#include <bits/stdc++.h>
using namespace std;
int main() {
  long long N;
  long long J;
  scanf("%lld%lld", &N, &J);
  long long mx, my;
  scanf("%lld%lld", &mx, &my);
  long long x = 0, y = 0;
  if (J % 2 == 0) {
    x = mx;
    y = my;
  } else {
    x = -mx;
    y = -my;
  }
  long long d = J / N;
  long long r = J % N;
  for (int i = 0; i < N; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    if ((d % 2 == 0) == (i < r)) {
      if ((i % 2 == (r + 1) % 2) == (i >= r)) {
        x -= 2 * a;
        y -= 2 * b;
      } else {
        x += 2 * a;
        y += 2 * b;
      }
    }
  }
  printf("%lld %lld\n", x, y);
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  long long N;
  long long J;
  scanf("%lld%lld", &N, &J);
  long long mx, my;
  scanf("%lld%lld", &mx, &my);
  long long x = 0, y = 0;
  if (J % 2 == 0) {
    x = mx;
    y = my;
  } else {
    x = -mx;
    y = -my;
  }
  long long d = J / N;
  long long r = J % N;
  for (int i = 0; i < N; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    if ((d % 2 == 0) == (i < r)) {
      if ((i % 2 == (r + 1) % 2) == (i >= r)) {
        x -= 2 * a;
        y -= 2 * b;
      } else {
        x += 2 * a;
        y += 2 * b;
      }
    }
  }
  printf("%lld %lld\n", x, y);
  return 0;
}

```