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;
double pi = acos(-1);
long long i, j;
long long mx = (1ll << 48) - 1, fb[100];
bool ff[100010];
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
long long C(long long n, long long m) {
  if (m < 0 || m > n) return 0;
  long long ans = 1;
  for (long long i = 1; i <= m; i++) ans = ans * (n - m + i) / i;
  return ans;
}
long long binaryPow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = ans % 998244353 * a;
    }
    a = a * a % 998244353;
    b >>= 1;
  }
  return ans % 998244353;
}
long long niyuan(long long a, long long b) {
  return (a * binaryPow(b, 998244353 - 2)) % 998244353;
}
void exgcd(int a, int b, int &d, int &x, int &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long a[200005];
void gg() {
  long long n, maxx = -(1ll << 40), k, p, d;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    maxx = max(maxx, a[i]);
  }
  d = maxx;
  maxx = -(1ll << 40);
  for (int i = 0; i < n; i++) {
    a[i] = d - a[i];
    maxx = max(maxx, a[i]);
  }
  d = maxx;
  if (k % 2 == 1)
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
  else
    for (int i = 0; i < n; i++) cout << d - a[i] << ' ';
  cout << '\n';
}
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) gg();
}