#include <bits/stdc++.h>
const int N = 1000005;
inline void fr(int &num) {
  num = 0;
  char c = getchar();
  int p = 1;
  while (c < '0' || c > '9') c == '-' ? p = -1, c = getchar() : c = getchar();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
  num *= p;
}
int n, a[N], b[N], x, y, o1[N], o2[N], cnt;
long long m;
int main() {
  fr(n), scanf("%lld", &m), fr(x), fr(y);
  for (register int i = 1; i <= n; ++i) fr(a[i]), fr(b[i]);
  register int T = 2;
  m %= (2 * n);
  while (T--)
    for (register int i = 1; i <= n; ++i)
      o1[cnt] = x, o2[cnt++] = y, x = (a[i] << 1) - x, y = (b[i] << 1) - y;
  printf("%d %d\n", o1[m], o2[m]);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int N = 1000005;
inline void fr(int &num) {
  num = 0;
  char c = getchar();
  int p = 1;
  while (c < '0' || c > '9') c == '-' ? p = -1, c = getchar() : c = getchar();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
  num *= p;
}
int n, a[N], b[N], x, y, o1[N], o2[N], cnt;
long long m;
int main() {
  fr(n), scanf("%lld", &m), fr(x), fr(y);
  for (register int i = 1; i <= n; ++i) fr(a[i]), fr(b[i]);
  register int T = 2;
  m %= (2 * n);
  while (T--)
    for (register int i = 1; i <= n; ++i)
      o1[cnt] = x, o2[cnt++] = y, x = (a[i] << 1) - x, y = (b[i] << 1) - y;
  printf("%d %d\n", o1[m], o2[m]);
  return 0;
}

```