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 JAVA):

Solution:
import java.util.Scanner;

public class B {

    public static void main(String[] args) {
        B solver = new B();
        solver.solve();
    }

    private void solve() {
        Scanner sc = new Scanner(System.in);
//        sc = new Scanner("5\n" +
//                "0 0\n" +
//                "0 4\n" +
//                "4 0\n" +
//                "4 4\n" +
//                "2 3\n");

        int n = sc.nextInt();
        int[] xs = new int[n];
        int[] ys = new int[n];

        for (int i = 0; i < n; i++) {
            xs[i] = sc.nextInt();
            ys[i] = sc.nextInt();
        }

        double s = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                long s1 = 0;
                long s2 = 0;

                for (int k = 0; k < n; k++) {
                    if (i == k || j == k) continue;

                    long s0 = (xs[k] - xs[j]) * (ys[j] - ys[i]) - (ys[k] - ys[j]) * (xs[j] - xs[i]);
                    if (s0 >= 0) {
                        s1 = Math.max(s1, s0);
                    } else {
                        s2 = Math.max(s2, -s0);
                    }
                }

                if (s1 > 0 && s2 > 0) {
                    s = Math.max(s, (s1 + s2) / 2.0);
                }
            }
        }

        System.out.println(s);
    }

}