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;
bool isPrime(long long a) {
  for (long long i = 2; i * i <= a; i++)
    if (a % i == 0) return false;
  return true;
}
bool isPowerOfTwo(long long a) {
  if (a == 1) return true;
  if (a & 1) return false;
  while (a) {
    a /= 2;
    if (a != 1 && a & 1) return false;
    if (a == 1) return true;
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (a < 0 || b < 0) {
    a = abs(a);
    b = abs(b);
  }
  if (a < b) swap(a, b);
  if (a % b == 0 || b == 0)
    return b;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long power(long long base, long long exponent) {
  long long ans = 1;
  for (long long i = 1; i <= exponent; i++) ans *= base;
  return ans;
}
bool isPowerOfX(long long x, long long value) {
  if (value == 1) return true;
  while (value) {
    value /= x;
    if (value % x && value != 1) return false;
  }
  return true;
}
bool allNeg(vector<long long> v) {
  int l = v.size();
  for (int i = 0; i < l; i++)
    if (v[i] >= 0) return false;
  return true;
}
bool allPos(vector<long long> v) {
  int l = v.size();
  for (int i = 0; i < l; i++)
    if (v[i] < 0) return false;
  return true;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    long long k;
    cin >> n >> k;
    long long mx = LONG_MIN;
    vector<long long> v;
    for (int i = 0; i < n; i++) {
      long long x;
      cin >> x;
      v.push_back(x);
      mx = max(mx, x);
    }
    if (allNeg(v) || allPos(v)) {
      long long mx2 = LONG_MIN;
      for (int i = 0; i < n; i++) {
        v[i] = mx - v[i];
        mx2 = max(mx2, v[i]);
      }
      if (k & 1) {
        for (int i = 0; i < n; i++) cout << v[i] << " ";
        cout << "\n";
        continue;
      } else if (!(k & 1)) {
        for (int i = 0; i < n; i++) v[i] = mx2 - v[i];
        for (int i = 0; i < n; i++) cout << v[i] << " ";
        cout << "\n";
        continue;
      }
    } else {
      long long mx2 = LONG_MIN;
      for (int i = 0; i < n; i++) {
        v[i] = mx - v[i];
        mx2 = max(mx2, v[i]);
      }
      if (k & 1) {
        for (int i = 0; i < n; i++) cout << v[i] << " ";
        cout << "\n";
        continue;
      } else if (!(k & 1)) {
        for (int i = 0; i < n; i++) {
          v[i] = mx2 - v[i];
        }
        for (int i = 0; i < n; i++) cout << v[i] << " ";
        cout << "\n";
        continue;
      }
    }
  }
  return 0;
}