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>
const int MOD = 1000000007;
using namespace std;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long powM(long long a, long long n, long long m) {
  if (!n) return 1;
  long long x = powM(a, n >> 1, m);
  x = x * x % m;
  if (n & 1) x = (x * a) % m;
  return x;
}
int myIntMin(int a, int b) { return a <= b ? a : b; }
long long int myllIntMin(long long int a, long long int b) {
  return a <= b ? a : b;
}
void myFactorial(long long int n, vector<long long int> &dp) {
  long long int f = 1;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    f = (f * i);
    dp[i] = f;
  }
}
void myFactorialMod(long long int n, vector<long long int> &dp) {
  long long int f = 1;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    f = (f * i) % MOD;
    dp[i] = f;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    long long int n, k;
    ;
    cin >> n >> k;
    vector<long long int> arr(n);
    long long int mx1 = INT_MIN;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
      mx1 = max(arr[i], mx1);
    }
    long long int mx2 = INT_MIN;
    for (int i = 0; i < n; ++i) {
      arr[i] = mx1 - arr[i];
      mx2 = max(mx2, arr[i]);
    }
    --k;
    bool flag = k % 2 == 0 ? true : false;
    if (flag) {
      for (int i = 0; i < n; ++i) cout << arr[i] << " ";
      cout << endl;
    } else {
      for (int i = 0; i < n; ++i) {
        cout << mx2 - arr[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}