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.awt.geom.Point2D;
import java.text.*;
import java.math.*;
import java.util.*;

public class Main implements Runnable {

	final String filename = "";

	long[] y;
	int c;
	int[] x;
	int n;

	boolean test(long d) {
		c = 0;
		long r = -1;
		y = new long[3];
		for (int j = 0; j < n; j++)
			if (x[j] > r) {
				if (c < 3)
					y[c] = 2 * x[j] + d;
				c++;
				r = x[j] + d;
			}
		return c <= 3;
	}

	public void solve() throws Exception {
		n = iread();
		x = new int[n];
		for (int i = 0; i < n; i++)
			x[i] = iread();
		Arrays.sort(x);
		long left = -1, right = 1000000001;
		while (left + 1 < right) {
			long d = (left + right) / 2;

			if (test(d))
				right = d;
			else
				left = d;
		}
		DecimalFormat df = new DecimalFormat("0.000000");
		test(right);
		out.write(df.format(right / 2.0) + "\n");
		for (int i = 0; i < 3; i++) {
			out.write(df.format(y[i] / 2.0) + "");
			if (i < 2)
				out.write(" ");
			else
				out.write("\n");
		}
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new BufferedWriter(new OutputStreamWriter(System.out));
			// in = new BufferedReader(new FileReader(filename+".in"));
			// out = new BufferedWriter(new FileWriter(filename+".out"));
			solve();
			out.flush();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public int iread() throws Exception {
		return Integer.parseInt(readword());
	}

	public double dread() throws Exception {
		return Double.parseDouble(readword());
	}

	public long lread() throws Exception {
		return Long.parseLong(readword());
	}

	BufferedReader in;

	BufferedWriter out;

	public String readword() throws IOException {
		StringBuilder b = new StringBuilder();
		int c;
		c = in.read();
		while (c >= 0 && c <= ' ')
			c = in.read();
		if (c < 0)
			return "";
		while (c > ' ') {
			b.append((char) c);
			c = in.read();
		}
		return b.toString();
	}

	public static void main(String[] args) {
		try {
			Locale.setDefault(Locale.US);
		} catch (Exception e) {

		}
		new Thread(new Main()).start();
		// new Thread(null, new Main(), "1", 1<<25).start();
	}
}