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 Main implements Runnable {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");
    private double[] x;
    private int n;
    private long t1;

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 256 * (1L << 20)).start();
    }

    public void run() {
        try {
            t1 = System.currentTimeMillis();
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            Locale.setDefault(Locale.US);
            solve();
            in.close();
            out.close();
            long t2 = System.currentTimeMillis();
            System.err.println("Time = " + (t2 - t1));
        } catch (Throwable t) {
            t.printStackTrace(System.err);
            System.exit(-1);
        }
    }

    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

    // solution

    private boolean can(double len) {
        double current = x[0];
        for (int i = 0; i < 3; i++) {
            int newIndex = Arrays.binarySearch(x, current + 2 * len);
            if (newIndex < 0) newIndex = -newIndex - 1;
            if (newIndex == n) {
                return true;
            }
            current = x[newIndex];
        }
        return false;
    }

    void solve() throws IOException {
        n = readInt();
        x = new double[n];
        for (int i = 0; i < n; i++) {
            x[i] = readDouble();
        }
        Arrays.sort(x);
        if (n <= 3) {
            out.println(0);
            for (int i = 0; i < 3; i++) {
                out.printf("%.6f ", x[Math.min(i, n - 1)]);
            }
            out.println();
        } else {
            double left = 0;
            double right = x[n - 1] - x[0];
            while (System.currentTimeMillis() - t1 < 1500) {
                double mid = (left + right) * .5;
                if (can(mid)) {
                    right = mid;
                } else {
                    left = mid;
                }
            }
            out.printf("%.6f\n", left);
            printAns(left + 1e-7);
            out.println();
        }
    }

    private void printAns(double len) {
        double current = x[0];
        for (int i = 0; i < 3; i++) {
            out.printf("%.6f ", current + len);
            int newIndex = Arrays.binarySearch(x, current + 2 * len);
            if (newIndex < 0) newIndex = -newIndex - 1;
            if (newIndex == n) {
                newIndex = n - 1;
            }
            current = x[newIndex];
        }
    }

}