#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
struct node {
  long long x, y;
};
node g[maxn], a[maxn];
int main() {
  std::ios::sync_with_stdio(false);
  long long n, j;
  while (cin >> n >> j) {
    cin >> g[0].x >> g[0].y;
    for (int i = 0; i < n; i++) {
      cin >> a[i].x >> a[i].y;
    }
    for (int i = 1; i < 2 * n; i++) {
      g[i].x = 2 * a[(i - 1) % n].x - g[i - 1].x;
      g[i].y = 2 * a[(i - 1) % n].y - g[i - 1].y;
    }
    cout << g[j % (2 * n)].x << " " << g[j % (2 * n)].y << endl;
  }
  return 0;
}

### 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;
const int maxn = 200005;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
struct node {
  long long x, y;
};
node g[maxn], a[maxn];
int main() {
  std::ios::sync_with_stdio(false);
  long long n, j;
  while (cin >> n >> j) {
    cin >> g[0].x >> g[0].y;
    for (int i = 0; i < n; i++) {
      cin >> a[i].x >> a[i].y;
    }
    for (int i = 1; i < 2 * n; i++) {
      g[i].x = 2 * a[(i - 1) % n].x - g[i - 1].x;
      g[i].y = 2 * a[(i - 1) % n].y - g[i - 1].y;
    }
    cout << g[j % (2 * n)].x << " " << g[j % (2 * n)].y << endl;
  }
  return 0;
}

```