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 <sstream>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>

using namespace std;

#define FOR(i,k,n) for(int i=(k); i<(int)(n); ++i)
#define REP(i,n) FOR(i,0,n)
#define FORIT(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

template<class T> void debug(T begin, T end){ for(T i = begin; i != end; ++i) cerr<<*i<<" "; cerr<<endl; }
inline bool valid(int x, int y, int W, int H){ return (x >= 0 && y >= 0 && x < W && y < H); }

typedef long long ll;
const int INF = 100000000;
const double EPS = 1e-8;
const int MOD = 1000000007;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

int sign(double r){ return (r < -EPS) ? -1 : (r > EPS) ? 1 : 0; }
bool eq(double a, double b){ return abs(b - a) < EPS; }

struct Point{
  double x, y;
  Point() {}
  Point(double x, double y) : x(x), y(y) {}
  Point operator+(const Point& b) const { return Point(x + b.x, y + b.y); }
  Point operator-(const Point& b) const { return Point(x - b.x, y - b.y); }
  Point operator*(const double b) const { return Point(x * b, y * b); }
  Point operator*(const Point& b) const { return Point(x * b.x - y * b.y, x * b.y + y * b.x); }
  Point operator/(const double b) const { return Point(x / b, y / b); }
  bool operator< (const Point& b) const { return x != b.x ? x < b.x : y < b.y; }
  bool operator==(const Point& b) const { return eq(x, b.x) && eq(y, b.y); }
  double norm(){ return x * x + y * y; }
  double arg(){ return atan2(x, y); }
  double abs(){ return sqrt(norm()); }
  Point rotate(double theta){ return Point(cos(theta) * x - sin(theta) * y, sin(theta) * x + cos(theta) * y); }
  Point rotate90(){ return Point(-y, x); }
};

ostream& operator<<(ostream& os, Point& p){ return os<<"("<<p.x<<","<<p.y<<")"; }
istream& operator>>(istream& is, Point& a){ return is>>a.x>>a.y; }
double dot(const Point& a, const Point& b){ return a.x * b.x + a.y * b.y; }
double cross(const Point& a, const Point& b){ return a.x * b.y - a.y * b.x; }
Point polor(double rho, double theta){ return Point(rho * cos(theta), rho * sin(theta)); }

struct Line : public vector<Point> {
  Line(const Point& a, const Point& b) {
    push_back(a); push_back(b);
  }
  Point vector() const {
    return back() - front();
  }
};
struct Circle : Point {
  double r;
  Circle(Point p, double r) : Point(p), r(r) { }
};
typedef vector<Point> Polygon;
Point curr(const Polygon& a, int x){ return a[x]; }
Point next(const Polygon& a, int x){ return a[(x + 1) % a.size()]; }
Point prev(const Polygon& a, int x){ return a[(x - 1 + a.size()) % a.size()]; }

