#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct p2d {
  T x, y;
  p2d(T x = 0, T y = 0) : x(x), y(y) {}
  p2d(const p2d &a) : x(a.x), y(a.y) {}
  p2d operator-(const p2d &a) const { return p2d(x - a.x, y - a.y); }
  p2d operator+(const p2d &a) const { return p2d(x + a.x, y + a.y); }
  p2d operator*(T k) const { return p2d(x * k, y * k); }
  p2d operator/(T k) const { return p2d(x / k, y / k); }
  void operator-=(const p2d &a) { x -= a.x, y -= a.y; }
  void operator+=(const p2d &a) { x += a.x, y += a.y; }
  void operator*=(T k) { x *= k, y *= k; }
  void operator/=(T k) { x /= k, y /= k; }
  bool operator==(const p2d &a) const { return x == a.x and y == a.y; }
  bool operator!=(const p2d &a) const { return x != a.x or y != a.y; }
  bool operator<(const p2d &a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
  T norm2() const { return x * x + y * y; }
  p2d ort() const { return p2d(-y, x); }
  T operator%(const p2d &a) const { return x * a.y - y * a.x; }
  T operator*(const p2d &a) const { return x * a.x + y * a.y; }
};
int n;
p2d<long long> A[100001];
int main() {
  ios::sync_with_stdio(0);
  long long q;
  cin >> n >> q;
  p2d<long long> M;
  cin >> M.x >> M.y;
  for (int i = 0; i < n; ++i) cin >> A[i].x >> A[i].y;
  long long div = (q / n);
  div -= div & 1;
  q -= div * n;
  for (int i = 0; i < q; ++i) M = A[i % n] * 2 - M;
  cout << M.x << " " << M.y << endl;
}

### Prompt

Generate 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>
using namespace std;
template <typename T>
struct p2d {
  T x, y;
  p2d(T x = 0, T y = 0) : x(x), y(y) {}
  p2d(const p2d &a) : x(a.x), y(a.y) {}
  p2d operator-(const p2d &a) const { return p2d(x - a.x, y - a.y); }
  p2d operator+(const p2d &a) const { return p2d(x + a.x, y + a.y); }
  p2d operator*(T k) const { return p2d(x * k, y * k); }
  p2d operator/(T k) const { return p2d(x / k, y / k); }
  void operator-=(const p2d &a) { x -= a.x, y -= a.y; }
  void operator+=(const p2d &a) { x += a.x, y += a.y; }
  void operator*=(T k) { x *= k, y *= k; }
  void operator/=(T k) { x /= k, y /= k; }
  bool operator==(const p2d &a) const { return x == a.x and y == a.y; }
  bool operator!=(const p2d &a) const { return x != a.x or y != a.y; }
  bool operator<(const p2d &a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
  T norm2() const { return x * x + y * y; }
  p2d ort() const { return p2d(-y, x); }
  T operator%(const p2d &a) const { return x * a.y - y * a.x; }
  T operator*(const p2d &a) const { return x * a.x + y * a.y; }
};
int n;
p2d<long long> A[100001];
int main() {
  ios::sync_with_stdio(0);
  long long q;
  cin >> n >> q;
  p2d<long long> M;
  cin >> M.x >> M.y;
  for (int i = 0; i < n; ++i) cin >> A[i].x >> A[i].y;
  long long div = (q / n);
  div -= div & 1;
  q -= div * n;
  for (int i = 0; i < q; ++i) M = A[i % n] * 2 - M;
  cout << M.x << " " << M.y << endl;
}

```