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;
int dest, i, a, n, t, q;
bool yes = false;
string s1, s2;
long long int max(long long int a, long long int b, long long int c) {
  return max(a, max(b, c));
}
struct comp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.first == b.first) return a.second < b.second;
    return a.first < b.first;
  }
};
bool check_pal(string &s) {
  int i = 0, j = s.length() - 1;
  while (i < j) {
    if (s[i] != s[j]) return false;
    i++, j--;
  }
  return true;
}
void swap(char &a, char &b) {
  char ch = a;
  a = b;
  b = ch;
  return;
}
int dfs(vector<vector<int> > &g, vector<int> &vis, vector<int> &co, int pos) {
  int mn = co[pos];
  vis[pos] = 1;
  for (int i = 0; i < g[pos].size(); i++) {
    if (!vis[g[pos][i]]) mn = min(dfs(g, vis, co, g[pos][i]), mn);
  }
  return mn;
}
void seive(int n, bitset<400000000UL> &isprime) {
  isprime[0] = 1;
  isprime[1] = 1;
  for (int i = 2; i < n + 1; i++) {
    if (!isprime[i]) {
      for (int j = 2; (j * i) < n + 1; j++) isprime[j * i] = 1;
    }
  }
}
const int N = int(1e6) + 40;
int A[N], ans = 0;
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n;
    cin >> k;
    vector<long long int> v(n);
    long long int x = INT_MIN;
    long long int mn = INT_MAX;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      x = max(v[i], x);
      mn = min(v[i], mn);
    }
    if (k == 0) {
      for (int i = 0; i < n; i++) cout << v[i] << " ";
      cout << "\n";
      continue;
    }
    vector<vector<long long int> > ans(2);
    for (int j = 0; j < n; j++) {
      long long int t = v[j];
      v[j] = x - t;
    }
    ans[0] = v;
    for (int j = 0; j < n; j++) {
      long long int t = v[j];
      v[j] = x - mn - t;
    }
    ans[1] = v;
    long long int ind = (k % 2 + 1) % 2;
    for (int i = 0; i < n; i++) cout << ans[ind][i] << " ";
    cout << "\n";
  }
  return 0;
}