Problem: Iahub has drawn a set of n points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral. 

Input

The first line contains integer n (4 ≤ n ≤ 300). Each of the next n lines contains two integers: xi, yi ( - 1000 ≤ xi, yi ≤ 1000) — the cartesian coordinates of ith special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide. 

Output

Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10 - 9.

Examples

Input

5
0 0
0 4
4 0
4 4
2 3


Output

16.000000

Note

In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int bfsz = 1 << 16;
char bf[bfsz + 5];
int rsz = 0;
int ptr = 0;
char gc() {
  if (rsz <= 0) {
    ptr = 0;
    rsz = (int)fread(bf, 1, bfsz, stdin);
    if (rsz <= 0) return EOF;
  }
  --rsz;
  return bf[ptr++];
}
void ga(char &c) {
  c = EOF;
  while (!isalpha(c)) c = gc();
}
int gs(char s[]) {
  int l = 0;
  char c = gc();
  while (isspace(c)) c = gc();
  while (c != EOF && !isspace(c)) {
    s[l++] = c;
    c = gc();
  }
  s[l] = '\0';
  return l;
}
template <class T>
bool gi(T &v) {
  v = 0;
  char c = gc();
  while (c != EOF && c != '-' && !isdigit(c)) c = gc();
  if (c == EOF) return false;
  bool neg = c == '-';
  if (neg) c = gc();
  while (isdigit(c)) {
    v = v * 10 + c - '0';
    c = gc();
  }
  if (neg) v = -v;
  return true;
}
struct triple {
  int x, y, z;
};
double x[3005], y[3005], res;
int n;
int main() {
  cin >> n;
  for (__typeof(1) i = 1; i <= n; i++) cin >> x[i] >> y[i];
  res = -1;
  for (__typeof(1) a = 1; a <= n - 1; a++)
    for (__typeof(a + 1) b = a + 1; b <= n; b++) {
      vector<double> tmp, t;
      double bb = (x[b] - x[a]) * (x[b] - x[a]) + (y[b] - y[a]) * (y[b] - y[a]),
             max1 = -1, max2 = -1;
      for (__typeof(1) k = 1; k <= n; k++)
        if (a != k && b != k) {
          double aa =
              (x[k] - x[a]) * (y[b] - y[a]) - (x[b] - x[a]) * (y[k] - y[a]);
          if (aa < 0) {
            tmp.push_back(fabs(aa) / sqrt(bb));
            if (max1 - fabs(aa) / sqrt(bb) < 0) max1 = fabs(aa) / sqrt(bb);
          }
          if (aa > 0) {
            if (max2 - fabs(aa) / sqrt(bb) < 0) max2 = fabs(aa) / sqrt(bb);
            t.push_back(fabs(aa) / sqrt(bb));
          }
        }
      if ((tmp.size() > 0) && (t.size() > 0)) {
        if ((res - sqrt(bb) * (max1 + max2) / 2.0) < 0)
          res = sqrt(bb) * (max1 + max2) / 2.0;
      }
    }
  printf("%.9lf\n", res);
}