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.io.*;
import java.util.*;

public class Solution
{ int n;
  int max=2*1000*1000*1000;
  int A[];
PrintWriter out;
BufferedReader in;
StringTokenizer ss;


   void dbg(String s){System.out.println(s);};

   String next_token() throws IOException
   {while (!ss.hasMoreTokens())ss=new StringTokenizer(in.readLine());
    return ss.nextToken();}

   Double _double() throws IOException
   {return Double.parseDouble (next_token());}

   int _int() throws IOException
   {return Integer.parseInt (next_token());}

   long _long() throws IOException
   {return Long.parseLong (next_token());}
   boolean pod(long m){long ob=0;int vus=0;
    for(int i=0;i<n&&vus<4;i++)
      if(A[i]>ob){vus++;ob=2*m+A[i];if(m+A[i]>max)ob=max;if(ob<0||ob>max)ob=max;}

    if(vus<=3)return true;
       return false;
   }



void run()throws IOException
{
in = new BufferedReader(new InputStreamReader(System.in));
out= new PrintWriter(System.out);
//String name = "file";12
//      in = new BufferedReader(new FileReader(name+".in"));
//      out = new PrintWriter(new File(name+".out"));
ss = new StringTokenizer("  ");


     n=_int();
     A=new int[n];
     for(int i=0;i<n;i++)A[i]=2*_int();
     Arrays.sort(A);
     int l=-1;int r=max;
     while(l+1<r){
         int m=(l+r)/2;
         if(pod(m))r=m;
         else{l=m;}
     }
    double vus[]=new double[3];long ob=0;long m=r;int kol=0;
    for(int i=0;i<n;i++)
          if(A[i]>ob){vus[kol]=A[i]+m;kol++;ob=2*m+A[i];if(r+A[i]>max)ob=max;if(ob<0||ob>max)ob=max;}
    double D=r;D=D/2;
    for(int i=0;i<3;i++)vus[i]=vus[i]/2;
    out.println(D);
    for(int i=0;i<3;i++)out.print(vus[i]+" ");


out.close();
}

public static void main(String[] args)throws Exception
{
try{new Solution().run();}catch (Exception e){int a[]=new int[2];a[1+3]=2;};
}
}