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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int n;
vector<long long> a(200013);
bool f(long long m) {
  long long l = (long long)(a[0] + 2 * m), r = (long long)(a[n - 1] - 2 * m);
  if (l >= r) {
    return 1;
  }
  if ((r - l) <= 2 * m) {
    return 1;
  }
  long long ll =
      a[int(upper_bound(a.begin(), a.begin() + n, l) - a.begin())] + 2 * m;
  long long rr = a[int(upper_bound(a.begin(), a.begin() + n, ll) - a.begin())];
  if (rr == 0) {
    return 1;
  }
  return (rr >= r);
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] *= 2;
  }
  sort(a.begin(), a.begin() + n);
  long long l = 0, r = 2000000000;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    if (!f(m)) {
      l = m;
    } else {
      r = m;
    }
  }
  if (!f(l)) l++;
  cout.setf(ios::fixed);
  cout.precision(6);
  cout << (double)l / 2.0 << endl
       << (double)(a[0] + l) / 2.0 << ' ' << (double)(a[n - 1] - l) / 2.0
       << ' ';
  long long ll = (long long)(a[0] + 2 * l);
  long long lll = a[int(upper_bound(a.begin(), a.begin() + n, ll) - a.begin())];
  cout << (double)(lll + l) / 2.0;
  return 0;
}