#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[110000];
pair<long long, long long> M;
int n;
long long k;
int main() {
  cin >> n >> k;
  k %= 2 * n;
  cin >> M.first >> M.second;
  for (int i = (0); i <= (n - 1); i++) scanf("%d%d", &p[i].first, &p[i].second);
  for (int i = (1); i <= (k); i++) {
    M.first = -M.first + 2 * p[(i - 1) % n].first;
    M.second = -M.second + 2 * p[(i - 1) % n].second;
  }
  cout << M.first << " " << M.second << endl;
  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;
pair<int, int> p[110000];
pair<long long, long long> M;
int n;
long long k;
int main() {
  cin >> n >> k;
  k %= 2 * n;
  cin >> M.first >> M.second;
  for (int i = (0); i <= (n - 1); i++) scanf("%d%d", &p[i].first, &p[i].second);
  for (int i = (1); i <= (k); i++) {
    M.first = -M.first + 2 * p[(i - 1) % n].first;
    M.second = -M.second + 2 * p[(i - 1) % n].second;
  }
  cout << M.first << " " << M.second << endl;
  return 0;
}

```