#include <bits/stdc++.h>
using namespace std;
template <class T>
int size(const T &a) {
  return (int)a.size();
}
template <class T>
T sqr(const T &a) {
  return a * a;
}
typedef long long int2d[3][3];
int2d a[100000], an, t, r;
void initA(int i) {
  int x, y;
  scanf("%d %d", &x, &y);
  a[i][0][0] = a[i][1][1] = -1;
  a[i][2][2] = 1;
  a[i][0][2] = 2 * x;
  a[i][1][2] = 2 * y;
}
void mult(int2d &a, int2d &b, int2d &r) {
  memset(t, 0, sizeof(t));
  for (int i = 0; i < 3; i++) {
    for (int k = 0; k < 3; k++) {
      for (int j = 0; j < 3; j++) {
        t[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  memcpy(r, t, sizeof(t));
}
int main() {
  int n;
  long long k;
  int x, y;
  int fk, rk;
  scanf("%d %lld %d %d", &n, &k, &x, &y);
  fk = k / n;
  rk = k % n;
  an[0][0] = an[1][1] = an[2][2] = 1;
  r[0][0] = r[1][1] = r[2][2] = 1;
  for (int i = 0; i < n; i++) {
    initA(i);
    mult(a[i], an, an);
    if (i < rk) {
      mult(a[i], r, r);
    }
  }
  while (fk) {
    if (fk & 1) {
      mult(r, an, r);
    }
    fk /= 2;
    mult(an, an, an);
  }
  printf("%lld %lld", r[0][2] + r[0][0] * x, r[1][2] + r[1][1] * 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>
using namespace std;
template <class T>
int size(const T &a) {
  return (int)a.size();
}
template <class T>
T sqr(const T &a) {
  return a * a;
}
typedef long long int2d[3][3];
int2d a[100000], an, t, r;
void initA(int i) {
  int x, y;
  scanf("%d %d", &x, &y);
  a[i][0][0] = a[i][1][1] = -1;
  a[i][2][2] = 1;
  a[i][0][2] = 2 * x;
  a[i][1][2] = 2 * y;
}
void mult(int2d &a, int2d &b, int2d &r) {
  memset(t, 0, sizeof(t));
  for (int i = 0; i < 3; i++) {
    for (int k = 0; k < 3; k++) {
      for (int j = 0; j < 3; j++) {
        t[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  memcpy(r, t, sizeof(t));
}
int main() {
  int n;
  long long k;
  int x, y;
  int fk, rk;
  scanf("%d %lld %d %d", &n, &k, &x, &y);
  fk = k / n;
  rk = k % n;
  an[0][0] = an[1][1] = an[2][2] = 1;
  r[0][0] = r[1][1] = r[2][2] = 1;
  for (int i = 0; i < n; i++) {
    initA(i);
    mult(a[i], an, an);
    if (i < rk) {
      mult(a[i], r, r);
    }
  }
  while (fk) {
    if (fk & 1) {
      mult(r, an, r);
    }
    fk /= 2;
    mult(an, an, an);
  }
  printf("%lld %lld", r[0][2] + r[0][0] * x, r[1][2] + r[1][1] * y);
  return 0;
}

```