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>
const double pi = acos(-1.0);
const double eps = 1e-9;
using namespace std;
double pos1, pos2, pos3;
bool ok(double d, vector<int> a) {
  pos1 = a[0] + d;
  int id = 0;
  for (int i = 0; i < a.size(); i++)
    if (a[i] > pos1 + d) {
      id = i;
      break;
    }
  pos2 = a[id] + d;
  for (int i = 0; i < a.size(); i++)
    if (a[i] > pos2 + d) {
      id = i;
      break;
    }
  pos3 = a[id] + d;
  for (int i = 0; i < a.size(); i++) {
    if (pos1 - d <= a[i] && a[i] <= pos1 + d) continue;
    if (pos2 - d <= a[i] && a[i] <= pos2 + d) continue;
    if (pos3 - d <= a[i] && a[i] <= pos3 + d) continue;
    return false;
  }
  return true;
}
void solve() {
  int n;
  cin >> n;
  set<int> st;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    st.insert(x);
  }
  n = st.size();
  vector<int> a;
  while (st.size() > 0) {
    a.push_back(*st.begin());
    st.erase(*st.begin());
  }
  double l = 0, r = a.back() + 1;
  vector<double> ans;
  for (int it = 0; it < 100; it++) {
    double mid = (l + r) * 0.5;
    if (ok(mid, a)) {
      r = mid;
      ans.clear();
      ans.push_back(pos1);
      ans.push_back(pos2);
      ans.push_back(pos3);
    } else
      l = mid;
  }
  cout << r << endl;
  for (auto x : ans) cout << x << " ";
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(10);
  int step;
  step = 1;
  for (int i = 1; i <= step; i++) {
    solve();
  }
}