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;
int main() {
  long long int t, a, b, c, d, i, j, k, l, n, m, x, y;
  cin >> t;
  while (t--) {
    cin >> n >> d;
    long long int s[n + 5];
    l = 0;
    for (i = 0; i < n; i++) {
      cin >> s[i];
    }
    vector<long long int> v, p;
    l = *max_element(s, s + n);
    for (i = 0; i < n; i++) {
      s[i] = l - s[i];
      v.push_back(s[i]);
    }
    k = *max_element(s, s + n);
    for (i = 0; i < n; i++) {
      s[i] = k - s[i];
      p.push_back(s[i]);
    }
    if (d % 2 != 0) {
      for (i = 0; i < n; i++) cout << v[i] << " ";
    } else if (d % 2 == 0) {
      for (i = 0; i < n; i++) cout << p[i] << " ";
    }
    v.clear();
    p.clear();
    cout << endl;
  }
}