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.*;
import static java.lang.Math.*;

public class Solution {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st;

    int n;
    TreeSet<Long> set = new TreeSet<Long>();

    long start;

    long f, s, t;

    boolean can(long d) {
        int cnt = 1;
        long cur = start;
        f = s = t = 0;
        while (true) {
            if (cnt == 4)
                break;
            if (cnt == 1)
                f = cur + d;
            else if (cnt == 2)
                s = cur + d;
            else if (cnt == 3)
                t = cur + d;
            Long next = set.higher(cur + 2L * d);
            if (next == null)
                break;
            cur = next;
            ++cnt;
        }
        return cnt <= 3;
    }

    void solve() throws IOException {
        n = ni();
        for (int i = 0; i < n; ++i) {
            int a = ni() * 2;
            set.add((long) a);
        }
        start = set.first();
        long l = 0, r = 2000000000L;
        while (l + 1 < r) {
            long m = (l + r) >> 1L;
            if (can(m))
                r = m;
            else
                l = m;
        }
        if (!can(l)) {
            l = r;
            can(l);
        }
        double d = (double) l / 2.0;
        double a = (double) f / 2.0;
        double b = (double) s / 2.0;
        double c = (double) t / 2.0;
        out.printf("%.6f\n", d);
        out.printf("%.6f %.6f %.6f", a, b, c);
    }

    public Solution() throws IOException {
        Locale.setDefault(Locale.US);
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

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

    int ni() throws IOException {
        return Integer.valueOf(ns());
    }

    long nl() throws IOException {
        return Long.valueOf(ns());
    }

    double nd() throws IOException {
        return Double.valueOf(ns());
    }

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