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;
vector<double> res;
bool check(vector<int> a, double x) {
  vector<double> ans;
  for (int i : a) {
    if (ans.size() == 0 || ans.back() + x < i) {
      ans.push_back(i + x);
    }
  }
  if (ans.size() > 3) return false;
  res = ans;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  double left = 0.0, right = 1000000000.0;
  while (abs(right - left) > 0.0000001) {
    double mid = (left + right) / 2;
    if (check(a, mid))
      right = mid;
    else
      left = mid;
  }
  printf("%.6lf\n", left);
  for (int i = 0; i < res.size(); i++) printf("%.6lf ", res[i]);
  for (int i = 0; i < 3 - res.size(); i++) printf("%.6lf ", res.back());
  printf("\n");
}