#include <bits/stdc++.h>
typedef struct punto {
  int x;
  int y;
} Punto;
int main() {
  long long int n;
  long long int k;
  scanf("%I64d %I64d", &n, &k);
  Punto p[n + 1];
  for (int i = 0; i < n + 1; ++i) scanf("%d %d", &p[i].x, &p[i].y);
  k = k % (n * 2);
  int x = 1;
  for (int i = 0; i < k; i++, x++) {
    long long int aux = 2 * p[x].x;
    p[0].x = aux - p[0].x;
    aux = 2 * p[x].y;
    p[0].y = aux - p[0].y;
    if (x == n) x = 0;
  }
  printf("%d %d", p[0].x, p[0].y);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
typedef struct punto {
  int x;
  int y;
} Punto;
int main() {
  long long int n;
  long long int k;
  scanf("%I64d %I64d", &n, &k);
  Punto p[n + 1];
  for (int i = 0; i < n + 1; ++i) scanf("%d %d", &p[i].x, &p[i].y);
  k = k % (n * 2);
  int x = 1;
  for (int i = 0; i < k; i++, x++) {
    long long int aux = 2 * p[x].x;
    p[0].x = aux - p[0].x;
    aux = 2 * p[x].y;
    p[0].y = aux - p[0].y;
    if (x == n) x = 0;
  }
  printf("%d %d", p[0].x, p[0].y);
  return 0;
}

```