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.math.*;
import java.io.*;

public class c {
	
	int n;
	int[] x;
	double[] place;

	private void solve() throws Exception {
		n = nextInt();
		x = new int[n];
		for (int i = 0; i < n; ++i)
			x[i] = nextInt();
		Arrays.sort(x);
		place = new double[3];
		double lo = 0, hi = 1000000000;
		for (int it = 0; it < 200; ++it) {
			double mid = (lo + hi) / 2.;
			if (canDo(mid))
				hi = mid;
			else
				lo = mid;
		}
		double ans = hi;
		//System.err.println(ans);
		out.printf("%.8f\n", ans);
		canDo(ans);
		Arrays.sort(place);
		for (double d : place)
			out.printf("%.6f ", d);
	}

	private boolean canDo(double d) {
		int cur = 0;
		for (int i = 0; i < 3; ++i) {
			if (cur >= n)
				place[i] = 42.;
			else
				place[i] = d + x[cur];
			while (cur < n && x[cur] < place[i] + d + 1e-10)
					++cur;
		}
		return cur == n;
	}

	public void run() {
		try {
			solve();
		} catch (Exception e) {
			NOO(e);
		} finally {
			out.close();
		}
	}

	PrintWriter out;
	BufferedReader in;
	StringTokenizer St;

	void NOO(Exception e) {
		e.printStackTrace();
		System.exit(1);
	}

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

	long nextLong() {
		return Long.parseLong(nextToken());
	}

	double nextDouble() {
		return Double.parseDouble(nextToken());
	}

	String nextToken() {
		while (!St.hasMoreTokens()) {
			try {
				String line = in.readLine();
				St = new StringTokenizer(line);
			} catch (Exception e) {
				NOO(e);
			}
		}
		return St.nextToken();
	}

	private c(String name) {
		try {
			in = new BufferedReader(new FileReader(name + ".in"));
			St = new StringTokenizer("");
			out = new PrintWriter(new FileWriter(name + ".out"));
		} catch (Exception e) {
			NOO(e);
		}
	}

	private c() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			St = new StringTokenizer("");
			out = new PrintWriter(System.out);
		} catch (Exception e) {
			NOO(e);
		}
	}

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

}