description: 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. 
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MOD = (int)1e9 + 7;
  const int FFTMOD = 119 << 23 | 1;
  const int INF = (int)1e9 + 23111992;
  const long long LINF = (long long)1e18 + 23111992;
  const long double PI = acos((long double)-1);
  const long double EPS = 1e-9;
  inline long long gcd(long long a, long long b) {
    long long r;
    while (b) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  inline long long fpow(long long n, long long k, int p = MOD) {
    long long r = 1;
    for (; k; k >>= 1) {
      if (k & 1) r = r * n % p;
      n = n * n % p;
    }
    return r;
  }
  template <class T>
  inline int chkmin(T& a, const T& val) {
    return val < a ? a = val, 1 : 0;
  }
  template <class T>
  inline int chkmax(T& a, const T& val) {
    return a < val ? a = val, 1 : 0;
  }
  inline unsigned long long isqrt(unsigned long long k) {
    unsigned long long r = sqrt(k) + 1;
    while (r * r > k) r--;
    return r;
  }
  inline long long icbrt(long long k) {
    long long r = cbrt(k) + 1;
    while (r * r * r > k) r--;
    return r;
  }
  inline void addmod(int& a, int val, int p = MOD) {
    if ((a = (a + val)) >= p) a -= p;
  }
  inline void submod(int& a, int val, int p = MOD) {
    if ((a = (a - val)) < 0) a += p;
  }
  inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
  inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
  inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
  inline int sign(long double x, long double y) { return sign(x - y); }
  mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
  inline int mrand() { return abs((int)mt()); }
  inline int mrand(int k) { return abs((int)mt()) % k; }
  struct point_t {
    int x, y;
    point_t() : x(0), y(0) {}
    point_t(int x, int y) : x(x), y(y) {}
    point_t(const point_t& rhs) : x(rhs.x), y(rhs.y) {}
    int operator<(const point_t& rhs) const {
      return make_pair(y, x) < make_pair(rhs.y, rhs.x);
    }
    int operator==(const point_t& rhs) const {
      return make_pair(y, x) == make_pair(rhs.y, rhs.x);
    }
    int operator!=(const point_t& rhs) const {
      return make_pair(y, x) != make_pair(rhs.y, rhs.x);
    }
    point_t operator-(const point_t& rhs) const {
      return point_t(x - rhs.x, y - rhs.y);
    }
  };
  long long cross(point_t a, point_t b) {
    return (long long)a.x * b.y - (long long)a.y * b.x;
  }
  long long area(point_t a, point_t b, point_t c) {
    return cross(a, b) + cross(b, c) + cross(c, a);
  }
  long long dist(point_t a, point_t b) {
    return (long long)(a.x - b.x) * (a.x - b.x) +
           (long long)(a.y - b.y) * (a.y - b.y);
  }
  void chemthan() {
    int n, k;
    cin >> n >> k;
    vector<point_t> pts(n);
    for (int i = (0); i < (n); ++i) cin >> pts[i].x >> pts[i].y;
    vector<vector<long long>> f(n, vector<long long>(n));
    for (int i = (0); i < (n); ++i) {
      long long sum = 0;
      for (int jj = (1); jj < (n); ++jj) {
        int u = (i + jj - 1) % n;
        int v = (u + 1) % n;
        sum += cross(pts[u], pts[v]);
        f[i][v] = sum;
      }
    }
    long long area = 0;
    for (int i = (0); i < (n); ++i) {
      int j = (i + 1) % n;
      area += cross(pts[i], pts[j]);
    }
    auto func = [&](int x, int y, int z, int t) {
      long long res = area;
      res -= f[x][y];
      res -= cross(pts[y], pts[x]);
      res -= f[z][t];
      res -= cross(pts[t], pts[z]);
      return res;
    };
    auto check = [&](long long mi) {
      vector<vector<pair<int, long long>>> dp(
          n, vector<pair<int, long long>>(n, make_pair(-1, 0)));
      function<pair<int, long long>(int, int)> calc = [&](int x, int y) {
        if (f[y][x] + cross(pts[x], pts[y]) < mi) {
          return make_pair(0, f[y][x] + cross(pts[x], pts[y]));
        }
        auto& res = dp[x][y];
        if (~res.first) return res;
        res = make_pair(0, f[y][x] + cross(pts[x], pts[y]));
        if (mi <= res.second) {
          res.first++;
          res.second = 0;
        }
        auto smart = [&](int x) { return x < n ? x : x - n; };
        {
          for (int i = smart(y + 1); i != x; i = smart(i + 1)) {
            auto tmp1 = calc(x, i);
            auto tmp2 = calc(i, y);
            auto tmp =
                make_pair(tmp1.first + tmp2.first,
                          tmp1.second + tmp2.second + cross(pts[x], pts[y]) +
                              cross(pts[y], pts[i]) + cross(pts[i], pts[x]));
            if (mi <= tmp.second) {
              tmp.first++;
              tmp.second = 0;
            }
            chkmax(res, tmp);
          }
        }
        return res;
      };
      return k + 1 <= calc(0, 1).first;
    };
    long long lo = 0, hi = area / (k + 1);
    while (lo < hi) {
      long long mi = lo + hi + 1 >> 1;
      if (check(mi)) {
        lo = mi;
      } else {
        hi = mi - 1;
      }
    }
    cout << (lo + hi >> 1) << "\n";
  }
  int32_t main(int32_t argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
      assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
      assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  