#include <bits/stdc++.h>
long long m, n, x[100010], y[100010], ax, ay, bx, by;
int main() {
  scanf("%lld %lld\n%lld %lld", &n, &m, &ax, &ay);
  for (long long i = 0; i < n; i++) scanf("%lld %lld", &x[i], &y[i]);
  m = m % (2 * n);
  for (long long i = 0; i < m; i++) {
    bx = x[i % n] * 2 - ax, by = y[i % n] * 2 - ay;
    ax = bx, ay = by;
  }
  printf("%lld %lld", ax, ay);
}

### Prompt

In CPP, your task is to solve 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>
long long m, n, x[100010], y[100010], ax, ay, bx, by;
int main() {
  scanf("%lld %lld\n%lld %lld", &n, &m, &ax, &ay);
  for (long long i = 0; i < n; i++) scanf("%lld %lld", &x[i], &y[i]);
  m = m % (2 * n);
  for (long long i = 0; i < m; i++) {
    bx = x[i % n] * 2 - ax, by = y[i % n] * 2 - ay;
    ax = bx, ay = by;
  }
  printf("%lld %lld", ax, ay);
}

```