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 long long nax = 2e5 + 7;
long long n;
vector<double> a;
bool check(double d, vector<double> &v) {
  v.push_back(a[0] + d);
  for (long long i = 1; i < n; ++i) {
    while (i < n && v.back() + d >= a[i]) i++;
    if (i < n) v.push_back(a[i] + d);
  }
  while (v.size() < 3) v.push_back(v.back() + 1e-6);
  if (v.size() <= 3) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  sort(a.begin(), a.end());
  a.resize(distance(a.begin(), unique(a.begin(), a.end())));
  n = a.size();
  double lo = 0, hi = 1e9, ans = 0;
  vector<double> res, temp;
  while (hi - lo >= 1e-6) {
    double mi = (hi + lo) / 2;
    temp.clear();
    if (check(mi, temp)) {
      hi = mi - 1e-6;
      res = temp;
      ans = mi;
    } else
      lo = mi + 1e-6;
  }
  cout << fixed << setprecision(6);
  cout << ans << "\n";
  for (auto it : res) cout << it << " ";
}