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.*;
import java.util.*;

public class C51 {
    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;
    int[] a;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.KOREA);
        new C51().run();
    }

    boolean check(double d) {
        int n = a.length;
        double s1s = a[0];
        double s1e = a[0] + 2 * d;
        int i = 0;
        while (i < n && a[i] <= s1e) {
            i++;
        }
        double s2s = i < n ? a[i] : 0;
        double s2e = s2s + 2 * d;
        while (i < n && a[i] < s2e) {
            i++;
        }
        double s3s = i < n ? a[i] : 0;
        double s3e = s3s + d * 2;
        while (i < n && a[i] < s3e) {
            i++;
        }
        return i >= n;
    }

    void restore(double d) {
        int n = a.length;
        double s1s = a[0];
        out.printf("%.6f ", (s1s + d));
        double s1e = a[0] + 2 * d;
        int i = 0;
        while (i < n && a[i] <= s1e) {
            i++;
        }
        double s2s = i < n ? a[i] : 0;
        out.printf("%.6f ", (s2s + d));
        double s2e = s2s + 2 * d;
        while (i < n && a[i] < s2e) {
            i++;
        }
        double s3s = i < n ? a[i] : 0;
        out.printf("%.6f ", (s3s + d));
        double s3e = s3s + d * 2;
        while (i < n && a[i] < s3e) {
            i++;
        }
    }

    void binsearch() {
        double l = 0;
        double r = 1000000000;
        for (int i = 0; i < 100; i++) {
            double m = (r + l) / 2;
            if (check(m)) {
                r = m;
            } else {
                l = m;
            }
        }
        out.printf("%.6f\n", r);
        restore(r);
    }

    public void solve() throws IOException {
        int n = nextInt();
        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        Arrays.sort(a);
        binsearch();
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
}