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;
vector<long long> v;
bool fun(double x) {
  long long req = 1;
  double l = v[0];
  for (long long i = 0; i < v.size(); i++) {
    if (v[i] - l <= 2 * x) {
    } else {
      req++;
      l = v[i];
    }
  }
  return req <= 3 ? true : false;
}
void solve() {
  long long n, x;
  cin >> n;
  set<long long> second;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    second.insert(x);
  }
  for (auto &x : second) v.push_back(x);
  if (v.size() <= 3) {
    cout << 0 << "\n";
    for (long long x : v) cout << x << " ";
    for (long long i = 0; i < 3 - v.size(); i++) cout << v.back() << " ";
    return;
  }
  double hi = v.back();
  double lo = 0.0;
  long long cnt = 100;
  while (cnt--) {
    double mid = (hi + lo) / 2;
    bool val = fun(mid);
    if (val) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  double ans[3];
  ans[0] = v[0] + lo;
  ans[2] = v.back() - lo;
  double yy = v[0] + 2 * lo;
  auto it = upper_bound(v.begin(), v.end(), (long long)(0.05 + v[0] + 2 * lo));
  ans[1] = v[it - v.begin()] + lo;
  cout << lo << "\n";
  cout << ans[0] << " " << ans[1] << " " << ans[2] << "\n";
  return;
}
int main() {
  std::cout << std::fixed;
  std::cout << std::setprecision(15);
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  while (t--) solve();
  return 0;
}