#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
/* attention
    long longのシフト演算には気をつけよう
    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも
    sizeは(int)とキャストしよう
    ごちゃごちゃ場合分けを考える前に全探索は考えましたか？
*/

const ll mod = 1e9+7;

void chmod(ll &M){
    if(M >= mod) M %= mod;
    else if(M < 0){
        M += (abs(M)/mod + 1)*mod;
        M %= mod;
    }
}


int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	double x1, x2, x3;
    double y1, y2, y3;
    cin >> x1 >> y1;
    cin >> x2 >> y2;
    cin >> x3 >> y3;

    double a = hypot(abs(x1-x2), abs(y1-y2));
    double b = hypot(abs(x2-x3), abs(y2-y3));
    double c = hypot(abs(x3-x1), abs(y3-y1));
    double len = max(a, max(b, c));

    double s = (a+b+c)/2;
    double area = sqrt(s*(s-a)*(s-b)*(s-c));
    double r = area/s;

    double left = 0, right = r;
    double mid;
    for(int i=0; i<10000000; ++i){
        mid = (left+right)/2;
        if(2*mid*r <= len*(r-mid)) left = mid;
        else right = mid;
    }

    printf("%.15lf\n", left);
	
	return 0;
}

### Prompt

Your task is to create a cpp solution to 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
/* attention
    long longのシフト演算には気をつけよう
    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも
    sizeは(int)とキャストしよう
    ごちゃごちゃ場合分けを考える前に全探索は考えましたか？
*/

const ll mod = 1e9+7;

void chmod(ll &M){
    if(M >= mod) M %= mod;
    else if(M < 0){
        M += (abs(M)/mod + 1)*mod;
        M %= mod;
    }
}


int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	double x1, x2, x3;
    double y1, y2, y3;
    cin >> x1 >> y1;
    cin >> x2 >> y2;
    cin >> x3 >> y3;

    double a = hypot(abs(x1-x2), abs(y1-y2));
    double b = hypot(abs(x2-x3), abs(y2-y3));
    double c = hypot(abs(x3-x1), abs(y3-y1));
    double len = max(a, max(b, c));

    double s = (a+b+c)/2;
    double area = sqrt(s*(s-a)*(s-b)*(s-c));
    double r = area/s;

    double left = 0, right = r;
    double mid;
    for(int i=0; i<10000000; ++i){
        mid = (left+right)/2;
        if(2*mid*r <= len*(r-mid)) left = mid;
        else right = mid;
    }

    printf("%.15lf\n", left);
	
	return 0;
}

```