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

public class Main{
    public static void main(String[] args) throws Exception{
        new Main().run();
    }

    private static final double EPS = 1e-9;

    void run() throws Exception{
        BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(sc.readLine());
        StringTokenizer st = new StringTokenizer(sc.readLine(), " ");
        int[] p = new int[n];
        for(int i = 0; i < n; i++)p[i] = Integer.parseInt((String)st.nextElement());
        Arrays.sort(p);
        double left = 0;
        double right = p[n-1];
        double ans = right+1;
        for(int o = 0; o < 60; o++){
            double m = (left+right)/2;
            if(ok(p, m)){
                ans = min(ans, m);
                right = m;
            }else{
                left = m;
            }
        }
        System.out.printf("%.6f\n", ans);
        int ind = 0;
        int count = 0;
        while(ind < p.length){
            double edge = p[ind] + ans*2;
            if(count > 0)System.out.print(" ");
            System.out.printf("%.6f", edge - ans);
            while(ind < p.length && p[ind] <= edge + EPS)ind++;
            count++;
        }
        while(count < 3){
            System.out.printf(" %.6f", 0.0);
            count++;
        }
        System.out.println();
    }

    boolean ok(int[] p, double r){
        int count = 0;
        int ind = 0;
        while(ind < p.length && count < 3){
            double edge = p[ind] + r*2;
            count++;
            while(ind < p.length && p[ind] <= edge + EPS)ind++;
        }
        return count <= 3 && ind == p.length;
    }
}