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:
"""
1392B - Omkar and Infinity Clock
https://codeforces.com/contest/1392/problem/B

#Solution Explanation
There's only two possible states the array can end up as. 
Which state it becomes after k turns is determined solely by the parity of k.
After the first move, the array will consists of all non-negative numbers (d−ai will never be negative because ai never exceeds d). 
After one turn, let's define x as max(ai) over all i. For any number ai, it will turn into x−ai. Because a zero will always exist after any one operation, x will be the maximum of the array in the next turn as well.
Then the value at index i will turn into x−(x−ai). This is just equal to ai!
Now that it's obvious that our array will enter a cycle with a period of 2,
we simply do the following: If k is odd, perform 1 operation. Otherwise perform 2 operations.
"""
for _ in range(int(input())):
	n,k=list(map(int,input().split()))
	array=[int(i) for i in input().split()]
	
	for j in range(2 if k%2==0 else 3):
		d=max(array)
		array=[d-i for i in array] 
	print(*array)