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 sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
void addEdge(vector<long long int> adj[], long long int u, long long int v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void DFSE(vector<long long int> adj[], vector<bool> &visited, long long int u) {
  long long int i;
  visited[u] = true;
  cout << u << " ";
  for (i = 0; i < adj[u].size(); i++) {
    if (visited[adj[u][i]] == false) {
      DFSE(adj, visited, adj[u][i]);
    }
  }
}
void DFS(vector<long long int> adj[], long long int u, long long int v) {
  long long int i;
  vector<bool> visited(v, false);
  for (i = 0; i < v; i++) {
    if (visited[i] == false) {
      DFSE(adj, visited, i);
    }
  }
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k, i, a;
    cin >> n >> k;
    vector<long long int> v;
    for (i = 0; i < n; i++) {
      cin >> a;
      v.push_back(a);
    }
    long long int m = *max_element(v.begin(), v.end());
    for (i = 0; i < n; i++) {
      v[i] = m - v[i];
    }
    k--;
    if (k % 2 == 0) {
      for (i = 0; i < n; i++) cout << v[i] << " ";
      cout << endl;
    } else {
      m = *max_element(v.begin(), v.end());
      for (i = 0; i < n; i++) {
        v[i] = m - v[i];
      }
      for (i = 0; i < n; i++) cout << v[i] << " ";
      cout << endl;
    }
  }
  return 0;
}