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 c51 {
	static int min;
	static int max;
	static int x[];
	static int n;
	static double ans[];

	static boolean can(double q) {
		int sum = 0;
		int next = 0;
		for (int i = 1; i < n; i++) {
			sum = x[i];
			if (sum - x[0] > 2 * q) {
				next = i;
				ans[0] = ((double) (x[next - 1] + x[0]) / 2);
				break;
			} else {
				if (i == n - 1) {
					ans[0] = (double) (max + min) / 2;
					ans[1] = (double) (max + min) / 2;
					ans[2] = (double) (max + min) / 2;
					next = n;
					return true;
				}
			}
		}
		sum = 0;
		int next1 = 0;
		if (next == n - 1) {
			ans[1] = x[n - 1];
			return true;
		}
		for (int i = next + 1; i < n; i++) {
			sum = x[i];
			if (sum - x[next] > 2 * q) {
				next1 = i;
				ans[1] = ((double) (x[next1 - 1] + x[next]) / 2);
				break;
			} else {
				if (i == n - 1) {
					ans[1] = (double) ((double) (x[n - 1] + x[next]) / 2);
					;
					ans[2] = (double) ((double) (x[n - 1] + x[next]) / 2);
					next1 = n;
					return true;
				}
			}
		}
		sum = 0;
		if (next1 == n - 1) {	
			ans[2] = x[n - 1];
			return true;
		}
		for (int i = next1 + 1; i < n; i++) {
			sum = x[i];
			if (sum - x[next1] > 2 * q) {
				return false;
			} else {
				if (i == n - 1) {
					ans[2] = ((double) (x[n - 1] + x[next1]) / 2);
				}
			}
		}
		return true;

	}

	public static void main(String[] args) throws FileNotFoundException {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		n = in.nextInt();
		x = new int[n];
		ans = new double[3];
		min = 1000000000;
		max = 0;
		for (int i = 0; i < n; i++) {
			x[i] = in.nextInt();
			max = Math.max(max, x[i]);
			min = Math.min(min, x[i]);
		}
		Arrays.sort(x);
		double l = 0;
		double r = max - min;
		for (int i = 0; i < 200; i++) {
			if (Math.abs(l - r) < 0.000001) {
				break;
			}
			double mid = (double) (l + r) / 2;
			if (can(mid)) {
				r = mid;
			} else {
				l = mid;
			}
		}
		if (n <= 3) {
			out.println(0);
			for (int i = 0; i < n; i++) {
				out.print(x[i] + " ");
			}
			for (int i = 0; i < 3 - n; i++) {
				out.print(x[n - 1] + " ");
			}
		} else {
			if (can(l)) {
				out.println(l);
			} else {
				boolean q = can(r);
				out.println(r);
			}
			out.print(ans[0] + " " + ans[1] + " " + ans[2]);
		}
		out.close();
	}
}