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.io.*;
import java.text.*;

public class C {
	static final double epsilon = 1E-7;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();

		int[] x = new int[n];
		for (int i = 0; i < n; i++)
			x[i] = sc.nextInt();
		Arrays.sort(x);

		double lo = 0, hi = x[x.length-1];
		List<Double> stationCoords = null;
		while (lo + epsilon <= hi) {
			double mid = lo + (hi - lo) / 2;
			stationCoords = getStationCoords(x, mid);
			if (stationCoords.size() <= 3) // If it's possible to cover
				hi = mid;
			else
				lo = mid;
		}

		stationCoords = getStationCoords(x, hi);

		DecimalFormat df = new DecimalFormat("0.000000");
		System.out.println(df.format(hi));
		for (int i = 0; i < 3; i++) {
			double coord =
				stationCoords.get(Math.min(i, stationCoords.size() - 1));
			System.out.print(df.format(coord));
			if (i < 2)
				System.out.print(" ");
		}
		System.out.println();
	}

	public static List<Double> getStationCoords(int[] x, double d) {
		List<Double> stations = new ArrayList<Double>();

		/*
		System.out.println("Generating coordinates for distance " + d);
		*/

		double curStation = -2*d; //This won't cover any houses
		for (int i = 0; i < x.length; i++) {
			/* Check if this house is NOT covered by the current station. */
			if (Math.abs(curStation - x[i]) > d + epsilon) {
				/* Make a new station, exactly d away from this one. */
				/*
				System.out.println(
						"Station " + curStation +
						" does not cover " + x[i]);
						*/
				curStation = x[i] + d;
				/*
				System.out.println(
						"Adding new station at " + curStation +
						" to cover " + x[i]);
						*/
				stations.add(curStation);

				if (stations.size() == 4)
					break;
			}
		}

		return stations;
	}
}