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;
int N;
int X[200002];
double T[3];
int f(double d) {
  int i = 0;
  int lst = X[i];
  T[0] = lst + d;
  while (i < N && X[i] <= T[0] + d) i++;
  if (i == N) return 1;
  lst = X[i];
  T[1] = lst + d;
  while (i < N && X[i] <= T[1] + d) i++;
  if (i == N) return 1;
  lst = X[i];
  T[2] = lst + d;
  while (i < N && X[i] <= T[2] + d) i++;
  if (i == N) return 1;
  return 0;
}
int main(int argc, char **argv) {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> X[i];
  sort(X, X + N);
  cout << fixed << setprecision(12);
  if (f(0)) {
    cout << 0.0 << endl;
    for (int i = 0; i < 3; i++) cout << T[i] << " ";
    cout << endl;
    return 0;
  }
  f(0.5);
  double l = 0.0, r = (double)1e9;
  for (int i = 0; i < 100; i++) {
    double m = (l + r) / 2;
    if (f(m))
      r = m;
    else
      l = m;
  }
  cout << r << endl;
  for (int i = 0; i < 3; i++) cout << T[i] << " ";
  cout << endl;
  return 0;
}