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 N;
int x[200000 + 5];
double pos[200000 + 5];
inline int f(double d) {
  double last = -1;
  int ret = 0;
  pos[0] = pos[1] = pos[2] = 0;
  for (int i = 0; i < N; i++) {
    if (x[i] > last + 1e-8) {
      last = x[i] + 2.0 * d;
      pos[ret] = last - d;
      ret++;
    }
  }
  return ret;
}
inline double bsearch() {
  double lo = 0.0;
  double hi = 1000111000.0;
  double mid;
  for (int k = 0; k < 100; k++) {
    mid = (lo + hi) * 0.5;
    if (f(mid) > 3) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  return mid;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &x[i]);
  }
  sort(x, x + N);
  double d = bsearch();
  printf("%.6lf\n", d);
  f(d + 1e-7);
  printf("%.6lf %.6lf %.6lf\n", pos[0], pos[1], pos[2]);
  return 0;
}