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>
const int INF = (int)1e9;
using namespace std;
const int MAX = 2e5 + 5;
const double eps = 1e-7;
int n;
double a[MAX];
double ans[4];
bool valid(double val) {
  ans[0] = a[0] + val;
  double pos = a[0] + val * 2.0;
  int cnt = 1;
  for (int i = 1; i < n && cnt <= 3; i++) {
    if (a[i] - pos >= eps) {
      ans[cnt++] = a[i] + val;
      pos = a[i] + val * 2.0;
    }
  }
  return cnt <= 3;
}
double bs() {
  double l = 0, h = 1e9;
  for (int i = 0; i < 100; i++) {
    double mid = l + (h - l) / 2.0;
    if (valid(mid)) {
      h = mid;
    } else
      l = mid;
  }
  return h;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  double res = bs();
  cout << fixed << setprecision(6) << res << '\n';
  valid(res);
  for (int i = 0; i < 3; i++) cout << fixed << setprecision(6) << ans[i] << " ";
  cout << '\n';
  return 0;
}