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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Echizen
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        private static final double eps = 1e-4;

        private double Get_area(double l, double a, double b, double c) {
            l = l / 2;
            return Math.sqrt(l * (l - a) * (l - b) * (l - c));
        }

        private double gcd(double a, double b) {
            if (b < eps) return a;
            return gcd(b, a % b);
        }

        private double dis(Point p1, Point p2) {
            return Math.sqrt((p1.x - p2.x) * (p1.x - p2.x) * 1.0 + (p1.y - p2.y) * (p1.y - p2.y) * 1.0);
        }

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            double x1, x2, x3, y1, y2, y3;
            double a, b, c;
            Point p1, p2, p3;
            x1 = in.nextDouble();
            y1 = in.nextDouble();
            x2 = in.nextDouble();
            y2 = in.nextDouble();
            x3 = in.nextDouble();
            y3 = in.nextDouble();
            p1 = new Point(x1, y1);
            p2 = new Point(x2, y2);
            p3 = new Point(x3, y3);
            a = dis(p1, p2);
            b = dis(p1, p3);
            c = dis(p2, p3);
            double l = a + b + c;
            double s = Get_area(l, a, b, c);
            double r = a * b * c / (4 * s);
            //余弦公式
            //cos [C]=(a^2+b^2-c^2)/(2*a*b); 再利用反函数得到相对应的弧度;
            double angle_a = Math.acos((b * b + c * c - a * a) / (2 * b * c));
            double angle_b = Math.acos((a * a + c * c - b * b) / (2 * a * c));
            double angle_c = Math.acos(-1) - angle_a - angle_b;
            //得到三个角的最大公约数
            double angle = gcd(gcd(angle_a, angle_b), angle_c);
            double n = Math.PI / angle;
            double sum = 0.5 * r * r * n * Math.sin(Math.PI * 2 / n);
            out.printf("%.8f\n", sum);
        }

        class Point {
            double x;
            double y;

            public Point() {
            }

            public Point(double x, double y) {
                this.x = x;
                this.y = y;
            }

        }

    }
}