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:
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF51C extends PrintWriter {
	CF51C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF51C o = new CF51C(); o.main(); o.flush();
	}

	double[] yy = new double[3];
	boolean check(int[] xx, int n, double d) {
		int i = 0;
		for (int h = 0; h < 3; h++) {
			if (i < n) {
				yy[h] = xx[i] + d;
				while (i < n && xx[i] <= yy[h] + d)
					i++;
			} else
				yy[h] = 0;
		}
		return i == n;
	}
	void main() {
		int n = sc.nextInt();
		int[] xx = new int[n];
		for (int i = 0; i < n; i++)
			xx[i] = sc.nextInt();
		xx = Arrays.stream(xx).boxed().sorted().mapToInt($->$).toArray();
		double lower = 0, upper = 2e9;
		while (upper - lower > 1e-7) {
			double d = (lower + upper) / 2;
			if (check(xx, n, d))
				upper = d;
			else
				lower = d;
		}
		check(xx, n, upper);
		printf("%f\n", upper);
		printf("%f %f %f\n", yy[0], yy[1], yy[2]);
	}
}