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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool sortin(const pair<long long int, long long int> &e,
            const pair<long long int, long long int> &f) {
  return (e.first < f.first);
}
bool POT(long long int x) { return x && (!(x & (x - 1))); }
long long int i, j, k, l, m, n, p, q, r, a, b, c, x, y, z, ts, mn = 1e18,
                                                               mod = 1e9 + 7;
long long int ar[250002], br[250002], xr[250002];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  cin >> ts;
  for (long long int test = 0; test < ts; test++) {
    cin >> n >> k;
    for (long long int i = 0; i <= n - 1; i++) cin >> ar[i];
    if (k % 2)
      k = 1;
    else
      k = 2;
    while (k--) {
      m = -1e18;
      for (long long int i = 0; i <= n - 1; i++) m = max(m, ar[i]);
      for (long long int i = 0; i <= n - 1; i++) ar[i] = m - ar[i];
    }
    for (long long int i = 0; i <= n - 1; i++) cout << ar[i], cout << ' ';
    cout << endl;
  }
}