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 long double epsilon = 1e-11;
int main() {
  int n;
  cin >> n;
  vector<double> v(n);
  for (int i = 0; i < n; i++) scanf("%lf", &v[i]);
  sort(v.begin(), v.end());
  double step = 2 * 1e9;
  double cur = 0;
  int idx;
  int cnt = 0;
  while (step > epsilon) {
    cur += step;
    double beg = v[0];
    idx = 0;
    cnt = 1;
    while (cnt < 4 && idx < n) {
      if (v[idx] - beg - epsilon < 2.0 * cur)
        idx++;
      else {
        beg = v[idx];
        cnt++;
        idx++;
      }
    }
    if (cnt < 4 && idx == n) cur -= step;
    step /= 2.0;
  }
  cur += 1e-8;
  vector<double> ans;
  double beg = v[0];
  ans.push_back(beg + cur);
  idx = 0;
  cnt = 1;
  while (cnt < 4 && idx < n) {
    if (idx < n && v[idx] - beg - 1e-6 < 2.0 * cur)
      idx++;
    else {
      beg = v[idx];
      ans.push_back(beg + cur);
      cnt++;
      idx++;
    }
  }
  while (ans.size() < 3) ans.push_back(0.0);
  printf("%.10lf\n%.10lf %.10lf %.10lf\n", cur, ans[0], ans[1], ans[2]);
  return 0;
}