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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int house[200020];
int n;
double a, b, c, len;
void _sort(int i, int j);
int main() {
  while (cin >> n) {
    int i, y, x;
    for (i = 0; i < n; i++) scanf("%d", &house[i]);
    sort(house, house + i);
    len = 10e9;
    for (int i = 0, j = 0; i < n; i++) {
      while (j < n - 1 && house[j] - house[i + 1] < house[n - 1] - house[j + 1])
        j++;
      _sort(i, j);
      _sort(i, j - 1);
    }
    printf("%.6lf\n%.6lf %.6lf %.6lf\n", len / 2.0, a, b, c);
  }
  return 0;
}
void _sort(int i, int j) {
  if (j < 0) j = 0;
  int z = house[i] - house[0];
  z = max(house[j] - house[min(i + 1, n - 1)], z);
  z = max(house[n - 1] - house[min(j + 1, n - 1)], z);
  if (z < len) {
    len = z;
    a = (house[i] + house[0]) / 2.0;
    b = (house[j] + house[min(i + 1, n - 1)]) / 2.0;
    c = (house[n - 1] + house[min(j + 1, n - 1)]) / 2.0;
  }
}