#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  sort(v.begin(), v.end());
  if (n < 4) {
    while (v.size() < 3) v.push_back(v[0]);
    cout << 0 << endl;
    cout << v[0] << ".000000 " << v[1] << ".000000 " << v[2] << ".000000"
         << endl;
    return 0;
  }
  int l = 0, r = n - 1;
  long long ans = 2000000000;
  long long p1 = -1, p2 = -1;
  long long firstHouse = v[0];
  long long lastHouse = v[n - 1];
  while (r - l > 1) {
    long long dl = v[l] - firstHouse;
    long long dr = lastHouse - v[r];
    long long dm = v[r - 1] - v[l + 1];
    long long temp = max(dm, max(dl, dr));
    if (temp < ans) {
      p1 = l;
      p2 = r;
      ans = temp;
    }
    if (v[l + 1] - v[l] + dl < v[r] - v[r - 1] + dr)
      ++l;
    else
      --r;
  }
  cout << ans / 2;
  if (ans % 2 == 1) cout << ".5";
  cout << endl;
  long long p3 = v[p2 - 1] + v[p1 + 1];
  p1 = v[p1] + firstHouse;
  p2 = lastHouse + v[p2];
  cout << p1 / 2;
  if (p1 % 2 == 1)
    cout << ".500000 ";
  else
    cout << ".000000 ";
  cout << p3 / 2;
  if (p3 % 2 == 1)
    cout << ".500000 ";
  else
    cout << ".000000 ";
  cout << p2 / 2;
  if (p2 % 2 == 1)
    cout << ".500000";
  else
    cout << ".000000";
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  sort(v.begin(), v.end());
  if (n < 4) {
    while (v.size() < 3) v.push_back(v[0]);
    cout << 0 << endl;
    cout << v[0] << ".000000 " << v[1] << ".000000 " << v[2] << ".000000"
         << endl;
    return 0;
  }
  int l = 0, r = n - 1;
  long long ans = 2000000000;
  long long p1 = -1, p2 = -1;
  long long firstHouse = v[0];
  long long lastHouse = v[n - 1];
  while (r - l > 1) {
    long long dl = v[l] - firstHouse;
    long long dr = lastHouse - v[r];
    long long dm = v[r - 1] - v[l + 1];
    long long temp = max(dm, max(dl, dr));
    if (temp < ans) {
      p1 = l;
      p2 = r;
      ans = temp;
    }
    if (v[l + 1] - v[l] + dl < v[r] - v[r - 1] + dr)
      ++l;
    else
      --r;
  }
  cout << ans / 2;
  if (ans % 2 == 1) cout << ".5";
  cout << endl;
  long long p3 = v[p2 - 1] + v[p1 + 1];
  p1 = v[p1] + firstHouse;
  p2 = lastHouse + v[p2];
  cout << p1 / 2;
  if (p1 % 2 == 1)
    cout << ".500000 ";
  else
    cout << ".000000 ";
  cout << p3 / 2;
  if (p3 % 2 == 1)
    cout << ".500000 ";
  else
    cout << ".000000 ";
  cout << p2 / 2;
  if (p2 % 2 == 1)
    cout << ".500000";
  else
    cout << ".000000";
}

```