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;
int n;
pair<double, double> p[301];
double a, b, c, den, len;
double area(int i) {
  double num = abs(a * p[i].first + b * p[i].second + c);
  return (num) / (2.0);
}
double f(int i, int j) {
  a = (p[j].second - p[i].second);
  b = (p[i].first - p[j].first);
  c = 0 - p[i].second * b - p[i].first * a;
  den = sqrtf(a * a + b * b);
  len = sqrtf((p[i].first - p[j].first) * (p[i].first - p[j].first) +
              (p[i].second - p[j].second) * (p[i].second - p[j].second));
  double m1 = -1, m2 = -1;
  for (int k = 0; k < n; ++k) {
    if (k == i or k == j) continue;
    long long temp = a * p[k].first + b * p[k].second + c;
    double flag = area(k);
    if (temp >= 0)
      m1 = max(m1, flag);
    else
      m2 = max(m2, flag);
  }
  if (m1 == -1 or m2 == -1) return 0;
  return m1 + m2;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    p[i] = {x, y};
  }
  double ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      ans = max(ans, f(i, j));
    }
  }
  printf("%0.10f\n", ans);
  return 0;
}