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;
template <class T>
inline T bigMod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modInverse(T a, T M) {
  return bigMod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T>
inline string int2String(T a) {
  ostringstream str;
  str << a;
  return str.str();
}
template <class T>
inline T string2int(string a) {
  stringstream str(a);
  T b;
  str >> b;
  return b;
}
template <class T>
inline void read(T& r) {
  r = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' or ch > '9') {
    f = (ch == '-' ? -1 : 1), ch = getchar();
  }
  while (ch >= '0' and ch <= '9') {
    r = r * 10 + ch - '0', ch = getchar();
  }
  r *= f;
}
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1, -2, -2, 2, 2, -1, -1, 1, 1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1, -1, 1, -1, 1, -2, 2, -2, 2};
long long a[200005], b[200005];
int main() {
  int t;
  long long n, k;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld %lld", &n, &k);
    long long mx = INT_MIN, mx1 = INT_MIN;
    for (int i = int(0); i < int(n); i++) {
      scanf("%lld", &a[i]);
      mx = max(mx, a[i]);
    }
    for (int i = int(0); i < int(n); i++) {
      b[i] = mx - a[i];
      mx1 = max(mx1, b[i]);
    }
    for (int i = int(0); i < int(n); i++) {
      a[i] = mx1 - b[i];
    }
    if (k % 2 == 0) {
      for (int i = int(0); i < int(n); i++) {
        printf("%d ", a[i]);
      }
    } else {
      for (int i = int(0); i < int(n); i++) {
        printf("%d ", b[i]);
      }
    }
    puts("");
  }
  return 0;
}