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.io.IOException;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Sol2 {

	public static boolean r(double x, double y) {
		if (Math.abs(x - y) < 0.000000001) {
			return false;
		}
		return true;
	}

	void run() throws IOException {
		Scanner sc = new Scanner(System.in);
		Comparator<double[]> com = new Comparator<double[]>() {
			public int compare(double[] a0, double[] a1) {
				if (a0[0] - a1[0] < 0) {
					return 1;
				}
				if (a0[0] - a1[0] > 0) {
					return -1;
				}

				return 0;
			}
		};
		int n = sc.nextInt();
		int[][] a = new int[n][2];
		for (int i = 0; i < n; i++) {
			a[i][0] = sc.nextInt();
			a[i][1] = sc.nextInt();
		}
		double[][] b = new double[n * (n - 1) / 2][6];
		int k = 0;
		for (int i = 0; i < n - 1; i++) {
			for (int j = i + 1; j < n; j++) {
				b[k][0] = Math.sqrt((a[i][0] - a[j][0]) * (a[i][0] - a[j][0])
						+ (a[i][1] - a[j][1]) * (a[i][1] - a[j][1]));
				b[k][1] = a[i][0] - a[j][0];
				b[k][2] = a[i][1] - a[j][1];
				b[k][3] = i;
				b[k][4] = j;

				k++;
			}
		}
		Arrays.sort(b, com);
		int i = 0;
		double max = 0;

		while ((i < k - 1) && (b[i][0] * b[i + 1][0] > max)) {
			int j = i + 1;
			while ((j < k) && (b[i][0] * b[j][0] > max)) {
				if (r(b[i][3], b[j][3]) && r(b[i][3], b[j][4])
						&& r(b[i][4], b[j][3]) && r(b[i][4], b[j][4])) {
					double ss = Math.atan2(b[i][1] * b[j][2] - b[i][2]
							* b[j][1], b[i][1] * b[j][1] + b[i][2] * b[j][2]);
					ss = Math.abs(ss);
					if (ss > Math.PI / 2) {
						ss = Math.PI-ss;
					}
					
					double s = Math.sin(ss) * b[i][0] * b[j][0];
					if (s > max) {
						max = s;
					}
				}
				j++;

			}
			i++;
		}
		System.out.println(max / 2);

	}

	public static void main(String[] args) throws IOException {
		new Sol2().run();
	}

}