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 Weapons19() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long int fun(long long int m, vector<long long int> &v) {
  long long int t = 0;
  long long int rq = 0;
  for (long long int i = 0; i < v.size(); i++) {
    if (v[i] > rq) {
      rq = v[i] + m;
      t++;
    }
  }
  if (t > 3) return 0;
  return 1;
}
int32_t main() {
  Weapons19();
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  map<long long int, long long int> z;
  for (auto a : v) z[a]++;
  if (z.size() <= 3) {
    cout << "0\n";
    for (auto a : z) cout << a.first << " ";
    for (long long int i = z.size(); i < 3; i++) cout << i + 10000034 << " ";
    return 0;
  }
  long long int s = 1, e = 1e10;
  while (s != e) {
    long long int m = (s + e) / 2;
    if (fun(m, v))
      e = m;
    else
      s = m + 1;
  }
  long double rt = s / 2.0;
  cout << fixed << setprecision(6) << rt << "\n";
  long long int rq = 0, t = 0;
  for (long long int i = 0; i < n; i++) {
    if (v[i] > rq) {
      t++;
      rq = v[i] + s;
      double p = v[i] + rt;
      cout << fixed << setprecision(6) << p << " ";
    }
  }
  while (t < 3) {
    t++;
    cout << t + 10000000 << " ";
  }
}