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;
const double eps = 1e-10;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<double> a(n);
  for (auto &t : a) {
    cin >> t;
  }
  sort(a.begin(), a.end());
  double tl = 0, tr = 1e9, st1 = 0, st2 = 0, st3 = 0;
  for (int i = 0; i < 300; i++) {
    double x = (tl + tr) / 2;
    st1 = 0;
    st2 = 0;
    st3 = 0;
    double lst = a[0] + x;
    st1 = a[0] + x;
    int k = 0, j = 0;
    while (k < n && j < 3) {
      while (k < n && a[k] <= lst + x) {
        k++;
      }
      if (k < n) {
        lst = a[k] + x;
        if (j == 0) {
          st2 = lst;
        }
        if (j == 1) {
          st3 = lst;
        }
      }
      j++;
    }
    if (k == n) {
      tr = x;
    } else {
      tl = x;
    }
  }
  cout << fixed << setprecision(6) << tr << '\n';
  double x = tr;
  st1 = 0;
  st2 = 0;
  st3 = 0;
  double lst = a[0] + x;
  st1 = a[0] + x;
  int k = 0, j = 0;
  while (k < n && j < 3) {
    while (k < n && a[k] <= lst + x) {
      k++;
    }
    if (k < n) {
      lst = a[k] + x;
      if (j == 0) {
        st2 = lst;
      }
      if (j == 1) {
        st3 = lst;
      }
    }
    j++;
  }
  cout << fixed << setprecision(6) << st1 << " " << st2 << " " << st3;
}