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 ld = long double;

constexpr ld eps = 1e-8;


namespace geometry2d {


using point = std::complex<long double>;
using polygon = std::vector<point>;


bool eq(long double a, long double b) {
    return (std::abs(a-b) < eps);
}

long double dot(point a, point b) {
    return std::real(std::conj(a) * b);
}

long double cross(point a, point b) {
    return std::imag(std::conj(a) * b);
}


class segment {
public:
    segment()
        : a(point(0, 0)),
          b(point(0, 0))
    {}
    segment(point a_, point b_)
        : a(a_), b(b_)
    {}

    point a, b;
};

class line {
public:
    line()
        : a(point(0, 0)), b(point(0, 0))
    {}
    line(point a_, point b_)
        : a(a_), b(b_)
    {}
    line(segment s)
        : a(s.a), b(s.b)
    {}

    point a, b;
};

class circle {
public:
    circle()
        : p(point(0, 0)), r(0)
    {}
    circle(point p_, long double r_)
        : p(p_), r(r_)
    {}

    point p;
    long double r;
};


int ccw(point a, point b, point c) {
    b -= a; c -= a;
    if(cross(b, c) > eps)           return 1;  // a -> b -> c : counterclockwise
    if(cross(b, c) < -eps)          return -1; // a -> b -> c : clockwise
    if(dot(b, c) < 0)               return 2;  // c -> a -> b : line
    if(std::norm(b) < std::norm(c)) return -2; // a -> b -> c : line
    return 0;                                  // a -> c -> b : line
}


bool isis_ll(line l, line m) {
    return !eq(cross(l.b - l.a, m.b - m.a), 0);
}

bool isis_ls(line l, segment s) {
    return (cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - s.a) < eps);
}

bool isis_lp(line l, point p) {
    return (std::abs(cross(l.b - p, l.a - p)) < eps);
}

bool isis_sp(segment s, point p) {
    return (std::abs(s.a - p) + std::abs(s.b - p) - std::abs(s.b - s.a) < eps);
}

bool isis_ss(segment s, segment t) {
    return ccw(s.a, s.b, t.a) * ccw(s.a, s.b, t.b) <= 0
        && ccw(t.a, t.b, s.a) * ccw(t.a, t.b, s.b) <= 0;
}


point proj(line l, point p) {
    long double t = dot(p - l.a, l.a - l.b) / std::norm(l.a - l.b);
    return l.a + t * (l.a - l.b);
}

point is_ll(line s, line t) {
    point sv = s.b - s.a, tv = t.b - t.a;
    assert(cross(sv, tv) != 0);
    return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}

long double dist_lp(line l, point p) {
    return std::abs(p - proj(l, p));
}

long double dist_ll(line l, line m) {
    return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}

long double dist_ls(line l, segment s) {
    return isis_ls(l, s) ? 0 : std::min(dist_lp(l, s.a), dist_lp(l, s.b));
}

long double dist_sp(segment s, point p) {
    point r = proj(line(s), p);
    return isis_sp(s, r) ? std::abs(r - p) : std::min(std::abs(s.a - p), std::abs(s.b - p));
}

long double dist_ss(segment s, segment t) {
    if(isis_ss(s, t)) return 0;
    long double d1 = std::min(dist_sp(s, t.a), dist_sp(s, t.b));
    long double d2 = std::min(dist_sp(t, s.a), dist_sp(t, s.b));
    return std::min(d1, d2);
}


std::vector<point> is_cc(circle const& c1, circle const& c2) {
    std::vector<point> res;
    long double d = std::abs(c1.p - c2.p);
    long double rc = (d * d + c1.r * c1.r - c2.r * c2.r) / (2 * d);
    long double dfr = c1.r * c1.r - rc * rc;
    if(std::abs(dfr) < eps) {
        dfr = 0.0;
    } else if(dfr < 0.0) {
        return res;
    }

    long double rs = std::sqrt(dfr);
    point diff = (c2.p - c1.p) / d;
    res.push_back(c1.p + diff * point(rc, rs));
    if(dfr != 0.0) {
        res.push_back(c1.p + diff * point(rc, -rs));
    }
    return res;
}

// 0 -> on, 1 -> in, 2 -> out
int is_in_polygon(polygon const& poly, point p) {
    int N = poly.size();
    ld sum = 0;
    for(int i=0; i<N; ++i) {
        point p1 = poly[i], p2 = poly[(i+1)%N];
        if(isis_sp(segment(p1, p2), p)) {
            return 0;
        }
        sum += arg((p2 - p) / (p1 - p));
    }
    return eq(sum, 0) ? 2 : 1;
}

point reflection(line const& l, point p) {
    auto pro = proj(l, p);
    return p + (pro - p) + (pro - p);
}

ld area(polygon const& p) {
    const int N = p.size();
    ld res = 0;
    for(int i=0; i<N; ++i) {
        res += cross(p[i], p[(i+1)%N]);
    }
    return res / 2;
}

// left side
polygon convex_cut(polygon const& p, line l) {
    const int N = p.size();
    polygon res;
    for(int i=0; i<N; ++i) {
        auto a = p[i], b = p[(i+1)%N];
        if(ccw(l.a, l.b, a) != -1) {
            res.push_back(a);
        }
        if(ccw(l.a, l.b, a) * ccw(l.a, l.b, b) < 0) {
            res.push_back(is_ll(line(a, b), l));
        }
    }
    return res;
}

line separate(point const& p1, point const& p2) {
    assert(p1 != p2);
    auto m = (p1 + p2) * point(0.5, 0);
    line res;
    res.a = m + (m - p1) * point(0, 1);
    res.b = m + (m - p1) * point(0, -1);
    return res;
}


} // namespace geometry2d

using namespace geometry2d;

int main() {
    int n;
    while(cin >> n, n) {
        polygon p(n);
        for(int i=0; i<n; ++i) {
            ld x, y;
            cin >> x >> y;
            p[i] = point(x, y);
        }
        ld lb = 0, ub = 1e9;
        while(abs(ub - lb) > eps) {
            auto p2 = p;
            ld m = (ub + lb) / 2;
            for(int i=0; i<n; ++i) {
                point a = p[i], b = p[(i+1)%n];
                point v = (b - a) * point(0, 1);
                v = v / abs(v) * m;
                a += v;
                b += v;
                p2 = convex_cut(p2, line(a, b));
            }
            if(p2.size() == 0) {
                ub = m;
            } else {
                lb = m;
            }
        }
        cout << fixed << setprecision(10) << lb << endl;
    }
}