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;
long long INF = 1LL << 60;
long long MOD = 1e9 + 7;
namespace output {
void pr(short x) { cout << x; }
void pr(signed x) { cout << x; }
void pr(long x) { cout << x; }
void pr(long long x) { cout << x; }
void pr(unsigned short x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(long double x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { cout << x; }
template <size_t sz>
void pr(const bitset<sz>& x) {
  cout << x;
}
template <class T>
void pr(const complex<T>& x) {
  cout << x;
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T1, class T2, class T3>
void pr(const tuple<T1, T2, T3>& x);
template <class T>
void pr(const T& x);
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("(", x.first, ", ", x.second, ")");
}
template <class T1, class T2, class T3>
void pr(const tuple<T1, T2, T3>& x) {
  pr("(", get<0>(x), ", ", get<1>(x), ", ", get<2>(x), ")");
}
template <class T>
void pr(const T& x) {
  pr("[");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("]");
}
void print() { pr("\n"); }
template <class T, class... Ts>
void print(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  print(ts...);
}
}  // namespace output
using namespace output;
template <class T1, class T2>
istream& operator>>(istream& is, pair<T1, T2>& x) {
  is >> x.first >> x.second;
  return is;
}
template <class T1, class T2, class T3>
istream& operator>>(istream& is, tuple<T1, T2, T3>& x) {
  is >> get<0>(x) >> get<1>(x) >> get<2>(x);
  return is;
}
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) is >> x;
  return is;
}
long long powa(long long base, long long exp) {
  long long res = 1;
  while (exp) {
    if (exp & 1) res = res * base % MOD;
    base = base * base % MOD;
    exp >>= 1;
  }
  return res;
}
bool mini(long long& a, long long b) { return b < a ? a = b, 1 : 0; }
bool maxi(long long& a, long long b) { return b > a ? a = b, 1 : 0; }
signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long _T;
  cin >> _T;
  for (long long caset = (1); caset < (_T + 1); ++caset) {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    cin >> a;
    for (long long _ = (0); _ < (k & 1 ? 1 : 2); ++_) {
      long long m = *max_element((a).begin(), (a).end());
      for (long long i = (0); i < (n); ++i) {
        a[i] = m - a[i];
      }
    }
    for (long long i = (0); i < (n); ++i) {
      cout << a[i] << ' ';
    }
    print();
  }
}