#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100 * 1000 + 10;
pair<long long, long long> m, a[MAXN];
int main() {
  ios::sync_with_stdio(false);
  long long n, w;
  cin >> n >> w;
  cin >> m.first >> m.second;
  for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
  w %= 2 * n;
  for (int i = 0; i < w; i++)
    m = pair<long long, long long>(2 * a[i % n].first - m.first,
                                   2 * a[i % n].second - m.second);
  cout << m.first << " " << m.second << endl;
  return 0;
}

### Prompt

Generate 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;
const int MAXN = 100 * 1000 + 10;
pair<long long, long long> m, a[MAXN];
int main() {
  ios::sync_with_stdio(false);
  long long n, w;
  cin >> n >> w;
  cin >> m.first >> m.second;
  for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
  w %= 2 * n;
  for (int i = 0; i < w; i++)
    m = pair<long long, long long>(2 * a[i % n].first - m.first,
                                   2 * a[i % n].second - m.second);
  cout << m.first << " " << m.second << endl;
  return 0;
}

```