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() {
  int N;
  while (scanf("%d", &N) != EOF) {
    set<double> st;
    double lowest = 1e12, highest = 0;
    for (int j = 0; j < (int)(N); ++j) {
      int x;
      scanf("%d", &x);
      st.insert((double)x);
      lowest = min(lowest, (double)x);
      highest = max(highest, (double)x);
    }
    double lo = 0, hi = 2e9;
    double center = lowest;
    for (int itr = 0; itr < (int)(80); ++itr) {
      double mid = (lo + hi) / 2.0;
      double eps = 1e-9;
      set<double>::iterator lower = st.lower_bound(lowest + mid * 2 + eps);
      set<double>::iterator upper = st.lower_bound(highest - mid * 2 - eps);
      if (lower == st.end() || upper == st.begin()) {
        hi = mid;
        continue;
      }
      upper--;
      if ((*upper) - (*lower) < mid * 2 + eps) {
        center = ((*upper) + (*lower)) / 2.0;
        hi = mid;
      } else
        lo = mid;
    }
    printf("%.6f\n", lo);
    printf("%.6f %.6f %.6f\n", lowest + lo, center, highest - lo);
  }
}