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;
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << endl; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
const int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
void INITIALIZE() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  int sqrtn = sqrt(n);
  for (long long i = 2; i <= sqrtn; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void Binary(long long x) {
  for (int i = 31; i >= 0; i--) {
    if ((1 << i) & x)
      cerr << "1";
    else
      cerr << "0";
  }
  cerr << "\n";
}
long long pow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    b >>= 1;
    a = (a * a) % 1000000007;
  }
  return res % 1000000007;
}
const int mxn = 2e5 + 5;
int n;
vector<int> a(mxn);
bool check(int x) {
  int tot = 0;
  int cur = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] > cur) {
      tot++;
      cur = a[i] + x;
    }
    if (tot > 3) return false;
  }
  return tot <= 3;
}
void solve() {
  cin >> n;
  a.resize(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  int l = 0, r = 1e9;
  int ans = 0;
  sort(a.begin(), a.end());
  while (l <= r) {
    int mid = l + (r - l) / 2;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << setprecision(6) << (double)1.0 * ans / 2.0 << "\n";
  int cur = 0;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] > cur) {
      cout << setprecision(6) << (double)a[i] + ans / 2.0 << " ";
      cur = a[i] + ans;
      cnt++;
    }
  }
  while (cnt < 3) {
    cnt++;
    cout << "0 ";
  }
}
int main() {
  INITIALIZE();
  int test;
  test = 1;
  while (test--) {
    solve();
  }
  return 0;
}