#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
void read(int& val) {
  int x = 0;
  int bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int maxn = 2e5 + 11;
const int mod = 1e9 + 7;
int a[maxn];
int n, k, t;
int main() {
  read(t);
  while (t--) {
    read(n);
    long long k;
    scanf("%lld", &k);
    for (int i = 1; i <= n; i++) read(a[i]);
    int mx = a[1];
    for (int i = 1; i <= n; i++) mx = max(mx, a[i]);
    for (int i = 1; i <= n; i++) {
      a[i] = mx - a[i];
    }
    k--;
    mx = a[1];
    for (int i = 1; i <= n; i++) {
      mx = max(mx, a[i]);
    }
    if (k & 1 || n == 1) {
      for (int i = 1; i <= n; i++) printf("%d ", mx - a[i]);
      puts("");
    } else {
      for (int i = 1; i <= n; i++) printf("%d ", a[i]);
      puts("");
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
void read(int& val) {
  int x = 0;
  int bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int maxn = 2e5 + 11;
const int mod = 1e9 + 7;
int a[maxn];
int n, k, t;
int main() {
  read(t);
  while (t--) {
    read(n);
    long long k;
    scanf("%lld", &k);
    for (int i = 1; i <= n; i++) read(a[i]);
    int mx = a[1];
    for (int i = 1; i <= n; i++) mx = max(mx, a[i]);
    for (int i = 1; i <= n; i++) {
      a[i] = mx - a[i];
    }
    k--;
    mx = a[1];
    for (int i = 1; i <= n; i++) {
      mx = max(mx, a[i]);
    }
    if (k & 1 || n == 1) {
      for (int i = 1; i <= n; i++) printf("%d ", mx - a[i]);
      puts("");
    } else {
      for (int i = 1; i <= n; i++) printf("%d ", a[i]);
      puts("");
    }
  }
  return 0;
}

```