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 int N = 4e5 + 5;
long long n, t, a[N], b, x, k;
set<pair<long long, long long> > s;
inline bool check(long long d) {
  long long p = 0;
  long long x = a[0] + d;
  while (a[p] <= x && p < n - 1) p++;
  long long x1 = a[p] + d;
  while (a[p] <= x1 && p < n - 1) p++;
  long long x2 = a[p] + d;
  while (a[p] <= x2 && p < n) p++;
  if (p >= n)
    return 1;
  else
    return 0;
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long r = 1e10;
  long long l = -1;
  while (r - l > 1) {
    long long mid = (l + r) / 2;
    if (check(mid) == 1) {
      r = mid;
    } else
      l = mid;
  }
  long long p = 0;
  cout << fixed << setprecision(6) << (long double)r / 2 << endl
       << (long double)a[0] + (long double)r / 2 << " ";
  while (a[p] <= a[0] + r && p < n - 1) p++;
  cout << fixed << setprecision(6) << (long double)a[x = p] + (long double)r / 2
       << " ";
  while (a[p] <= a[x] + r && p < n - 1) p++;
  cout << fixed << setprecision(6) << (long double)a[p] + (long double)r / 2
       << " ";
  return 0;
}