#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
template <class T>
inline void gmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void gmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
const int N = 1e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647,
          ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567,
          ms127 = 2139062143;
const double PI = acos(-1.0), eps = 1e-8;
void fre() {
  freopen("/Users/luras/Desktop/in.txt", "r", stdin);
  freopen("/Users/luras/Desktop/out.txt", "w", stdout);
}
const int INF = 1e9;
int casenum, casei;
long long n, m;
long long ax[N], ay[N], x0, Y0;
long long sumx, sumy;
int main() {
  while (~scanf("%lld%lld", &n, &m)) {
    scanf("%lld%lld", &x0, &Y0);
    for (int i = 0; i < n; i++) {
      scanf("%lld%lld", &ax[i], &ay[i]);
    }
    m = m % (2 * n);
    sumx = 0;
    sumy = 0;
    for (int i = m - 1; i >= 0; i--) {
      if ((m - i) & 1) {
        sumx += ax[i % n] * 2;
        sumy += ay[i % n] * 2;
      } else {
        sumx -= ax[i % n] * 2;
        sumy -= ay[i % n] * 2;
      }
    }
    if (m & 1) {
      sumx -= x0;
      sumy -= Y0;
    } else {
      sumx += x0;
      sumy += Y0;
    }
    printf("%lld %lld\n", sumx, sumy);
  }
  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>
#pragma comment(linker, "/STACK:102400000,102400000")
template <class T>
inline void gmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void gmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
const int N = 1e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647,
          ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567,
          ms127 = 2139062143;
const double PI = acos(-1.0), eps = 1e-8;
void fre() {
  freopen("/Users/luras/Desktop/in.txt", "r", stdin);
  freopen("/Users/luras/Desktop/out.txt", "w", stdout);
}
const int INF = 1e9;
int casenum, casei;
long long n, m;
long long ax[N], ay[N], x0, Y0;
long long sumx, sumy;
int main() {
  while (~scanf("%lld%lld", &n, &m)) {
    scanf("%lld%lld", &x0, &Y0);
    for (int i = 0; i < n; i++) {
      scanf("%lld%lld", &ax[i], &ay[i]);
    }
    m = m % (2 * n);
    sumx = 0;
    sumy = 0;
    for (int i = m - 1; i >= 0; i--) {
      if ((m - i) & 1) {
        sumx += ax[i % n] * 2;
        sumy += ay[i % n] * 2;
      } else {
        sumx -= ax[i % n] * 2;
        sumy -= ay[i % n] * 2;
      }
    }
    if (m & 1) {
      sumx -= x0;
      sumy -= Y0;
    } else {
      sumx += x0;
      sumy += Y0;
    }
    printf("%lld %lld\n", sumx, sumy);
  }
  return 0;
}

```