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, i;
double l, r, m, a, b, c, x[300000];
bool Less(double k) {
  a = b = c = x[0] + k;
  for (i = 0; i < n; i++) {
    if (a + k < x[i]) break;
  }
  b = x[i] + k;
  for (; i < n; i++) {
    if (b + k < x[i]) break;
  }
  c = x[i] + k;
  for (; i < n; i++) {
    if (c + k < x[i]) break;
  }
  if (i == n) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  double t = clock();
  cin >> n;
  for (i = 0; i < n; i++) cin >> x[i];
  sort(x, x + n);
  l = 0;
  r = 1000000000;
  while (clock() - t < 930.0) {
    m = (l + r) / 2.0;
    if (Less(m))
      l = m;
    else
      r = m;
  }
  Less(r);
  cout << fixed << setprecision(6) << l << endl;
  cout << fixed << setprecision(6) << a << " " << b << " " << c;
  return 0;
}