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
import bisect as bi
import math
from collections import defaultdict as dd
input=sys.stdin.readline
sys.setrecursionlimit(10**5)
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
s1="abcdefghijklmnopqrstuvwxyz"
res=[];ans=[]
def solve(targ,ans,res,l):
    if(targ<0):
        return []
    if(targ==0):
        if(ans<res):
            ans=res
        return ans
    for i in range(25,-1,-1):
        if targ-l[i]>=0:
            x=res.append(s1[i])
            if len(solve((targ-l[i]),ans,x,l))>0:
                return ans
    return []
    
for _ in range(inin()):
    n,k=cin()
    l=ain()
    sarr=[0]*n;ma=max(l);tarr=[0]*n
    for i in range(n):
        sarr[i]=ma-l[i]
    ma2=max(sarr)
    for i in range(n):
        tarr[i]=ma2-sarr[i]
    if(k%2==1):
        print(*sarr)
    else:
        print(*tarr)