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;
bool isCovered(std::vector<long long>& houses, double d) {
  double left_upper = houses[0] + 2 * d;
  double right_lower = houses[houses.size() - 1] - 2 * d;
  std::vector<long long>::iterator lower, upper;
  lower = upper_bound(houses.begin(), houses.end(), left_upper);
  upper = lower_bound(houses.begin(), houses.end(), right_lower);
  upper = upper - 1;
  return ((*upper - *lower) <= 2 * d);
}
int main() {
  int N;
  cin >> N;
  std::vector<long long> houses;
  for (int i = 0; i < N; i++) {
    int input;
    cin >> input;
    houses.push_back(input);
  }
  sort(houses.begin(), houses.end());
  double low, high, mid;
  low = 0;
  high = double(houses[houses.size() - 1] - houses[0]) / 6.0;
  while (high - low > 1e-6) {
    mid = (high + low) / 2;
    if (isCovered(houses, mid)) {
      high = mid;
    } else {
      low = mid;
    }
  }
  mid = (high + low) / 2;
  cout << fixed << showpoint << setprecision(6) << mid << endl;
  double mid_base;
  long long left_upper = long(houses[0] + 2 * mid + 0.5);
  long long right_lower = long(houses[houses.size() - 1] - 2 * mid + 0.5);
  std::vector<long long>::iterator lower, upper;
  lower = upper_bound(houses.begin(), houses.end(), left_upper);
  upper = lower_bound(houses.begin(), houses.end(), right_lower);
  if (upper != houses.begin()) upper = upper - 1;
  if (lower == houses.end()) lower = upper;
  mid_base = double(*upper + *lower) / 2.0;
  cout << double(houses[0] + mid) << " " << mid_base << " "
       << double(houses[houses.size() - 1] - mid) << endl;
  return 0;
}