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.util.Arrays;
import java.util.Scanner;


public class C51 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] x = new int[n];
		for (int i = 0; i < n; i++) {
			x[i] = in.nextInt();
		}
		Arrays.sort(x);
		int x0 = x[0];
		for (int i = 0; i < n; i++) {
			x[i] -= x0;
			x[i] *= 2;
		}
//		System.out.println(Arrays.toString(x));
		int L = -1;
		int R = x[n - 1];
//		System.out.println(R);
		while (R - L > 1) {
			int M = (L + R) / 2;
			int counter = 1;
			int left = 0;
			for (int i = 0; i < n; i++) {
				if (x[i] - left > 2 * M) {
					left = x[i];
					counter++;
				}
			}
//			System.out.println(M + " " + counter);
			if (counter <= 3) {
				R = M;
			} else {
				L = M;
			}
		}
		System.out.println(R % 2 == 0? R / 2 : R / 2 + ".5");
		int left = 0;
		int counter = 0;
		for (int i = 0; i < n; i++) {
			if (x[i] - left > 2 * R) {
				System.out.print(((x[i - 1] + left) % 4 == 0? (x[i - 1] + left) / 4 + x0 : (x[i - 1] + left) / 4 + x0 + ".5") + " ");
				left = x[i];
				counter++;
			}
		}
		for (;counter < 3; counter++) {
			System.out.print(((x[n - 1] - R) % 2 == 0?  (x[n - 1] - R) / 2 + x0 : (x[n - 1] - R) / 2 + x0 + ".5")  + " ");
		}
	}
}