#include <bits/stdc++.h>
using namespace std;
struct point {
  long long x, y;
};
point sym(point a, point center) {
  point ans;
  ans.x = 2 * center.x - a.x;
  ans.y = 2 * center.y - a.y;
  return ans;
}
int main() {
  point m;
  int n;
  long long j;
  scanf("%d%I64d", &n, &j);
  scanf("%I64d%I64d", &m.x, &m.y);
  point term[100005];
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &term[i].x, &term[i].y);
  }
  for (int i = 0; i < j % (2 * n); i++) {
    m = sym(m, term[i % n]);
  }
  printf("%I64d %I64d\n", m.x, m.y);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 sym(point a, point center) {
  point ans;
  ans.x = 2 * center.x - a.x;
  ans.y = 2 * center.y - a.y;
  return ans;
}
int main() {
  point m;
  int n;
  long long j;
  scanf("%d%I64d", &n, &j);
  scanf("%I64d%I64d", &m.x, &m.y);
  point term[100005];
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &term[i].x, &term[i].y);
  }
  for (int i = 0; i < j % (2 * n); i++) {
    m = sym(m, term[i % n]);
  }
  printf("%I64d %I64d\n", m.x, m.y);
  return 0;
}

```