#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x, y;
  Point() {}
  Point(int pX, int pY) : x(pX), y(pY) {}
};
Point a[100010], m[200010];
int main() {
  int i, n;
  long long k;
  while (scanf("%d%I64lld", &n, &k) != EOF) {
    scanf("%d%d", &m[0].x, &m[0].y);
    for (i = 1; i <= n; i++) {
      scanf("%d%d", &a[i - 1].x, &a[i - 1].y);
      m[i] = Point(2 * a[i - 1].x - m[i - 1].x, 2 * a[i - 1].y - m[i - 1].y);
    }
    for (i = n + 1; i <= 2 * n; i++)
      m[i] = Point(2 * a[(i - 1) % n].x - m[i - 1].x,
                   2 * a[(i - 1) % n].y - m[i - 1].y);
    k = k % (2 * n);
    printf("%d %d\n", m[k].x, m[k].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 {
  int x, y;
  Point() {}
  Point(int pX, int pY) : x(pX), y(pY) {}
};
Point a[100010], m[200010];
int main() {
  int i, n;
  long long k;
  while (scanf("%d%I64lld", &n, &k) != EOF) {
    scanf("%d%d", &m[0].x, &m[0].y);
    for (i = 1; i <= n; i++) {
      scanf("%d%d", &a[i - 1].x, &a[i - 1].y);
      m[i] = Point(2 * a[i - 1].x - m[i - 1].x, 2 * a[i - 1].y - m[i - 1].y);
    }
    for (i = n + 1; i <= 2 * n; i++)
      m[i] = Point(2 * a[(i - 1) % n].x - m[i - 1].x,
                   2 * a[(i - 1) % n].y - m[i - 1].y);
    k = k % (2 * n);
    printf("%d %d\n", m[k].x, m[k].y);
  }
  return 0;
}

```