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 sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		for(int i=0; i<n; i++){
			a[i] = sc.nextInt();
		}
		
		Arrays.sort(a);
		
		int low=0, high=(int)1e9;
		
		while(true){
			int m = (low+high)/2;
			int base=0;
			int count=1;
			for(int i=0;i<n;i++){
				if(a[i]-a[base]>m){
					count++;
					base=i;
				}
			}
			
			if(count<=3){
				high=m;
			}
			else{
				low=m+1;
			}
						
			if(low==high){
				break;
			}
		}
		
		int result = low;
		System.out.println(result/2.0);
		
		int base = 0;
        int count = 1;
        for (int i = 0; i < n; i++){
            if (a[i] - a[base] > result){
                System.out.print(a[base] + result / 2.0 + " ");
                base = i;
                count++;
            }
        }
        System.out.print(a[n - 1] - result / 2.0 + " ");
        for (int i = 0; i < 3 - count; i++)
            System.out.print(0.0 + " ");
	
	}
}