#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:136777216")
using namespace std;
int bits(int x) { return x == 0 ? 0 : 1 + bits(x & (x - 1)); }
const double PI = acos(-1.0);
const double eps = 1e-9;
const int INF = 1000000000;
string nextString() {
  char buf[1000000];
  scanf("%s", buf);
  return buf;
}
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
  long long n, j;
  scanf("%I64d %I64d\n", &n, &j);
  long long x0, y0;
  scanf("%I64d %I64d\n", &x0, &y0);
  j %= 2 * n;
  vector<pair<int, int> > in(n);
  for (int(i) = 0; (i) < (n); (i)++)
    scanf("%d %d\n", &in[i].first, &in[i].second);
  long long cur_x = x0, cur_y = y0;
  for (int(i) = 0; (i) < (j); (i)++) {
    cur_x = 2LL * (long long)in[i % n].first - cur_x;
    cur_y = 2LL * (long long)in[i % n].second - cur_y;
  }
  printf("%I64d %I64d\n", cur_x, cur_y);
  return 0;
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:136777216")
using namespace std;
int bits(int x) { return x == 0 ? 0 : 1 + bits(x & (x - 1)); }
const double PI = acos(-1.0);
const double eps = 1e-9;
const int INF = 1000000000;
string nextString() {
  char buf[1000000];
  scanf("%s", buf);
  return buf;
}
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
  long long n, j;
  scanf("%I64d %I64d\n", &n, &j);
  long long x0, y0;
  scanf("%I64d %I64d\n", &x0, &y0);
  j %= 2 * n;
  vector<pair<int, int> > in(n);
  for (int(i) = 0; (i) < (n); (i)++)
    scanf("%d %d\n", &in[i].first, &in[i].second);
  long long cur_x = x0, cur_y = y0;
  for (int(i) = 0; (i) < (j); (i)++) {
    cur_x = 2LL * (long long)in[i % n].first - cur_x;
    cur_y = 2LL * (long long)in[i % n].second - cur_y;
  }
  printf("%I64d %I64d\n", cur_x, cur_y);
  return 0;
}

```