#include <bits/stdc++.h>
using namespace std;
vector<int> Ax;
vector<int> Ay;
long long Mx;
long long My;
int main() {
  int n;
  long long obj;
  cin >> n >> obj >> Mx >> My;
  Ax = Ay = vector<int>(n);
  for (unsigned i = 0; i < n; i++) {
    cin >> Ax[i] >> Ay[i];
  }
  if (obj > 2 * n) {
    obj = obj % ((long long)(2 * n));
  }
  long long currJ = 0;
  while (currJ < obj) {
    for (unsigned i = 0; (i < n) and (currJ < obj); i++, currJ++) {
      Mx = (Ax[i] << 1) - Mx;
      My = (Ay[i] << 1) - My;
    }
  }
  cout << Mx << ' ' << My << endl;
}

### Prompt

Create a solution in Cpp for 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;
vector<int> Ax;
vector<int> Ay;
long long Mx;
long long My;
int main() {
  int n;
  long long obj;
  cin >> n >> obj >> Mx >> My;
  Ax = Ay = vector<int>(n);
  for (unsigned i = 0; i < n; i++) {
    cin >> Ax[i] >> Ay[i];
  }
  if (obj > 2 * n) {
    obj = obj % ((long long)(2 * n));
  }
  long long currJ = 0;
  while (currJ < obj) {
    for (unsigned i = 0; (i < n) and (currJ < obj); i++, currJ++) {
      Mx = (Ax[i] << 1) - Mx;
      My = (Ay[i] << 1) - My;
    }
  }
  cout << Mx << ' ' << My << endl;
}

```