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 long long int mod = 1e9 + 7;
ostream& operator<<(ostream& o, vector<long long int> s) {
  for (long long int i = 0; i < (long long int)s.size(); i++) o << s[i] << " ";
  o << "\n";
  return o;
}
long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (a > mod) a -= mod;
  return c;
}
void sub(long long int& a, long long int b) {
  a = a - b;
  if (a < 0) a += mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int exp(long long int a, long long int pow) {
  return (pow == 0 ? 1
                   : (pow & 1 ? mul(a, exp(mul(a, a), pow / 2))
                              : exp(mul(a, a), pow / 2)));
}
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
void solve(long long int test) {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (long long int& x : v) cin >> x;
  vector<long long int> v1, v2;
  long long int mx = *max_element(v.begin(), v.end());
  for (long long int x : v) v1.push_back(mx - x);
  mx = *max_element(v1.begin(), v1.end());
  for (long long int x : v1) v2.push_back(mx - x);
  if (k & 1)
    cout << v1;
  else
    cout << v2;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  for (long long int i = 0; i < t; i++) solve(i + 1);
}