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:
from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')): 
    #------------------Sublime--------------------------------------#
    sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
    def I():return (int(input()))
    def In():return(map(int,input().split()))
else:
    #------------------PYPY FAst I/o--------------------------------#
    def I():return (int(stdin.readline()))
    def In():return(map(int,stdin.readline().split()))
#sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d


def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_right(a, x)
    if i != len(a):
        return i
    else:            
        return -1
def main():
    try:
        n,k=In()
        l=list(In())
        t=max(l)
        ma=-1
        for x in range(n):
            l[x]=t-l[x]
            ma=max(ma,l[x])
        #print(l)
        k-=1
        if k&1==1 and k>=1:
            #print(ma)
            for x in range(n):
                l[x]=ma-l[x]
            print(*l)
        else:
            print(*l)
    except: 
        pass
        
M = 998244353
P = 1000000007
Inf=1000000000000000000
if __name__ == '__main__':
    for _ in range(I()):main()
    #for _ in range(1):main()