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 inf = 1e18 + 5;
const long long nax = 2e5 + 5;
long long n;
long long arr[nax];
bool visited[nax];
bool check(double mid) {
  long long cnt = 0;
  double ch = 0;
  for (long long i = 0; i < n; i++) {
    if (arr[i] > ch) {
      cnt++;
      ch = arr[i] + mid;
    }
  }
  if (cnt <= 3) return 1;
  return 0;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  double st = 0.0, en = 1e9;
  double ans = 0.0;
  while (en - st > 0.0000001) {
    double mid = st + (en - st) / 2;
    if (check(mid) == 1) {
      ans = mid;
      en = mid;
    } else {
      st = mid;
    }
  }
  double result = ans / 2;
  cout << fixed << setprecision(6) << result << '\n';
  double ch = 0;
  vector<double> res;
  for (long long i = 0; i < n; i++) {
    if (arr[i] > ch) {
      double bh = arr[i] + ans / 2;
      res.push_back(bh);
      ch = arr[i] + ans;
    }
  }
  long long si = res.size();
  long long hey = 3 - si;
  if (si < 3) {
    double fuck = res[0];
    while (hey) {
      hey--;
      res.push_back(fuck);
    }
  }
  sort(res.begin(), res.end());
  for (auto it : res) {
    cout << fixed << setprecision(6) << it << " ";
  }
  return 0;
}