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 MainClass
{
    public static void main(String[] args)throws IOException
    {
        Reader in = new Reader();
        int n = in.nextInt();
        long[] A = new long[n];
        for (int i=0;i<n;i++)   A[i] = in.nextLong();
        new MergeSortLong().sort(A, 0, n - 1);
        double l = 0.0d;
        double r = 1000000000.0d;
        int noOfIterations = 80;
        double ans = -1;
        while (noOfIterations > 0)
        {
            double mid = 0.5 * (l + r);
            if (check(A, mid))
            {
                ans = mid;
                r = mid;
            }
            else
            {
                l = mid;
            }
            noOfIterations--;
        }
        System.out.println(ans);
        double pos1 = A[0] + ans;
        double pos2 = pos1;
        double pos3 = pos1;
        int idx1 = returnIndex(A, A[0] + 2 * ans);
        if (idx1 != n)
        {
            pos2 = A[idx1] + ans;
            int idx2 = returnIndex(A, A[idx1] + 2 * ans);
            if (idx2 != n)
                pos3 = A[idx2] + ans;
        }
        System.out.println(pos1 + " " + pos2 + " " + pos3);

    }
    public static boolean check(long[] A, double d)
    {
        int n = A.length;
        double x1 = A[0] + 2 * d;
        int idx1 = returnIndex(A, x1);
        if (idx1 == n)
            return true;
        int idx2 = returnIndex(A, A[idx1] + 2 * d);
        if (idx2 == n)
            return true;
        int idx3 = returnIndex(A, A[idx2] + 2 * d);
        if (idx3 == n)
            return true;
        return false;
    }
    public static int returnIndex(long[] A, double val)
    {
        for(int i=0;i<A.length;i++)
        {
            if(A[i] > val)
                return i;
        }
        return A.length;
    }
}
class Reader
{
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader()
    {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException
    {
        din = new DataInputStream(new FileInputStream(file_name));
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException
    {
        byte[] buf = new byte[64]; // line length
        int cnt = 0, c;
        while ((c = read()) != -1)
        {
            if (c == '\n')
                break;
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException
    {
        int ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do
        {
            ret = ret * 10 + c - '0';
        }  while ((c = read()) >= '0' && c <= '9');

        if (neg)
            return -ret;
        return ret;
    }

    public long nextLong() throws IOException
    {
        long ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');
        if (neg)
            return -ret;
        return ret;
    }

    public double nextDouble() throws IOException
    {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();

        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');

        if (c == '.')
        {
            while ((c = read()) >= '0' && c <= '9')
            {
                ret += (c - '0') / (div *= 10);
            }
        }

        if (neg)
            return -ret;
        return ret;
    }

    private void fillBuffer() throws IOException
    {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1)
            buffer[0] = -1;
    }

    private byte read() throws IOException
    {
        if (bufferPointer == bytesRead)
            fillBuffer();
        return buffer[bufferPointer++];
    }

    public void close() throws IOException
    {
        if (din == null)
            return;
        din.close();
    }
}
class MergeSortLong
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(long arr[], int l, int m, int r) {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        long L[] = new long[n1];
        long R[] = new long[n2];

        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(long arr[], int l, int r) {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
}