#include <bits/stdc++.h>
using namespace std;
void run() {
  int n;
  scanf("%d", &n);
  long long ind;
  cin >> ind;
  long long x, y;
  cin >> x >> y;
  vector<int> px(n), py(n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &px[i], &py[i]);
  ind = ind % (2 * n);
  for (int i = (0); i < (ind); ++i) {
    int j = i % n;
    x = 2 * px[j] - x;
    y = 2 * py[j] - y;
  }
  cout << x << " " << y << endl;
}
int main() {
  run();
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
void run() {
  int n;
  scanf("%d", &n);
  long long ind;
  cin >> ind;
  long long x, y;
  cin >> x >> y;
  vector<int> px(n), py(n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &px[i], &py[i]);
  ind = ind % (2 * n);
  for (int i = (0); i < (ind); ++i) {
    int j = i % n;
    x = 2 * px[j] - x;
    y = 2 * py[j] - y;
  }
  cout << x << " " << y << endl;
}
int main() {
  run();
  return 0;
}

```