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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    public static void main (String[] args) throws IOException
    {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

        double[][] points = new double[3][2];
        double[] sides = new double[3];
        final double MARGIN = 1e-4;

        // Gets input
        for (int i=0; i<3; ++i) {
            String[] line = reader.readLine().split(" ");
            for (int j=0; j<2; ++j) {
                points[i][j] = Double.parseDouble(line[j]);
            }
        }

        // Gets side lengths
        for (int i=0; i<3; ++i) {
            sides[i] = distForm(points[i], points[(i+1)%3]);
        }

        double halfCircumference = (sides[0] + sides[1] + sides[2]) / 2;

        double radius = (sides[0] * sides[1] * sides[2]) / (2 * Math.sqrt(halfCircumference * (halfCircumference - sides[0]) * (halfCircumference - sides[1]) * (halfCircumference - sides[2]))) / 2;

        // Gets angles
        double angle1 = cosineLaw(sides[0], sides[1], sides[2]);
        double angle2 = cosineLaw(sides[1], sides[0], sides[2]);
        double angle3 = cosineLaw(sides[2], sides[0], sides[1]);

        // Gets number of sides
        int numSides;
        for(numSides = 2; numSides <= 99; numSides++) {
            double side1Val = numSides * angle1;
            double side2Val = numSides * angle2;
            double side3Val = numSides * angle3;
            if(Math.abs(side1Val - Math.round(side1Val)) < MARGIN && Math.abs(side2Val - Math.round(side2Val)) < MARGIN && Math.abs(side3Val - Math.round(side3Val)) < MARGIN) {
                break;
            }
        }

        // Finds and outputs area
        double area = numSides * .5  * Math.pow(radius, 2) * Math.sin(Math.PI * 2 / numSides);
        System.out.println(String.format("%.7f", area));
    }

    // Finds distance between two points
    static double distForm(double[] first, double[] second) {
        return Math.sqrt(Math.pow(second[0] - first[0], 2) + Math.pow(second[1] - first[1], 2));
    }

    // Finds triangle angles
    static double cosineLaw(double A, double B, double C){
        return Math.acos((Math.pow(B, 2)+Math.pow(C, 2)-Math.pow(A, 2))/(2*B*C)) / Math.PI;
    }
}