Problem: The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: "Where is the most distant point from the sea?" The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.

In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.



Input

The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.


n
x1    y1
.
.
.
xn    yn


Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.

n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi) - (xi+1, yi+1) (1 ≤ i ≤ n - 1) and the line segment (xn, yn) - (x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.

You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.

The last dataset is followed by a line containing a single zero.

Output

For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces.

The answer should not have an error greater than 0.00001 (10-5 ). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.

Example

Input

4
0 0
10000 0
10000 10000
0 10000
3
0 0
10000 0
7000 1000
6
0 40
100 20
250 40
250 70
100 90
0 70
3
0 0
10000 10000
5000 5001
0


Output

5000.000000
494.233641
34.542948
0.353553

Solution (in CPP):

Solution:
/*
LA3890 Most Distant Point from the Sea
题意：大海中央有一个凸n边形的小岛，求岛上离海最远的点，输出其距离
思路：
	二分答案，对于小岛的每条边，作一条相应的直线，若最后形成的半平面非空，
	那么可以满足。 
*/
#include <bits/stdc++.h>
using namespace std;
const double eps=1e-6;
const double PI=acos(-1);

struct Point		//定义点or向量 
{
	double x,y;
	Point ( double x=0,double y=0 ) : x(x),y(y) {}
};
typedef Point Vector;		//区分点和向量，因为结构体一样 

Vector operator + ( Vector A,Vector B )
{
	return Vector( A.x+B.x,A.y+B.y );
} 

Vector operator - ( Point A,Point B )
{
	return Vector( A.x-B.x,A.y-B.y );
}

Vector operator * ( Vector A,double b )
{
	return Vector( A.x*b,A.y*b );
}

Vector operator / ( Vector A,double b )
{
	return Vector( A.x/b,A.y/b );
}

bool operator < ( const Point &a,const Point &b )
{
	return a.x<b.x || ( a.x==b.x && a.y<b.y );
}

int dcmp( double x )
{
	if ( fabs( x )<eps ) return 0;
	else return x<0 ? -1 : 1;
}

bool operator == ( const Point &a,const Point &b )
{
	return dcmp( a.x-b.x )==0 && dcmp( a.y-b.y )==0;
}

double Dot( Vector A,Vector B )		//求点积 
{
	return A.x*B.x+A.y*B.y;
}

double Length( Vector A )		//求向量的模 
{
	return sqrt( Dot( A,A ) );
}

double Cross( Vector A,Vector B )		//根据向量求叉积 
{
	return A.x*B.y-A.y*B.x;
}

Vector Normal( Vector A )		//左转九十度长度归一，注意不能是零向量
{
	double l=Length( A );
	return Vector( -A.y/l,A.x/l );
}

struct Line2		//有向直线
{
	Point P; Vector v; double ang;
	Line2() {}
	Line2( Point P,Vector v ) :P(P),v(v) {ang=atan2( v.y,v.x );}
	bool operator <( const Line2 &L ) const
	{
		return ang<L.ang;
	}
};

bool Onleft( Line2 L,Point p )		//点p在有向直线左边 
{
	return Cross( L.v,p-L.P ) >0;
}

Point GetLineIntersection( Line2 a,Line2 b )		//两直线交点 
{
	Vector u=a.P-b.P;
	double t=Cross( b.v,u )/Cross( a.v,b.v );
	return a.P+a.v*t;
}

int HalfPlaneIntersection( Line2 *L,int n,Point *poly )
{
	sort( L,L+n );
	int first,last; Point *p=new Point[n]; Line2 *q=new Line2[n];
	q[first=last=0]=L[0];
	for ( int i=1; i<n; i++ )
	{
		while ( first<last && !Onleft( L[i],p[last-1] ) ) last--;
		while ( first<last && !Onleft( L[i],p[first] ) ) first++;
		q[++last]=L[i];
		if ( fabs( Cross( q[last].v,q[last-1].v ) )<eps )
		{
			last--;
			if ( Onleft( q[last],L[i].P ) ) q[last]=L[i];
		}
		if ( first<last ) p[last-1]=GetLineIntersection( q[last-1],q[last] );
	}
	while (first<last && !Onleft( q[first],p[last-1] ) ) last--;
	if ( last-first<=1 ) return 0;
	p[last]=GetLineIntersection( q[last],q[first] );
	int m=0;
	for ( int i=first; i<=last; i++ )
		poly[m++]=p[i];
	return m;
}

Point p[210],poly[210];
Vector v[210],v2[210];
Line2 L[210];
int n;

int main()
{
	while ( scanf( "%d",&n ) && n )
	{
		int m;
		for ( int i=0,x,y; i<n; i++ )
		{
			scanf( "%d%d",&x,&y ); p[i]=Point( x,y );
		}
		
		for ( int i=0; i<n; i++ )
			v[i]=p[(i+1)%n]-p[i],v2[i]=Normal( v[i] );
		double left=0,right=20010;
		while ( dcmp( right-left) )
		{
			double mid=(left+right)/2;
			for ( int i=0; i<n; i++ )
				L[i]=Line2( p[i]+v2[i]*mid,v[i] );
			m=HalfPlaneIntersection( L,n,poly );
			if ( !m ) right=mid;
			else left=mid;
		}
		
		printf( "%.6lf\n",left );
	}
}