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:

for _ in range(int(input())):
    n,k = map(int,input().split())
    a = list(map(int,input().split()))
    # a.sort()
    zero =  a.count(0)
    mx = max(a)
    # if _!=2: continue
    b = a[:]
    c = a[:]
    while k:
        k-=1
        mx = max(c)
        for i in range(n):
            c[i] = mx-c[i]
        if c==a:
            break
        a = b[:]
        b = c[:]
        # print(*b)
    # if not zero:
    #     k-=1
    #     for i in range(n):
    #         a[i] = mx-a[i]
    # mx = max(a)
    # b = a[:]
    # for i in range(n):
    #     b[i] = mx-b[i]
    k+=1
    par = k%2
    if par:
        print(*c)
    else:
        print(*b)