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 <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

/* 基本要素 */
typedef complex<double> Point;
typedef pair<Point, Point> Line;
typedef vector<Point> VP;
const double EPS = 1e-6; // 許容誤差^2
const double INF = 1e9;
#define X real()
#define Y imag()
// #define LE(n,m) ((n) < (m) + EPS)
#define LE(n,m) ((n) - (m) < EPS)
// #define GE(n,m) ((n) + EPS > (m))
#define GE(n,m) (EPS > (m) - (n))
#define EQ(n,m) (abs((n)-(m)) < EPS)

// 内積　dot(a,b) = |a||b|cosθ
double dot(Point a, Point b) {
  return a.X*b.X + a.Y*b.Y;
}

// 外積　cross(a,b) = |a||b|sinθ
double cross(Point a, Point b) {
  return a.X*b.Y - a.Y*b.X;
}

// 点の進行方向
int ccw(Point a, Point b, Point c) {
  b -= a;  c -= a;
  if (cross(b,c) >  EPS) return +1;  // counter clockwise
  if (cross(b,c) < -EPS) return -1;  // clockwise
  if (dot(b,c)   < -EPS) return +2;  // c--a--b on line
  if (norm(b) < norm(c)) return -2;  // a--b--c on line or a==b
  return 0;                          // a--c--b on line or a==c or b==c
}

// 直線と点
bool isecLP(Point a1, Point a2, Point b) {
  return abs(ccw(a1, a2, b)) != 1;
  // return EQ(cross(a2-a1, b-a1), 0); と等価
}

// 直線と直線
bool isecLL(Point a1, Point a2, Point b1, Point b2) {
  return !isecLP(a2-a1, b2-b1, 0) || isecLP(a1, b1, b2);
}

// 線分と点
bool isecSP(Point a1, Point a2, Point b) {
  return !ccw(a1, a2, b);
}

// 線分と線分
bool isecSS(Point a1, Point a2, Point b1, Point b2) {
  return ccw(a1, a2, b1)*ccw(a1, a2, b2) <= 0 &&
  ccw(b1, b2, a1)*ccw(b1, b2, a2) <= 0;
}

// 点pの直線aへの射影点を返す
Point proj(Point a1, Point a2, Point p) {
  return a1 + dot(a2-a1, p-a1)/norm(a2-a1) * (a2-a1);
}

double distSP(Point a1, Point a2, Point p) {
  Point r = proj(a1, a2, p);
  if (isecSP(a1, a2, r)) return abs(r-p);
  //= !ccw(a1, a2, r)
  return min(abs(a1-p), abs(a2-p));
}

// 2直線の交点
Point crosspointLL(Point a1, Point a2, Point b1, Point b2) {
  double d1 = cross(b2-b1, b1-a1);
  double d2 = cross(b2-b1, a2-a1);
  if (EQ(d1, 0) && EQ(d2, 0)) return a1;  // same line
  assert(!EQ(d2, 0)); // 交点がない
  return a1 + d1/d2 * (a2-a1);
}

// 凸多角形の内部判定　O(n)
// 点が領域内部なら1、境界上なら2、外部なら0を返す
int inConvex(Point p, const VP& ps) {
  int n = ps.size();
  int dir = ccw(ps[0], ps[1], p);
  rep (i, n) {
    int ccwc = ccw(ps[i], ps[(i + 1) % n], p);
    if (!ccwc) return 2;  // 線分上に存在
    if (ccwc != dir) return 0;
  }
  return 1;
}

int main(){
    int n;
    while(cin >>n,n){
        VP p(n);
        rep(i,n){
            int x,y;
            cin >>x >>y;
            p[i] = Point(x,y);
        }

        auto D = [&](Point z){
            double ret = INF;
            rep(i,n){
                ret = min(ret, distSP(p[i],p[(i+1)%n],z));
            }
            // printf(" (%f,%f) -> %f\n",z.X,z.Y,ret);
            return ret;
        };

        auto check = [&](double d){
            vector<Line> l;
            rep(i,n){
                Point p1 = p[i], p2 = p[(i+1)%n];
                double vx = p1.X-p2.X, vy = p1.Y-p2.Y;
                double sz = sqrt(vx*vx + vy*vy);

                for(int j=-1; j<=1; j+=2){
                    Point q1(p1.X+vy*d/sz*j, p1.Y-vx*d/sz*j);
                    Point q2(p2.X+vy*d/sz*j, p2.Y-vx*d/sz*j);
                    l.pb({q1,q2});
                }
            }

            int L = l.size();
            rep(i,L)rep(j,i){
                if(!isecLL(l[i].fi, l[i].se, l[j].fi, l[j].se)) continue;

                Point c = crosspointLL(l[i].fi, l[i].se, l[j].fi, l[j].se);
                if(inConvex(c, p) != 1) continue;

                if(GE(D(c),d)) return true;
            }
            return false;
        };

        double l=0,r=10000;
        rep(loop,50){
            double m = (l+r)/2;
            if(check(m)) l=m;
            else r=m;
        }
        printf("%.10f\n", l);
    }
    return 0;
}