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;
int32_t main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  double x[n];
  for (long long i = 0; i < n; i++) {
    cin >> x[i];
  }
  sort(x, x + n);
  double l = 0, r = INT_MAX;
  double ans = INT_MAX;
  vector<double> res;
  while (r - l > 0.00000001) {
    double mid = (l + r) / 2.0;
    double t1 = mid + x[0], t2, t3;
    long long a = upper_bound(x, x + n, t1 + mid) - x;
    if (a == n) {
      ans = mid;
      r = mid;
      continue;
    }
    t2 = mid + x[a];
    a = upper_bound(x, x + n, t2 + mid) - x;
    if (a == n) {
      ans = mid;
      r = mid;
      continue;
    }
    t3 = mid + x[a];
    a = upper_bound(x, x + n, t3 + mid) - x;
    if (a == n) {
      ans = mid;
      r = mid;
    } else {
      l = mid;
    }
  }
  cout << fixed << setprecision(6) << ans << endl;
  double t1, t2, t3;
  t1 = ans + x[0];
  long long a = upper_bound(x, x + n, t1 + ans) - x;
  if (a == n) {
    t2 = t1;
    t3 = t1;
    cout << fixed << setprecision(6) << t1 << " " << t2 << " " << t3;
    return 0;
  }
  t2 = ans + x[a];
  a = upper_bound(x, x + n, t2 + ans) - x;
  if (a == n) {
    t3 = t2;
    cout << fixed << setprecision(6) << t1 << " " << t2 << " " << t3;
    return 0;
  }
  t3 = ans + x[a];
  cout << fixed << setprecision(6) << t1 << " " << t2 << " " << t3;
}