#include <bits/stdc++.h>
using namespace std;
namespace IO {
void setIn(string s) { freopen(s.c_str(), "r", stdin); }
void setOut(string s) { freopen(s.c_str(), "w", stdout); }
void setIO(string s = "") {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  if (s.size()) {
    setIn(s + ".inp");
    setOut(s + ".out");
  } else {
  }
}
}  // namespace IO
using namespace IO;
namespace Function {
template <typename T1, typename T2>
void amax(T1 &a, T2 b) {
  assert(!(typeid(a).name() == typeid(int).name() &&
           typeid(b).name() == typeid(long long).name()));
  if (a < b) a = b;
}
template <typename T1, typename T2>
void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
template <typename T>
long long sqr(T x) {
  return 1LL * x * x;
}
template <typename T1, typename T2>
long long gcd(T1 a, T2 b) {
  return (b == 0 ? a : gcd(b, a % b));
}
template <typename T1, typename T2>
long long lcm(T1 a, T2 b) {
  return 1LL * a / gcd(a, b) * b;
}
}  // namespace Function
using namespace Function;
namespace Output {
void print(int x) { cout << x << "\n"; }
void print(unsigned int x) { cout << x << "\n"; }
void print(long unsigned int x) { cout << x << "\n"; }
void print(long long x) { cout << x << "\n"; }
void print(unsigned long long x) { cout << x << "\n"; }
void print(float x) { cout << x << "\n"; }
void print(double x) { cout << x << "\n"; }
void print(long double x) { cout << x << "\n"; }
void print(char x) { cout << x << "\n"; }
void print(const char *x) { cout << x << "\n"; }
void print(string x) { cout << x << "\n"; }
void print(bool x) { cout << x << "\n"; }
template <class T, class... Ts>
void print(T t, Ts... ts) {
  cout << t << " ";
  print(ts...);
}
template <typename T1, typename T2>
void print(pair<T1, T2> a) {
  print(a.first, a.second);
}
template <typename T>
void print(T a) {
  for (auto it : a) {
    print(it);
  }
}
template <class T, class... Ts>
void prine(T t, Ts... ts) {
  print(t, ts...);
  exit(0);
}
}  // namespace Output
using namespace Output;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
const int INF = 1e9;
const long long INFL = 1e18;
const int MOD = 1e9 + 7;
const int N = 1e3 + 10;
long long x[N], y[N];
long long dis(int i, int j) { return sqr(x[i] - x[j]) + sqr(y[i] - y[j]); }
int main() {
  setIO();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
  }
  double res = INFL;
  for (int i = 0; i < n; i++) {
    long long S = 1LL * (x[(i + 1) % n] - x[i]) * (y[i] + y[(i + 1) % n]);
    S += 1LL * (x[(i + 2) % n] - x[(i + 1) % n]) *
         (y[(i + 2) % n] + y[(i + 1) % n]);
    S += 1LL * (x[i] - x[(i + 2) % n]) * (y[i] + y[(i + 2) % n]);
    amin(res, abs(S) / sqrt(dis(i, (i + 2) % n)));
  }
  cout << fixed << setprecision(6);
  print(res / 2);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
namespace IO {
void setIn(string s) { freopen(s.c_str(), "r", stdin); }
void setOut(string s) { freopen(s.c_str(), "w", stdout); }
void setIO(string s = "") {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  if (s.size()) {
    setIn(s + ".inp");
    setOut(s + ".out");
  } else {
  }
}
}  // namespace IO
using namespace IO;
namespace Function {
template <typename T1, typename T2>
void amax(T1 &a, T2 b) {
  assert(!(typeid(a).name() == typeid(int).name() &&
           typeid(b).name() == typeid(long long).name()));
  if (a < b) a = b;
}
template <typename T1, typename T2>
void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
template <typename T>
long long sqr(T x) {
  return 1LL * x * x;
}
template <typename T1, typename T2>
long long gcd(T1 a, T2 b) {
  return (b == 0 ? a : gcd(b, a % b));
}
template <typename T1, typename T2>
long long lcm(T1 a, T2 b) {
  return 1LL * a / gcd(a, b) * b;
}
}  // namespace Function
using namespace Function;
namespace Output {
void print(int x) { cout << x << "\n"; }
void print(unsigned int x) { cout << x << "\n"; }
void print(long unsigned int x) { cout << x << "\n"; }
void print(long long x) { cout << x << "\n"; }
void print(unsigned long long x) { cout << x << "\n"; }
void print(float x) { cout << x << "\n"; }
void print(double x) { cout << x << "\n"; }
void print(long double x) { cout << x << "\n"; }
void print(char x) { cout << x << "\n"; }
void print(const char *x) { cout << x << "\n"; }
void print(string x) { cout << x << "\n"; }
void print(bool x) { cout << x << "\n"; }
template <class T, class... Ts>
void print(T t, Ts... ts) {
  cout << t << " ";
  print(ts...);
}
template <typename T1, typename T2>
void print(pair<T1, T2> a) {
  print(a.first, a.second);
}
template <typename T>
void print(T a) {
  for (auto it : a) {
    print(it);
  }
}
template <class T, class... Ts>
void prine(T t, Ts... ts) {
  print(t, ts...);
  exit(0);
}
}  // namespace Output
using namespace Output;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
const int INF = 1e9;
const long long INFL = 1e18;
const int MOD = 1e9 + 7;
const int N = 1e3 + 10;
long long x[N], y[N];
long long dis(int i, int j) { return sqr(x[i] - x[j]) + sqr(y[i] - y[j]); }
int main() {
  setIO();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
  }
  double res = INFL;
  for (int i = 0; i < n; i++) {
    long long S = 1LL * (x[(i + 1) % n] - x[i]) * (y[i] + y[(i + 1) % n]);
    S += 1LL * (x[(i + 2) % n] - x[(i + 1) % n]) *
         (y[(i + 2) % n] + y[(i + 1) % n]);
    S += 1LL * (x[i] - x[(i + 2) % n]) * (y[i] + y[(i + 2) % n]);
    amin(res, abs(S) / sqrt(dis(i, (i + 2) % n)));
  }
  cout << fixed << setprecision(6);
  print(res / 2);
  return 0;
}

```