#include <bits/stdc++.h>
using namespace std;
int x[100005];
int y[100005];
int main() {
  long long j;
  long long n, mx, my;
  cin >> n >> j >> mx >> my;
  for (int i = (0); i < (n); ++i) scanf("%d%d", &x[i], &y[i]);
  j %= 2 * n;
  if (j <= n) {
    for (int i = (0); i < (j); ++i) {
      mx = 2 * x[i] - mx;
      my = 2 * y[i] - my;
    }
  } else {
    reverse(x, x + n);
    reverse(y, y + n);
    for (int i = (0); i < (2 * n - j); ++i) {
      mx = 2 * x[i] - mx;
      my = 2 * y[i] - my;
    }
  }
  cout << mx << ' ' << my;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 x[100005];
int y[100005];
int main() {
  long long j;
  long long n, mx, my;
  cin >> n >> j >> mx >> my;
  for (int i = (0); i < (n); ++i) scanf("%d%d", &x[i], &y[i]);
  j %= 2 * n;
  if (j <= n) {
    for (int i = (0); i < (j); ++i) {
      mx = 2 * x[i] - mx;
      my = 2 * y[i] - my;
    }
  } else {
    reverse(x, x + n);
    reverse(y, y + n);
    for (int i = (0); i < (2 * n - j); ++i) {
      mx = 2 * x[i] - mx;
      my = 2 * y[i] - my;
    }
  }
  cout << mx << ' ' << my;
}

```