#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, x, y, i;
  long long int j;
  pair<int, int> m;
  cin >> n >> j;
  cin >> m.first >> m.second;
  int xsum = 0, ysum = 0;
  j %= (2 * n);
  int r1, r2;
  if (j >= n) {
    r1 = j - n + 1;
    r2 = n;
  } else {
    r1 = 1;
    r2 = j;
  }
  for (i = 1; i <= n; ++i) {
    cin >> x >> y;
    if (r1 <= i && i <= r2) {
      if (i % 2) {
        xsum += 2 * x;
        ysum += 2 * y;
      } else {
        xsum -= 2 * x;
        ysum -= 2 * y;
      }
    }
  }
  xsum -= m.first;
  ysum -= m.second;
  if ((j & 1) == 0) {
    xsum *= -1;
    ysum *= -1;
  }
  cout << xsum << " " << ysum << endl;
}

### Prompt

Develop a solution in Cpp 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;
int main() {
  int n, x, y, i;
  long long int j;
  pair<int, int> m;
  cin >> n >> j;
  cin >> m.first >> m.second;
  int xsum = 0, ysum = 0;
  j %= (2 * n);
  int r1, r2;
  if (j >= n) {
    r1 = j - n + 1;
    r2 = n;
  } else {
    r1 = 1;
    r2 = j;
  }
  for (i = 1; i <= n; ++i) {
    cin >> x >> y;
    if (r1 <= i && i <= r2) {
      if (i % 2) {
        xsum += 2 * x;
        ysum += 2 * y;
      } else {
        xsum -= 2 * x;
        ysum -= 2 * y;
      }
    }
  }
  xsum -= m.first;
  ysum -= m.second;
  if ((j & 1) == 0) {
    xsum *= -1;
    ysum *= -1;
  }
  cout << xsum << " " << ysum << endl;
}

```