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>
template <class T>
using min_q = std::priority_queue<T, std::vector<T>, std::greater<T>>;
int comp_double(double a, double b) {
  if (fabs(a - b) <= (1e-10)) return 0;
}
using namespace std;
int is_C_S(char f) {
  if (f >= 'a' && f <= 'z') return (f - 'a') + 1;
  return (f - 'A') + 27;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, -1, 1, 1, -1, -1, 1};
template <class T>
T fast_power(T a, T p) {
  if (p == 0) return 1;
  return (((p & 1) ? a : 1) * fast_power(a * a, p / 2));
}
double _3_pos[4];
int h_p[200010];
int n;
bool check(double d) {
  int l = 0;
  int cnt = 0;
  while (l < n) {
    double val = h_p[l] + d;
    while (h_p[l] <= val + d && l < n) l++;
    _3_pos[cnt++] = val;
    if (l >= n) return 1;
    if (cnt == 3) return 0;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < (int)n; i++) cin >> h_p[i];
  sort(h_p, h_p + n);
  double lo = 0, hi = 1e10;
  for (int i = 0; i < (int)1000; i++) {
    double mid = (lo + hi) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid;
  }
  cout << fixed << setprecision(6) << hi << "\n";
  for (int i = 0; i < (int)3; i++) {
    cout << fixed << setprecision(6) << _3_pos[i] << " ";
  }
  cout << endl;
  return 0;
}