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

public class Pr51C {
	FastScanner in;
	PrintWriter out;

	final double eps = 1e-7;

	public void solve() throws IOException {
		int n = in.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
		}
		Arrays.sort(a);
		double l = 0;
		double r = (a[n - 1] - a[0]) / 3.0 + 1;
		while (r - l > eps) {
			double m = (l + r) / 2;
			int id = 0;
			int last = 0;
			for (int i = 0; i < 3; i++) {
				while (id < n && a[id] <= a[last] + 2 * m) {
					id++;
				}
				last = id;
			}
			if (id < n) {
				l = m;
			} else {
				r = m;
			}
		}
		out.printf("%.6f%n", r);
		int id = 0;
		int last = 0;
		for (int i = 0; i < 3; i++) {
			while (id < n && a[id] <= a[last] + 2 * r) {
				id++;
			}
			out.printf("%.6f ", last >= n ? (a[n-1]+r) : (a[last] + r));
			last = id;
		}

	}

	public void run() {
		try {
			in = new FastScanner(System.in);
			out = new PrintWriter(System.out, true);
			solve();
			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public FastScanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

	public static void main(String[] arg) {
		Locale.setDefault(Locale.US);
		new Pr51C().run();
	}
}