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.*;


public class Main {

    final int maxn = 1 << 17;

    int n;
    int x[] = new int[maxn], y[] = new int[maxn];

    public void Run() {
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        for (int i = 0; i < n; i++) {
            x[i] = in.nextInt();
            y[i] = in.nextInt();
        }

        double ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                double A = y[i] - y[j];
                double B = x[j] - x[i];
                double C = -A * x[i] - B * y[i];
                double d = Math.sqrt(A * A + B * B);
                A /= d; B /= d; C /= d;
                d = Math.sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j])) / 2.0;

                double a1 = -1, a2 = -1;
                for (int k = 0; k < n; k++) if (k != i && k != j) {
                    if (A * x[k] + B * y[k] + C > 0) {
                        a1 = Math.max(a1, (A * x[k] + B * y[k] + C) * d);
                    }
                    if (A * x[k] + B * y[k] + C < 0) {
                        a2 = Math.max(a2, -(A * x[k] + B * y[k] + C) * d);
                    }
                }
                if (a1 > 0 && a2 > 0) ans = Math.max(ans, a1 + a2);
            }
        }
        System.out.println(ans);
    }

    public static void main(String[] args) {
        new Main().Run();
    }
}