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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
long long power(long long b, long long e, long long m) {
  if (b == 0) return 0;
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (b == 0) return 0;
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
bool isPowerOfTwo(long long x) { return (x && !(x & (x - 1))); }
bool isPrime(long long n) {
  if (n == 1) {
    return false;
  }
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  if (n == 1) {
    cout << 0;
    return;
  }
  long long d = *max_element((v).begin(), (v).end());
  vector<long long> w(n);
  for (long long i = 0; i < n; i++) {
    w[i] = d - v[i];
  }
  long long e = *max_element((w).begin(), (w).end());
  if (k & 1) {
    for (auto i : w) cout << i << " ";
  } else {
    for (auto i : w) cout << e - i << " ";
  }
}
int main() {
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
    cout << "\n";
  }
}