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:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cassert>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <complex>
#include <stack>
#include <queue>
#include <list>
#include <set>
#include <map>
#include <bitset>
#include <functional>
#include <iterator>

using namespace std;

#define dump(n) cout<<"# "<<#n<<'='<<(n)<<endl
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define peri(i,a,b) for(int i=int(b);i-->int(a);)
#define rep(i,n) repi(i,0,n)
#define per(i,n) peri(i,0,n)
#define iter(c) __typeof__((c).begin())
#define foreach(i,c) for(iter(c) i=(c).begin();i!=(c).end();++i)
#define all(c) (c).begin(),(c).end()
#define mp make_pair

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<string> vs;
typedef pair<int,int> pii;

const int INFTY=1<<29;
const double EPS=1e-9;

template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& p){
	return os<<'('<<p.first<<','<<p.second<<')';
}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& a){
	os<<'[';
	rep(i,a.size()) os<<(i?" ":"")<<a[i];
	return os<<']';
}

const double PI=acos(-1);

int Signum(double x){
	return abs(x)<EPS?0:x<0?-1:1;
}

struct Point{
	double x,y;
	Point(){}
	Point(double x,double y):x(x),y(y){}
};
bool operator==(Point a,Point b){
	return abs(a.x-b.x)<EPS && abs(a.y-b.y)<EPS;
}
bool operator!=(Point a,Point b){
	return abs(a.x-b.x)>=EPS || abs(a.y-b.y)>=EPS;
}
Point operator+(Point a,Point b){
	return Point(a.x+b.x,a.y+b.y);
}
Point operator-(Point a,Point b){
	return Point(a.x-b.x,a.y-b.y);
}
Point operator*(double c,Point p){
	return Point(c*p.x,c*p.y);
}
Point operator/(Point p,double c){
	return Point(p.x/c,p.y/c);
}
double Abs(Point p){
	return sqrt(p.x*p.x+p.y*p.y);
}
double Abs2(Point p){
	return p.x*p.x+p.y*p.y;
}
double Dot(Point a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point Rot(Point p,double t){
	return Point(cos(t)*p.x-sin(t)*p.y,sin(t)*p.x+cos(t)*p.y);
}

struct Line{
	Point pos,dir;
	Line(){}
	Line(Point p,Point d):pos(p),dir(d){}
	Line(double px,double py,double dx,double dy):pos(px,py),dir(dx,dy){}
};
typedef Line Segment;
Point InterLL(Line a,Line b){
	if(abs(Cross(a.dir,b.dir))<EPS)
		return a.pos;
	return a.pos+Cross(b.pos-a.pos,b.dir)/Cross(a.dir,b.dir)*a.dir;
}
Point InterLS(Line l,Segment s){
	return InterLL(s,l);
}

int CCW(Point a,Point b,Point c){
	Point d1=b-a,d2=c-a;
	if(int sign=Signum(Cross(d1,d2)))
		return sign;	// 1:ccw,-1:cw
	if(Dot(d1,d2)<-EPS)
		return -2;		// c-a-b
	if(Abs2(d1)<Abs2(d2)-EPS)
		return 2;		// a-b-c
	return 0;			// a-c-b
}

vector<Point> ConvexCut(const vector<Point>& ps,Line l){
	int n=ps.size();
	vector<Point> res;
	rep(i,n){
		int c1=CCW(l.pos,l.pos+l.dir,ps[i]);
		int c2=CCW(l.pos,l.pos+l.dir,ps[(i+1)%n]);
		if(c1!=-1)
			res.push_back(ps[i]);
		if(c1*c2==-1)
			res.push_back(InterLS(l,Segment(ps[i],ps[(i+1)%n]-ps[i])));
	}
	return res;
}

vector<Point> calc(const vector<Point>& ps,double x)
{
	vector<Point> res=ps;
	rep(i,ps.size()){
		Point dir=ps[(i+1)%ps.size()]-ps[i];
		Point pos=ps[i]+x*Rot(dir/Abs(dir),PI/2);
		res=ConvexCut(res,Line(pos,dir));
	}
	return res;
}

int main()
{
	for(int n;cin>>n,n;){
		vector<Point> ps(n);
		rep(i,n) cin>>ps[i].x>>ps[i].y;
		
		double lo=0,hi=1e6;
		rep(_,50){
			double mi=(lo+hi)/2;
			if(calc(ps,mi).size()==0)
				hi=mi;
			else
				lo=mi;
		}
		printf("%f\n",lo);
	}
	
	return 0;
}