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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double dist(double x, double y, double xx, double yy) {
  x -= xx;
  y -= yy;
  return sqrt(x * x + y * y);
}
double geta(double d1, double d2, double d3) {
  return acos((d2 * d2 + d3 * d3 - d1 * d1) / (2.0 * d2 * d3));
}
double ab(double x) { return x >= 0 ? x : -x; }
int ch(double a, double phi) {
  return ab(a / phi - (int)(a / phi + 0.5)) < 1e-5;
}
int main() {
  double x1, y1, x2, y2, x3, y3;
  cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
  double d1 = dist(x1, y1, x2, y2);
  double d2 = dist(x1, y1, x3, y3);
  double d3 = dist(x2, y2, x3, y3);
  double a1 = geta(d1, d2, d3);
  double a2 = geta(d1, d3, d2);
  double a3 = geta(d2, d3, d1);
  double pi = 2.0 * acos(0.0);
  int n;
  for (n = 3; n <= 100; n++) {
    double phi = pi / n;
    if (ch(a1, phi) && ch(a2, phi) && ch(a3, phi)) {
      break;
    }
  }
  double p = (d1 + d2 + d3) / 2.0;
  double s = sqrt(p * (p - d1) * (p - d2) * (p - d3));
  double r = (d1 * d2 * d3) / (4.0 * s);
  double res = n * 0.5 * r * r * sin(2 * pi / n);
  printf("%.10lf\n", res);
  return 0;
}