#include <bits/stdc++.h>
using namespace std;
const int N = 205, M = 55;
int n, m;
long long ans, mx[M], f[N][N][M];
vector<int> g[N], r[N];
queue<int> q[N];
struct P {
  int x, y;
  double k;
  P(int xx = 0, int yy = 0) {
    x = xx;
    y = yy;
  }
  inline P operator-(const P& b) const { return P(x - b.x, y - b.y); }
  inline bool operator<(const P& b) const { return k < b.k; }
  inline long long operator*(const P& b) const {
    return 1ll * x * b.y - 1ll * y * b.x;
  }
} a[N], p[N];
inline bool lf(P a, P b, P c) { return (b - a) * (c - b) > 0; }
inline void add(int x, int y) {
  while (!q[x].empty() && lf(p[q[x].front()], p[x], p[y]))
    add(q[x].front(), y), q[x].pop();
  r[x].push_back(y);
  g[y].push_back(x);
  q[y].push(x);
}
inline void init(int n) {
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    r[i].clear();
  }
  for (int i = 1; i <= n; i++)
    while (!q[i].empty()) q[i].pop();
  for (int i = 1; i < n; i++) add(i, i + 1);
}
inline void dp(int n) {
  memset(f, 0, sizeof(f));
  for (int i = n; i; i--) {
    for (int j = 0; j < m; j++) mx[j] = 0;
    reverse(g[i].begin(), g[i].end());
    int x = r[i].size() - 1;
    for (int t = 0; t < g[i].size(); t++) {
      int j = g[i][t];
      f[j][i][1] = p[j] * p[i];
      for (int k = 2; k <= m; k++)
        if (mx[k - 1]) f[j][i][k] = mx[k - 1] + p[j] * p[i];
      for (; ~x && lf(p[j], p[i], p[r[i][x]]); x--)
        for (int k = 1; k < m; k++)
          if (mx[k] < f[i][r[i][x]][k])
            mx[k] = f[i][r[i][x]][k], f[j][i][k + 1] = mx[k] + p[j] * p[i];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < r[i].size(); j++) ans = max(ans, f[i][r[i][j]][m - 2]);
}
inline void sol(int id) {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (i ^ id)
      if (a[i].x > a[id].x || (a[i].x == a[id].x && a[i].y > a[id].y))
        p[++cnt] = a[i] - a[id], p[cnt].k = atan2(p[cnt].y, p[cnt].x);
  sort(p + 1, p + cnt + 1);
  init(cnt);
  dp(cnt);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].x, &a[i].y);
  for (int i = 1; i <= n; i++) sol(i);
  cout << ans / 2 << "." << (ans & 1 ? "50" : "00") << endl;
  return 0;
}
