=[0]*r;
    for i in range(r):mat[i]=iia();
    return r,c,mat;
def prefix_sum(n,arr):
    for i in range(1,n):arr[i]+=arr[i-1]
    return arr;
def binomial(n, k):
    if 0 <= k <= n:
        ntok = 1;ktok = 1
        for t in range(1, min(k, n - k) + 1):ntok *= n;ktok *= t;n -= 1
        return ntok // ktok
    else:return 0
def divisors(n):
    res = [];
    for i in range(1,ceil(sqrt(n))+1):
        if n%i == 0:
            if i==n//i:res.append(i)
            else:res.append(i);res.append(n//i)
    return res;
# code start from here-

for _ in range(ii1()):
    n ,k= iia()
    arr = iia()

    a=max(arr)
    for i in range(n):
        arr[i]=a-arr[i]
    if k==1:print(*arr)
    else:
        k-=1
        if k%2==1:
            a=max(arr)
            for i in range(n):
                arr[i]=a-arr[i]
            print(*arr)
        else:
            print(*arr)