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;
double power(double x, long long int y) {
  double temp;
  if (y == 0) return 1;
  temp = fmod(power(x, y / 2), 1000000007);
  if (y % 2 == 0)
    return fmod(temp * temp, 1000000007);
  else {
    if (y > 0)
      return fmod(x * temp * temp, 1000000007);
    else
      return fmod((temp * temp) / x, 1000000007);
  }
}
long long int isprime(long long int n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  long long int x = gcd(a, b);
  return (a * b / x);
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (auto &i : v) cin >> i;
  ;
  long long int mx = *max_element(v.begin(), v.end()), mx1 = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    v[i] = mx - v[i];
    if (v[i] > mx1) mx1 = v[i];
  }
  if (k % 2 == 0) {
    for (long long int i = 0; i < n; i++) v[i] = mx1 - v[i];
  }
  for (auto i : v) cout << i << " ";
  ;
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}