#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long N, j;
  cin >> N >> j;
  long long resx, resy;
  cin >> resx >> resy;
  if (j % 2 == 1) {
    resx = -resx;
    resy = -resy;
  }
  for (long long i = 1; i <= N; i++) {
    long long x, y;
    cin >> x >> y;
    long long cnt = j / N + (j % N >= i);
    if (cnt % 2 == 0) continue;
    long long other = (j - 1) % N + 1;
    if ((other < i && (other + N - i) % 2 == 0) ||
        (other >= i && (other - i) % 2 == 0)) {
      resx += 2 * x;
      resy += 2 * y;
    } else {
      resx -= 2 * x;
      resy -= 2 * y;
    }
  }
  cout << resx << ' ' << resy << '\n';
  return 0;
}

### Prompt

Please formulate 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 long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long N, j;
  cin >> N >> j;
  long long resx, resy;
  cin >> resx >> resy;
  if (j % 2 == 1) {
    resx = -resx;
    resy = -resy;
  }
  for (long long i = 1; i <= N; i++) {
    long long x, y;
    cin >> x >> y;
    long long cnt = j / N + (j % N >= i);
    if (cnt % 2 == 0) continue;
    long long other = (j - 1) % N + 1;
    if ((other < i && (other + N - i) % 2 == 0) ||
        (other >= i && (other - i) % 2 == 0)) {
      resx += 2 * x;
      resy += 2 * y;
    } else {
      resx -= 2 * x;
      resy -= 2 * y;
    }
  }
  cout << resx << ' ' << resy << '\n';
  return 0;
}

```