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>
int main() {
  int n;
  std::cin >> n;
  auto comp = [](const double a, const double b) -> bool {
    return a < b - (1e-9);
  };
  std::set<double, decltype(comp)> set(comp);
  for (int i = 0; i < n; ++i) {
    double a = 0.;
    scanf("%lf", &a);
    set.insert(a);
  }
  double l = 0.;
  double r = 2. * 1e+9;
  double x = 0.;
  double oldX = 0.;
  for (int i = 0; i < 256; ++i) {
    x = (l + r) * 0.5;
    if (fabs(oldX - x) < 1e-9) break;
    oldX = x;
    auto it_1 = set.upper_bound((*set.begin() + 2. * x));
    if (it_1 != set.end()) {
      auto it_2 = set.upper_bound((*it_1 + 2. * x));
      if (it_2 != set.end()) {
        auto it_3 = set.upper_bound((*it_2 + 2. * x));
        if (it_3 != set.end())
          l = x;
        else
          r = x;
      } else
        r = x;
    } else
      r = x;
  }
  std::cout.precision(6);
  std::cout << std::fixed << x << '\n';
  std::cout << std::fixed << *set.begin() + x << ' ';
  auto it_1 = set.upper_bound(round(*set.begin() + 2. * x));
  if (it_1 == set.end()) {
    std::cout << std::fixed << *set.begin() + x << ' ';
    std::cout << std::fixed << *set.begin() + x << ' ';
  } else {
    std::cout << std::fixed << *it_1 + x << ' ';
    auto it_2 = set.upper_bound(round(*it_1 + 2. * x));
    if (it_2 == set.end())
      std::cout << std::fixed << *it_1 + x;
    else
      std::cout << std::fixed << *it_2 + x;
  }
  return 0;
}