Problem: A prisoner wants to escape from a prison. The prison is represented by the interior of the convex polygon with vertices P_1, P_2, P_3, …, P_{n+1}, P_{n+2}, P_{n+3}. It holds P_1=(0,0), P_{n+1}=(0, h), P_{n+2}=(-10^{18}, h) and P_{n+3}=(-10^{18}, 0).

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The prison walls P_{n+1}P_{n+2}, P_{n+2}P_{n+3} and P_{n+3}P_1 are very high and the prisoner is not able to climb them. Hence his only chance is to reach a point on one of the walls P_1P_2, P_2P_3,..., P_{n}P_{n+1} and escape from there. On the perimeter of the prison, there are two guards. The prisoner moves at speed 1 while the guards move, remaining always on the perimeter of the prison, with speed v.

If the prisoner reaches a point of the perimeter where there is a guard, the guard kills the prisoner. If the prisoner reaches a point of the part of the perimeter he is able to climb and there is no guard there, he escapes immediately. Initially the prisoner is at the point (-10^{17}, h/2) and the guards are at P_1. 

Find the minimum speed v such that the guards can guarantee that the prisoner will not escape (assuming that both the prisoner and the guards move optimally).

Notes:

  * At any moment, the guards and the prisoner can see each other. 
  * The "climbing part" of the escape takes no time. 
  * You may assume that both the prisoner and the guards can change direction and velocity instantly and that they both have perfect reflexes (so they can react instantly to whatever the other one is doing). 
  * The two guards can plan ahead how to react to the prisoner movements. 

Input

The first line of the input contains n (1 ≤ n ≤ 50).

The following n+1 lines describe P_1, P_2,..., P_{n+1}. The i-th of such lines contain two integers x_i, y_i (0≤ x_i, y_i≤ 1,000) — the coordinates of P_i=(x_i, y_i).

It is guaranteed that P_1=(0,0) and x_{n+1}=0. The polygon with vertices P_1,P_2,..., P_{n+1}, P_{n+2}, P_{n+3} (where P_{n+2}, P_{n+3} shall be constructed as described in the statement) is guaranteed to be convex and such that there is no line containing three of its vertices.

Output

Print a single real number, the minimum speed v that allows the guards to guarantee that the prisoner will not escape. Your answer will be considered correct if its relative or absolute error does not exceed 10^{-6}.

Examples

Input


2
0 0
223 464
0 749


Output


1


Input


3
0 0
2 2
2 4
0 6


Output


1.0823922


Input


4
0 0
7 3
7 4
5 7
0 8


Output


1.130309669


Input


5
0 0
562 248
460 610
281 702
206 723
0 746


Output


1.148649561


Input


7
0 0
412 36
745 180
747 184
746 268
611 359
213 441
0 450


Output


1.134745994

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
int read() {
  char c;
  while ((c = getchar()) < '-')
    ;
  if (c == '-') {
    int x = (c = getchar()) - '0';
    while ((c = getchar()) >= '0') x = x * 10 + c - '0';
    return -x;
  }
  int x = c - '0';
  while ((c = getchar()) >= '0') x = x * 10 + c - '0';
  return x;
}
const int N = 50 + 5;
int sgn(const double &x, const double &eps = 1e-9) {
  return x < -eps ? -1 : x > eps;
}
double sqr(const double &x) { return x * x; }
struct Point {
  double x, y;
  Point unit() {
    double len = sqrt(sqr(x) + sqr(y));
    assert(sgn(len));
    return {x / len, y / len};
  }
} p[N];
int n;
Point operator-(const Point &a, const Point &b) {
  return {a.x - b.x, a.y - b.y};
}
Point operator+(const Point &a, const Point &b) {
  return {a.x + b.x, a.y + b.y};
}
Point operator*(const Point &a, const double &b) { return {a.x * b, a.y * b}; }
double dis(const Point &a, const Point &b) {
  return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}
double sum_len[N];
pair<double, double> mx[N][N];
double F(int i, int j, double vi) {
  Point p0 = p[i] + (p[i + 1] - p[i]).unit() * vi;
  double l;
  Point d;
  if (j < i) {
    l = sum_len[i] - sum_len[j + 1] + vi;
    d = (p[j] - p[j + 1]).unit();
    p0 = p[j + 1] - p0;
  } else {
    l = sum_len[j] - sum_len[i] - vi;
    d = (p[j + 1] - p[j]).unit();
    p0 = p[j] - p0;
  }
  double len = dis(p[j], p[j + 1]);
  auto F = [&](const double &v) -> double {
    double mu = sqr(d.x * v + p0.x) + sqr(d.y * v + p0.y);
    if (sgn(mu) <= 0) return 1;
    return (v + l) / sqrt(mu);
  };
  double ans = max(F(0), F(len));
  Point p3 = p0 - d * l;
  double a = sqr(p3.x) + sqr(p3.y), b = d.x * p3.x + d.y * p3.y;
  if (sgn(b)) {
    double v = -a / b - l;
    if (sgn(v) > 0 && sgn(len - v) > 0) chmax(ans, F(v));
  }
  return ans;
}
void init_mx(int i, int j) {
  double l = 0, r = dis(p[i], p[i + 1]);
  for (int tmp = 1; tmp <= 200; ++tmp) {
    double m1 = (l * 2 + r) / 3, m2 = (l + 2 * r) / 3;
    if (F(i, j, m1) > F(i, j, m2))
      r = m2;
    else
      l = m1;
  }
  mx[i][j].first = l;
  mx[i][j].second = F(i, j, l);
}
pair<double, double> check(const double &v, int i, int j) {
  if (mx[i][j].second < v) return {1, 0};
  pair<double, double> ans;
  double l = 0, r = mx[i][j].first;
  for (int tmp = 0; tmp <= 50; ++tmp) {
    double mid = (l + r) / 2;
    if (F(i, j, mid) >= v)
      r = mid;
    else
      l = mid;
  }
  ans.first = l;
  l = mx[i][j].first;
  r = dis(p[i], p[i + 1]);
  for (int tmp = 0; tmp <= 50; ++tmp) {
    double mid = (l + r) / 2;
    if (F(i, j, mid) >= v)
      l = mid;
    else
      r = mid;
  }
  ans.second = l;
  return ans;
}
bool check(const double &v) {
  for (int i = 1; i <= n; ++i) {
    vector<pair<double, double>> seg;
    for (int j = 1; j <= i - 1; ++j) seg.push_back(check(v, i, j));
    for (int j = i + 1; j <= n; ++j) {
      auto p1 = check(v, i, j);
      for (auto p0 : seg)
        if (max(p0.first, p1.first) < min(p0.second, p1.second)) return 1;
    }
  }
  return 0;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n + 1; ++i) {
    p[i].x = read();
    p[i].y = read();
  }
  for (int i = 1; i <= n; ++i)
    sum_len[i + 1] = sum_len[i] + dis(p[i], p[i + 1]);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (j != i) init_mx(i, j);
  double l = 1, r = 2000;
  for (int tmp = 1; tmp <= 50; ++tmp) {
    double mid = (l + r) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  printf("%.9f\n", l);
}