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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;


public class Contest48_C {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        String []s= in.readLine().split(" ");
        int [] inp = new int [n];
        for (int i = 0; i < inp.length; i++) {
            inp[i] = Integer.parseInt(s[i]);
        }
        double min = 0;
        double max = 1000000000;
        double e = 1e-9;
        Arrays.sort(inp);
        int a = 0;
        double d11 = 0;
        double d22 = 0;
        double d33 = 0;
        double d = 0;
        while(a<100){
            a++;
            d = min + ((max-min)/2);
            double d1 = inp[0]+d;
            double d2 = d1;
            double d3 = d1;
            boolean flag = true;
            boolean flag2 = true;
            boolean can = true;
            for (int i = 1; i < inp.length; i++) {
                if(inp[i] + e >d1+d && flag && flag2) {
                    flag = false;
                    d2 = inp[i] + d;
                }else if( inp[i] + e > d2+d && !flag && flag2) {
                    d3 = inp[i] + d;
                    flag2 = false;
                }else if(inp[i] + e > d3+d && !flag && !flag2) {
                    can = false;
                    break;
                }
            }
            if(can){
                d11 = d1;
                d22 = d2;
                d33 = d3;
                max = d;
            }else {
                min = d;
            }
        }
        System.out.println(min);
        System.out.println(d11+" "+d22+" "+d33);
    }
}