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>
const double pi = std::acos(-1);
using namespace std;
const long long int mod = 1000000000 + 7;
long long int power(long long int x, long long int y, long long int md) {
  long long int res = 1;
  x = x % md;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    y = y >> 1;
    x = (x * x) % md;
  }
  return res;
}
long long int m_mul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  return (a * b + mod) % mod;
}
long long int m_add(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  return (a + b + mod) % mod;
}
long long int spf[(long long int)1e7 + 2] = {0};
void siv() {
  spf[1] = 1;
  for (long long int i = 1; i < 1e7 + 2; i++) spf[i] = i;
  for (long long int i = 2; i < 1e7 + 2; i++) spf[i] = 2, i++;
  for (long long int i = 3; i * i < (long long int)1e7 + 2; i++) {
    if (spf[i] == i)
      for (long long int j = i * i; j < (long long int)1e7 + 2; j = j + i)
        if (spf[j] == j) spf[j] = i;
  }
}
int main() {
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int n, k;
    cin >> n >> k;
    long long int arr[n], brr[n];
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
      brr[i] = arr[i];
    }
    sort(brr, brr + n, greater<long long int>());
    long long int e = 0;
    long long int ans = 0;
    if (k % 2 == 1 || n == 1)
      ans = brr[0];
    else {
      ans = brr[n - 1];
    }
    for (long long int i = 0; i < n; i++) {
      if (k % 2 == 1)
        arr[i] = ans - arr[i];
      else
        arr[i] = arr[i] - ans;
    }
    for (long long int i = 0; i < n; i++) cout << arr[i] << " ";
    cout << '\n';
  }
  return 0;
}