#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > mull(const vector<vector<long long> >& a,
                                const vector<vector<long long> >& b) {
  vector<vector<long long> > res(3, vector<long long>(3, 0));
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++) res[i][j] += a[i][k] * b[k][j];
  return res;
}
void pow(vector<vector<long long> >& a, long long b) {
  vector<vector<long long> > tmp = a;
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++) a[i][j] = (i == j);
  for (; b;) {
    if (b & 1ll) a = mull(a, tmp);
    tmp = mull(tmp, tmp);
    b >>= 1LL;
  }
}
void gen(vector<vector<long long> >& res, int x, int y) {
  res[0][0] = res[1][1] = -1;
  res[0][2] = 2 * x;
  res[1][2] = 2 * y;
  res[2][2] = 1;
}
vector<vector<long long> > m[100100];
int main() {
  int n;
  long long M;
  cin >> n >> M;
  long long x0, y0;
  cin >> x0 >> y0;
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    m[i].resize(3);
    m[i][0].resize(3);
    m[i][1].resize(3);
    m[i][2].resize(3);
    gen(m[i], x, y);
  }
  vector<vector<long long> > tmp;
  tmp.resize(3);
  for (int i = 0; i < 3; i++) {
    tmp[i].resize(3);
    tmp[i][i] = 1;
  }
  for (int i = 0; i < n; i++) {
    tmp = mull(m[i], tmp);
    long long x, y;
    x = tmp[0][0] * x0 + tmp[0][1] * y0 + tmp[0][2];
    y = tmp[1][0] * x0 + tmp[1][1] * y0 + tmp[1][2];
  }
  pow(tmp, M / n);
  M %= n;
  for (int i = 0; i < M; i++) tmp = mull(m[i], tmp);
  long long x, y;
  x = tmp[0][0] * x0 + tmp[0][1] * y0 + tmp[0][2];
  y = tmp[1][0] * x0 + tmp[1][1] * y0 + tmp[1][2];
  cout << x << " " << y << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
vector<vector<long long> > mull(const vector<vector<long long> >& a,
                                const vector<vector<long long> >& b) {
  vector<vector<long long> > res(3, vector<long long>(3, 0));
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++) res[i][j] += a[i][k] * b[k][j];
  return res;
}
void pow(vector<vector<long long> >& a, long long b) {
  vector<vector<long long> > tmp = a;
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++) a[i][j] = (i == j);
  for (; b;) {
    if (b & 1ll) a = mull(a, tmp);
    tmp = mull(tmp, tmp);
    b >>= 1LL;
  }
}
void gen(vector<vector<long long> >& res, int x, int y) {
  res[0][0] = res[1][1] = -1;
  res[0][2] = 2 * x;
  res[1][2] = 2 * y;
  res[2][2] = 1;
}
vector<vector<long long> > m[100100];
int main() {
  int n;
  long long M;
  cin >> n >> M;
  long long x0, y0;
  cin >> x0 >> y0;
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    m[i].resize(3);
    m[i][0].resize(3);
    m[i][1].resize(3);
    m[i][2].resize(3);
    gen(m[i], x, y);
  }
  vector<vector<long long> > tmp;
  tmp.resize(3);
  for (int i = 0; i < 3; i++) {
    tmp[i].resize(3);
    tmp[i][i] = 1;
  }
  for (int i = 0; i < n; i++) {
    tmp = mull(m[i], tmp);
    long long x, y;
    x = tmp[0][0] * x0 + tmp[0][1] * y0 + tmp[0][2];
    y = tmp[1][0] * x0 + tmp[1][1] * y0 + tmp[1][2];
  }
  pow(tmp, M / n);
  M %= n;
  for (int i = 0; i < M; i++) tmp = mull(m[i], tmp);
  long long x, y;
  x = tmp[0][0] * x0 + tmp[0][1] * y0 + tmp[0][2];
  y = tmp[1][0] * x0 + tmp[1][1] * y0 + tmp[1][2];
  cout << x << " " << y << endl;
  return 0;
}

```