#include <bits/stdc++.h>
using namespace std;
template <class T>
int sgn(T x) {
  return (x > 0) - (x < 0);
}
template <class T>
struct Point {
  T x, y;
  explicit Point(T _x = 0, T _y = 0) : x(_x), y(_y) {}
  bool operator<(Point p) const { return tie(x, y) < tie(p.x, p.y); }
  bool operator==(Point p) const { return tie(x, y) == tie(p.x, p.y); }
  Point operator+(Point p) const { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) const { return Point(x - p.x, y - p.y); }
  Point operator*(T d) const { return Point(x * d, y * d); }
  Point operator/(T d) const { return Point(x / d, y / d); }
  T dot(Point p) const { return x * p.x + y * p.y; }
  T cross(Point p) const { return x * p.y - y * p.x; }
  T cross(Point a, Point b) const { return (a - *this).cross(b - *this); }
  T dist2() const { return x * x + y * y; }
  double dist() const { return sqrt((double)dist2()); }
  double angle() const { return atan2(y, x); }
  Point unit() const { return *this / dist(); }
  Point perp() const { return Point(-y, x); }
  Point normal() const { return perp().unit(); }
  Point rotate(double a) const {
    return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
  }
  friend ostream& operator<<(ostream& os, Point p) {
    return os << "(" << p.x << "," << p.y << ")";
  }
};
template <class T>
T polygonArea2(vector<Point<T>>& v) {
  T a = v.back().cross(v[0]);
  for (int i = 0; i < ((int)(v).size() - 1); ++i) a += v[i].cross(v[i + 1]);
  return a;
}
using ll = int64_t;
using Point = Point<ll>;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, K;
  cin >> n >> K;
  vector<Point> polygon(n);
  for (Point& p : polygon) {
    cin >> p.x >> p.y;
  }
  ll s = 0;
  ll e = 1e17;
  vector<vector<ll>> area(n, vector<ll>(n));
  for (int j = 0; j < n; j++) {
    for (int i = j - 1; i >= 0; i--) {
      vector<Point> pts;
      for (int k = i; k <= j; k++) {
        pts.push_back(polygon[k]);
      }
      area[i][j] = polygonArea2(pts);
    }
  }
  while (s + 1 < e) {
    ll thresh = (s + e) / 2;
    vector<vector<pair<int, ll>>> dp(n, vector<pair<int, ll>>(n, {-1e8, 0}));
    for (int j = 0; j < n; j++) {
      for (int i = j - 1; i >= 0; i--) {
        dp[i][j] = {0, area[i][j]};
        if (area[i][j] >= thresh) dp[i][j] = {1, 0};
        for (int k = i + 1; k < j; k++) {
          int np = dp[i][k].first + dp[k][j].first;
          ll narea = (polygon[k] - polygon[i]).cross(polygon[j] - polygon[i]) +
                     dp[i][k].second + dp[k][j].second;
          if (narea >= thresh) {
            narea = 0;
            np += 1;
          }
          dp[i][j] = max(dp[i][j], {np, narea});
        }
      }
    }
    if (dp[0][n - 1].first >= K + 1) {
      s = thresh;
    } else {
      e = thresh;
    }
  }
  cout << s << '\n';
}

### Prompt

Please formulate a CPP solution to the following problem:

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>
using namespace std;
template <class T>
int sgn(T x) {
  return (x > 0) - (x < 0);
}
template <class T>
struct Point {
  T x, y;
  explicit Point(T _x = 0, T _y = 0) : x(_x), y(_y) {}
  bool operator<(Point p) const { return tie(x, y) < tie(p.x, p.y); }
  bool operator==(Point p) const { return tie(x, y) == tie(p.x, p.y); }
  Point operator+(Point p) const { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) const { return Point(x - p.x, y - p.y); }
  Point operator*(T d) const { return Point(x * d, y * d); }
  Point operator/(T d) const { return Point(x / d, y / d); }
  T dot(Point p) const { return x * p.x + y * p.y; }
  T cross(Point p) const { return x * p.y - y * p.x; }
  T cross(Point a, Point b) const { return (a - *this).cross(b - *this); }
  T dist2() const { return x * x + y * y; }
  double dist() const { return sqrt((double)dist2()); }
  double angle() const { return atan2(y, x); }
  Point unit() const { return *this / dist(); }
  Point perp() const { return Point(-y, x); }
  Point normal() const { return perp().unit(); }
  Point rotate(double a) const {
    return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
  }
  friend ostream& operator<<(ostream& os, Point p) {
    return os << "(" << p.x << "," << p.y << ")";
  }
};
template <class T>
T polygonArea2(vector<Point<T>>& v) {
  T a = v.back().cross(v[0]);
  for (int i = 0; i < ((int)(v).size() - 1); ++i) a += v[i].cross(v[i + 1]);
  return a;
}
using ll = int64_t;
using Point = Point<ll>;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, K;
  cin >> n >> K;
  vector<Point> polygon(n);
  for (Point& p : polygon) {
    cin >> p.x >> p.y;
  }
  ll s = 0;
  ll e = 1e17;
  vector<vector<ll>> area(n, vector<ll>(n));
  for (int j = 0; j < n; j++) {
    for (int i = j - 1; i >= 0; i--) {
      vector<Point> pts;
      for (int k = i; k <= j; k++) {
        pts.push_back(polygon[k]);
      }
      area[i][j] = polygonArea2(pts);
    }
  }
  while (s + 1 < e) {
    ll thresh = (s + e) / 2;
    vector<vector<pair<int, ll>>> dp(n, vector<pair<int, ll>>(n, {-1e8, 0}));
    for (int j = 0; j < n; j++) {
      for (int i = j - 1; i >= 0; i--) {
        dp[i][j] = {0, area[i][j]};
        if (area[i][j] >= thresh) dp[i][j] = {1, 0};
        for (int k = i + 1; k < j; k++) {
          int np = dp[i][k].first + dp[k][j].first;
          ll narea = (polygon[k] - polygon[i]).cross(polygon[j] - polygon[i]) +
                     dp[i][k].second + dp[k][j].second;
          if (narea >= thresh) {
            narea = 0;
            np += 1;
          }
          dp[i][j] = max(dp[i][j], {np, narea});
        }
      }
    }
    if (dp[0][n - 1].first >= K + 1) {
      s = thresh;
    } else {
      e = thresh;
    }
  }
  cout << s << '\n';
}

```