#include <bits/stdc++.h>
using namespace std;
int n;
long long j;
long long ix, iy;
int mx[100100], my[100100];
inline void reflect(int in, long long ax, long long ay, long long& bx,
                    long long& by) {
  bx = ax + ((mx[in] - ax) << 1);
  by = ay + ((my[in] - ay) << 1);
  return;
}
int main() {
  cin >> n >> j;
  int i, cnt = 0;
  long long a, b;
  memset((mx), (0), sizeof((mx)));
  memset((my), (0), sizeof((my)));
  j = j % ((long long)(2 * n));
  cin >> ix >> iy;
  for (i = 0; i < n; i++) {
    cin >> mx[i] >> my[i];
  }
  a = ix;
  b = iy;
  while (cnt < j) {
    for (i = 0; i < n && cnt < j; i++) {
      reflect(i, ix, iy, a, b);
      ix = a;
      iy = b;
      cnt++;
    }
  }
  cout << ix << " " << iy << "\n";
  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 n;
long long j;
long long ix, iy;
int mx[100100], my[100100];
inline void reflect(int in, long long ax, long long ay, long long& bx,
                    long long& by) {
  bx = ax + ((mx[in] - ax) << 1);
  by = ay + ((my[in] - ay) << 1);
  return;
}
int main() {
  cin >> n >> j;
  int i, cnt = 0;
  long long a, b;
  memset((mx), (0), sizeof((mx)));
  memset((my), (0), sizeof((my)));
  j = j % ((long long)(2 * n));
  cin >> ix >> iy;
  for (i = 0; i < n; i++) {
    cin >> mx[i] >> my[i];
  }
  a = ix;
  b = iy;
  while (cnt < j) {
    for (i = 0; i < n && cnt < j; i++) {
      reflect(i, ix, iy, a, b);
      ix = a;
      iy = b;
      cnt++;
    }
  }
  cout << ix << " " << iy << "\n";
  return 0;
}

```