#include <bits/stdc++.h>
using namespace std;
const int N = 100006;
struct data {
  int x, y;
} P[N], M[N];
int n;
long long J;
int main() {
  scanf("%d%I64d", &n, &J);
  scanf("%d%d", &M[0].x, &M[0].y);
  for (int i = 0; i < n; i++) scanf("%d%d", &P[i].x, &P[i].y);
  for (int i = 1; i < n; i++) {
    M[i].x = 2 * P[i - 1].x - M[i - 1].x;
    M[i].y = 2 * P[i - 1].y - M[i - 1].y;
  }
  int x = 2 * P[n - 1].x - M[n - 1].x;
  int y = 2 * P[n - 1].y - M[n - 1].y;
  long long cnt = J / n;
  int id = J % n;
  if (cnt % 2 == 1) M[0].x = x, M[0].y = y;
  for (int i = 1; i <= id; i++) {
    M[i].x = 2 * P[i - 1].x - M[i - 1].x;
    M[i].y = 2 * P[i - 1].y - M[i - 1].y;
  }
  printf("%d %d\n", M[id].x, M[id].y);
  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;
const int N = 100006;
struct data {
  int x, y;
} P[N], M[N];
int n;
long long J;
int main() {
  scanf("%d%I64d", &n, &J);
  scanf("%d%d", &M[0].x, &M[0].y);
  for (int i = 0; i < n; i++) scanf("%d%d", &P[i].x, &P[i].y);
  for (int i = 1; i < n; i++) {
    M[i].x = 2 * P[i - 1].x - M[i - 1].x;
    M[i].y = 2 * P[i - 1].y - M[i - 1].y;
  }
  int x = 2 * P[n - 1].x - M[n - 1].x;
  int y = 2 * P[n - 1].y - M[n - 1].y;
  long long cnt = J / n;
  int id = J % n;
  if (cnt % 2 == 1) M[0].x = x, M[0].y = y;
  for (int i = 1; i <= id; i++) {
    M[i].x = 2 * P[i - 1].x - M[i - 1].x;
    M[i].y = 2 * P[i - 1].y - M[i - 1].y;
  }
  printf("%d %d\n", M[id].x, M[id].y);
  return 0;
}

```