#include <bits/stdc++.h>
using namespace std;
const int inf(1010101010);
int n;
long long j;
int x[100005], y[100005];
int main() {
  cin >> n >> j;
  int thex, they;
  cin >> thex >> they;
  for (int i(0); i < (n); i++) cin >> x[i] >> y[i];
  long long resx(0), resy(0);
  for (int i(0); i < (n); i++) {
    long long times((j + n - 1 - i) / n);
    if (times % 2) {
      if (i % 2 == 0) resx += 2 * x[i], resy += 2 * y[i];
      if (i % 2 == 1) resx -= 2 * x[i], resy -= 2 * y[i];
    }
  }
  if (j % 2 == 0) resx *= -1, resy *= -1;
  if (j % 2 == 0) resx += thex, resy += they;
  if (j % 2 == 1) resx -= thex, resy -= they;
  cout << resx << " " << resy << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int inf(1010101010);
int n;
long long j;
int x[100005], y[100005];
int main() {
  cin >> n >> j;
  int thex, they;
  cin >> thex >> they;
  for (int i(0); i < (n); i++) cin >> x[i] >> y[i];
  long long resx(0), resy(0);
  for (int i(0); i < (n); i++) {
    long long times((j + n - 1 - i) / n);
    if (times % 2) {
      if (i % 2 == 0) resx += 2 * x[i], resy += 2 * y[i];
      if (i % 2 == 1) resx -= 2 * x[i], resy -= 2 * y[i];
    }
  }
  if (j % 2 == 0) resx *= -1, resy *= -1;
  if (j % 2 == 0) resx += thex, resy += they;
  if (j % 2 == 1) resx -= thex, resy -= they;
  cout << resx << " " << resy << endl;
}

```