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 INF = 0x7fffffff;
const double eps = 1e-3;
int dcmp(double x) {
  if (fabs(x) < eps)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
const double PI = acos(-1);
const double TWO_PI = PI * 2;
double NormalizeAngle(double rad, double center = PI) {
  return rad - TWO_PI * floor((rad + PI - center) / TWO_PI);
}
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);
}
Point operator*(const Point& A, double p) { return Point(A.x * p, A.y * p); }
Point operator/(const Point& A, double p) { return Point(A.x / p, A.y / p); }
bool operator<(const Point& a, const Point& b) {
  return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator==(const Point& a, const Point& b) {
  return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double angle(Point v) { return atan2(v.y, v.x); }
double Dot(const Point& A, const Point& B) { return A.x * B.x + A.y * B.y; }
double Dist(const Point& A, const Point& B) {
  return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
double Dist2(const Point& A, const Point& B) {
  return (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y);
}
double Cross(const Point& A, const Point& B) { return A.x * B.y - A.y * B.x; }
double Length(Point A) { return sqrt(Dot(A, A)); }
double Angle(Point A, Point B) { return atan2(Cross(A, B), Dot(A, B)); }
double Area2(Point A, Point B, Point C) { return Cross(B - A, C - A); }
Point Rotate(Point A, double rad) {
  return Point(A.x * cos(rad) - A.y * sin(rad),
               A.x * sin(rad) + A.y * cos(rad));
}
Point Normal(const Point& A) {
  double L = Length(A);
  return Point(-A.y / L, A.x / L);
}
struct Line {
  Point p;
  Point v;
  double ang;
  Line() {}
  Line(Point p, Point v) : p(p), v(v) { ang = atan2(v.y, v.x); }
  bool operator<(const Line& L) const { return ang < L.ang; }
  Point point(double t) { return p + v * t; }
};
bool OnLeft(const Line& L, const Point& p) { return Cross(L.v, p - L.p) > 0; }
Point GetLineIntersection(const Line& a, const Line& b) {
  Point u = a.p - b.p;
  double t = Cross(b.v, u) / Cross(a.v, b.v);
  return a.p + a.v * t;
}
double x, y;
vector<Point> V;
int main(void) {
  cin.sync_with_stdio(false);
  for (int i = (0); i < (3); i++) {
    cin >> x >> y;
    V.push_back(Point(x, y));
  }
  Line l1 = Line((V[0] + V[1]) / 2, Normal(V[1] - V[0]));
  Line l2 = Line((V[1] + V[2]) / 2, Normal(V[2] - V[1]));
  Point C = GetLineIntersection(l1, l2);
  double r = Length(C - V[1]);
  double a[5];
  a[0] = fabs(Angle(V[1] - C, V[0] - C)) / (2 * PI);
  a[1] = fabs(Angle(V[2] - C, V[1] - C)) / (2 * PI);
  a[2] = fabs(Angle(V[0] - C, V[2] - C)) / (2 * PI);
  cerr << a[0] << ' ' << a[1] << ' ' << a[2] << endl;
  int n;
  for (int t = (3); t < (101); t++) {
    int good = 0;
    for (int i = (0); i < (3); i++) {
      double b1 = a[i] * t;
      double c1 = floor(b1);
      if (dcmp(b1 - c1) == 0 || dcmp(b1 - c1 - 1) == 0) {
        good++;
      }
      if (t == 38) {
        cerr << b1 << ' ' << c1 << endl;
      }
    }
    if (good == 3) {
      n = t;
      break;
    }
  }
  cerr << "n = " << n << ", r = " << r << endl;
  double ans = r * r * sin(2 * PI / n) * n / 2.0;
  cout << fixed << setprecision(15) << ans << endl;
  return 0;
}