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;
long double x[200005];
vector<long double> v;
bool solve(long double in) {
  v.clear();
  long double p = x[0] + in;
  int cnt = 2;
  v.push_back(p);
  for (int i = 1; i < n && cnt >= 0; i++) {
    if (x[i] - p > in) {
      p = x[i] + in;
      v.push_back(p);
      cnt--;
    }
  }
  return cnt >= 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i];
  sort(x, x + n);
  long double low = 0, mid, high = x[n - 1];
  for (int i_i = 0; i_i < 100; i_i++) {
    mid = (low + high) / 2.0;
    if (solve(mid) == 1)
      high = mid;
    else
      low = mid;
  }
  cout << fixed << setprecision(6) << mid << endl;
  cout << fixed << setprecision(6) << v[0] << " "
       << v[min(1, (int)v.size() - 1)] << " " << v[min(2, (int)v.size() - 1)];
  return 0;
}