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 eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
template <class T>
T sqr(T x) {
  return x * x;
}
const int nmax = 300300;
int n;
double a[nmax];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) scanf("%lf", &a[i]);
  sort(a, a + n);
  double l, r;
  l = 0, r = 1e10;
  for (int i = 0; i < (int)(100); i++) {
    double m = (l + r) / 2;
    double now = a[0] - 1;
    int res = 0;
    for (int j = 0; j < (int)(n); j++)
      if (a[j] > now) {
        res++;
        now = a[j] + m * 2;
      }
    if (res <= 3)
      r = m;
    else
      l = m;
  }
  printf("%0.9lf\n", l);
  double now = a[0] - 1;
  vector<double> w;
  for (int j = 0; j < (int)(n); j++) {
    if (a[j] > now) {
      w.push_back(a[j] + l);
      now = a[j] + l * 2;
    }
  }
  while (w.size() < 3) w.push_back(0);
  sort(w.begin(), w.end());
  for (int i = 0; i < (int)(w.size()); i++) printf("%0.9lf ", w[i]);
  puts("");
  return 0;
}