Problem: You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.

You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.

Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.

Input

The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.

The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).

Output

Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.

Examples

Input

4
0 0
0 1
1 1
1 0


Output

0.3535533906


Input

6
5 0
10 0
12 -4
10 -8
5 -8
3 -4


Output

1.0000000000

Note

Here is a picture of the first sample

<image>

Here is an example of making the polygon non-convex.

<image>

This is not an optimal solution, since the maximum distance we moved one point is  ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most  ≈ 0.3535533906.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct point {
  double x, y;
  point(double aa, double bb) : x(aa), y(bb) {}
};
double sq(double x) { return x * x; }
double dist(point p1, point p2) {
  return sqrt(sq(p1.x - p2.x) + sq(p1.y - p2.y));
}
double dist2(point v, point w) { return sqrt(v.x - w.x) + sqrt(v.y - w.y); }
double pDistance(double x, double y, double x1, double y1, double x2,
                 double y2) {
  double A = x - x1;
  double B = y - y1;
  double C = x2 - x1;
  double D = y2 - y1;
  double dot = A * C + B * D;
  double len_sq = C * C + D * D;
  double param = -1;
  if (len_sq != 0) param = dot / len_sq;
  double xx, yy;
  if (param < 0) {
    xx = x1;
    yy = y1;
  } else if (param > 1) {
    xx = x2;
    yy = y2;
  } else {
    xx = x1 + param * C;
    yy = y1 + param * D;
  }
  double dx = x - xx;
  double dy = y - yy;
  return sqrt(dx * dx + dy * dy);
}
int n;
const int n_ = 1001;
vector<point> P;
struct vec {
  double x, y;
  vec(double _x, double _y) : x(_x), y(_y) {}
};
vec toVec(point a, point b) { return vec(b.x - a.x, b.y - a.y); }
double cross(vec a, vec b) { return a.x * b.y - a.y * b.x; }
bool ccw(point p, point q, point r) {
  return cross(toVec(p, q), toVec(p, r)) > 0;
}
bool isConvex() {
  int sz = (int)P.size();
  if (sz <= 3) return false;
  bool isLeft = ccw(P[0], P[1], P[2]);
  for (int i = 1; i < sz - 1; i++)
    if (ccw(P[i], P[i + 1], P[(i + 2) == sz ? 1 : i + 2]) != isLeft)
      return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i <= n - 1; i++) {
    double a, b;
    cin >> a >> b;
    P.push_back(point(a, b));
  }
  double res = 1e11;
  for (int i = 0; i <= n - 1; i++) {
    double here = pDistance(P[i].x, P[i].y, P[(i - 1) == -1 ? n - 1 : i - 1].x,
                            P[(i - 1) == -1 ? n - 1 : i - 1].y,
                            P[(i + 1) == n ? 0 : i + 1].x,
                            P[(i + 1) == n ? 0 : i + 1].y) /
                  2;
    res = min(res, here);
  }
  cout << setprecision(7) << fixed << res << "\n";
}