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.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class c {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        HashSet<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < n; i++) {
            set.add(sc.nextInt());
        }
        Integer[] houses = new Integer[set.size()];
        set.toArray(houses);
        Arrays.sort(houses);
        double l = 0;
        double h = 2e9;
        double mid;
        
        for(int _=0;_<100;_++) {
            mid = l + (h - l) / 2;
            double min = houses[0];
            double max = houses[0] + 2 * mid;
            int count = 1;
            boolean cor = true;
            for (int i = 0; i < houses.length; i++) {
                if (houses[i] <= max+1e-9 && houses[i]+1e-9 >= min)
                    continue;
                else {
                    if (count == 3) {
                        cor = false;
                        break;

                    }
                    count++;
                    min = houses[i];
                    max = min + 2 * mid;
                }
            }
            if (cor) {
                h = mid;
            } else
                l = mid;
        }
        double min = houses[0];
        double max = houses[0] + 2 * l;
        System.out.println(l);
        System.out.print(houses[0]+l+" ");
        int count=1;
        for (int i = 0; i < houses.length; i++) {
            if (houses[i] <= max+1e-7 && houses[i]+1e-7 >= min)
                continue;
            else {
                System.out.print((houses[i]+l)+" ");
                count++;
                min = houses[i];
                max = min + 2 * l;
            }
        }
        if(count<3) {
            for(int i=count+1;i<=3;i++)
                System.out.print(houses[houses.length-1]+" ");
        }
    }
}