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;
const int N = 5e5 + 5;
int n;
int ar[N];
bool valid(double g) {
  double last = -2e9, station = 0;
  for (int i = 1; i <= n; i++) {
    if (last + g * 2 < ar[i]) {
      last = ar[i];
      station++;
    }
  }
  return station <= 3;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
  }
  sort(ar + 1, ar + n + 1);
  int counter = 400;
  double lef = 0, R = 2e9;
  while (counter--) {
    double mid = lef + (R - lef) / 2;
    if (valid(mid)) {
      R = mid;
    } else {
      lef = mid;
    }
  }
  cout << fixed << setprecision(6) << R << '\n';
  double last = -2e9;
  vector<double> p;
  for (int i = 1; i <= n; i++) {
    if (last + 2 * R < ar[i]) {
      last = ar[i];
      p.push_back(ar[i] + R);
    }
  }
  for (int i = 0; i < 3; i++) cout << p[i] << " ";
  return 0;
}