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.*;
import java.util.*;
import java.math.*;

public class C {

  private double x0, x1, x2, x3, y0, y1, y2, y3, r, x12, x22, x32, y12, y22, y32, z2;
  
  private double diff(double a, double b) {
    return Math.abs(a - b);
  }

  private double angle(double x1, double y1, double x2, double y2) {
    return Math.acos(((x1 - x0) * (x2 - x0) + (y1 - y0) * (y2 - y0)) /
                     (Math.hypot((x1 - x0), (y1 - y0)) * Math.hypot((x2 - x0), (y2 - y0))));
  }

  public void run() {
    x1 = Double.parseDouble(sc.next()); x12 = x1 * x1;
    y1 = Double.parseDouble(sc.next()); y12 = y1 * y1;
    x2 = Double.parseDouble(sc.next()); x22 = x2 * x2;
    y2 = Double.parseDouble(sc.next()); y22 = y2 * y2;
    x3 = Double.parseDouble(sc.next()); x32 = x3 * x3;
    y3 = Double.parseDouble(sc.next()); y32 = y3 * y3;

    z2 = 2 * ((y3 - y1) * x2 + (x1 - x3) * y2 - x1 * y3 + x3 * y1);
    x0 = ((x12 + y12) * (y2 - y3) + (x22 + y22) * (y3 - y1) + (x32 + y32) * (y1 - y2)) / z2;
    y0 = ((x12 + y12) * (x3 - x2) + (x22 + y22) * (x1 - x3) + (x32 + y32) * (x2 - x1)) / z2;
    r = Math.hypot(x1 - x0, y1 - y0);

    int n, minN = 3;
    double div1, div2, div3;
    double minDiff = Double.MAX_VALUE;

    for (n = 3; n < 101; n++) {
      div1 = angle(x1, y1, x2, y2) / (2 * Math.PI / n);
      div2 = angle(x1, y1, x3, y3) / (2 * Math.PI / n);
      div3 = angle(x2, y2, x3, y3) / (2 * Math.PI / n);

      div1 = new Double(div1).equals(Double.NaN) ? 0 : div1;
      div2 = new Double(div2).equals(Double.NaN) ? 0 : div2;
      div3 = new Double(div3).equals(Double.NaN) ? 0 : div3;

      double diff1 = diff(div1, Math.round(div1));
      double diff2 = diff(div2, Math.round(div2));
      double diff3 = diff(div3, Math.round(div3));

      double diff = Math.max(Math.max(diff1, diff2), diff3);

      if (diff < minDiff) { minDiff = diff; minN = n; }
    }

    pw.println((minN / 2.0 * r * r * Math.sin((2.0 * Math.PI / minN))));
  }

  public static Scanner sc;
  public static PrintWriter pw;

  public static void main(String[] args) throws Exception {
    sc = new Scanner(new BufferedInputStream(System.in));
    pw = new PrintWriter(new BufferedOutputStream(System.out));
    
    new C().run();

    pw.close();
  }

}