#include <bits/stdc++.h>
using namespace std;
void run() {
  long long n, idx;
  cin >> n >> idx;
  long long mx, my;
  vector<long long> xx(n), yy(n);
  cin >> mx >> my;
  for (int i = 0; i < (n); ++i) cin >> xx[i] >> yy[i];
  idx %= (n << 1);
  for (int i = 0; i < (idx); ++i) {
    int ii = (i % n);
    mx = (xx[ii] << 1) - mx;
    my = (yy[ii] << 1) - my;
  }
  cout << mx << " " << my << endl;
}
int main() {
  run();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You are given the following points with integer coordinates on the plane: M0, A0, A1, ..., An - 1, where n is odd number. Now we define the following infinite sequence of points Mi: Mi is symmetric to Mi - 1 according <image> (for every natural number i). Here point B is symmetric to A according M, if M is the center of the line segment AB. Given index j find the point Mj.

Input

On the first line you will be given an integer n (1 ≤ n ≤ 105), which will be odd, and j (1 ≤ j ≤ 1018), where j is the index of the desired point. The next line contains two space separated integers, the coordinates of M0. After that n lines follow, where the i-th line contain the space separated integer coordinates of the point Ai - 1. The absolute values of all input coordinates will not be greater then 1000.

Output

On a single line output the coordinates of Mj, space separated.

Examples

Input

3 4
0 0
1 1
2 3
-5 3


Output

14 0


Input

3 1
5 5
1000 1000
-1000 1000
3 100


Output

1995 1995

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void run() {
  long long n, idx;
  cin >> n >> idx;
  long long mx, my;
  vector<long long> xx(n), yy(n);
  cin >> mx >> my;
  for (int i = 0; i < (n); ++i) cin >> xx[i] >> yy[i];
  idx %= (n << 1);
  for (int i = 0; i < (idx); ++i) {
    int ii = (i % n);
    mx = (xx[ii] << 1) - mx;
    my = (yy[ii] << 1) - my;
  }
  cout << mx << " " << my << endl;
}
int main() {
  run();
  return 0;
}

```