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.io.*;
import java.util.*;

public class ThreeBaseStations {

    static boolean check(int[] pos, int d) {
        int j = 0;
        for (int i = 0; i < 3 && j < pos.length; ++i) {
            double l = pos[j];
            double r = l + d;
            while (j < pos.length && pos[j] <= r) {
                ++j;
            }
        }
        
        return j == pos.length;// true if d can cover the range
    }
    
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        
        int N = in.nextInt();
        int[] pos = new int[N];
        
        for (int i = 0; i < N; ++i)
            pos[i] = in.nextInt();
        
        Arrays.sort(pos);

        int lo = 0;
        int hi = pos[N - 1] - pos[0];
        while (lo < hi) {
            int mid = (lo + hi) / 2;
            if (check(pos, mid)) {
                hi = mid;
            } else {
                lo = mid + 1;
            }
        }
        
        int d = lo;
        double[] t = {pos[0], pos[0], pos[0]};
        int j = 0;
        for (int i = 0; i < 3 && j < pos.length; ++i) {
            int l = pos[j];
            double r = l + d;
            int rr = l;
            while (j < pos.length && pos[j] <= r) {
                rr = pos[j];
                ++j;
            }
            //System.out.println(l + " " + rr + " " + r + " " + d);
            t[i] = (l + rr) / 2.0;
        }
        Arrays.sort(t);
        
        System.out.format(Locale.ENGLISH, "%.6f\n", d / 2.0);
        System.out.format(Locale.ENGLISH, "%.6f %.6f %.6f\n", t[0], t[1], t[2]);
    }
    

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = new StringTokenizer("");
        }

        public String next() {
            try {
                while (!tokenizer.hasMoreTokens())
                    tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            
            return tokenizer.nextToken();
        }
        
        public String nextLine() {
            String line = null;
            try {
                tokenizer = new StringTokenizer("");
                line =  reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        
        public double nextDouble() {
            return Double.parseDouble(next());
        }
        
        public long nextLong() {
            return Long.parseLong(next());
        }
        
        boolean hasNext() {
            while (!tokenizer.hasMoreTokens()) {
                String line = nextLine();
                if (line == null) return false;
                tokenizer = new StringTokenizer(line);
            }
            return true;
        }
    }
}