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.*;

public class CF51C {
    private static final double eps = 1e-7;
    private static double[] base;
    private static int[] x;
    private static int n;

    private static boolean ok (double d) {
        base = new double[3];
        int j = 0;
        for (int i = 0; i < 3 && j < n; ++i, ++j) {
            base[i] = x[j];
            while (j+1 < n && x[j+1] - base[i] < 2*(d+eps))
                ++ j;
            base[i] = (base[i] + x[j]) / 2;
        }
        return j == n;
    }

    public static void main (String[] args) {
        Scanner scan = new Scanner(System.in);
        n = scan.nextInt();
        x = new int[n];
        for (int i = 0; i < n; ++i)
            x[i] = scan.nextInt();
        Arrays.sort(x);

        double low = 0, high = 1e9, mid = 5e8;
        boolean lastok = false;
        while (high - low >= eps || !lastok) {
            mid = (low + high) / 2;
            lastok = ok(mid);
            if (lastok)
                high = mid;
            else
                low = mid;
        }

        System.out.printf("%.6f\n", mid);
        System.out.printf("%.6f %.6f %.6f\n",
            base[0], base[1], base[2]);
    }
}