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 = 200005, INF = 2000000000000000000;
long double pi = 3.1415926535897932, minn = 1e-9;
long long power(long long a, long long b, long long p) {
  if (a == 0) return 0;
  long long res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(std::numeric_limits<double>::max_digits10);
  ;
  long long n;
  cin >> n;
  vector<long double> v;
  for (long long i = 0; i < n; i++) {
    long double a;
    cin >> a;
    v.push_back(a);
  }
  sort((v).begin(), (v).end());
  long double l = 0, r = v[n - 1];
  long double x, y, z, ans;
  while ((r - l) >= minn) {
    long double m = (l + r) / 2.0;
    long double a1 = v[0] + m, a2, a3;
    long long f = 0;
    auto next = upper_bound((v).begin(), (v).end(), a1 + m);
    if (next == v.end()) {
      a2 = v[n - 1] - m;
      a3 = v[n - 1] - m;
    } else {
      long long pos = (long long)(next - v.begin());
      a2 = v[pos] + m;
      next = upper_bound((v).begin(), (v).end(), a2 + m);
      if (next == v.end()) {
        a3 = v[n - 1] - m;
      } else {
        pos = (long long)(next - v.begin());
        a3 = v[pos] + m;
        if (v[n - 1] > (a3 + m)) {
          f = -1;
        }
      }
    }
    if (f == -1)
      l = m;
    else {
      ans = m;
      x = a1;
      y = a2;
      z = a3;
      r = m;
    }
  }
  cout << ans << "\n" << x << " " << y << " " << z;
}