#include <bits/stdc++.h>
using namespace std;
struct Matrix {
  int data[2][2];
  void init() { memset(data, 0, sizeof(data)); }
  Matrix operator*(const Matrix &o) {
    Matrix ret;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < 2; j++) {
        ret.data[i][j] = 0;
        for (int k = 0; k < 2; k++) {
          ret.data[i][j] += data[i][k] * o.data[k][j];
        }
      }
    }
    return ret;
  }
  static Matrix power(Matrix o, long long n) {
    Matrix ret = o;
    n--;
    while (n) {
      if (n & 1) {
        ret = ret * o;
      }
      o = o * o;
      n >>= 1;
    }
    return ret;
  }
};
int n;
long long K;
int a[100005], b[100005];
Matrix meta;
Matrix matrix[100005];
int gao(int p, int num[100005]) {
  for (int i = 0; i < n; i++) {
    matrix[i].init();
    matrix[i].data[0][0] = -1;
    matrix[i].data[1][0] = 2 * num[i];
    matrix[i].data[1][1] = 1;
    if (i) {
      matrix[i] = matrix[i - 1] * matrix[i];
    }
  }
  meta.init();
  meta.data[0][0] = p;
  meta.data[0][1] = 1;
  long long div = K / n;
  int mod = K % n;
  if (div) {
    meta = meta * Matrix::power(matrix[n - 1], div);
  }
  if (mod) {
    meta = meta * matrix[mod - 1];
  }
  return meta.data[0][0];
}
int main() {
  while (scanf("%d", &n) != EOF) {
    cin >> K;
    int mx, my;
    scanf("%d %d", &mx, &my);
    for (int i = 0; i < n; i++) {
      scanf("%d %d", &a[i], &b[i]);
    }
    int x = gao(mx, a);
    int y = gao(my, b);
    printf("%d %d\n", x, y);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
struct Matrix {
  int data[2][2];
  void init() { memset(data, 0, sizeof(data)); }
  Matrix operator*(const Matrix &o) {
    Matrix ret;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < 2; j++) {
        ret.data[i][j] = 0;
        for (int k = 0; k < 2; k++) {
          ret.data[i][j] += data[i][k] * o.data[k][j];
        }
      }
    }
    return ret;
  }
  static Matrix power(Matrix o, long long n) {
    Matrix ret = o;
    n--;
    while (n) {
      if (n & 1) {
        ret = ret * o;
      }
      o = o * o;
      n >>= 1;
    }
    return ret;
  }
};
int n;
long long K;
int a[100005], b[100005];
Matrix meta;
Matrix matrix[100005];
int gao(int p, int num[100005]) {
  for (int i = 0; i < n; i++) {
    matrix[i].init();
    matrix[i].data[0][0] = -1;
    matrix[i].data[1][0] = 2 * num[i];
    matrix[i].data[1][1] = 1;
    if (i) {
      matrix[i] = matrix[i - 1] * matrix[i];
    }
  }
  meta.init();
  meta.data[0][0] = p;
  meta.data[0][1] = 1;
  long long div = K / n;
  int mod = K % n;
  if (div) {
    meta = meta * Matrix::power(matrix[n - 1], div);
  }
  if (mod) {
    meta = meta * matrix[mod - 1];
  }
  return meta.data[0][0];
}
int main() {
  while (scanf("%d", &n) != EOF) {
    cin >> K;
    int mx, my;
    scanf("%d %d", &mx, &my);
    for (int i = 0; i < n; i++) {
      scanf("%d %d", &a[i], &b[i]);
    }
    int x = gao(mx, a);
    int y = gao(my, b);
    printf("%d %d\n", x, y);
  }
  return 0;
}

```