#include <bits/stdc++.h>
using namespace std;
int M[200005][2], a[200005][2];
void slove() {
  long long n, j;
  cin >> n >> j >> M[0][0] >> M[0][1];
  for (int i = (0); i < (n); i++) cin >> a[i][0] >> a[i][1];
  for (int i = (1); i < (2 * n); i++) {
    M[i][0] = 2 * a[(i - 1) % n][0] - M[i - 1][0];
    M[i][1] = 2 * a[(i - 1) % n][1] - M[i - 1][1];
  }
  cout << M[j % (2 * n)][0] << ' ' << M[j % (2 * n)][1] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  slove();
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int M[200005][2], a[200005][2];
void slove() {
  long long n, j;
  cin >> n >> j >> M[0][0] >> M[0][1];
  for (int i = (0); i < (n); i++) cin >> a[i][0] >> a[i][1];
  for (int i = (1); i < (2 * n); i++) {
    M[i][0] = 2 * a[(i - 1) % n][0] - M[i - 1][0];
    M[i][1] = 2 * a[(i - 1) % n][1] - M[i - 1][1];
  }
  cout << M[j % (2 * n)][0] << ' ' << M[j % (2 * n)][1] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  slove();
  return 0;
}

```