#include <bits/stdc++.h>
using namespace std;
pair<long long int, long long int> a[100001];
pair<long long int, long long int> ans[200001];
int main() {
  long long int n, x;
  long long int m0, m1;
  cin >> n >> x >> m0 >> m1;
  for (int j = 0; j < n; j++) cin >> a[j].first >> a[j].second;
  ans[0].first = m0;
  ans[0].second = m1;
  for (int j = 1; j < 2 * n; j++) {
    int ind = (j - 1) % n;
    ans[j].first = 2 * a[ind].first - ans[j - 1].first;
    ans[j].second = 2 * a[ind].second - ans[j - 1].second;
  }
  x = x % (2 * n);
  cout << ans[x].first << " " << ans[x].second << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
pair<long long int, long long int> a[100001];
pair<long long int, long long int> ans[200001];
int main() {
  long long int n, x;
  long long int m0, m1;
  cin >> n >> x >> m0 >> m1;
  for (int j = 0; j < n; j++) cin >> a[j].first >> a[j].second;
  ans[0].first = m0;
  ans[0].second = m1;
  for (int j = 1; j < 2 * n; j++) {
    int ind = (j - 1) % n;
    ans[j].first = 2 * a[ind].first - ans[j - 1].first;
    ans[j].second = 2 * a[ind].second - ans[j - 1].second;
  }
  x = x % (2 * n);
  cout << ans[x].first << " " << ans[x].second << endl;
  return 0;
}

```