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;
const long double pi = acosl(-1);
const long double radian = 180 / pi;
const long double eps = 1e-12;
inline int sgn(const long double &x) {
  if (fabsl(x) < eps)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
inline int sgn(const long long &x) {
  if (x == 0)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
inline long long sqr(const long long &x) { return x * x; }
inline long double refresh(const long double &x) {
  if (x < -1 + eps)
    return -1;
  else if (x > 1 - eps)
    return 1;
  else
    return x;
}
class Point {
 public:
  long long x, y;
  Point(void) : x(0), y(0) {}
  Point(const long long &x, const long long &y) : x(x), y(y) {}
  long double length(void) const { return sqrtl(sqr(x) + sqr(y)); }
  long long length2(void) const { return sqr(x) + sqr(y); }
  long double distance(const Point &rhs) const {
    return sqrtl(sqr(rhs.x - x) + sqr(rhs.y - y));
  }
  long double distance2(const Point &rhs) const {
    return sqr(rhs.x - x) + sqr(rhs.y - y);
  }
  Point conjugate(void) const { return Point(x, -y); }
  Point normal(const bool &lft = true) const {
    return lft ? Point(-y, x) : Point(y, -x);
  }
  long long vect(const Point &lhs, const Point &rhs) const {
    return (lhs - *this) * (rhs - *this);
  }
  long long scal(const Point &lhs, const Point &rhs) const {
    return (lhs - *this) & (rhs - *this);
  }
  long double distance_to_line(const Point &p, const Point &q) const {
    return fabsl(p.vect(q, *this) / p.distance(q));
  }
  long double distance_to_segment(const Point &p, const Point &q) const {
    if (p.scal(q, *this) < 0)
      return p.distance(*this);
    else if (q.scal(p, *this) < 0)
      return q.distance(*this);
    else
      return this->distance_to_line(p, q);
  }
  bool on_line(const Point &p, const Point &q) const {
    return sgn(p.vect(q, *this)) == 0;
  }
  bool on_segment(const Point &p, const Point &q) const {
    return on_line(p, q) && sgn((p.x - x) * (q.x - x)) <= 0 &&
           sgn((p.y - y) * (q.y - y)) <= 0;
  }
  Point operator+(const Point &rhs) const {
    return Point(x + rhs.x, y + rhs.y);
  }
  Point operator-(const Point &rhs) const {
    return Point(x - rhs.x, y - rhs.y);
  }
  Point operator*(const long long &rhs) const {
    return Point(x * rhs, y * rhs);
  }
  Point operator/(const long long &rhs) const {
    return Point(x / rhs, y / rhs);
  }
  long long operator*(const Point &rhs) const { return x * rhs.y - y * rhs.x; }
  long long operator&(const Point &rhs) const { return x * rhs.x + y * rhs.y; }
  Point operator<<(const Point &rhs) const {
    return Point(x * rhs.x - y * rhs.y, x * rhs.y + y * rhs.x);
  }
  long double operator^(const Point &rhs) const {
    int s = sgn(*this * rhs);
    long double a = acosl(
        refresh((long double)(*this & rhs) / this->length() / rhs.length()));
    return s < 0 ? -a : a;
  }
  bool operator==(const Point &rhs) const { return x == rhs.x && y == rhs.y; }
  inline int quadrant(void) const {
    if (x == 0 && y == 0)
      return 0;
    else if (x < 0 && y < 0)
      return 1;
    else if (x == 0 && y < 0)
      return 2;
    else if (x > 0 && y < 0)
      return 3;
    else if (x > 0 && y == 0)
      return 4;
    else if (x > 0 && y > 0)
      return 5;
    else if (x == 0 && y > 0)
      return 6;
    else if (x < 0 && y > 0)
      return 7;
    else
      return 8;
  }
  bool operator<(const Point &rhs) const {
    int lq = quadrant(), rq = rhs.quadrant();
    if (lq != rq)
      return lq < rq;
    else {
      int s = sgn(*this * rhs);
      return s != 0 ? s > 0 : sgn(length2() - rhs.length2()) < 0;
    }
  }
  bool on_left(const Point &rhs, const bool &inclusive = true) const {
    long long s = *this * rhs, t = *this & rhs;
    return s == 0 ? t >= 0 || inclusive : s > 0;
  }
  bool on_right(const Point &rhs, const bool &inclusive = true) const {
    long long s = *this * rhs, t = *this & rhs;
    return s == 0 ? t >= 0 || inclusive : s < 0;
  }
};
const int maxn = 1000;
int n;
Point p[maxn];
int main(void) {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i].x >> p[i].y;
  long double ans = 1e20;
  for (int i = 0; i < n; ++i)
    ans =
        min(ans, p[i].distance_to_line(p[(i + n - 1) % n], p[(i + 1) % n]) / 2);
  cout << fixed << setprecision(20);
  cout << ans << endl;
  return 0;
}