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;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
string blank = "";
const int N = 2e5 + 10;
vector<double> arr(N);
int n;
vector<double> can(double mid) {
  vector<double> ans;
  int i = arr[0];
  ans.push_back(min(arr[n - 1], i + mid));
  i = *upper_bound(arr.begin(), arr.begin() + n, i + (mid * 2));
  if (i == 0) i = arr[n - 1];
  ans.push_back(min(arr[n - 1], i + mid));
  i = *upper_bound(arr.begin(), arr.begin() + n, i + (mid * 2));
  if (i == 0) i = arr[n - 1];
  ans.push_back(min(arr[n - 1], i + mid));
  if (ans[2] + mid < arr[n - 1]) ans.clear();
  return ans;
}
void solve() {
  cin >> n;
  for (long long i = (0); i < (n); i++) cin >> arr[i];
  sort(arr.begin(), arr.begin() + n);
  double l = 0, r = arr[n - 1] - arr[0], mid;
  int rep = 100;
  vector<double> ans;
  while (rep--) {
    mid = l + ((r - l) / 2.0);
    vector<double> res = can(mid);
    if (!res.empty())
      r = mid, ans = res;
    else
      l = mid;
  }
  cout << fixed << setprecision(6) << l << '\n';
  for (long long i = (0); i < (3); i++)
    cout << fixed << setprecision(6) << ans[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}