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.Arrays;
import java.util.Scanner;

public class B168 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        double eps = 1e-7;
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        Arrays.sort(a);
        double l = 0;
        double r = a[n - 1] - a[0];
        while (r - l > eps) {
            int st = a[0];
            int cnt = 1;
            double m = (l + r) / 2;
            boolean flag = true;
            for (int i = 0; i < n; i++) {
                if (a[i] > st + 2 * m) {
                    if (cnt != 3) {
                        cnt++;
                        st = a[i];
                    } else {
                        flag = false;
                        break;
                    }
                }
            }
            if (flag) {
                r = m;
            } else {
                l = m;
            }
        }
        System.out.printf("%.6f\n", r);
        double st = a[0];
        int cnt = 1;
        System.out.printf("%.6f ", (st + r));
        for (int i = 0; i < n; i++) {
            if (a[i] > st + 2 * r) {
                st = a[i];
                System.out.printf("%.6f ", (st + r));
                cnt++;
            }
        }
        while (cnt != 3) {
            System.out.println(0);
            cnt++;
        }
        in.close();
    }

}