#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, val;
  cin >> n >> val;
  vector<pair<long long, long long>> m(2 * n + 1), a(2 * n + 1);
  cin >> m[0].first >> m[0].second;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  val %= 2 * n;
  for (long long i = 1; i <= val + 1; i++) {
    m[i].first = 2 * a[(i - 1) % n].first - m[i - 1].first;
    m[i].second = 2 * a[(i - 1) % n].second - m[i - 1].second;
  }
  cout << m[val].first << " " << m[val].second << "\n";
  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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, val;
  cin >> n >> val;
  vector<pair<long long, long long>> m(2 * n + 1), a(2 * n + 1);
  cin >> m[0].first >> m[0].second;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  val %= 2 * n;
  for (long long i = 1; i <= val + 1; i++) {
    m[i].first = 2 * a[(i - 1) % n].first - m[i - 1].first;
    m[i].second = 2 * a[(i - 1) % n].second - m[i - 1].second;
  }
  cout << m[val].first << " " << m[val].second << "\n";
  return 0;
}

```