#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
void __never(int a) { printf("\nOPS %d", a); }
int n;
long long J;
pair<int, int> m0;
pair<int, int> t[300000];
void init() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void sol() {
  cin >> n >> J;
  cin >> m0.first >> m0.second;
  for (int(i) = (0); (i) <= (n - 1); ++(i)) {
    scanf("%d %d", &t[i].first, &t[i].second);
  }
  for (int(i) = (n); (i) <= (n + n - 1); ++(i)) t[i] = t[i - n];
  n += n;
  J %= n;
  for (int(i) = (1); (i) <= (J); ++(i)) {
    m0.first = t[i - 1].first + t[i - 1].first - m0.first;
    m0.second = t[i - 1].second + t[i - 1].second - m0.second;
  }
  cout << m0.first << " " << m0.second;
}
int main() {
  sol();
  return 0;
}

### Prompt

Develop a solution in CPP 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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
void __never(int a) { printf("\nOPS %d", a); }
int n;
long long J;
pair<int, int> m0;
pair<int, int> t[300000];
void init() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void sol() {
  cin >> n >> J;
  cin >> m0.first >> m0.second;
  for (int(i) = (0); (i) <= (n - 1); ++(i)) {
    scanf("%d %d", &t[i].first, &t[i].second);
  }
  for (int(i) = (n); (i) <= (n + n - 1); ++(i)) t[i] = t[i - n];
  n += n;
  J %= n;
  for (int(i) = (1); (i) <= (J); ++(i)) {
    m0.first = t[i - 1].first + t[i - 1].first - m0.first;
    m0.second = t[i - 1].second + t[i - 1].second - m0.second;
  }
  cout << m0.first << " " << m0.second;
}
int main() {
  sol();
  return 0;
}

```