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;
bool valid(vector<double> &v, double &r) {
  int low = upper_bound(v.begin(), v.end(), *v.begin() + 2 * r) - v.begin(),
      high =
          lower_bound(v.begin(), v.end(), *(--v.end()) - 2 * r) - v.begin() - 1;
  if (low >= high) {
    return true;
  } else {
    return v[low] + 2 * r >= v[high];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  double x;
  set<double> s;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x;
    s.insert(x);
  }
  vector<double> v(s.begin(), s.end());
  cout << fixed << setprecision(6);
  if (v.size() <= 3) {
    cout << 0 << endl;
    for (int i = 0; i < 3; ++i) {
      cout << v[i % v.size()] << ' ';
    }
  } else {
    double l = 0, r = (*(--v.end()) - *v.begin()) / 3, m;
    int tr = 1e5;
    while (r - l > 1e-8 && --tr) {
      m = l + (r - l) / 2;
      if (valid(v, m)) {
        r = m;
      } else {
        l = m;
      }
    }
    int low = upper_bound(v.begin(), v.end(), *v.begin() + 2 * r) - v.begin(),
        high = lower_bound(v.begin(), v.end(), *(--v.end()) - 2 * r) -
               v.begin() - 1;
    cout << m << endl;
    cout << *v.begin() + m << ' ' << v[low] + (v[high] - v[low]) / 2 << ' '
         << *(--v.end()) - m;
  }
  return 0;
}