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 int INF = 1ll << 30;
const long long LINF = 1ll << 60;
bool isprime(long long a) {
  if (a == 2) {
    return 1;
  }
  if (!(a & 1)) {
    return 0;
  }
  for (long long i = 3; i * i <= a; i += 2) {
    if (a % i == 0) {
      return 0;
    }
  }
  return 1;
}
bool isPerfectSquare(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second < b.second)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    long long unsigned k;
    cin >> n >> k;
    long long a[n + 1], b[n + 1];
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
      b[i] = a[i];
    }
    long long flag = 0, cnt = 0;
    long long mx = *max_element(a, a + n);
    for (long long i = 0; i < n; ++i) a[i] = mx - a[i];
    long long mx1 = *max_element(a, a + n);
    for (long long i = 0; i < n; ++i) b[i] = mx1 - a[i];
    for (long long i = 0; i < n; ++i)
      if (mx == a[i]) cnt++;
    if (cnt == n) {
      for (long long i = 0; i < n; ++i) cout << 0 << " ";
      cout << endl;
    } else {
      if (k % 2 == 1)
        for (long long i = 0; i < n; ++i) cout << a[i] << " ";
      else if (k % 2 == 0)
        for (long long i = 0; i < n; ++i) cout << b[i] << " ";
      cout << endl;
    }
  }
}