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<int> ans;
int makeElement(string str);
int makeRow(string str) {
  str = str.substr(4, str.size() - 9);
  int cur = 0;
  bool fnd = true;
  while (fnd) {
    fnd = false;
    for (int i = 0; i < str.size(); i++) {
      if (str.substr(i, 5) == "</td>") {
        cur += makeElement(str.substr(0, i + 5));
        fnd = true;
        str = str.substr(i + 5, 23424);
        break;
      }
    }
  }
  return cur;
}
void DFS(string str) {
  str = str.substr(7, str.size() - 15);
  int cur = 0;
  bool fnd = true;
  while (fnd) {
    fnd = false;
    for (int i = 0; i < str.size(); i++) {
      if (str.substr(i, 5) == "</tr>") {
        cur += makeRow(str.substr(0, i + 5));
        fnd = true;
        str = str.substr(i + 5, 23424);
        break;
      }
    }
  }
  ans.push_back(cur);
}
int makeElement(string str) {
  str = str.substr(4, str.size() - 9);
  if (str == "") return 1;
  DFS(str);
  return 1;
}
const int INF = 0x7f7f7f7f;
const double EPS = 1e-8;
int s[211111];
int main() {
  int n;
  cin >> n;
  double mns = INF;
  double mxs = -INF;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x < mns) mns = x;
    if (x > mxs) mxs = x;
    s[i] = x;
  }
  sort(s, s + n);
  double mn = 0;
  double mx = INF;
  double bst[3];
  for (int i = 0; i < 110; i++) {
    double md = (mx + mn) * 0.5;
    bool can = true;
    double pos[3];
    int last = 0;
    for (int u = 0; u < n; u++) {
      if (u == 0 || pos[last - 1] + md + EPS < s[u]) {
        if (last == 3) {
          can = false;
          break;
        }
        pos[last++] = s[u] + md;
      }
    }
    if (can) {
      memcpy(bst, pos, sizeof bst);
      mx = md;
    } else {
      mn = md;
    }
  }
  cout << fixed << setprecision(6) << (mn + mx) * 0.5 << endl;
  for (int i = 0; i < 3; i++) cout << fixed << setprecision(6) << bst[i] << ' ';
  return 0;
}