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 fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
double EPS = 0.0000001;
long long n, m, u, v;
vector<double> a;
map<int, int> mp1;
bool valid(double x) {
  double a1 = a[0], l = 0;
  for (int i = 0; i < 3; i++) {
    l = a1 + x * 2.0;
    if (l >= a.back()) return 1;
    vector<double>::iterator it = upper_bound((a).begin(), (a).end(), l);
    if (it == a.end()) return 1;
    a1 = *it;
  }
  return (l >= a.back());
}
double bs(double st, double en) {
  while (fabs(en - st) > EPS) {
    double mid = st + (en - st) / 2;
    if (valid(mid))
      en = mid;
    else
      st = mid;
  }
  return en;
}
int main(void) {
  fast();
  cin >> n;
  double z;
  for (int i = 0; i < n; i++) {
    cin >> z;
    if (!mp1[(int)z]) {
      a.push_back(z);
      mp1[(int)z]++;
    }
  }
  sort((a).begin(), (a).end());
  double sol = bs(0, 1e9 + 1.0);
  cout << fixed << setprecision(6) << sol << "\n";
  double a1 = a[0], l = 0.0;
  for (int i = 0; i < 3; i++) {
    cout << fixed << setprecision(6) << min(a1 + sol, a.back()) << " ";
    l = a1 + sol + sol;
    vector<double>::iterator it = upper_bound((a).begin(), (a).end(), l);
    if (it >= a.end()) it--;
    a1 = *it;
  }
}