#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> arr[100005];
int main() {
  int n;
  long long j;
  scanf("%d %lld", &n, &j);
  vector<pair<long long, long long> > v;
  pair<long long, long long> m;
  scanf("%lld %lld", &m.first, &m.second);
  pair<long long, long long> ans = m;
  for (int i = int(1); i < int(n + 1); ++i) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    v.emplace_back(a, b);
  }
  for (int i = int(0); i < int(j % (2 * n)); ++i) {
    pair<long long, long long> cur;
    cur.first = v[i % n].first * 2 - ans.first;
    cur.second = v[i % n].second * 2 - ans.second;
    ans = cur;
  }
  printf("%lld %lld\n", ans.first, ans.second);
  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;
pair<long long, long long> arr[100005];
int main() {
  int n;
  long long j;
  scanf("%d %lld", &n, &j);
  vector<pair<long long, long long> > v;
  pair<long long, long long> m;
  scanf("%lld %lld", &m.first, &m.second);
  pair<long long, long long> ans = m;
  for (int i = int(1); i < int(n + 1); ++i) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    v.emplace_back(a, b);
  }
  for (int i = int(0); i < int(j % (2 * n)); ++i) {
    pair<long long, long long> cur;
    cur.first = v[i % n].first * 2 - ans.first;
    cur.second = v[i % n].second * 2 - ans.second;
    ans = cur;
  }
  printf("%lld %lld\n", ans.first, ans.second);
  return 0;
}

```