//a->b->c
int ccw(Point a, Point b, Point c) {
  b = b - a; c = c - a;
  if (cross(b, c) > EPS)   return +1; // 反時計回り
  if (cross(b, c) < -EPS)   return -1; // 時計回り
  if (dot(b, c) < 0)     return +2; // c--a--b の順番で一直線上
  if (b.norm() < c.norm()) return -2; // a--b--c の順番で一直線上
  return 0;                         // a--c--b の順番で一直線上
}
// 二直線の交差判定(平行でない)
bool intersectLL(const Line& l, const Line& m) {
  return abs(cross(l.vector(), m.vector())) > EPS;
}
// 二直線の同一判定
bool equalLL(const Line& l, const Line& m){
  return abs(cross(l.vector(), m[0] - l[0])) < EPS;
}
// 直線と線分の交差判定(同一直線上にある場合も含む)
bool intersectLS(const Line& l, const Line& s) {
  return cross(l.vector(), s[0] - l[0]) * cross(l.vector(), s[1] - l[0]) < EPS; //直線lについて、線分sの端点が異なる側にある
}
// 直線と点の交差判定
bool intersectLP(const Line& l, const Point& p) {
  return abs(ccw(l[0], l[1], p)) != 1;
}
// 二つの線分の交差判定
bool intersectSS(const Line& s, const Line& t) {
  return ccw(s[0], s[1], t[0]) * ccw(s[0], s[1], t[1]) <= 0 &&
         ccw(t[0], t[1], s[0]) * ccw(t[0], t[1], s[1]) <= 0;
}
// 線分と点の交差判定
bool intersectSP(const Line& s, const Point& p) {
  return ccw(s[0], s[1], p) == 0;
}
// 円と点の交差判定(EPS注意)
bool intersectCP(const Circle& c, const Point& p){
  return (c - p).abs() < c.r + EPS;
}
// 二つの円の交差判定(EPS注意)
bool intersectCC(const Circle& c, const Circle& d){
  return (c - d).abs() < c.r + d.r + EPS && (c - d).abs() > abs(c.r - d.r) - EPS;
}
// 射影
Point projection(const Line& l, const Point& p){
  double t = dot(p - l[0], l.vector()) / l.vector().norm();
  return l[0] + l.vector() * t;
}
// 反射
Point reflection(const Line& l, const Point& p){
  return p + (projection(l, p) - p) * 2;
}
// 直線と点の距離
double distanceLP(const Line& l, const Point& p){
  return (p - projection(l, p)).abs();
}
// 二直線の距離
double distanceLL(const Line& l, const Line& m){
  return intersectLL(l, m) ? 0 : distanceLP(l, m[0]);
}
// 直線と線分の距離
double distanceLS(const Line& l, const Line& s){
  if(intersectLS(l, s)) return 0;
  return min(distanceLP(l, s[0]), distanceLP(l, s[1]));
}
// 線分と点の距離
double distanceSP(const Line& s, const Point& p){
  const Point r = projection(s, p);
  if(intersectSP(s, r)) return (r - p).abs(); //垂線が線分に交わるとき
  return min((s[0] - p).abs(), (s[1] - p).abs());
}
// 二つの線分の距離
double distanceSS(const Line& s, const Line& t){
  if(intersectSS(s, t)) return 0;
  return min(min(distanceSP(s, t[0]), distanceSP(s, t[1])),
             min(distanceSP(t, s[0]), distanceSP(t, s[1])));
}
// 二直線の交点
Point crosspointLL(const Line& l, const Line& m){
  double d = cross(m.vector(), l.vector());
  if(abs(d) < EPS) assert(false); // 直線が交わらない
  return l[0] + l.vector() * cross(m.vector(), m[1] - l[0]) * (1.0 / d);
}

double area(const Polygon& P) {
  double A = 0;
  REP(i, P.size()){
    A += cross(curr(P, i), next(P, i));
  }
  return A / 2.0;
}

Polygon convex_cut(const Polygon& P, const Line& l){
  Polygon Q;
  REP(i, P.size()){
    Point A = curr(P, i), B = next(P, i);
    if(ccw(l[0], l[1], A) != -1) Q.push_back(A); //Aが直線lの右側でない
    if(ccw(l[0], l[1], A) * ccw(l[0], l[1], B) < 0)
      Q.push_back(crosspointLL(l, Line(A, B)));
  }
  return Q;
}
int main(){
  int N;
  while(cin>>N && N){
    Polygon P(N);
    REP(i, N) cin>>P[i];
    double lb = 0, ub = 1e6;
    REP(iter, 50){
      double mid = (lb + ub) / 2;
      Polygon Q = P;
      //cout<<"mid is "<<mid<<endl;
      REP(i, N){
        Line l(curr(P, i), next(P, i));
        //cout<<"l is "<<l[0]<<" "<<l[1]<<endl;
        Point u = l.vector().rotate90() / l.vector().abs();
        Line m(l[0] + u * mid, l[1] + u * mid);
        //cout<<"m is "<<m[0]<<" "<<m[1]<<endl;
        Q = convex_cut(Q, m);
        if(Q.empty()) break;
        //cout<<"Q is "; REP(i, Q.size()) cout<<Q[i]<<" "; cout<<endl;
      }
      if(Q.empty()){
        ub = mid;
      }else{
        lb = mid;
      }
    }
    printf("%.6lf\n", ub);
  }
  return 0;
}