#include <bits/stdc++.h>
using namespace std;
int n, Mx, My, i, sumx, sumy;
int Ax[100010], Ay[100010];
long long j;
int main() {
  cin >> n >> j;
  cin >> Mx >> My;
  j = j % (2 * n);
  for (i = 0; i < n; ++i) {
    cin >> Ax[i] >> Ay[i];
  }
  for (i = j - 1; i >= 0; i -= 2) {
    sumx += Ax[i % n];
    sumy += Ay[i % n];
  }
  for (i = j - 2; i >= 0; i -= 2) {
    sumx -= Ax[i % n];
    sumy -= Ay[i % n];
  }
  if (j % 2 == 0)
    cout << sumx * 2 + Mx << " " << sumy * 2 + My << endl;
  else
    cout << sumx * 2 - Mx << " " << sumy * 2 - My << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 n, Mx, My, i, sumx, sumy;
int Ax[100010], Ay[100010];
long long j;
int main() {
  cin >> n >> j;
  cin >> Mx >> My;
  j = j % (2 * n);
  for (i = 0; i < n; ++i) {
    cin >> Ax[i] >> Ay[i];
  }
  for (i = j - 1; i >= 0; i -= 2) {
    sumx += Ax[i % n];
    sumy += Ay[i % n];
  }
  for (i = j - 2; i >= 0; i -= 2) {
    sumx -= Ax[i % n];
    sumy -= Ay[i % n];
  }
  if (j % 2 == 0)
    cout << sumx * 2 + Mx << " " << sumy * 2 + My << endl;
  else
    cout << sumx * 2 - Mx << " " << sumy * 2 - My << endl;
  return 0;
}

```