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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
long long dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long dx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const long long maxn = 1e5;
const long long OO = 1e18;
const long long mod = 1e9 + 7;
const long double PI = acos(-1);
void maxself(long long &a, long long b) { a = max(a, b); }
void minself(long long &a, long long b) { a = min(a, b); }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    long long mx = -OO;
    for (long long i = 0; i < n; i++) cin >> v[i], mx = max(mx, v[i]);
    if (k % 2)
      k = 1;
    else
      k = 2;
    while (k--) {
      long long nmx = -OO;
      for (long long i = 0; i < n; i++) {
        v[i] = mx - v[i];
        nmx = max(nmx, v[i]);
      }
      mx = nmx;
    }
    for (auto x : v) cout << x << ' ';
    cout << "\n";
  }
  return 0;
}