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 mod = 1000000007;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long _a, long long _b) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
long long modInverse(long long x) { return powermodm(x, mod - 2, mod); }
long long fact[100005];
long long ncrmodm(long long n, long long r, long long m) {
  return (fact[n] * (modInverse(fact[n - r]) * modInverse(fact[r])) % m) % m;
}
void init_fact() {
  fact[0] = 0;
  for (long long i = 1; i < 100001; i++) fact[i] = fact[i - 1] * i;
}
void write(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
void show(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
vector<vector<long long> > vec;
vector<long long> visited;
void dfs(int x, int p = -1) {
  visited[x] = 1;
  for (int i = 0; i < vec[x].size(); i++) {
    if (p == vec[x][i]) continue;
    if (!visited[vec[x][i]]) {
      visited[vec[x][i]] = 1;
      dfs(vec[x][i], x);
    }
  }
}
bool prime[100000];
void seive() {
  int n = 100000;
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  long long ma = -INT_MAX, mi = INT_MAX;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    ma = max(ma, a[i]);
    mi = min(mi, a[i]);
  }
  if (k % 2) {
    for (long long i = 0; i < n; i++) {
      cout << ma - a[i] << " ";
    }
    cout << endl;
  } else {
    for (long long i = 0; i < n; i++) {
      cout << -mi + a[i] << " ";
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}