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 int MAXN = 1e9 + 7;
const int INF = 1e9 + 7;
const signed long long MOD = 1e9 + 7;
using namespace std;
bool comp1(signed long long a, signed long long b) { return (sin(a) < sin(b)); }
signed long long binpow(signed long long a, signed long long b) {
  signed long long res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
signed long long phi(signed long long n) {
  signed long long result = n;
  for (signed long long i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  if (n > 1) result -= result / n;
  return result;
}
signed long long gcd(signed long long a, signed long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  for (size_t jj = 0; jj < t; jj++) {
    signed long long n, k;
    cin >> n >> k;
    vector<signed long long> mas(n);
    signed long long mx = -1e9 - 47;
    for (size_t i = 0; i < n; i++) {
      cin >> mas[i];
      mx = max(mx, mas[i]);
    }
    signed long long mx1 = -1e9 - 47;
    for (size_t i = 0; i < n; i++) {
      mas[i] = mx - mas[i];
      mx1 = max(mx1, mas[i]);
    }
    k--;
    if (k % 2) {
      for (size_t i = 0; i < n; i++) {
        mas[i] = mx1 - mas[i];
      }
    }
    for (size_t i = 0; i < n; i++) {
      cout << mas[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}