#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 100;
struct point {
  long long x, y;
  point(){};
  point(long long a, long long b) : x(a), y(b){};
} rem[N];
point solve(point a, point b) {
  long long fri = abs(a.x - b.x);
  long long sec = abs(a.y - b.y);
  return point((a.x > b.x) ? b.x - fri : b.x + fri,
               (a.y > b.y) ? b.y - sec : b.y + sec);
}
signed main() {
  long long n, m;
  cin >> n >> m;
  ;
  long long x, y;
  cin >> x >> y;
  point ans = point(x, y);
  for (long long i = 0; i < n; i++) {
    cin >> x >> y;
    rem[i] = point(x, y);
  }
  long long cnt = 0;
  m = m % (2 * n);
  while (m--) {
    ans = solve(ans, rem[cnt]);
    cnt = (cnt + 1) % n;
  }
  cout << ans.x << " " << ans.y << endl;
}

### 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 long long N = 1e5 + 100;
struct point {
  long long x, y;
  point(){};
  point(long long a, long long b) : x(a), y(b){};
} rem[N];
point solve(point a, point b) {
  long long fri = abs(a.x - b.x);
  long long sec = abs(a.y - b.y);
  return point((a.x > b.x) ? b.x - fri : b.x + fri,
               (a.y > b.y) ? b.y - sec : b.y + sec);
}
signed main() {
  long long n, m;
  cin >> n >> m;
  ;
  long long x, y;
  cin >> x >> y;
  point ans = point(x, y);
  for (long long i = 0; i < n; i++) {
    cin >> x >> y;
    rem[i] = point(x, y);
  }
  long long cnt = 0;
  m = m % (2 * n);
  while (m--) {
    ans = solve(ans, rem[cnt]);
    cnt = (cnt + 1) % n;
  }
  cout << ans.x << " " << ans.y << endl;
}

```