Problem: Snuke received a triangle as a birthday present. The coordinates of the three vertices were (x_1, y_1), (x_2, y_2), and (x_3, y_3).

He wants to draw two circles with the same radius inside the triangle such that the two circles do not overlap (but they may touch). Compute the maximum possible radius of the circles.

Constraints

* 0 ≤ x_i, y_i ≤ 1000
* The coordinates are integers.
* The three points are not on the same line.

Input

The input is given from Standard Input in the following format:


x_1 y_1
x_2 y_2
x_3 y_3


Output

Print the maximum possible radius of the circles. The absolute error or the relative error must be at most 10^{-9}.

Examples

Input

0 0
1 1
2 0


Output

0.292893218813


Input

3 1
1 5
4 9


Output

0.889055514217

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<algorithm>
#include<utility>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>
#include<climits>
#include<set>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)
using namespace std;
typedef pair<int, int> pii;
typedef long long int ll;
typedef pair<ll, ll> pll;
//int dx[4] = { 1,0,0,-1 };
//int dy[4] = { 0,1,-1,0 };
#define MOD 1000000007
#define ARRAY_MAX 1005

const int INF = 1e9 + 7;


int main(){


	ios::sync_with_stdio(false);
	cin.tie(0);

	vector<double> x(3), y(3);
	for (int i = 0; i < 3; i++) {
		cin >> x[i] >> y[i];
	}

	double maxi = 0;

	double len_sum = 0;

	for (int i = 0; i < 3; i++)
	{
		double len = (x[i % 3] - x[(i + 1) % 3])*(x[i % 3] - x[(i + 1) % 3]) + (y[i % 3] - y[(i + 1) % 3]) * (y[i % 3] - y[(i + 1) % 3]);
		len_sum += sqrt(len);
		maxi = max(maxi,len);
	}

	maxi = sqrt(maxi);

	double x1 = x[1] - x[0], x2 = x[2] - x[0];
	double y1 = y[1] - y[0], y2 = y[2] - y[0];

	double s = abs(x1 * y2 - y1 * x2) / 2;

	double r = 2 * s / (len_sum);

	double ans = (r*maxi) / (2 * r + maxi);

	printf("%.12f\n", ans);

	return 0;
}