Problem: Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input

The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output

Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples

Input

0.000000 0.000000
1.000000 1.000000
0.000000 1.000000


Output

1.00000000

Solution (in JAVA):

Solution:
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

import java.util.Scanner;

/**
 *
 * @author IR_HAM
 */
public class AncientBerlandCircus
{
    public static void main(String [] args)
    {
        Scanner in = new Scanner(System.in);
        double [] x = new double[3];
        double [] y = new double[3];
        
        for(int i = 0; i < x.length; i++)
        {
            x[i] = in.nextDouble();
            y[i] = in.nextDouble();
        }
        
        double [] z = new double[3];
        z[0] = length(x[0], y[0], x[1], y[1]);
        z[1] = length(x[1], y[1], x[2], y[2]);
        z[2] = length(x[2], y[2], x[0], y[0]);
        
        for(int i = 0; i < z.length; i++)
        {
            int k = i;
            for(int j = i; j < z.length; j++)
                if(z[k] > z[j])
                    k = j;
            
            double w = z[i];
            z[i] = z[k];
            z[k] = w;
        }
        
        double r = z[0] * z[1] * z[2] / (4 * area(z));
        
        z[0] = Math.acos(1 - sqr(z[0]) / (2 * sqr(r)));
        z[1] = Math.acos(1 - sqr(z[1]) / (2 * sqr(r)));
        z[2] = 2 * Math.PI - (z[0] + z[1]);
        
        int n = optimalN(z[0], z[1], z[2]);
        //System.out.println(r + " " + z[0] + " " + z[1] + " " + z[2] + " " + n);
        double poly_area = polygonArea(n, r);
        System.out.println(String.format("%.6f", poly_area));
        /*System.out.println("\n");
        for(int i = 3; i <= 100; i++)
            System.out.println(i + " " + polygonArea(i, r));*/
    }
    
    public static double polygonArea(int n, double r)
    {
        return n * sqr(r) * Math.sin(360.0 * Math.PI / (180 * n)) / 2;
    }
    
    public static int optimalN(double a, double b, double c)
    {
        int m = 100;
        int n = 3;
        
        double error = 0.00001;
        for(int i = 3; i <= m; i++)
        {
            double d = 2 * Math.PI / i;
            
            double e = a / d;
            double f = b / d;
            double g = c / d;
            
            int n1 = round(e);
            int n2 = round(f);
            int n3 = round(g);
            
            int t = n1 + n2 + n3;
            
            double epsilon = (abs(n1 * d - a) + abs(n2 * d - b) + abs(n3 * d - c)) / d;
            epsilon = Double.parseDouble(String.format("%.013f", epsilon));
            /*System.out.println(i + " " + (n1 + n2 + n3) + " \\ " + d + " / "
                    + " | " + n1 + " <-> " + e + " <-> " + n1 * d + " <-> " + a + " <--> " + a * 180 / Math.PI
                    + " | " + n2 + " <-> " + f + " <-> " + n2 * d + " <-> " + b + " <--> " + b * 180 / Math.PI
                    + " | " + n3 + " <-> " + g + " <-> " + n3 * d + " <-> " + c + " <--> " + c * 180 / Math.PI
                    + " | " + " ====> " + epsilon + " | " + error);*/
            if(t == i && epsilon < error)
            {
                n = i;
                break;
            }
        }
         
        return n;
    }
    
    public static double length(double x1, double y1, double x2, double y2)
    {
        return Math.sqrt((sqr(x1 - x2) + sqr(y1 - y2)));
    }
    
    public static double sqr(double x)
    {
        return x * x;
    }
    
    public static double area(double [] z)
    {
        double a = Math.acos((sqr(z[2]) + sqr(z[1]) - sqr(z[0])) / (2 * z[1] * z[2]));
        a = z[1] * z[2] * Math.sin(a) / 2;
        
        return a;
    }
    
    public static double abs(double x)
    {
        if(x >= 0)
            return x;
        else
            return -x;
    }
    
    public static int round(double x)
    {
        return (int) (Double.parseDouble(String.format("%.0f", x)));
    }
}
//-13.242302 -45.014124 -33.825369 51.083964 84.512928 -55.134407
//17.288379 68.223317 48.776683 71.688379 23.170559 106.572762
//25.428124 39.407248 17.868098 39.785933 11.028461 43.028890