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;
using ll = long long;
using ld = long double;
using pi = pair<int, int>;
using vi = vector<int>;
using si = set<int>;
const ld PI = acos(-1.l);
const vi dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return 31 - __builtin_clz(x); }
template <class T>
void remDup(vector<T>& v) {
  sort(begin(v), end(v));
  v.erase(unique(begin(v), end(v)), end(v));
}
template <class A, class B>
void re(pair<A, B>& p);
template <class A>
void re(vector<A>& v);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(ld& d) {
  string t;
  re(t);
  d = stold(t);
}
template <class H, class... T>
void re(H& h, T&... t) {
  re(h);
  re(t...);
}
template <class A, class B>
void re(pair<A, B>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& x) {
  for (auto& a : x) re(a);
}
string to_string(char c) { return string(1, c); }
string to_string(const char* s) { return (string)s; }
string to_string(string s) { return s; }
string to_string(bool b) { return to_string((int)b); }
template <class A, class B>
string to_string(pair<A, B> p);
template <class T>
string to_string(T v) {
  bool fst = 1;
  string res = "";
  for (const auto& x : v) {
    if (!fst) {
      res += " ";
    }
    fst = 0;
    res += to_string(x);
  }
  return res;
}
template <class A, class B>
string to_string(pair<A, B> p) {
  return to_string(p.first) + " " + to_string(p.second);
}
template <class T>
void pr(T x) {
  cout << to_string(x);
}
template <class H, class... T>
void pr(const H& h, const T&... t) {
  pr(h);
  pr(t...);
}
void ps() { pr("\n"); }
template <class H, class... T>
void ps(const H& h, const T&... t) {
  pr(h);
  if (sizeof...(t)) pr(" ");
  ps(t...);
}
void dbg() { cerr << "]" << '\n'; }
template <class H, class... T>
void dbg(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  dbg(t...);
}
ld K(ld a) { return a * a; }
struct Point {
  ld x, y;
  friend void re(Point& p) { cin >> p.x >> p.y; }
  friend string to_string(Point p) {
    return "{" + to_string(p.x) + ", " + to_string(p.y) + "}";
  }
  Point operator+(Point b) { return Point{x + b.x, y + b.y}; }
  Point operator-(Point b) { return Point{x - b.x, y - b.y}; }
  Point operator*(ld mul) { return Point{x * mul, y * mul}; }
  ld operator*(Point b) { return x * b.y - y * b.x; }
  ld dot(Point b) { return x * b.x + y * b.y; }
  ld len() { return sqrt(K(x) + K(y)); }
  Point lenTo(ld to) { return *this * (to / len()); }
  ld dist(Point& b) { return (*this - b).len(); }
};
struct Line {
  Point one, two;
  ld dist(Point p) { return abs((p - one) * (p - two)) / one.dist(two); }
};
void solve() {
  int n;
  re(n);
  vector<Point> v(n);
  re(v);
  ld res = 1e15;
  for (int i = (0); i < (n); ++i) {
    int j = (i + 1) % n, k = (i + 2) % n;
    ckmin(res, Line{v[i], v[k]}.dist(v[j]));
  }
  cout << fixed << setprecision(10) << res / 2 << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ll T = 1;
  while (T--) solve();
}