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.lang.*;
import java.io.*;
import java.util.*;
import java.math.*;

public class Solution implements Runnable{
    private static BufferedReader br = null;
    private static PrintWriter out = null;
    private static StringTokenizer stk = null;
    
    public static void main(String[] args) {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        (new Thread(new Solution())).start();
    }
    
    private void loadLine() {
        try {
            stk = new StringTokenizer(br.readLine());
        }
        catch (IOException e) {
            e.printStackTrace();
        }
    }
    
    private String nextLine() {
        try {
            return br.readLine();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }
    
    private Integer nextInt() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Integer.parseInt(stk.nextToken());
    }
    
    private Long nextLong() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Long.parseLong(stk.nextToken());
    }
    
    private String nextWord() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return (stk.nextToken());
    }
    
    private Double nextDouble() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Double.parseDouble(stk.nextToken());
    }
    
    boolean enough(double d) {
        //first
        int start = 0;
        for (int i = 1; i < arr.length; ++i) {
            if (arr[i] - arr[start] > 2*d) {
                start = i;
                break;
            }
        }
        
        if (start == 0) {
            return true;
        }
        
        int tmp = start;
        for (int i = start+1; i < arr.length; ++i) {
            if (arr[i] - arr[start] > 2*d) {
                start = i;
                break;
            }
        }
        
        if (start == tmp) {
            return true;
        }
        
        tmp = start;
        for (int i = start+1; i < arr.length; ++i) {
            if (arr[i] - arr[start] > 2*d) {
                start = i;
                break;
            }
        }
        
        if (start == tmp) {
            return true;
        }
        return false;
    }
    
    int[] arr;
    public void run() {
        int n = nextInt();
        arr = new int[n];
        for (int i = 0; i < n; ++i) {
            arr[i] = nextInt();
        }
        Arrays.sort(arr);
        
        double l = 0.0;
        double r = 2e9;
        for (int iter = 0; iter < 100; ++iter) {
            double mid = (l+r)/2;
            if (enough(mid)) {
                r = mid;
            }
            else {
                l = mid;
            }
        }
        out.printf(Locale.ENGLISH, "%.6f\n", r);
        double t1, t2, t3;
        double d = r;
        int start = 0;
        t1 = 1.0*arr[start]+d;
        for (int i = 1; i < arr.length; ++i) {
            if (arr[i] - arr[start] > 2*d) {
                start = i;
                break;
            }
        }
        
        
        int tmp = start;
        t2 = 1.0*arr[start] + d;
        for (int i = start+1; i < arr.length; ++i) {
            if (arr[i] - arr[start] > 2*d) {
                start = i;
                break;
            }
        }
        
        tmp = start;
        t3 = 1.0*arr[start]+d;
        for (int i = start+1; i < arr.length; ++i) {
            if (arr[i] - arr[start] > 2*d) {
                start = i;
                break;
            }
        }
        
        out.printf(Locale.ENGLISH, "%.6f %.6f %.6f\n", t1, t2, t3);
        out.flush();
    }
}