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;
template <typename T>
inline T Abs(T a) {
  return a > 0 ? a : -a;
}
int x[1000000];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  sort(x, x + n);
  double l = 0;
  double r = x[n - 1];
  double x1, x2, x3;
  double m;
  while (r - l > 1e-6) {
    m = (l + r) / 2.0;
    x1 = x[0] + m;
    x2 = x1;
    x3 = x1;
    int next = upper_bound(x, x + n, (int)(x[0] + 2 * m)) - x;
    if (next != n) {
      x2 = x[next] + m;
      next = upper_bound(x, x + n, (int)(x[next] + 2 * m)) - x;
      x3 = x2;
      if (next != n) {
        x3 = x[next] + m;
      }
    }
    if (x3 + m >= x[n - 1])
      r = m;
    else
      l = m;
  }
  m = r;
  x1 = x[0] + m;
  x2 = x1;
  x3 = x1;
  int next = upper_bound(x, x + n, (int)(x[0] + 2 * m)) - x;
  if (next != n) {
    x2 = x[next] + m;
    next = upper_bound(x, x + n, (int)(x[next] + 2 * m)) - x;
    x3 = x2;
    if (next != n) {
      x3 = x[next] + m;
    }
  }
  printf("%.6lf\n", m);
  printf("%.6lf %.6lf %.6lf", x1, x2, x3);
  return 0;
}