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:
/**
 * Created by IntelliJ IDEA.
 * User: aircube
 * Date: 11.04.11
 * Time: 20:09
 * To change this template use File | Settings | File Templates.
 */

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

public class BasicStation {

    void solve() throws IOException {
        int n = nextInt();
        int[] x = new int[n];
        for (int i = 0; i < n; ++i) {
            x[i] = nextInt();
        }
        Arrays.sort(x);
        int L = 0, R = 1000000000, res = R;
        while (L <= R) {
            int m = (L + R) >> 1;
            if (check(m, x, new int[4])) {
                res = Math.min(res, m);
                R = m - 1;
            } else {
                L = m + 1;
            }
        }
        if (res % 2 == 0) {
            out.println(res / 2 + ".000000");
        } else {
            out.println(res / 2 + ".500000");
        }
        int c[] = new int[4];
        check(res, x, c);
        for (int i = 0; i < 3; ++i) {
            if (res % 2 == 0) {
                out.print((x[Math.min(c[i], n - 1)] + res / 2)+ ".000000 ");
            } else {
                out.print((x[Math.min(c[i], n - 1)] + res / 2)+ ".500000 ");
            }
        }
    }

    boolean check(int d, int [] x, int p[]) {
        p[0] = 0;
        for (int i = 0; i < 3; ++i) {
            p[i + 1] = getPos(d, p[i], x);
        }
        return p[3] == x.length;
    }
    int getPos(int d, int start, int [] x) {
        int c = start;
        while (c < x.length && x[c] - x[start] <= d) ++ c;
        return c;
    }
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;

    public static void main(String[] args) throws IOException {
        new BasicStation().run();
    }

    void run() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        st = null;
        out = new PrintWriter(System.out);
        solve();
        br.close();
        out.close();
    }

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

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

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

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