#include <bits/stdc++.h>
using namespace std;
struct Point {
  long long x, y;
  Point(long long x = 0, long long y = 0) : x(x), y(y) {}
  Point operator+(Point a) { return Point(x + a.x, y + a.y); }
  Point operator-(Point a) { return Point(x - a.x, y - a.y); }
} a[100111];
int main() {
  ios ::sync_with_stdio(false);
  cout << (fixed) << setprecision(6);
  int n;
  long long j;
  while (cin >> n >> j) {
    j %= 2 * n;
    Point M;
    cin >> M.x >> M.y;
    for (int i = 0, _a = (n); i < _a; i++) cin >> a[i].x >> a[i].y;
    if (j == 0) {
      cout << M.x << ' ' << M.y << endl;
      continue;
    }
    for (int x = (1), _b = (j); x <= _b; x++) {
      M = a[(x - 1) % n] + a[(x - 1) % n] - M;
      if (x == j) cout << M.x << ' ' << M.y << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
struct Point {
  long long x, y;
  Point(long long x = 0, long long y = 0) : x(x), y(y) {}
  Point operator+(Point a) { return Point(x + a.x, y + a.y); }
  Point operator-(Point a) { return Point(x - a.x, y - a.y); }
} a[100111];
int main() {
  ios ::sync_with_stdio(false);
  cout << (fixed) << setprecision(6);
  int n;
  long long j;
  while (cin >> n >> j) {
    j %= 2 * n;
    Point M;
    cin >> M.x >> M.y;
    for (int i = 0, _a = (n); i < _a; i++) cin >> a[i].x >> a[i].y;
    if (j == 0) {
      cout << M.x << ' ' << M.y << endl;
      continue;
    }
    for (int x = (1), _b = (j); x <= _b; x++) {
      M = a[(x - 1) % n] + a[(x - 1) % n] - M;
      if (x == j) cout << M.x << ' ' << M.y << endl;
    }
  }
  return 0;
}

```