Problem: Barney has finally found the one, a beautiful young lady named Lyanna. The problem is, Lyanna and Barney are trapped in Lord Loss' castle. This castle has shape of a convex polygon of n points. Like most of castles in Demonata worlds, this castle has no ceiling.

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Barney and Lyanna have an escape plan, but it requires some geometry knowledge, so they asked for your help.

Barney knows that demons are organized and move in lines. He and Lyanna want to wait for the appropriate time so they need to watch for the demons. Each of them wants to stay in a point inside the castle (possibly on edges or corners), also they may stay in the same position. They both want to pick a real number r and watch all points in the circles with radius r around each of them (these two circles may overlap).

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We say that Barney and Lyanna are watching carefully if and only if for every edge of the polygon, at least one of them can see at least one point on the line this edge lies on, thus such point may not be on the edge but it should be on edge's line. Formally, each edge line should have at least one common point with at least one of two circles.

The greater r is, the more energy and focus they need. So they asked you to tell them the minimum value of r such that they can watch carefully.

Input

The first line of input contains a single integer n (3 ≤ n ≤ 300) — the number of castle polygon vertices.

The next n lines describe the polygon vertices in counter-clockwise order. i-th of them contains two integers xi and yi (|xi|, |yi| ≤ 104) — the coordinates of i-th point of the castle. It is guaranteed that given points form a convex polygon, in particular, any three of them do not line on the same line.

Output

In the first line print the single number r — minimum radius of guys' watching circles.

In the second line print the pair of coordinates of point where Barney should stay.

In the third line print the pair of coordinates of point where Lyanna should stay.

Points should lie inside the polygon.

Coordinates may not be integers. If there are multiple answers print any of them.

Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 6.

