Problem: The New Vasjuki village is stretched along the motorway and that's why every house on it is characterized by its shift relative to some fixed point — the xi coordinate. The village consists of n houses, the i-th house is located in the point with coordinates of xi.

TELE3, a cellular communication provider planned to locate three base stations so as to provide every house in the village with cellular communication. The base station having power d located in the point t provides with communication all the houses on the segment [t - d, t + d] (including boundaries).

To simplify the integration (and simply not to mix anything up) all the three stations are planned to possess the equal power of d. Which minimal value of d is enough to provide all the houses in the village with cellular communication.

Input

The first line contains an integer n (1 ≤ n ≤ 2·105) which represents the number of houses in the village. The second line contains the coordinates of houses — the sequence x1, x2, ..., xn of integer numbers (1 ≤ xi ≤ 109). It is possible that two or more houses are located on one point. The coordinates are given in a arbitrary order.

Output

Print the required minimal power d. In the second line print three numbers — the possible coordinates of the base stations' location. Print the coordinates with 6 digits after the decimal point. The positions of the stations can be any from 0 to 2·109 inclusively. It is accepted for the base stations to have matching coordinates. If there are many solutions, print any of them.

Examples

Input

4
1 2 3 4


Output

0.500000
1.500000 2.500000 3.500000


Input

3
10 20 30


Output

0
10.000000 20.000000 30.000000


Input

5
10003 10004 10001 10002 1


Output

0.500000
1.000000 10001.500000 10003.500000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct iii {
  int a, b, c;
  iii() {}
  iii(int a, int b, int c) : a(a), b(b), c(c) {}
  bool operator<(const iii& e) const {
    return a < e.a || a == e.a && (b < e.b || b == e.b && c < e.c);
  }
};
template <typename T>
T sq(const T& x) {
  return x * x;
}
template <typename T>
void minl(T& a, const T& b) {
  if (a > b) a = b;
}
template <typename T>
void maxl(T& a, const T& b) {
  if (a < b) a = b;
}
template <typename T>
void usort(vector<T>& a) {
  sort((a).begin(), (a).end());
  a.erase(unique((a).begin(), (a).end()), a.end());
}
template <typename T>
int index(const vector<T>& a, const T& x) {
  return lower_bound((a).begin(), (a).end(), x) - a.begin();
}
pair<int, int> operator+(const pair<int, int>& a, const pair<int, int>& b) {
  return pair<int, int>(a.first + b.first, a.second + b.second);
}
pair<int, int> operator-(const pair<int, int>& a, const pair<int, int>& b) {
  return pair<int, int>(a.first - b.first, a.second - b.second);
}
long long cross(const pair<int, int>& a, const pair<int, int>& b) {
  return (long long)a.first * b.second - (long long)b.first * a.second;
}
int n;
double a[200000];
double pos[3];
bool check(double d) {
  int j = 0;
  double p = a[0] + d + d;
  for (int i = 0; i < 3; i++) {
    pos[i] = p - d;
    for (;; j++) {
      if (j >= n) return true;
      if (a[j] > p) break;
    }
    p = a[j] + d + d;
  }
  return false;
}
void york() {
  double lf = 0.0, rt = 1e9;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lf", a + i);
  sort(a, a + n);
  int rep = 85;
  while (rep--) {
    double mid = (lf + rt) / 2;
    if (check(mid))
      rt = mid;
    else
      lf = mid;
  }
  printf("%.6f\n", rt);
  printf("%.6f %.6f %.6f\n", pos[0], pos[1], pos[2]);
}
int main() {
  york();
  return 0;
}