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("-O3")
using namespace std;
const long long N = 1e6 + 5;
long long mod = 1e9 + 7;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long pow1(long long n, long long p) {
  if (p == 0) return 1;
  long long x = pow1(n, p / 2);
  x = (x * x) % mod;
  if (p % 2 == 0)
    return x;
  else
    return (x * n) % mod;
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
bool compare(const pair<long long, long long> &p1,
             const pair<long long, long long> &p2) {
  if (p1.first < p2.first) return true;
  if (p1.first == p2.first) return p1.second < p2.second;
  return false;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n, m, k, f = 0, flag = 0;
    cin >> n >> k;
    vector<long long> a(n + 1);
    long long mx = -1e18;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      mx = (mx > a[i] ? mx : a[i]);
    }
    if (k == 0) {
      for (long long i = 0; i < n; i++) cout << a[i] << " ";
      cout << "\n";
    } else if (k >= 1) {
      for (long long i = 0; i < n; i++) {
        a[i] = mx - a[i];
      }
      mx = -1e18;
      for (long long i = 0; i < n; i++) mx = (mx > a[i] ? mx : a[i]);
      k--;
      if (k % 2 == 0) {
        for (long long i = 0; i < n; i++) cout << a[i] << " ";
        cout << "\n";
      } else {
        vector<long long> b(n + 1);
        for (long long i = 0; i < n; i++) {
          b[i] = mx - a[i];
        }
        for (long long i = 0; i < n; i++) cout << b[i] << " ";
        cout << "\n";
      }
    }
  }
  return 0;
}