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.Arrays;
import java.util.Scanner;

public class Main {

    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; ++i) {
            a[i] = 2*in.nextInt();
        }
        Arrays.sort(a);
        
        
        long[] cent = new long[3];
        
        long low = 0, high = 5000000000L;
        
        while(low != high) {
            long mid = low+(high-low)/2;   //No +1 if ...,false,true,...
            
            boolean P = true;
            long[] newCent = new long[3];            
            newCent[0] = a[0]+mid;
            
            int t = 0;
            for(int i = 0; i < a.length; ++i) {
                if(a[i] > newCent[t]+mid) {
                    if(t >= 2) {
                        P = false;
                        break;
                    }
                    newCent[++t] = a[i]+mid;
                }
            }
            
            if(P) {
                high = mid;
                cent = newCent;
                //System.out.println(mid+" "+Arrays.toString(cent));

            }
            else low = mid+1;      //low = mid+1 if ...,false,true,...
            //INVARIANT: Solution is in {low,...,high}
        }
     
        

        Arrays.sort(cent);
        
        System.out.println(String.format("%6f",0.5d*low));

        for(int i = 0; i < cent.length; ++i) {
            
            System.out.print(String.format("%6f",0.5d*cent[i]));
            if(i != cent.length-1) System.out.print(" ");
        }
    }

}