#include <bits/stdc++.h>
using namespace std;
double PI = acos(-1);
struct point {
  long long x, y;
  int cs;
  point(long long x = 0, long long y = 0) {
    this->x = x;
    this->y = y;
  }
  bool operator==(point& p) { return p.x == x && p.y == y; }
  bool operator<(const point& p) const {
    if (y == p.y) return x < p.x;
    return y < p.y;
  }
  point rotate(double alpha) {
    return point(x * cos(alpha) - y * sin(alpha),
                 x * sin(alpha) + y * cos(alpha));
  }
} M[200010], A[100010];
struct vecto {
  long long x, y;
  vecto(long long x = 0, long long y = 0) {
    this->x = x;
    this->y = y;
  }
  bool operator==(vecto& v) { return v.x == x && v.y == y; }
  bool operator<(const vecto& v) const {
    if (v.x == x) return y < v.y;
    return x < v.x;
  }
};
vecto toVec(point a, point b) { return vecto(b.x - a.x, b.y - a.y); }
long long crossProduct(vecto a, vecto b) { return a.x * b.y - a.y * b.x; }
int main() {
  ios::sync_with_stdio(0);
  int n;
  long long k;
  cin >> n >> k;
  cin >> M[0].x >> M[0].y;
  for (int i = (0); i <= (n - 1); i++) cin >> A[i].x >> A[i].y;
  for (int i = (1); i <= (2 * n - 1); i++) {
    M[i].x = 2 * A[(i - 1 + n) % n].x - M[i - 1].x;
    M[i].y = 2 * A[(i - 1 + n) % n].y - M[i - 1].y;
  }
  k %= (2 * n);
  cout << M[k].x << " " << M[k].y;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
double PI = acos(-1);
struct point {
  long long x, y;
  int cs;
  point(long long x = 0, long long y = 0) {
    this->x = x;
    this->y = y;
  }
  bool operator==(point& p) { return p.x == x && p.y == y; }
  bool operator<(const point& p) const {
    if (y == p.y) return x < p.x;
    return y < p.y;
  }
  point rotate(double alpha) {
    return point(x * cos(alpha) - y * sin(alpha),
                 x * sin(alpha) + y * cos(alpha));
  }
} M[200010], A[100010];
struct vecto {
  long long x, y;
  vecto(long long x = 0, long long y = 0) {
    this->x = x;
    this->y = y;
  }
  bool operator==(vecto& v) { return v.x == x && v.y == y; }
  bool operator<(const vecto& v) const {
    if (v.x == x) return y < v.y;
    return x < v.x;
  }
};
vecto toVec(point a, point b) { return vecto(b.x - a.x, b.y - a.y); }
long long crossProduct(vecto a, vecto b) { return a.x * b.y - a.y * b.x; }
int main() {
  ios::sync_with_stdio(0);
  int n;
  long long k;
  cin >> n >> k;
  cin >> M[0].x >> M[0].y;
  for (int i = (0); i <= (n - 1); i++) cin >> A[i].x >> A[i].y;
  for (int i = (1); i <= (2 * n - 1); i++) {
    M[i].x = 2 * A[(i - 1 + n) % n].x - M[i - 1].x;
    M[i].y = 2 * A[(i - 1 + n) % n].y - M[i - 1].y;
  }
  k %= (2 * n);
  cout << M[k].x << " " << M[k].y;
  return 0;
}

```