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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  std::cin >> n;
  std::vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    std::cin >> a[i];
  }
  std::sort(a.begin(), a.end());
  auto check = [&](long long dist) {
    std::vector<long long> ans;
    for (int l = 0, r = 0; l < n; l = r) {
      while (r < n && a[r] - a[l] <= dist) {
        r++;
      }
      ans.push_back((a[r - 1] + a[l]));
    }
    return ans;
  };
  long long l = 0, r = 2e9;
  while (l != r) {
    long long mid = (l + r) / 2;
    if ((int)check(mid).size() > 3) {
      l = mid + 1;
    } else {
      r = mid;
    }
  }
  auto ans = check(l);
  std::cout << l / 2 << '.' << l % 2 * 5 << "00000\n";
  while ((int)ans.size() < 3) {
    ans.push_back(1);
  }
  for (int i = 0; i < (int)ans.size(); i++) {
    std::cout << ans[i] / 2 << '.' << ans[i] % 2 * 5 << "00000"
              << (i + 1 == (int)ans.size() ? '\n' : ' ');
  }
}