#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:20000000")
string problem_name = "a";
void init() { freopen((problem_name + ".in").c_str(), "rt", stdin); }
int x[200100], y[200100];
long long xr[200100], yr[200100];
int main() {
  int n;
  long long j;
  scanf("%d%I64d", &n, &j);
  for (int i = 0; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  xr[0] = x[0];
  yr[0] = y[0];
  for (int i = 1; i <= 2 * n + 15; i++) {
    xr[i] = x[(i - 1) % n + 1] * 2 - xr[i - 1];
    yr[i] = y[(i - 1) % n + 1] * 2 - yr[i - 1];
  }
  n *= 2;
  printf("%I64d %I64d\n", xr[j % n], yr[j % n]);
  return 0;
}

### Prompt

Generate 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;
#pragma comment(linker, "/STACK:20000000")
string problem_name = "a";
void init() { freopen((problem_name + ".in").c_str(), "rt", stdin); }
int x[200100], y[200100];
long long xr[200100], yr[200100];
int main() {
  int n;
  long long j;
  scanf("%d%I64d", &n, &j);
  for (int i = 0; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  xr[0] = x[0];
  yr[0] = y[0];
  for (int i = 1; i <= 2 * n + 15; i++) {
    xr[i] = x[(i - 1) % n + 1] * 2 - xr[i - 1];
    yr[i] = y[(i - 1) % n + 1] * 2 - yr[i - 1];
  }
  n *= 2;
  printf("%I64d %I64d\n", xr[j % n], yr[j % n]);
  return 0;
}

```