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;
static char inp[1024], *sde = inp + 1024;
static const char *sp = sde;
struct __X {
  bool d[256], dd[256], kk = 1, s = 1;
  __X() {
    for (int i = 48; i < 58; d[i++] = 1)
      ;
    d[45] = dd[32] = dd[10] = dd[13] = dd[9] = 1;
  }
  __attribute__((always_inline)) inline void __ra() {
    size_t r = sde - sp;
    if (r < 24) {
      memcpy(inp, sp, r);
      size_t sz = fread(inp + r, 1, 1024 - r, stdin);
      sp = inp;
      sde = inp + r + sz;
      if (sde != inp + 1024) *sde = 0;
    }
    if (sp == sde) kk = 0;
  }
  __attribute__((always_inline)) inline char P() {
    if (!(sde - sp)) __ra();
    return *(sp++);
  }
  __attribute__((always_inline)) inline __X &operator,(char &x) {
    if (kk)
      for (x = P(); kk && dd[x]; x = P())
        ;
    return *this;
  }
  __attribute__((always_inline)) inline __X &operator,(char *x) {
    if (kk) {
      char _t = P();
      for (; kk && dd[_t]; _t = P())
        ;
      if (_t)
        for (; _t && !dd[_t]; _t = P()) *x++ = _t;
      *x = 0;
    }
    return *this;
  }
  __attribute__((always_inline)) inline __X &operator,(string &x) {
    x = "";
    if (kk) {
      char _t = P();
      for (; kk && dd[_t]; _t = P())
        ;
      if (_t)
        for (; _t && !dd[_t]; _t = P()) x += _t;
    }
    return *this;
  }
  template <class T>
  __attribute__((always_inline)) inline __X &operator,(T &x) {
    if (kk) {
      x = s = 0;
      __ra();
      while (!d[*sp]) __ra(), ++sp;
      if (*sp == '-')
        s = 1;
      else
        x = (*sp & 15);
      ++sp;
      while (d[*sp]) x = (x << 1) + (x << 3) + (*(sp++) & 15);
      if (s) x = -x;
    }
    return *this;
  }
} in;
int main() {
  int t, n, a[200000], i, v;
  int k;
  in, t;
  while (t--) {
    in, n, k;
    for (i = 0; i < n; ++i) in, a[i];
    if (k & 1) {
      v = *max_element(a, a + n);
      for (i = 0; i < n; ++i) a[i] = v - a[i];
    } else {
      v = *max_element(a, a + n);
      for (i = 0; i < n; ++i) a[i] = v - a[i];
      v = *max_element(a, a + n);
      for (i = 0; i < n; ++i) a[i] = v - a[i];
    }
    for (i = 0; i < n; ++i) cout << a[i] << ' ';
    cout << '\n';
  }
}