#include <iostream>
#include <complex>
#include <cstdio>
#include <utility>
using namespace std;
typedef complex<double> P;
typedef pair<P,P> L;
#define X real()
#define Y imag()

double dot(P a, P b){
  return a.X * b.X + a.Y * b.Y;
}
double cross(P a, P b){
  return a.X * b.Y - a.Y * b.X;
}

L angle_bisector(P a, P b, P c){
  //角abcをの二等分線
  //a, b or b, cが一致するとゼロ徐算が発生しREになる
  P p = b + (a-b)/abs(a-b) + (c-b)/abs(c-b);
  return L(b,p);
}

P intersection(L a, L b){
  //直線aと直線bの交点
  return a.first + cross(b.second-b.first,a.first-b.first)/(-cross(b.second-b.first,a.second-b.first)+cross(b.second-b.first,a.first-b.first))*(a.second-a.first);
}

double p_to_s_dist(P a, L s){
  return abs(cross(s.first-s.second,a-s.second))/abs(s.first-s.second);
}

int main(){
  P p[3];
  double x, y;
  for(int i = 0; i < 3; ++i){
    cin >> x >> y;
    p[i] = P(x,y);
  }
  L bisectors[3];
  for(int i = 0; i < 3; ++i){
    bisectors[i] = angle_bisector(p[(i+2)%3],p[i],p[(i+1)%3]);
  }
  P inner_center = intersection(bisectors[0],bisectors[1]);
  double ans = 0.0;
  for(int i = 0; i < 3; ++i){
    P a = p[i], b = inner_center;
    L l = L(p[i],p[(i+1)%3]);
    for(int j = 0; j < 10000; ++j){
      P m = (a+b)/2.0;
      double d1 = p_to_s_dist(m,l)*2;
      P k = intersection(bisectors[(i+1)%3], L(m,m+p[(i+1)%3]-p[i]));
      double d2 = abs(k-m);
      if(d1 < d2){
	a = m;
      }else{
	b = m;
      }
    }
    ans = max(ans,p_to_s_dist((a+b)/2.0,l));
  }
  printf("%.12f\n",ans);
  return 0;
}




### Prompt

In cpp, your task is to solve the following 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

### Response

```cpp
#include <iostream>
#include <complex>
#include <cstdio>
#include <utility>
using namespace std;
typedef complex<double> P;
typedef pair<P,P> L;
#define X real()
#define Y imag()

double dot(P a, P b){
  return a.X * b.X + a.Y * b.Y;
}
double cross(P a, P b){
  return a.X * b.Y - a.Y * b.X;
}

L angle_bisector(P a, P b, P c){
  //角abcをの二等分線
  //a, b or b, cが一致するとゼロ徐算が発生しREになる
  P p = b + (a-b)/abs(a-b) + (c-b)/abs(c-b);
  return L(b,p);
}

P intersection(L a, L b){
  //直線aと直線bの交点
  return a.first + cross(b.second-b.first,a.first-b.first)/(-cross(b.second-b.first,a.second-b.first)+cross(b.second-b.first,a.first-b.first))*(a.second-a.first);
}

double p_to_s_dist(P a, L s){
  return abs(cross(s.first-s.second,a-s.second))/abs(s.first-s.second);
}

int main(){
  P p[3];
  double x, y;
  for(int i = 0; i < 3; ++i){
    cin >> x >> y;
    p[i] = P(x,y);
  }
  L bisectors[3];
  for(int i = 0; i < 3; ++i){
    bisectors[i] = angle_bisector(p[(i+2)%3],p[i],p[(i+1)%3]);
  }
  P inner_center = intersection(bisectors[0],bisectors[1]);
  double ans = 0.0;
  for(int i = 0; i < 3; ++i){
    P a = p[i], b = inner_center;
    L l = L(p[i],p[(i+1)%3]);
    for(int j = 0; j < 10000; ++j){
      P m = (a+b)/2.0;
      double d1 = p_to_s_dist(m,l)*2;
      P k = intersection(bisectors[(i+1)%3], L(m,m+p[(i+1)%3]-p[i]));
      double d2 = abs(k-m);
      if(d1 < d2){
	a = m;
      }else{
	b = m;
      }
    }
    ans = max(ans,p_to_s_dist((a+b)/2.0,l));
  }
  printf("%.12f\n",ans);
  return 0;
}




```