#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
#pragma GCC target("avx", "avx2")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 205;
template <typename T>
int sign(T x) {
  if (x < 0) return -1;
  if (x == 0) return 0;
  return 1;
}
struct Vec {
  long long x, y;
  Vec(long long _x = 0, long long _y = 0) { x = _x, y = _y; }
} a[N];
Vec operator+(Vec x, Vec y) { return Vec(x.x + y.x, x.y + y.y); }
Vec operator-(Vec x, Vec y) { return Vec(x.x - y.x, x.y - y.y); }
long long Cross(Vec x, Vec y) { return x.x * y.y - x.y * y.x; }
int n, m;
pair<long long, long long> dp[N][N];
bool check(long long area) {
  for (int len = 3; len <= n; len++) {
    for (int i = 1, j = len; j <= n; i++, j++) {
      dp[i][j] = {-1e9, 0};
      for (int k = i + 1; k <= j - 1; k++) {
        pair<long long, long long> f = {dp[i][k].first + dp[k][j].first,
                                        dp[i][k].second + dp[k][j].second +
                                            Cross(a[k] - a[i], a[j] - a[i])};
        if (f.second >= area) f.first++, f.second = 0;
        ckmax(dp[i][j], f);
      }
    }
  }
  return dp[1][n].first >= m;
}
int main() {
  n = read(), m = read() + 1;
  for (int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
  long long l = 0, r = 1e18;
  while (l < r) {
    long long mid = l + r + 1 >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  print(l, '\n');
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given a strictly convex polygon with n vertices.

You will make k cuts that meet the following conditions: 

  * each cut is a segment that connects two different nonadjacent vertices; 
  * two cuts can intersect only at vertices of the polygon. 



Your task is to maximize the area of the smallest region that will be formed by the polygon and those k cuts.

Input

The first line contains two integers n, k (3 ≤ n ≤ 200, 0 ≤ k ≤ n-3).

The following n lines describe vertices of the polygon in anticlockwise direction. The i-th line contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^8) — the coordinates of the i-th vertex.

It is guaranteed that the polygon is convex and that no two adjacent sides are parallel.

Output

Print one integer: the maximum possible area of the smallest region after making k cuts multiplied by 2.

Examples

Input


8 4
-2 -4
2 -2
4 2
1 5
0 5
-4 4
-5 0
-5 -1


Output


11


Input


6 3
2 -2
2 -1
1 2
0 2
-2 1
-1 0


Output


3

Note

In the first example, it's optimal to make cuts between the following pairs of vertices: 

  * (-2, -4) and (4, 2), 
  * (-2, -4) and (1, 5), 
  * (-5, -1) and (1, 5), 
  * (-5, 0) and (0, 5). 

<image> Points (-5, -1), (1, 5), (0, 5), (-5, 0) determine the smallest region with double area of 11. 

In the second example, it's optimal to make cuts between the following pairs of vertices: 

  * (2, -1) and (0, 2), 
  * (2, -1) and (1, 0), 
  * (-1, 0) and (0, 2). 

<image> Points (2, -2), (2, -1), (-1, 0) determine one of the smallest regions with double area of 3. 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
#pragma GCC target("avx", "avx2")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 205;
template <typename T>
int sign(T x) {
  if (x < 0) return -1;
  if (x == 0) return 0;
  return 1;
}
struct Vec {
  long long x, y;
  Vec(long long _x = 0, long long _y = 0) { x = _x, y = _y; }
} a[N];
Vec operator+(Vec x, Vec y) { return Vec(x.x + y.x, x.y + y.y); }
Vec operator-(Vec x, Vec y) { return Vec(x.x - y.x, x.y - y.y); }
long long Cross(Vec x, Vec y) { return x.x * y.y - x.y * y.x; }
int n, m;
pair<long long, long long> dp[N][N];
bool check(long long area) {
  for (int len = 3; len <= n; len++) {
    for (int i = 1, j = len; j <= n; i++, j++) {
      dp[i][j] = {-1e9, 0};
      for (int k = i + 1; k <= j - 1; k++) {
        pair<long long, long long> f = {dp[i][k].first + dp[k][j].first,
                                        dp[i][k].second + dp[k][j].second +
                                            Cross(a[k] - a[i], a[j] - a[i])};
        if (f.second >= area) f.first++, f.second = 0;
        ckmax(dp[i][j], f);
      }
    }
  }
  return dp[1][n].first >= m;
}
int main() {
  n = read(), m = read() + 1;
  for (int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
  long long l = 0, r = 1e18;
  while (l < r) {
    long long mid = l + r + 1 >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  print(l, '\n');
  return 0;
}

```