#include <bits/stdc++.h>
using namespace std;
int n;
long long j;
int xs, ys, px[100010], py[100010];
long long solve(int x0, int* px) {
  long long C = 0;
  for (int i = 1; i <= (int)(n); i++)
    if (i & 1)
      C += px[i];
    else
      C -= px[i];
  C *= 2;
  long long r = j / n;
  if (r & 1) x0 = C - x0;
  long long rest = j % n;
  long long res = x0;
  int i = 1;
  while (rest--) {
    res = 2 * px[i] - res;
    i++;
  }
  return res;
}
int main() {
  cin >> n >> j;
  scanf("%d %d", &xs, &ys);
  for (int i = 1; i <= (int)(n); i++) scanf("%d %d", &px[i], &py[i]);
  long long xa = solve(xs, px);
  long long ya = solve(ys, py);
  cout << xa << " " << ya << endl;
}

### Prompt

Please formulate a Cpp solution 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>
using namespace std;
int n;
long long j;
int xs, ys, px[100010], py[100010];
long long solve(int x0, int* px) {
  long long C = 0;
  for (int i = 1; i <= (int)(n); i++)
    if (i & 1)
      C += px[i];
    else
      C -= px[i];
  C *= 2;
  long long r = j / n;
  if (r & 1) x0 = C - x0;
  long long rest = j % n;
  long long res = x0;
  int i = 1;
  while (rest--) {
    res = 2 * px[i] - res;
    i++;
  }
  return res;
}
int main() {
  cin >> n >> j;
  scanf("%d %d", &xs, &ys);
  for (int i = 1; i <= (int)(n); i++) scanf("%d %d", &px[i], &py[i]);
  long long xa = solve(xs, px);
  long long ya = solve(ys, py);
  cout << xa << " " << ya << endl;
}

```