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;
const int mm = 2e5 + 5;
const int oo = 1e9;
int d[mm], n;
double len = oo, a, b, c;
void jud(int x, int y) {
  if (y < 0) y = 0;
  int z;
  z = d[x] - d[0];
  z = max(d[y] - d[min(x + 1, n - 1)], z);
  z = max(d[n - 1] - d[min(y + 1, n - 1)], z);
  if (z < len) {
    len = z;
    a = d[x] + d[0];
    b = d[y] + d[min(x + 1, n - 1)];
    c = d[n - 1] + d[min(y + 1, n - 1)];
  }
}
int main() {
  int i, j;
  while (scanf("%d", &n) != EOF) {
    len = oo;
    for (i = 0; i < n; i++) scanf("%d", &d[i]);
    sort(d, d + n);
    for (i = 0, j = 0; i < n; i++) {
      while (j < n - 1 && d[n - 1] - d[j + 1] > d[j] - d[i + 1]) ++j;
      jud(i, j);
      jud(i, j - 1);
    }
    printf("%.6lf\n%.6lf %.6lf %.6lf", len / 2.0, a / 2.0, b / 2.0, c / 2.0);
  }
  return 0;
}