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;
using ll = long long;
using ull = unsigned long long;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
const ll N = 1e5 + 10;
const ll INF = 1e10;
const ll M = 1e3 + 1;
const ll mod = 1e9 + 7;
const ll L = 31;
ll solve() {
  ll n, k;
  cin >> n >> k;
  ll arr[n];
  ll mx = -1e10, mi = 1e10;
  for (ll i = 0; i < n; i++) {
    cin >> arr[i];
    mx = max(mx, arr[i]);
    mi = min(mi, arr[i]);
  }
  for (ll i = 0; i < n; i++) {
    arr[i] = mx - arr[i];
  }
  k--;
  if ((k % 2) == 1) {
    for (ll i = 0; i < n; i++) {
      arr[i] = (mx - mi) - arr[i];
    }
  }
  for (ll i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << '\n';
  return 0;
}
signed main() {
  ll cas;
  cin >> cas;
  for (ll i = 0; i < cas; i++) {
    solve();
  }
  return 0;
}