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.*;
public class maximus {
	static double pos[]=new double[3];
public static boolean check(long [] array,double mid){
	int t=0;
	long cnt=array[0];
	for(int i=0;i<array.length;i++){
		if(array[i]-cnt<=2*mid)continue;
		else {
		if(t<=2)
		pos[t]=cnt+mid;
		cnt=array[i];
		t++;	
		}
	}
	if(t<=2)pos[2]=cnt+mid;
	if(t<=2)return true;
	return false;
}	
public static void main(String []args){
	Scanner in=new Scanner(System.in);
	int n=in.nextInt();
	long array[]=new long[n];
	for(int i=0;i<n;i++)array[i]=in.nextLong();
	Arrays.sort(array);
	double high=1e9;
	double low=0;
	for(int i=1;i<=100;i++){
	double mid=(high+low)/2;
	boolean flag=check(array,mid);
	if(flag)high=mid;
	else low=mid;	
		//System.out.println(flag+" "+low+" "+high);
	}
	System.out.println(low);
	System.out.print(pos[0]+" "+pos[1]+" "+pos[2]);
}
}