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;
double pt[301][2];
double mx[301][301][2];
double area(int a, int b, int c) {
  double res = pt[a][0] * (pt[b][1] - pt[c][1]) +
               pt[b][0] * (pt[c][1] - pt[a][1]) +
               pt[c][0] * (pt[a][1] - pt[b][1]);
  return res / 2.0;
}
int main() {
  int n;
  while (cin >> n) {
    double best = 0;
    for (int i = 0; i < n; i++) cin >> pt[i][0] >> pt[i][1];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) mx[i][j][0] = mx[i][j][1] = -100000000000.0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (j == i) continue;
        for (int k = 0; k < n; k++) {
          if (i == k || k == j) continue;
          double tmp = area(i, j, k);
          if (tmp < 0) {
            tmp = -tmp;
            if (tmp > mx[i][j][1]) mx[i][j][1] = tmp;
          } else {
            if (tmp > mx[i][j][0]) mx[i][j][0] = tmp;
          }
        }
      }
    }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        if (mx[i][j][0] + mx[i][j][1] > best) {
          best = mx[i][j][0] + mx[i][j][1];
        }
      }
    printf("%.10lf\n", best);
  }
  return 0;
}