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;
template <class T, class U>
inline void relax_min(T &a, U b) {
  a = min(a, b);
}
template <class T, class U>
inline void relax_max(T &a, U b) {
  a = max(a, b);
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    in >> a[i];
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    out << a[i] << ' ';
  return out;
}
int next_int() {
  int(tmp);
  scanf("%d", &(tmp));
  return tmp;
}
long long next_ll() {
  long long(tmp);
  scanf("%lld", &(tmp));
  return tmp;
}
long long modPow(long long a, long long b, long long MOD) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1)
      ret = ((((((long long)(ret) * (long long)(a))) % ((MOD))) + ((MOD))) %
             ((MOD)));
    a = ((((((long long)(a) * (long long)(a))) % ((MOD))) + ((MOD))) % ((MOD)));
  }
  return ret;
}
const long long EXACTLY_ONE_HUNDRED = 100;
const long long EXACTLY_ONE_THOUSAND = 1000;
const long long EXACTLY_ONE_MILLION = 1000 * 1000;
const long long EXACTLY_ONE_BILLION =
    EXACTLY_ONE_THOUSAND * EXACTLY_ONE_MILLION;
const int MAXN1 = (int)(EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN2 = (int)(2 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN3 = (int)(3 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN4 = (int)(4 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN5 = (int)(5 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN = MAXN3;
const int INF = (int)EXACTLY_ONE_BILLION + 7;
const int INF2 = 1 << 30;
const long long INFLL = 4LL * EXACTLY_ONE_BILLION * EXACTLY_ONE_BILLION + 7;
const long long INFLL2 = 1LL << 60;
const long double Pi = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = (int)EXACTLY_ONE_BILLION + 7;
struct Vector {
  int id;
  long long x, y;
  explicit Vector(long long _x = 0, long long _y = 0, long long _id = 0) {
    x = _x;
    y = _y;
    id = _id;
  }
  Vector operator+(Vector const &a) { return Vector(x + a.x, y + a.y); }
  Vector operator-(Vector const &a) { return Vector(x - a.x, y - a.y, id); }
  long long operator*(Vector const &a) { return x * a.x + y * a.y; }
  Vector operator*(long double a) { return Vector(x * a, y * a); }
  long long operator%(Vector const &a) { return x * a.y - a.x * y; }
  Vector ort() { return Vector(-y, x); }
  bool operator<(Vector const &a) const {
    if (y == a.y) return (x < a.x);
    return (y < a.y);
  }
  bool operator==(Vector const &a) const { return (x == a.x && y == a.y); }
  long double len() { return sqrt((long double)x * x + y * y); }
  long long sqr_len() { return x * x + y * y; }
  Vector dir() { return Vector(x, y); }
};
ostream &operator<<(ostream &out, const Vector &a) {
  out << a.x << " " << a.y;
  return out;
}
istream &operator>>(istream &in, Vector &a) {
  in >> a.x >> a.y;
  return in;
}
int main() {
  int(n);
  scanf("%d", &(n));
  vector<int> x(n), y(n);
  for (decltype(0)((IT_FOR_SCANNING)) = (0); ((IT_FOR_SCANNING)) < ((((n))));
       ++((IT_FOR_SCANNING)))
    scanf("%d%d", &(((x))[IT_FOR_SCANNING]), &(((y))[IT_FOR_SCANNING]));
  for (decltype(0)((i)) = (0); ((i)) < ((2)); ++((i))) {
    x.push_back(x[i]);
    y.push_back(y[i]);
  }
  long double ans = INFLL;
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    Vector A(x[i], y[i]), B(x[i + 1], y[i + 1]), C(x[i + 2], y[i + 2]);
    Vector AC = C - A, AB = B - A;
    updmin(ans, abs(AC % AB / AC.len()));
  }
  printf("%.30Lf\n", ans / 2);
  return 0;
}