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:
def myFun(arr):
    d = max(arr)
    for i in range(len(arr)):
        arr[i] = d - arr[i]
    return arr

t = int(input())
for p in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    if n == 1:
        print(0)
    else:
        if k == 1:
            print(*myFun(a))
        elif k == 2:
            print(*myFun(myFun(a)))
        else:
            count = 0
            for i in a:
                if i<0:
                    a = myFun(a)
                    count += 1
                    break
            if 0 not in a:
                a = myFun(a)
                count += 1
            if (k-count)%2 == 0:
                print(*a)
            else:
                print(*myFun(a))