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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> x(n);
  for (long long i = 0; i < n; i++) cin >> x[i];
  sort(x.begin(), x.end());
  long long l = 0, r = 2e9;
  long double ans;
  while (l <= r) {
    long long mid = (l + r) / 2;
    auto it1 = upper_bound(x.begin(), x.end(), mid + x[0]);
    if (it1 != x.end()) {
      auto it2 = upper_bound(x.begin(), x.end(), mid + *it1);
      if (it2 != x.end()) {
        auto it3 = upper_bound(x.begin(), x.end(), mid + *it2);
        if (it3 != x.end()) {
          l = mid + 1;
        } else {
          ans = mid;
          r = mid - 1;
        }
      } else {
        ans = mid;
        r = mid - 1;
      }
    } else {
      ans = mid;
      r = mid - 1;
    }
  }
  printf("%0.15Lf\n", ans / 2);
  printf("%0.15Lf ", x[0] + ans / 2);
  auto it1 = upper_bound(x.begin(), x.end(), x[0] + (long long)ans);
  if (it1 != x.end()) {
    printf("%0.15Lf ", *it1 + ans / 2);
    auto it2 = upper_bound(x.begin(), x.end(), (long long)ans + *it1);
    if (it2 != x.end()) {
      printf("%0.15Lf ", *it2 + ans / 2);
    } else {
      printf("%0.15Lf ", x[0] + ans / 2);
    }
  } else {
    printf("%0.15Lf ", x[0] + ans / 2);
    printf("%0.15Lf ", x[0] + ans / 2);
  }
  return 0;
}