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.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	private static final double eps = 1e-8;
	private static int n;
	private static double[] x;

	public static void main(String[] args) throws IOException {
		InputStream fin = System.in;
//		fin = new FileInputStream(new File("in.txt"));
		Scanner cin = new Scanner(fin);

		while (cin.hasNextInt()) {
			n = cin.nextInt();
			x = new double[n];

			for (int i = 0; i < n; i++)
				x[i] = cin.nextDouble();

			Arrays.sort(x);
			if (n <= 3) {
				System.out.printf("%.6f\n", 0.0);
				for (int i = 0; i < 3; i++) {
					if (i < n)
						System.out.printf("%.6f ", x[i]);
					else
						System.out.printf("%.6f ", x[0]);
				}
				System.out.println();
				continue;
			}

			double l = 0, r = x[n - 1] - x[0], mid = 0;

			for (int i = 0; i < 100; i++) {
				mid = (l + r) / 2.0;
				if (doit(mid))
					r = mid;
				else
					l = mid;
			}

			mid = r;
			System.out.printf("%.6f\n", mid);
			r = x[0] - 1;
			int ret = 0;
			for (int i = 0; i < n; i++) {
				if (x[i] > r + eps) {
					r = x[i] + 2 * mid;
					System.out.printf("%.6f ", (x[i] + mid));
					ret++;
				}
			}
			for (int i = ret; i < 3; i++)
				System.out.printf("%.6f ", x[0]);
			System.out.println();
		}
	}

	private static boolean doit(double d) {
		int ret = 0;
		double r = x[0] - 1;
		for (int i = 0; ret <= 3 && i < n; i++) {
			if (x[i] > r + eps) {
				r = x[i] + 2 * d;
				ret++;
			}
		}
		return ret <= 3;
	}
}