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.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Zyflair Griffane
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		PandaScanner in = new PandaScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		C solver = new C();
		solver.solve(1, in, out);
		out.close();
	}
}

class C implements BinarySearchable{
    TreeSet<Long> positions = new TreeSet<>();
    long[] placed = new long[3];
    public void solve(int testNumber, PandaScanner in, PrintWriter out) {
        int n = in.nextInt();
        for (int i = 0; i < n; i++) {
            positions.add((long) in.nextInt() << 1);
        }
        int res = BinarySearchUtil.minIntSearch(0, 1_000_000_000, this);
        solves(res);
        Arrays.sort(placed);
        out.printf("%.6f\n", res / 2.0);
        out.printf("%.6f %.6f %.6f\n", placed[0] / 2.0, placed[1] / 2.0, placed[2] / 2.0);
    }
    public boolean solves(int distance) {
        Long at = Long.MIN_VALUE;
        for (int i = 0; i < 3; i++) {
            at = positions.higher(at);
            if (at == null) {
                return true;
            }
            placed[i] = at + distance;
            at = positions.floor(placed[i] + distance);
        }
        return at == positions.last();
    }
}

interface BinarySearchable {
    public boolean solves(int mid);
}

class PandaScanner {
    public BufferedReader br;
    public StringTokenizer st;
    public InputStream in;

    public PandaScanner(InputStream in) {
        br = new BufferedReader(new InputStreamReader(this.in = in));
    }

    public String nextLine() {
        try {
            return br.readLine();
        }
        catch (Exception e) {
            return null;
        }
    }

    public String next() {
        if (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(nextLine().trim());
            return next();
        }
        return st.nextToken();
    }

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

    }

class BinarySearchUtil {

    public static int minIntSearch(int low, int high, BinarySearchable solver) {
        int res = high;
        while (low <= high) {
            int mid = low + ((high - low) >> 1);
            if (solver.solves(mid)) {
                res = mid;
                high = mid - 1;
            }
            else {
                low = mid + 1;
            }
        }
        return res;
    }

    }