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>
typedef struct {
  double x, y;
} t_coord;
double polygon_center_angle(int n) { return 2 * 3.14159265358979323846 / n; }
double center_angle(t_coord center, t_coord point) {
  return atan2(point.x - center.x, point.y - center.y);
}
t_coord move(const t_coord point, const t_coord vector) {
  t_coord res;
  res.x = point.x + vector.x;
  res.y = point.y + vector.y;
  return res;
}
t_coord calc_center(const t_coord p[3]) {
  t_coord res;
  double d = 2 * ((p[0].x * (p[1].y - p[2].y)) + (p[1].x * (p[2].y - p[0].y)) +
                  (p[2].x * (p[0].y - p[1].y)));
  res.x =
      ((((p[0].x) * (p[0].x)) + ((p[0].y) * (p[0].y))) * (p[1].y - p[2].y) +
       (((p[1].x) * (p[1].x)) + ((p[1].y) * (p[1].y))) * (p[2].y - p[0].y) +
       (((p[2].x) * (p[2].x)) + ((p[2].y) * (p[2].y))) * (p[0].y - p[1].y)) /
      d;
  res.y =
      ((((p[0].x) * (p[0].x)) + ((p[0].y) * (p[0].y))) * (p[2].x - p[1].x) +
       (((p[1].x) * (p[1].x)) + ((p[1].y) * (p[1].y))) * (p[0].x - p[2].x) +
       (((p[2].x) * (p[2].x)) + ((p[2].y) * (p[2].y))) * (p[1].x - p[0].x)) /
      d;
  return res;
}
double check(double angle, double divisor_angle) {
  double tmp = angle / divisor_angle;
  int factor = tmp + 0.5;
  double res = fabs(factor - tmp);
  return res;
}
int main() {
  t_coord p[3], center;
  double angle[3], angle_diff[2], poly_min_angle, diff;
  int min_poly;
  for (int i = 0; i < 3; ++i) scanf("%lf %lf", &p[i].x, &p[i].y);
  center = calc_center(p);
  for (int i = 0; i < 3; ++i) angle[i] = center_angle(center, p[i]);
  for (int i = 0; i < 2; ++i) {
    angle_diff[i] = angle[i] - angle[i + 1];
    if (angle_diff[i] < 0.) angle_diff[i] += 2 * 3.14159265358979323846;
  }
  for (int i = 3; i <= 100; ++i) {
    poly_min_angle = polygon_center_angle(i);
    for (int j = 0; j < 2; ++j) {
      diff = check(angle_diff[j], poly_min_angle);
      if (diff > 1e-5) break;
    }
    if (diff < 1e-5) {
      min_poly = i;
      break;
    }
  }
  double r = sqrt(((p[0].x - center.x) * (p[0].x - center.x)) +
                  ((p[0].y - center.y) * (p[0].y - center.y)));
  double area =
      min_poly * (cos(poly_min_angle / 2) * r) * (sin(poly_min_angle / 2) * r);
  printf("%f", area);
  return 0;
}