#include <bits/stdc++.h>
using namespace std;
double a[200000];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int count1 = 1, count2 = 2;
  if (n == 1) {
    cout << 0 << endl << a[0] << ' ' << a[0] << ' ' << a[0];
    return 0;
  }
  if (n == 2) {
    cout << 0 << endl << a[0] << ' ' << a[0] << ' ' << a[1];
    return 0;
  }
  if (n == 3) {
    cout << 0 << endl << a[0] << ' ' << a[1] << ' ' << a[2];
    return 0;
  }
  double r = (a[n - 1] - a[count2]);
  bool doit = false;
  do {
    doit = false;
    if (count2 < n - 1) {
      if ((a[count2] - a[count1]) <= r) {
        count2++;
        doit = true;
      }
    }
    if (!doit && count2 - count1 > 1) {
      if ((a[count1] - a[0]) <= r) {
        count1++;
        doit = true;
      }
    }
    if (!doit && count2 < n - 1) {
      if ((a[count1] - a[0]) <= r && (a[count2] - a[count1 + 1]) <= r) {
        count1++;
        count2++;
        doit = true;
      }
    }
    r = max((a[count1 - 1] - a[0]),
            max((a[count2 - 1] - a[count1]), (a[n - 1] - a[count2])));
  } while (doit);
  r /= 2;
  printf("%.6f\n%.6f %.6f %.6f", r, a[0] + r, a[count1] + r, a[count2] + r);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
double a[200000];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int count1 = 1, count2 = 2;
  if (n == 1) {
    cout << 0 << endl << a[0] << ' ' << a[0] << ' ' << a[0];
    return 0;
  }
  if (n == 2) {
    cout << 0 << endl << a[0] << ' ' << a[0] << ' ' << a[1];
    return 0;
  }
  if (n == 3) {
    cout << 0 << endl << a[0] << ' ' << a[1] << ' ' << a[2];
    return 0;
  }
  double r = (a[n - 1] - a[count2]);
  bool doit = false;
  do {
    doit = false;
    if (count2 < n - 1) {
      if ((a[count2] - a[count1]) <= r) {
        count2++;
        doit = true;
      }
    }
    if (!doit && count2 - count1 > 1) {
      if ((a[count1] - a[0]) <= r) {
        count1++;
        doit = true;
      }
    }
    if (!doit && count2 < n - 1) {
      if ((a[count1] - a[0]) <= r && (a[count2] - a[count1 + 1]) <= r) {
        count1++;
        count2++;
        doit = true;
      }
    }
    r = max((a[count1 - 1] - a[0]),
            max((a[count2 - 1] - a[count1]), (a[n - 1] - a[count2])));
  } while (doit);
  r /= 2;
  printf("%.6f\n%.6f %.6f %.6f", r, a[0] + r, a[count1] + r, a[count2] + r);
  return 0;
}

```