#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
long long int power(long long int a, long long int b,
                    long long int m = 1000000007) {
  long long int ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1LL * a * ans) % m;
    b >>= 1;
    a = (1LL * a * a) % m;
  }
  return ans;
}
long long int ncr(long long int n, long long int r) {
  long long int res = 1;
  if (r > n - r) r = n - r;
  for (long long int i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
bool check(double d, vector<double> &a) {
  double f = a.front();
  for (long long int i = 0; i < 3; i++) {
    auto it = lower_bound(a.begin(), a.end(), f + 2 * d);
    if (it == a.end())
      return true;
    else {
      if (*it > f + 2 * d)
        f = *it;
      else {
        it++;
        if (it == a.end())
          return true;
        else
          f = *it;
      }
    }
  }
  return false;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int tt = 1;
  long long int ii = 0;
  while (tt-- && ++ii) {
    long long int n;
    cin >> n;
    set<long long int> temp;
    for (long long int i = 0; i < n; i++) {
      long long int x;
      cin >> x;
      temp.insert(x);
    }
    vector<double> a;
    for (auto x : temp) a.emplace_back(x);
    double lo = 0, hi = a.back();
    while ((hi - lo) > 1e-7) {
      double mid = (hi + lo) / 2;
      if (check(mid, a))
        hi = mid;
      else
        lo = mid;
    }
    double val = hi;
    cout << fixed << setprecision(7) << val << "\n";
    double f = a.front();
    for (long long int i = 0; i < 3; i++) {
      cout << f + val << " ";
      auto it = lower_bound(a.begin(), a.end(), f + 2 * val);
      if (*it - (f + 2 * val) > 1e-7)
        f = *it;
      else {
        it++;
        if (it == a.end())
          f = a.back();
        else
          f = *it;
      }
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
long long int power(long long int a, long long int b,
                    long long int m = 1000000007) {
  long long int ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1LL * a * ans) % m;
    b >>= 1;
    a = (1LL * a * a) % m;
  }
  return ans;
}
long long int ncr(long long int n, long long int r) {
  long long int res = 1;
  if (r > n - r) r = n - r;
  for (long long int i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
bool check(double d, vector<double> &a) {
  double f = a.front();
  for (long long int i = 0; i < 3; i++) {
    auto it = lower_bound(a.begin(), a.end(), f + 2 * d);
    if (it == a.end())
      return true;
    else {
      if (*it > f + 2 * d)
        f = *it;
      else {
        it++;
        if (it == a.end())
          return true;
        else
          f = *it;
      }
    }
  }
  return false;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int tt = 1;
  long long int ii = 0;
  while (tt-- && ++ii) {
    long long int n;
    cin >> n;
    set<long long int> temp;
    for (long long int i = 0; i < n; i++) {
      long long int x;
      cin >> x;
      temp.insert(x);
    }
    vector<double> a;
    for (auto x : temp) a.emplace_back(x);
    double lo = 0, hi = a.back();
    while ((hi - lo) > 1e-7) {
      double mid = (hi + lo) / 2;
      if (check(mid, a))
        hi = mid;
      else
        lo = mid;
    }
    double val = hi;
    cout << fixed << setprecision(7) << val << "\n";
    double f = a.front();
    for (long long int i = 0; i < 3; i++) {
      cout << f + val << " ";
      auto it = lower_bound(a.begin(), a.end(), f + 2 * val);
      if (*it - (f + 2 * val) > 1e-7)
        f = *it;
      else {
        it++;
        if (it == a.end())
          f = a.back();
        else
          f = *it;
      }
    }
  }
  return 0;
}

```