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

public class C {
	public static void main(String[] args) throws Exception{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.valueOf(in.readLine().trim());
		int[] V = new int[N];
		StringTokenizer stok = new StringTokenizer(in.readLine());
		for(int i = 0; i < N;i++)
			V[i] = Integer.valueOf(stok.nextToken())*2;
		Arrays.sort(V);
		long low = 0;
		long high = 1000000000000L;
		ans = new long[3];
		while(low < high){
			long mid = (low+high)/2;
			if(pos(mid, V)){
				high = mid;
			}else{
				low = mid+1;
			}
		}
		pos(high, V);
		System.out.format("%.06f\n",high/2.0);
		System.out.format("%.06f %.06f %.06f\n", ans[0]/2.0, ans[1]/2.0, ans[2]/2.0);
	}
	static long[] ans;
	private static boolean pos(long D, int[] V) {
		int at = 0;
		for(int v:V){
			if(at > 0 && abs(ans[at-1]-v) <= D)
				continue;
			if(at == 3)
				return false;
			ans[at++] = v+D;
		}
		return true;
	}
}