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 long N = 2e5 + 10;
long long n, ar[N];
vector<long double> vec;
bool check(long double x) {
  vec.clear();
  long long cnt = 1, indx = 0;
  long double last = (long double)ar[0] + 2.0 * x;
  vec.push_back((long double)ar[0] + x);
  while (indx < n) {
    while (indx < n && (long double)ar[indx] <= last) indx++;
    if (indx < n) {
      cnt++;
      last = (long double)ar[indx] + 2.0 * x;
      vec.push_back((long double)ar[indx] + x);
    }
  }
  return cnt < 4;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> ar[i];
  sort(ar, ar + n);
  if (n < 4) {
    cout << 0 << endl;
    for (long long i = 0; i < 3; i++) cout << ar[i] << ' ';
    return 0;
  }
  long double st = 0.0, ed = (long double)1e9;
  for (long long i = 0; i < 100; i++) {
    long double mid = (long double)(st + ed) / 2.0;
    if (check(mid))
      ed = mid;
    else
      st = mid;
  }
  check(ed);
  while (vec.size() < 3) vec.push_back(0.0);
  cout << fixed << setprecision(10) << ed << '\n';
  for (long double x : vec) cout << x << ' ';
}