#include <bits/stdc++.h>
using namespace std;
struct st {
  int x, y;
  bool operator<(const st o) const { return y < o.y || (y == o.y && x < o.x); }
};
long long a[1000000][2];
long long n, m, x, y;
long long M[1000000][2];
int main() {
  cin >> n >> m;
  cin >> M[0][0] >> M[0][1];
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(2); j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i < 2 * n; i++) {
    for (int j = 0; j < (int)(2); j++) {
      M[i][j] = 2 * (a[(i - 1) % n][j]) - M[i - 1][j];
    }
  }
  cout << M[m % (2 * n)][0] << " " << M[m % (2 * n)][1];
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
struct st {
  int x, y;
  bool operator<(const st o) const { return y < o.y || (y == o.y && x < o.x); }
};
long long a[1000000][2];
long long n, m, x, y;
long long M[1000000][2];
int main() {
  cin >> n >> m;
  cin >> M[0][0] >> M[0][1];
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(2); j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i < 2 * n; i++) {
    for (int j = 0; j < (int)(2); j++) {
      M[i][j] = 2 * (a[(i - 1) % n][j]) - M[i - 1][j];
    }
  }
  cout << M[m % (2 * n)][0] << " " << M[m % (2 * n)][1];
  return 0;
}

```