#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 1e2;
pair<long long, long long> A[MAX], B[MAX << 1];
pair<long long, long long> getNext(pair<long long, long long> a,
                                   pair<long long, long long> b) {
  long long x = 2 * b.first - a.first;
  long long y = 2 * b.second - a.second;
  return make_pair(x, y);
}
int main() {
  int n;
  long long k;
  scanf("%d %I64d", &n, &k);
  scanf("%I64d %I64d", &B[0].first, &B[0].second);
  for (int i = 0; i < (n); i++) scanf("%I64d %I64d", &A[i].first, &A[i].second);
  for (int i = 0; i < (2 * n); i++) B[i + 1] = getNext(B[i], A[i % n]);
  printf("%I64d %I64d\n", B[k % (2 * n)].first, B[k % (2 * n)].second);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MAX = 1e5 + 1e2;
pair<long long, long long> A[MAX], B[MAX << 1];
pair<long long, long long> getNext(pair<long long, long long> a,
                                   pair<long long, long long> b) {
  long long x = 2 * b.first - a.first;
  long long y = 2 * b.second - a.second;
  return make_pair(x, y);
}
int main() {
  int n;
  long long k;
  scanf("%d %I64d", &n, &k);
  scanf("%I64d %I64d", &B[0].first, &B[0].second);
  for (int i = 0; i < (n); i++) scanf("%I64d %I64d", &A[i].first, &A[i].second);
  for (int i = 0; i < (2 * n); i++) B[i + 1] = getNext(B[i], A[i % n]);
  printf("%I64d %I64d\n", B[k % (2 * n)].first, B[k % (2 * n)].second);
  return 0;
}

```