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;
long long powr(long long b, long long m) {
  long long ans = 1;
  while (m > 0) {
    if (m % 2 == 1) {
      ans = (ans * b) % 1000000007;
    }
    b = (b * b) % 1000000007;
    m = m / 2;
  }
  return ans;
}
long long gcd1(long long a, long long b) {
  if (a == 0)
    return b;
  else {
    return gcd1(b % a, a);
  }
}
long long lcm(long long a, long long b) { return (a * b) / gcd1(a, b); }
long long mod1(long long a) {
  if (a < 0)
    return -1 * a;
  else {
    return a;
  }
}
bool isPrime(int num) {
  if (num == 1) return 0;
  for (int i = 2; i * i <= (num); i++) {
    if (num % i == 0) {
      return false;
      break;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int f1;
  cin >> f1;
  for (int y3 = 0; y3 < f1; y3++) {
    long long n, m;
    cin >> n >> m;
    long long a[n], count = 0, b[n], c[n], max1 = -1000000000000,
                    max2 = -100000000000;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      max1 = max(max1, a[i]);
    }
    for (int i = 0; i < n; i++) {
      b[i] = max1 - a[i];
      max2 = max(max2, b[i]);
    }
    for (int i = 0; i < n; i++) {
      c[i] = max2 - b[i];
    }
    if (m % 2) {
      for (long long i = 0; i < n; i++) cout << b[i] << " ";
    } else {
      for (long long i = 0; i < n; i++) cout << c[i] << " ";
    }
    cout << "\n";
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}