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.text.DecimalFormat;
import java.util.Arrays;

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

public class round48C {

    static int N;
    static int [] locations;
    static DecimalFormat df = new DecimalFormat("0.000000");
    public static boolean valid(double d){
        double cur = locations[0] + 2 * d;
        int pos = 0;
        int i;
        for(i = 1 ; i < locations.length ; ++i){
            if(locations[i] > cur){
                pos = i; break;
            }
        }
        cur = locations[pos] + 2 * d;
        for(i = pos + 1 ; i < locations.length ; ++i){
            if(locations[i] > cur){
                pos = i; break;
            }
        }
        cur = locations[pos] + 2 * d;
        for(i = pos + 1 ; i < locations.length ; ++i){
            if(locations[i] > cur){
                pos = i; break;
            }
        }
        if(i == locations.length)return true;
        else return false;
    }
    public static double [] getLocations(double d){
        double [] ans = new double [3];
        ans[0] = locations[0] + d;
        double cur = locations[0] + 2 * d;
        int i;
        int pos = 0;
        for(i = 1 ; i < locations.length ; ++i){
            if(locations[i] > cur){
                pos = i; break;
            }
        }
        ans[1] = locations[pos] + d;
        cur = locations[pos] + 2 * d;
        for(i = pos + 1 ; i < locations.length ; ++i){
            if(locations[i] > cur){
                pos = i; break;
            }
        }
        ans[2] = locations[pos] + d;
        return ans;
    }
    public static void main(String[] args)throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        N = parseInt(br.readLine());
        String [] use = br.readLine().split(" ");
        locations = new int [N];
        for(int i = 0 ; i < use.length ; ++i){
            locations[i] = parseInt(use[i]);
        }
        Arrays.sort(locations);
        //binary search
        double lo = 0.0;
        double hi = 1000000000;
        for(int i = 0 ; i < 200 ; ++i){
            double mid = lo + ((hi - lo) / 2.0);
            if(valid(mid)){
                hi = mid;
            }else{
                lo = mid;
            }
        }
        System.out.println(df.format(lo));
        double [] locs = getLocations(parseDouble(df.format(lo)));
        for(double d : locs){
            System.out.print(df.format(d) + " ");
        }
    }
}