Examples

Input

4
-41 67
-16 20
25 25
-36 85


Output

0
-16 20
-36 85


Input

7
-7 54
-5 31
-2 17
20 19
32 23
34 27
26 57


Output

2.9342248
32.019503 23.0390067
-6.929116 54.006444

Note

In the first example guys can stay in opposite corners of the castle.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int RLEN = 1 << 20 | 1;
inline char gc() {
  static char ibuf[RLEN], *ib, *ob;
  (ib == ob) && (ob = (ib = ibuf) + fread(ibuf, 1, RLEN, stdin));
  return (ib == ob) ? EOF : *ib++;
}
inline int read() {
  char ch = gc();
  int res = 0;
  bool f = 1;
  while (!isdigit(ch)) f ^= ch == '-', ch = gc();
  while (isdigit(ch)) res = (res + (res << 2) << 1) + (ch ^ 48), ch = gc();
  return f ? res : -res;
}
inline long long readll() {
  char ch = gc();
  long long res = 0;
  bool f = 1;
  while (!isdigit(ch)) f ^= ch == '-', ch = gc();
  while (isdigit(ch)) res = (res + (res << 2) << 1) + (ch ^ 48), ch = gc();
  return f ? res : -res;
}
inline char readchar() {
  char ch = gc();
  while (isspace(ch)) ch = gc();
  return ch;
}
inline int readstring(char *s) {
  int top = 0;
  char ch = gc();
  while (isspace(ch)) ch = gc();
  while (!isspace(ch) && ch != EOF) s[++top] = ch, ch = gc();
  s[top + 1] = '\0';
  return top;
}
template <typename tp>
inline void chemx(tp &a, tp b) {
  a = max(a, b);
}
template <typename tp>
inline void chemn(tp &a, tp b) {
  a = min(a, b);
}
const long double eps = 1e-12;
inline int sign(long double x) {
  return (x > eps) ? 1 : (x < -eps ? -1 : 0);
  return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
}
struct pt {
  long double x, y;
  pt(long double _x = 0, long double _y = 0) : x(_x), y(_y) {}
  friend inline pt operator+(const pt &a, const pt &b) {
    return pt(a.x + b.x, a.y + b.y);
  }
  friend inline pt operator-(const pt &a, const pt &b) {
    return pt(a.x - b.x, a.y - b.y);
  }
  friend inline long double operator*(const pt &a, const pt &b) {
    return a.x * b.y - a.y * b.x;
  }
  friend inline long double operator^(const pt &a, const pt &b) {
    return a.x * b.x - a.y * b.y;
  }
  friend inline pt operator*(const pt &a, const long double &b) {
    return pt(a.x * b, a.y * b);
  }
  friend inline pt operator/(const pt &a, const long double &b) {
    return pt(a.x / b, a.y / b);
  }
  inline long double ang() { return atan2(y, x); }
  inline long double dis() const { return sqrt(x * x + y * y); }
  pt perp() { return pt(-y, x) / dis(); }
};
struct line {
  pt s, t;
  long double rad;
  line() {}
  line(pt _s, pt _t) { s = _s, t = _t, rad = (t - s).ang(); }
  void write() {
    cout << s.x << " " << s.y << " " << t.x << " " << t.y << " " << rad << '\n';
  }
  friend inline bool operator<(const line &a, const line &b) {
    return sign(a.rad - b.rad) ? a.rad < b.rad
                               : sign((a.t - a.s) * (b.t - a.s)) > 0;
  }
  inline line move(const long double &k) const {
    pt tmp = pt(k * sin(rad), -k * cos(rad));
    return line(s + tmp, t + tmp);
  }
};
inline pt Inter(const line &a, const line &b) {
  long double s1 = (b.t - a.s) * (b.s - a.s), s2 = (b.t - b.s) * (a.t - b.s);
  long double k = (sign(s1 + s2)) ? (s1) / (s1 + s2) : 0;
  return a.s + (a.t - a.s) * k;
}
const int N = 605;
pt p[N], ans1, ans2, res1, res2;
int n, cnt, hd, tl, id[N], ps[N], tot;
bool vis[N];
line t[N], q[N], st[N], tmp[N];
inline bool comp(const line &a, const line &b, const line &c) {
  return sign((Inter(a, b) - c.s) * (c.t - c.s)) >= 0;
}
inline bool half_inter() {
  int m = 0;
  for (int i = 1; i <= cnt; i++) {
    if (sign(q[i].rad - q[i - 1].rad)) m++;
    q[m] = q[i];
  }
  cnt = m;
  if (cnt == 1) {
    st[1] = st[2] = q[1];
    return true;
  }
  hd = 1, tl = 2, st[1] = q[1], st[2] = q[2];
  for (int i = 3; i <= cnt; i++) {
    while (hd < tl && comp(st[tl - 1], st[tl], q[i])) tl--;
    while (hd < tl && comp(st[hd], st[hd + 1], q[i])) hd++;
    st[++tl] = q[i];
  }
  while (hd < tl && comp(st[tl - 1], st[tl], st[hd])) tl--;
  while (hd < tl && comp(st[hd], st[hd + 1], st[tl])) hd++;
  return (tl - hd) >= 2;
}
inline line rev(line x) { return line(x.t, x.s); }
inline bool judge(int l, int r, pt &ans) {
  cnt = 0;
  if (l + 1 >= r) {
    ans = Inter(t[l], t[r]);
    return true;
  }
  for (int i = 1; i <= n; i++) vis[ps[i]] = 1;
  for (int i = l; i <= r; i++) vis[ps[(i - 1) % n + 1 + n]] = 1;
  for (int i = 1; i <= tot; i++)
    if (vis[i]) q[++cnt] = tmp[id[i]], vis[i] = 0;
  bool c = half_inter();
  if (!c) return false;
  ans = Inter(st[hd], st[hd + 1]);
  return true;
}
inline bool check(long double k, int fg = 0) {
  tot = 0;
  for (int i = 1; i <= n; i++) tmp[++tot] = t[i];
  for (int i = 1; i <= n; i++) tmp[++tot] = rev(t[i].move(-k));
  sort(id + 1, id + tot + 1,
       [](const int &i, const int &j) { return tmp[i] < tmp[j]; });
  for (int i = 1; i <= tot; i++) ps[id[i]] = i;
  for (int i = 1, j = 1; i <= n; i++) {
    while (j < n + i - 1 && judge(i, j + 1, ans1)) j++;
    if (judge(j + 1, n + i - 1, ans2)) return true;
  }
  return false;
}
int main() {
  n = read();
  for (int i = 1; i <= n * 2; i++) id[i] = i;
  for (int i = 1; i <= n; i++) p[i].x = read(), p[i].y = read();
  for (int i = 1; i < n; i++) t[i] = line(p[i], p[i + 1]);
  t[n] = line(p[n], p[1]);
  for (int i = 1; i <= n; i++) t[i + n] = t[i];
  long double l = 0, r = 1e5;
  while (l + eps < r) {
    long double mid = (l + r) * 0.5;
    if (check(mid))
      r = mid, res1 = ans1, res2 = ans2;
    else
      l = mid;
  }
  printf("%.10lf\n", (double)l);
  printf("%.10lf %.10lf\n", (double)res1.x, (double)res1.y);
  printf("%.10lf %.10lf\n", (double)res2.x, (double)res2.y);
  return 0;
}