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("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
#pragma GCC optimize("Os")
using namespace std;
using ll = long long int;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
using vll = vector<ll>;
using vi = vector<int>;
const ll mod = (ll)(1e9) + 7LL;
const ll M = 988244353LL;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ld pi = acos(-1);
ll gcd(ll a, ll b) { return (b ? gcd(b, a % b) : a); }
ll P(ll B, ll power, ll modulo) {
  ll ans = 1LL;
  while (power > 0LL) {
    if (power % 2LL == 1LL) {
      ans = (ans * B) % modulo;
    }
    B = (B * B) % modulo;
    power /= 2LL;
  }
  return ans;
}
bool isPrime(ll n) {
  if (n <= 1LL) {
    return false;
  }
  if (n <= 3LL) {
    return true;
  }
  if (n % 2 == 0LL || n % 3 == 0LL) {
    return false;
  }
  for (ll i = 5LL; (i * i) <= n; i += 6LL) {
    if (n % i == 0LL || n % (i + 2LL) == 0LL) {
      return false;
    }
  }
  return true;
}
void vok() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
}
const int mxN = int(1e5) + 100;
void solve() {
  ll n, k;
  cin >> n >> k;
  vll v(n);
  for (auto &x : v) {
    cin >> x;
  }
  ll maxi = *max_element((v).begin(), (v).end());
  for (int i = 0; i < n; i++) {
    v[i] = maxi - v[i];
  }
  vll poss1 = v;
  if (k == 1) {
    for (auto x : poss1) {
      cout << x << " ";
    }
    cout << "\n";
    return;
  }
  maxi = *max_element((v).begin(), (v).end());
  for (int i = 0; i < n; i++) {
    poss1[i] = maxi - poss1[i];
  }
  vll poss2 = poss1;
  for (int i = 0; i < n; i++) {
    poss2[i] = maxi - poss2[i];
  }
  if (k % 2) {
    for (auto x : poss2) {
      cout << x << " ";
    }
    cout << "\n";
  } else {
    for (auto x : poss1) {
      cout << x << " ";
    }
    cout << "\n";
  }
}
int main() {
  vok();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}