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;
const int white = 0, gray = 1, black = 2;
const int inf = 0x0fffffff;
const int Size = 10000;
char buffer[Size];
const double eps = 10e-6;
bool eq(double a, double b) { return fabs(a - b) < eps; }
bool lt(double a, double b) { return a < b && !eq(a, b); }
bool le(double a, double b) { return a < b || eq(a, b); }
struct Point {
  double x, y;
  Point(double x = 0., double y = 0.) : x(x), y(y) {}
};
Point operator+(const Point &a, const Point &b) {
  return Point(a.x + b.x, a.y + b.y);
}
Point operator-(const Point &a, const Point &b) {
  return Point(a.x - b.x, a.y - b.y);
}
double Cross(const Point &a, const Point &b) { return a.x * b.y - a.y * b.x; }
struct Seg {
  Point a, b;
  Seg(const Point &a = Point(), const Point &b = Point()) : a(a), b(b) {}
};
double Det(double a, double b, double c, double d) { return a * d - b * c; }
int CrossLine(const Seg &a, const Seg &b, Point &res) {
  Point p1 = a.b - a.a;
  double a1 = p1.y, b1 = -p1.x, c1 = -Cross(a.a, p1);
  Point p2 = b.b - b.a;
  double a2 = p2.y, b2 = -p2.x, c2 = -Cross(b.a, p2);
  double d = Det(a1, b1, a2, b2);
  if (eq(d, 0.)) return 0;
  double x = -Det(c1, b1, c2, b2) / d;
  double y = -Det(a1, c1, a2, c2) / d;
  res = Point(x, y);
  return 1;
}
Point operator*(const Point &a, double k) { return Point(a.x * k, a.y * k); }
double Dist(const Point &a) {
  double d = a.x * a.x + a.y * a.y;
  if (eq(d, 0.)) return 0.;
  return sqrt(d);
}
Point Norm(const Point &a) {
  double d = 1 / Dist(a);
  return a * d;
}
double gcd(double a, double b) {
  if (fabs(b) < eps) return a;
  return gcd(b, fmod(a, b));
}
double Scalar(const Point &a, const Point &b) { return a.x * b.x + a.y * b.y; }
double GetAngle(const Point &a, const Point &b) {
  return acos(Scalar(a, b) / (Dist(a) * Dist(b)));
}
double pi = acos(-1.);
int Solution(int nTest) {
  Point ar[3];
  for (int i = 0; i < 3; i++) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    ar[i] = Point(x, y);
  }
  Point p01 = ar[1] - ar[0], p02 = ar[2] - ar[0];
  Point p01n(p01.y, -p01.x), p02n(p02.y, -p02.x);
  Point t1 = ar[0] + (p01 * 0.5);
  Point t2 = ar[0] + (p02 * 0.5);
  Seg m1(t1, t1 + p01n);
  Seg m2(t2, t2 + p02n);
  Point c1, c2, c3;
  if (CrossLine(m1, m2, c1) == 0) throw "exc";
  Point p12 = ar[2] - ar[1];
  Point p12n = Point(p12.y, -p12.x);
  Point t3 = ar[1] + p12 * 0.5;
  Seg m3(t3, t3 + p12n);
  CrossLine(m1, m3, c3);
  CrossLine(m2, m3, c2);
  Point c = (c1 + c2 + c3) * (1. / 3.);
  double a1 = fabs(GetAngle(ar[0] - c, ar[1] - c)),
         a2 = fabs(GetAngle(ar[0] - c, ar[2] - c));
  double a3 = fabs(GetAngle(ar[2] - c, ar[1] - c));
  for (int n = 3; n <= 100; n++) {
    double g = (2. * pi) / (double)n;
    int state = 0;
    for (int j = 1; j < n; j++) {
      if (eq((double)j * g, a1)) state++;
      if (eq((double)j * g, a2)) state++;
      if (eq((double)j * g, a3)) state++;
    }
    if (state == 3) {
      double r = Dist(c - ar[0]);
      double s1 = r * r * sin(g) / 2.;
      double s = s1 * n;
      printf("%.7lf\n", s);
      return 0;
    }
  }
  return 0;
}
int main() {
  int i = 0, n = 99999;
  while (i < n && Solution(i)) {
    i++;
  }
  return 0;
}