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 long double EPS = 1e-4;
const int inf = 1e9;
const long double PI = acos(-1);
int mod = (int)998244353;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
const int a228 = 18;
const long long kekmod = 1791791791;
const long long bestmod = 1148822869;
const long long secmod = (int)1e9 + 113;
vector<long long> mods = {kekmod, bestmod, mod, MOD9, 1000000007};
vector<long long> hashpows = {29, 31, 37, 43, 47, 53, 179, 229};
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count() +
            228 + 'i' + 'q' + 1337 + 1488);
long long MOD = mods[rnd() % mods.size()];
long long hashpow = hashpows[rand() % hashpows.size()];
long double hypot(long double a, long double b, long double c, long double d) {
  return sqrt(((a - c) * (a - c)) + ((b - d) * (b - d)));
}
long double gcd(long double a, long double b) {
  while (abs(b) > EPS) {
    a -= floor(a / b) * b;
    swap(a, b);
  }
  return a;
}
long double getang(long double a, long double b, long double c) {
  return acos(-(c * c - a * a - b * b) / (2 * a * b));
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  srand(time(NULL));
  long double a, b, c;
  long double x1, y1, x2, y2, x3, y3;
  cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
  a = hypot(x1, y1, x2, y2);
  b = hypot(x2, y2, x3, y3);
  c = hypot(x3, y3, x1, y1);
  long double a1 = getang(a, b, c);
  long double a2 = getang(b, c, a);
  long double a3 = getang(a, c, b);
  long double mere = gcd(gcd(a1, a2), a3);
  int n = (int)((PI + EPS) / mere);
  long double R = c / sin(a1) / 2;
  long double tang = PI * 2 / n;
  long double ans = n * (0.5 * R * R * sin(tang));
  cout << fixed << setprecision(20) << ans;
}