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;
const int N = 2e5 + 5;
double x, y;
int n;
vector<int> v(N);
bool ok(double d) {
  x = 0;
  y = 0;
  double last = v[0] + d;
  for (int i = 1; i < n; i++) {
    if (v[i] > last + d) {
      if (x == 0) {
        x = v[i] + d;
        last = v[i] + d;
      } else if (y == 0) {
        y = v[i] + d;
        last = v[i] + d;
      } else
        return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(6);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.begin() + n);
  double st = 0, ed = 2e9;
  double ans1, ans2;
  for (double sz = (ed - st) / 2; sz > 1e-6; sz *= 0.5) {
    if (ok(sz + st)) {
      ans1 = x;
      ans2 = y;
    } else
      st += sz;
  }
  cout << st << endl << v[0] + st << " " << ans1 << " " << ans2 << endl;
  return 0;
}