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.*;
import static java.lang.Math.*;

public class ProblemC_48 {
    
    final boolean ONLINE_JUDGE=System.getProperty("ONLINE_JUDGE")!=null;
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok=new StringTokenizer("");
    
    void init() throws FileNotFoundException{
        if (ONLINE_JUDGE){
            in=new BufferedReader(new InputStreamReader(System.in));
            out =new PrintWriter(System.out);
        }
        else{
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        }
    }
    
    String readString() throws IOException{
        while(!tok.hasMoreTokens()){
            tok=new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
    
    int readInt() throws IOException{
        return Integer.parseInt(readString());
    }
    
    public static void main(String[] args){
        new ProblemC_48().run();
    }
    
    public void run(){
        try{
            long t1=System.currentTimeMillis();
            init();
            solve();
            out.close();
            long t2=System.currentTimeMillis();
            System.err.println("Time = "+(t2-t1));
        }catch (Exception e){
            e.printStackTrace(System.err);
            System.exit(-1);
        }
    }
    
    void solve() throws IOException{
        int n=readInt();
        int[] a=new int[n];
        for (int i=0; i<n; i++){
            a[i]=readInt()*2;
        }
        Arrays.sort(a);
        int k=a.length;
        for (int i=1; i<n; i++){
            if (a[i]==a[i-1]){
                a[i-1]=-1;
                k--;
            }
        }
        double[] x=new double[k];
        k=0;
        for (int i=0; i<n; i++){
            if (a[i]!=-1){
                x[k++]=a[i];
            }
        }
        long left=0;
        long right=2000000000;
        double x1=0, x2=0, x3=0;
        for (; right-left>0; ){
            long r=(left+right)/2;
            x1=x[0]+r;
            if (x1+r>=x[x.length-1]){
                right=(long)ceil(r);
                continue;
            }
            int i2=Arrays.binarySearch(x, x1+r);
            if (i2<0){
                i2=-(i2+1);
            }else{
                i2++;
            }
            if (i2>=x.length){
                right=(long)ceil(r);
                continue;
            }
            x2=x[i2]+r;
            if (x2+r>=x[x.length-1]){
                right=(long)ceil(r);
                continue;
            }
            int i3=Arrays.binarySearch(x, x2+r);
            if (i3<0){
                i3=-(i3+1);
            }else{
                i3++;
            }
            if (i3>=x.length){
                right=(long)ceil(r);
                continue;
            }
            x3=x[i3]+r;
            if (x3+r>=x[x.length-1]){
                right=(long)ceil(r);
                continue;
            }
            if (left==r){
                left=r+1;
            }else{
                left=r;
            }
        }
        double r=(double)(left+right)/2;
        x1=x[0]+r;
        int i2=Arrays.binarySearch(x, x1+r);
        if (i2<0){
            i2=-(i2+1);
        }else{
            i2++;
        }
        if (i2>=x.length){
            i2=x.length-1;
        }
        x2=x[i2]+r;
        int i3=Arrays.binarySearch(x, x2+r);
        if (i3<0){
            i3=-(i3+1);
        }else{
            i3++;
        }
        if (i3>=x.length){
            i3=x.length-1;
        }
        x3=x[i3]+r;
        out.printf(Locale.US, "%.6f", (double)(left+right)/4);
        out.println();
        out.printf(Locale.US, "%.6f", x1/2);
        out.print(" ");
        out.printf(Locale.US, "%.6f", x2/2);
        out.print(" ");
        out.printf(Locale.US, "%.6f", x3/2);
    }       
}