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> house;
double ans[3];
bool check(double d) {
  double sta[3];
  int pos = 0;
  sta[0] = (double)house[0] + d;
  for (int i = 1; i < house.size(); i++) {
    if (house[i] > sta[pos] + d) {
      if (pos == 2) return false;
      pos++;
      sta[pos] = house[i] + d;
    }
  }
  ans[0] = sta[0];
  ans[1] = sta[1];
  ans[2] = sta[2];
  return true;
}
int main() {
  int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    house.push_back(x);
  }
  sort(house.begin(), house.end());
  double lo = 0.0, hi = 1000000000.0;
  while (hi - lo > 1e-7) {
    double mid = lo + 0.5 * (hi - lo);
    if (check(mid)) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  cout << fixed << setprecision(6) << lo << endl;
  cout << fixed << setprecision(6) << ans[0] << " " << ans[1] << " " << ans[2]
       << endl;
  return 0;
}