#include <bits/stdc++.h>
int X[100001], Y[100001];
int main() {
  int n, i, d, x0, y0, ansx, ansy, sign;
  long long m;
  scanf("%d %lld", &n, &m);
  scanf("%d %d", &x0, &y0);
  for (i = 0; i < n; i++) scanf("%d %d", &X[i], &Y[i]);
  if ((m / n) % 2)
    d = m % n + n;
  else
    d = m % n;
  sign = -1;
  ansx = x0;
  ansy = y0;
  for (i = 0; i < d; i++) {
    ansx = ansx * sign + X[i % n] * 2;
    ansy = ansy * sign + Y[i % n] * 2;
  }
  printf("%d %d\n", ansx, ansy);
  return 0;
}

### Prompt

Your task is to create 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>
int X[100001], Y[100001];
int main() {
  int n, i, d, x0, y0, ansx, ansy, sign;
  long long m;
  scanf("%d %lld", &n, &m);
  scanf("%d %d", &x0, &y0);
  for (i = 0; i < n; i++) scanf("%d %d", &X[i], &Y[i]);
  if ((m / n) % 2)
    d = m % n + n;
  else
    d = m % n;
  sign = -1;
  ansx = x0;
  ansy = y0;
  for (i = 0; i < d; i++) {
    ansx = ansx * sign + X[i % n] * 2;
    ansy = ansy * sign + Y[i % n] * 2;
  }
  printf("%d %d\n", ansx, ansy);
  return 0;
}

```