#include <bits/stdc++.h>
using namespace std;
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int infi = 1e9;
const long long Linfi = (long long)9e18;
const long long MOD = 1000000007;
const int c1 = 11;
const int c2 = 37;
int n;
long long m;
pair<long long, long long> A[200005];
pair<long long, long long> M[200005];
void Init() {}
void solve() {
  m = m % (2 * n);
  for (int i = 1; i <= m; i++) {
    M[i].first = 2 * A[(i - 1) % n].first - M[i - 1].first;
    M[i].second = 2 * A[(i - 1) % n].second - M[i - 1].second;
  }
  cout << M[m].first << " " << M[m].second << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  cin >> M[0].first >> M[0].second;
  for (int i = 0; i <= n - 1; i++) cin >> A[i].first >> A[i].second;
  solve();
  return 0;
}

### Prompt

Your task is to create 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>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int infi = 1e9;
const long long Linfi = (long long)9e18;
const long long MOD = 1000000007;
const int c1 = 11;
const int c2 = 37;
int n;
long long m;
pair<long long, long long> A[200005];
pair<long long, long long> M[200005];
void Init() {}
void solve() {
  m = m % (2 * n);
  for (int i = 1; i <= m; i++) {
    M[i].first = 2 * A[(i - 1) % n].first - M[i - 1].first;
    M[i].second = 2 * A[(i - 1) % n].second - M[i - 1].second;
  }
  cout << M[m].first << " " << M[m].second << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  cin >> M[0].first >> M[0].second;
  for (int i = 0; i <= n - 1; i++) cin >> A[i].first >> A[i].second;
  solve();
  return 0;
}

```