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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
typedef double ld;
typedef complex<ld> Point;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
const ld eps = 1e-6;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;

bool eq(ld a, ld b) {
	return (abs(a - b) < eps);
}
//内積
ld dot(Point a, Point b) {
	return real(conj(a)*b);
}
//外積
ld cross(Point a, Point b) {
	return imag(conj(a)*b);
}

//3点の位置関係
int ccw(Point a, Point b, Point c) {
	b -= a; c -= a;
	if (cross(b, c) > eps)return 1;//a,b,cが反時計回り
	if (cross(b, c) < -eps)return -1;//a,b,cが時計回り
	if (dot(b, c) < 0)return 2;//c,a,bの順に一直線
	if (norm(b) < norm(c))return -2;//a,b,cの順に一直線
	return 0;//a,c,bの順に一直線
}

//直線にするなら十分通い２点を端点とすればよい
class Line {
public:
	Point a, b;
};

//2直線の交差判定
bool isis_ll(Line l, Line m) {
	return !eq(cross(l.b - l.a, m.b - m.a), 0);
}

//直線と直線の交点
//平行な２直線に対しては使うな！！！！
Point is_ll(Line s, Line t) {
	Point sv = s.b - s.a; Point tv = t.b - t.a;
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}

//点から直線への垂線の足
Point proj(Line l, Point p) {
	ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}

//直線と点の距離
ld dist_lp(Line l, Point p) {
	return abs(p - proj(l, p));
}

int n;
void solve() {
	vector<Point> v(n);
	rep(i, n) {
		ld x, y; cin >> x >> y;
		v[i] = { x,y };
	}
	v.push_back(v[0]);
	vector<Line> l(n);
	vector<LDP> d(n);
	rep(i, n) {
		l[i] = { v[i],v[i + 1] };
		ld dx = real(v[i + 1]) - real(v[i]);
		ld dy = imag(v[i + 1]) - imag(v[i]);
		ld r = pow(dx, 2) + pow(dy, 2); r = sqrt(r);
		d[i] = { dx / r,dy / r };
	}
	ld ans = 0;
	rep(i, n) {
		rep(j, n) {
			Rep(k, i+ 1, n) {
				if (i == j || j == k)continue;
				if (!isis_ll(l[i], l[j]) || !isis_ll(l[j], l[k]))continue;
				Point a = is_ll(l[i], l[j]), b = is_ll(l[j], l[k]);
				ld vx = (d[j].first - d[i].first) / 2.0;
				ld vy = (d[j].second - d[i].second) / 2.0;
				Point sa = { real(a) + vx,imag(a) + vy };
				vx = (d[k].first - d[j].first) / 2.0;
				vy = (d[k].second - d[j].second) / 2.0;
				Point sb = { real(b) + vx,imag(b) + vy };
				Line la = { a,sa }, lb = { b,sb };
				Point m = is_ll(la, lb);
				ld mi = INF;
				int cnt = 0;
				rep(x, n) {
					cnt += ccw(v[x], v[x + 1], m);
					mi = min(mi, dist_lp(l[x], m));
				}
				//cout << mi << endl;
				//ans = max(ans, mi);
				cnt = abs(cnt);
				if (cnt == n) {
					ans = max(ans, mi);
				}
			}
		}
	}
	cout << ans << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(9);
	while (cin >>n,n) {
		solve();
	}
	//stop
	return 0;
}