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.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.math.*;
import static java.lang.Math.*;

public class Main implements Runnable {

    boolean in(double left, double right, double c) {
        return c >= left && c <= right;
    }

    boolean can(int[] a, double c, double[] wh) {
        int n = a.length;
        wh[0] = a[0] + c;
        wh[1] = wh[0];
        wh[2] = a[a.length - 1] - c;
        double minBad = Double.MAX_VALUE;
        double maxBad = Double.MIN_VALUE;
        for(int i = 0; i < n; ++i) {
            if (!in(wh[0] - c, wh[0] + c, a[i]) && !in(wh[2] - c, wh[2] + c, a[i])) {
                minBad = Math.min(minBad, a[i]);
                maxBad = Math.max(maxBad, a[i]);
            }
        }
        if (minBad == Double.MAX_VALUE)
            return true;
        if (maxBad - minBad <= 2 * c) {
            wh[1] = (maxBad + minBad) / 2;
            return true;
        }
        return false;
    }

    public void solve() throws IOException {
        int n = nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; ++i)
            a[i] = nextInt();
        Arrays.sort(a);
        double l = 0, r = a[a.length - 1] - a[0];
        double[] wh = new double[3];
        for(int i = 0; i < 100; ++i) {
            double c = (l + r) / 2;
            if (can(a, c, wh))
                r = c;
            else
                l = c;
        }
        can(a, r, wh);
        pw.printf(Locale.US, "%1.8f\n%1.8f %1.8f %1.8f", r, wh[0], wh[1], wh[2]);
    }

    static final String filename = "A";
    static final boolean fromConsole = true;

    public void run() {
        try {
            if (!fromConsole) {
                in = new BufferedReader(new FileReader(filename + ".in"));
                pw = new PrintWriter(filename + ".out");
            } else {
                in = new BufferedReader(new InputStreamReader(System.in));
                pw = new PrintWriter(System.out);
            }
            st = new StringTokenizer("");
            long st = System.currentTimeMillis();
            solve();
            //pw.printf("\nWorking time: %d ms\n", System.currentTimeMillis() - st);
            pw.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    private StringTokenizer st;
    private BufferedReader in;
    private PrintWriter pw;

    boolean hasNext() throws IOException {
        while (!st.hasMoreTokens()) {
            String line = in.readLine();
            if (line == null) {
                return false;
            }
            st = new StringTokenizer(line);
        }
        return st.hasMoreTokens();
    }

    String next() throws IOException {
        return hasNext() ? st.nextToken() : null;
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    BigInteger nextBigInteger() throws IOException {
        return new BigInteger(next());
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public static void main(String[] args) {
        new Thread(new Main()).start();
    }
}