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 JAVA):

Solution:
import java.util.*;

public class Main
{
  static double arr[];
  public static void main(String[] arg)
  {
        Scanner pd=new Scanner(System.in);
        int n=pd.nextInt();
        arr=new double[n];
        for(int i=0;i<n;i++)
            arr[i]=pd.nextLong();
        Arrays.sort(arr);
        double s=0;
        double e=arr[arr.length-1];
        double ans[]=new double[3];
        double print=-1;
        while(e-s>=1e-7)
        {
            double mid=(s+e)/2;
            double tmp[]=possible(mid);
            if(tmp!=null)
            {
                print=mid;
                ans[0]=tmp[0];
                ans[1]=tmp[1];
                ans[2]=tmp[2];
                e=mid;
            }
            else
            {
                s=mid;
            }
        }
        Arrays.sort(ans);
        System.out.printf("%.6f\n",print);
        System.out.printf("%.6f %.6f %.6f\n",ans[0],ans[1],ans[2]);
  }
  public static double[] possible(double m)
  {
    double tmp[]=new double[3];
    tmp[0]=arr[0]+m;
    int k=0;
    for(int i=1;i<arr.length;i++)
    {
        if(Math.abs(arr[i]-tmp[k])-(1e-9)>m)
        {
            if(k==2)return null;
            tmp[++k]=arr[i]+m;
        }
    }
    
    return tmp;
  }
}