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;
void print(vector<long long> v) {
  long long n = v.size();
  for (long long i = 0; i < n; i++) cout << v[i] << " ";
  cout << '\n';
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b != 0) {
    if (b & 1) res *= a;
    a = a * a;
    b /= 2;
  }
  return res;
}
bool check(double mid, vector<double> &v) {
  double x = v[0];
  long long cnt = 0;
  while (true) {
    auto it = upper_bound(v.begin(), v.end(), x + 2 * mid);
    cnt++;
    if (it == v.end()) break;
    x = *it;
  }
  if (cnt > 3) return false;
  return true;
}
double binary(vector<double> v) {
  double low = double(0), high = double(1000000000), mid;
  double err = 0.000001;
  while (high - low > err) {
    mid = low + 0.5 * (high - low);
    if (check(mid, v)) {
      high = mid;
    } else {
      low = mid;
    }
  }
  return high;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  vector<double> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  double d = binary(v);
  double x = v[0];
  vector<double> ans;
  ans.push_back(x + d);
  while (true) {
    auto it = upper_bound(v.begin(), v.end(), x + 2 * d);
    if (it == v.end()) break;
    x = *it;
    ans.push_back(x + d);
  }
  while (ans.size() < 3) {
    ans.push_back(v[n - 1]);
  }
  cout << fixed << setprecision(12);
  cout << d << '\n';
  for (long long i = 0; i < 3; i++) cout << ans[i] << " ";
  return 0;
}