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;
void shikhar7s(long long int cas) {
  long long int n, i;
  cin >> n;
  long long int a;
  vector<long long int> v;
  for (i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  sort(v.begin(), v.end());
  long long int st = 0, end = 1000000000, mid, ans, j, d, c = 0;
  while (st <= end) {
    j = 0;
    c = 0;
    mid = (st + end) / 2;
    while (j < n && c < 3) {
      d = v[j] + (2 * mid);
      j = upper_bound(v.begin(), v.end(), d) - v.begin();
      c++;
    }
    if (d < v[n - 1])
      st = mid + 1;
    else {
      ans = mid;
      end = mid - 1;
    }
  }
  double an = ans - (0.5);
  j = 0;
  c = 0;
  while (j < n && c < 3) {
    d = v[j] + (2 * an);
    j = upper_bound(v.begin(), v.end(), d) - v.begin();
    c++;
  }
  if (d < v[n - 1]) an += 0.5;
  cout << setprecision(12) << an << "\n";
  j = 0;
  c = 0;
  long long int l;
  while (j < n && c < 3) {
    d = v[j] + (2 * an);
    cout << setprecision(12) << v[j] + an << " ";
    l = j;
    j = upper_bound(v.begin(), v.end(), d) - v.begin();
    c++;
  }
  while (c < 3) {
    cout << setprecision(12) << v[l] + an << " ";
    c++;
  }
}
signed main() {
  std::ios_base::sync_with_stdio(false);
  long long int t = 1;
  long long int cas = 1;
  while (cas <= t) {
    shikhar7s(cas);
    cas++;
  }
  return 0;
}