#include <bits/stdc++.h>
int main() {
  int mat[100001][2];
  long long j;
  int i, n, mx, my;
  scanf("%d %I64d", &n, &j);
  j %= 2 * n;
  scanf("%d %d", &mx, &my);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &mat[i][0], &mat[i][1]);
  }
  for (i = 1; i <= j; i++) {
    mx = 2 * mat[(i - 1) % n][0] - mx;
    my = 2 * mat[(i - 1) % n][1] - my;
  }
  printf("%d %d\n", mx, my);
  return 0;
}

### Prompt

Create a solution in CPP for 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>
int main() {
  int mat[100001][2];
  long long j;
  int i, n, mx, my;
  scanf("%d %I64d", &n, &j);
  j %= 2 * n;
  scanf("%d %d", &mx, &my);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &mat[i][0], &mat[i][1]);
  }
  for (i = 1; i <= j; i++) {
    mx = 2 * mat[(i - 1) % n][0] - mx;
    my = 2 * mat[(i - 1) % n][1] - my;
  }
  printf("%d %d\n", mx, my);
  return 0;
}

```