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 double pi = acos(-1.0);
const long long MOD = 1000007;
const double eps = 1E-8;
long long a[200010];
long long n;
double solve(double l, double r) {
  if (r - l < eps) return r;
  double d = (l + r) / 2;
  long long counter = 0;
  double last = -1;
  for (long long i = 0; i <= n - 1; i++)
    if (a[i] > last) {
      ++counter;
      last = a[i] + 2 * d;
    }
  if (counter > 3)
    return solve(d, r);
  else
    return solve(l, d);
}
int main() {
  cin >> n;
  for (long long i = 0; i <= n - 1; i++) cin >> a[i];
  sort(a, a + n);
  double d = solve(0, a[n - 1]);
  printf("%.6lf\n", d);
  double last = -1;
  int counter = 0;
  double temp;
  for (long long i = 0; i <= n - 1; i++) {
    if (a[i] > last) {
      ++counter;
      temp = a[i] + d;
      last = a[i] + d * 2;
      printf("%.6lf", temp);
      if (counter == 3)
        cout << endl;
      else
        cout << " ";
    }
  }
  while (counter < 3) {
    ++counter;
    printf("%.6lf", temp);
    if (counter == 3)
      cout << endl;
    else
      cout << " ";
  }
}