#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> M;
pair<long long, long long> q[100005];
pair<long long, long long> get(pair<long long, long long> a, int id) {
  a.first = q[id].first * 2 - a.first;
  a.second = q[id].second * 2 - a.second;
  return a;
}
int main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  scanf("%lld%lld", &M.first, &M.second);
  m %= (n << 1);
  for (int i = 0; i < n; i++) scanf("%lld%lld", &q[i].first, &q[i].second);
  for (int i = 1; i <= m; i++) M = get(M, (i - 1) % n);
  printf("%lld %lld", M.first, M.second);
}

### Prompt

Your challenge is to write 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;
pair<long long, long long> M;
pair<long long, long long> q[100005];
pair<long long, long long> get(pair<long long, long long> a, int id) {
  a.first = q[id].first * 2 - a.first;
  a.second = q[id].second * 2 - a.second;
  return a;
}
int main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  scanf("%lld%lld", &M.first, &M.second);
  m %= (n << 1);
  for (int i = 0; i < n; i++) scanf("%lld%lld", &q[i].first, &q[i].second);
  for (int i = 1; i <= m; i++) M = get(M, (i - 1) % n);
  printf("%lld %lld", M.first, M.second);
}

```