#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
const int INF = 0x7fffffff;
const int MOD = 1e9 + 7;
long long px[MAXN];
long long py[MAXN];
long long mx, my;
int main() {
  long long n, j;
  scanf("%lld", &(n)), scanf("%lld", &(j));
  scanf("%lld", &(mx)), scanf("%lld", &(my));
  for (int i = 0; i < n; i++) scanf("%lld", &(px[i])), scanf("%lld", &(py[i]));
  j = j % (n * 2);
  for (long long i = 0; i < j; i++) {
    mx = 2 * px[i % n] - mx;
    my = 2 * py[i % n] - my;
  }
  printf("%lld %lld", mx, my);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1e6 + 10;
const int INF = 0x7fffffff;
const int MOD = 1e9 + 7;
long long px[MAXN];
long long py[MAXN];
long long mx, my;
int main() {
  long long n, j;
  scanf("%lld", &(n)), scanf("%lld", &(j));
  scanf("%lld", &(mx)), scanf("%lld", &(my));
  for (int i = 0; i < n; i++) scanf("%lld", &(px[i])), scanf("%lld", &(py[i]));
  j = j % (n * 2);
  for (long long i = 0; i < j; i++) {
    mx = 2 * px[i % n] - mx;
    my = 2 * py[i % n] - my;
  }
  printf("%lld %lld", mx, my);
  return 0;
}

```