#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int n;
long long id;
int a[2][maxn], b[2];
int main() {
  scanf("%d%I64d", &n, &id);
  scanf("%d%d", &b[0], &b[1]);
  for (int i = (1), _b = (n); i <= _b; ++i) scanf("%d%d", &a[0][i], &a[1][i]);
  for (int k = (0), _b = (2); k < _b; ++k) {
    int first = 1;
    int ans = 0;
    if ((id / n) % 2)
      for (int i = (n), _b = (1); i >= _b; --i)
        ans += first * a[k][i], first = -first;
    first = 1;
    if ((id % n) % 2) ans = -ans;
    for (int i = (id % n), _b = (1); i >= _b; --i)
      ans += first * a[k][i], first = -first;
    ans <<= 1;
    if (id % 2)
      ans -= b[k];
    else
      ans += b[k];
    printf("%d ", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int maxn = 1e5 + 7;
int n;
long long id;
int a[2][maxn], b[2];
int main() {
  scanf("%d%I64d", &n, &id);
  scanf("%d%d", &b[0], &b[1]);
  for (int i = (1), _b = (n); i <= _b; ++i) scanf("%d%d", &a[0][i], &a[1][i]);
  for (int k = (0), _b = (2); k < _b; ++k) {
    int first = 1;
    int ans = 0;
    if ((id / n) % 2)
      for (int i = (n), _b = (1); i >= _b; --i)
        ans += first * a[k][i], first = -first;
    first = 1;
    if ((id % n) % 2) ans = -ans;
    for (int i = (id % n), _b = (1); i >= _b; --i)
      ans += first * a[k][i], first = -first;
    ans <<= 1;
    if (id % 2)
      ans -= b[k];
    else
      ans += b[k];
    printf("%d ", ans);
  }
  return 0;
}

```