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;
const double PI = acos(-1), EPS = 1e-7;
const int dx[] = {-1, 0, 0, 1, -1, 1, 1, -1};
const int dy[] = {0, -1, 1, 0, 1, 1, -1, -1};
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return ((x / gcd(x, y)) * y); }
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long r = power(x, y / 2);
  return r * r * power(x, y % 2);
}
void file() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
struct edge {
  int from, to, w;
  edge(int from, int to, int w) : from(from), to(to), w(w) {}
  bool operator<(const edge &e) const { return w > e.w; }
};
long long t, n;
void change(vector<long long> &v) {
  long long mx = LLONG_MIN;
  for (int i = 0; i < (n); ++i) mx = max(v[i], mx);
  for (int i = 0; i < (n); ++i) {
    v[i] = mx - v[i];
  }
}
int main() {
  file();
  int a;
  cin >> a;
  while (a--) {
    bool f = 0;
    cin >> n >> t;
    vector<long long> v(n);
    for (int i = 0; i < (n); ++i) cin >> v[i];
    if (n == 1) {
      cout << "0" << '\n';
      f = 1;
    } else if (t & 1)
      change(v);
    else {
      change(v);
      change(v);
    }
    if (f == 0) {
      for (int i = 0; i < (n); ++i) cout << v[i] << " ";
      cout << '\n';
    }
  }
}