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:
import java.util.Scanner;


public class Main {
	public static final double eps=1e-3;
	public static point intersect(tline a,tline b)
	{
		double x=(a.b-b.b)/(b.a-a.a);
		double y=a.a*x+a.b;
		point p=new point(x,y);
		return p;
	}
	public static point rotate(point mid,point p,double ang)
	{
//		x' = x*cos(a) - y*sin(a), y' = y*cos(a) + x*sin(a) )
		double x=p.x-mid.x;
		double y=p.y-mid.y;
		double nx=x*Math.cos(ang)-y*Math.sin(ang);
		double ny=y*Math.cos(ang)+x*Math.sin(ang);
		nx+=mid.x;
		ny+=mid.y;
		point ans=new point(nx,ny);
		return ans;
	}
	public static boolean samepoint(point i,point j)
	{
		if(Math.abs(i.x-j.x)<eps&&Math.abs(i.y-j.y)<eps)
		{
			return true;
		}
		return false;
	}
	public static double calculate(int n,double r)
	{
		double ang=2*Math.PI/n;
		double ans=0;
		ans=r*r*Math.sin(ang)/2;
		return ans*n;
	}
	public static double distance(point i,point j)
	{
		double ans=Math.sqrt((i.x-j.x)*(i.x-j.x)+(i.y-j.y)*(i.y-j.y));
		return ans;
	}
	public static void main(String[] args)
	{
		Scanner cin=new Scanner(System.in);
		while(cin.hasNextDouble())
		{
			point p[]=new point[3];
			for(int flag=0;flag<3;flag++)
			{
				double x,y;
				x=cin.nextDouble();
				y=cin.nextDouble();
				p[flag]=new point(x,y);
			}
			point mid1=new point();
			point mid2=new point();
			if(Math.abs(p[1].y-p[0].y)<eps)
			{
				point tem=p[1];
				p[1]=p[2];
				p[2]=tem;
			}
			if(Math.abs(p[2].y-p[1].y)<eps)
			{
				point tem=p[0];
				p[0]=p[1];
				p[1]=tem;
			}

			mid1.x=(p[0].x+p[1].x)/2;
			mid1.y=(p[0].y+p[1].y)/2;
			mid2.x=(p[2].x+p[1].x)/2;
			mid2.y=(p[2].y+p[1].y)/2;
			double k=-(p[1].x-p[0].x)/(p[1].y-p[0].y);
			tline l1=new tline(k,mid1.y-k*mid1.x);
			k=-(p[2].x-p[1].x)/(p[2].y-p[1].y);
			tline l2=new tline(k,mid2.y-k*mid2.x);
			point interpoint=intersect(l1,l2);
			point lastpoint=new point();
			point newpoint=new point();
			for(int n=3;n<=100;n++)
			{
				lastpoint=p[0];
				int count=0;
				for(int nn=0;nn<n;nn++)
				{
					newpoint=rotate(interpoint,lastpoint,2*Math.PI/n);
					if(samepoint(newpoint,p[1])||samepoint(newpoint,p[2]))
					{
						count++;
					}
					lastpoint=newpoint;
				}
				if(count>=2)
				{
					System.out.println(calculate(n,distance(interpoint,p[0])));
					break;
				}
			}
		}
	}
}

class point
{
	point()
	{
		
	}
	point(double x,double y)
	{
		this.x=x;
		this.y=y;
	}
	double x,y;
}
class line
{
	point a,b;
	line()
	{
		
	}
	line(point a,point b)
	{
		this.a=a;
		this.b=b;
	}
}
class tline
{
	double a,b;
	tline(double a,double b)
	{
		this.a=a;
		this.b=b;
	}
}