#include <bits/stdc++.h>
using namespace std;
long long ansx, ansy, j, k;
int n, i, mx, my, x[100005], y[100005], s, m, l;
int main() {
  cin >> n >> j;
  scanf("%d %d", &mx, &my);
  for (i = 0; i < n; i++) scanf("%d %d", &x[i], &y[i]);
  s = (j & 1) ? -1 : 1;
  k = j / n;
  ansx = s * mx;
  ansy = s * my;
  if (k & 1) {
    m = ((j % n) & 1) ? -1 : 1;
    for (l = n - 1; l >= j % n; l--) {
      ansx += 2 * m * x[l];
      ansy += 2 * m * y[l];
      m *= -1;
    }
  } else if (j % n > 0) {
    m = 1;
    for (l = j % n - 1; l >= 0; l--) {
      ansx += 2 * m * x[l];
      ansy += 2 * m * y[l];
      m *= -1;
    }
  }
  cout << ansx << " " << ansy << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
long long ansx, ansy, j, k;
int n, i, mx, my, x[100005], y[100005], s, m, l;
int main() {
  cin >> n >> j;
  scanf("%d %d", &mx, &my);
  for (i = 0; i < n; i++) scanf("%d %d", &x[i], &y[i]);
  s = (j & 1) ? -1 : 1;
  k = j / n;
  ansx = s * mx;
  ansy = s * my;
  if (k & 1) {
    m = ((j % n) & 1) ? -1 : 1;
    for (l = n - 1; l >= j % n; l--) {
      ansx += 2 * m * x[l];
      ansy += 2 * m * y[l];
      m *= -1;
    }
  } else if (j % n > 0) {
    m = 1;
    for (l = j % n - 1; l >= 0; l--) {
      ansx += 2 * m * x[l];
      ansy += 2 * m * y[l];
      m *= -1;
    }
  }
  cout << ansx << " " << ansy << endl;
  return 0;
}

```