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;
long long int n, m, k, ab, ac, bc, ratioa;
long long int a[200005];
vector<double> res;
double chk(double mid) {
  res.clear();
  double x = a[0] + mid;
  res.push_back(x);
  long long int ans = 1;
  for (long long int i = 1; i < n; i++) {
    if (a[i] <= x + mid)
      continue;
    else {
      ans++;
      x = a[i] + mid;
      res.push_back(x);
    }
  }
  if (ans <= 3) {
    return true;
  }
  return false;
}
void solve() {
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n == 1) {
    cout << 0 << "\n";
    cout << a[0] << " " << a[0] << " " << a[0] << "\n";
    return;
  } else if (n == 2) {
    cout << 0 << "\n";
    cout << a[0] << " " << a[0] << " " << a[1] << "\n";
    return;
  }
  sort(a, a + n);
  vector<double> v;
  double l = 0.0, h = (double)a[n - 1], mid, ans = -1, o = 100;
  while (o--) {
    mid = l + (h - l) / 2;
    if (chk(mid)) {
      v = res;
      ans = mid;
      h = mid;
    } else {
      l = mid;
    }
  }
  cout << fixed << setprecision(9) << ans << "\n";
  if (v.size() == 3) {
    for (auto it : v) {
      cout << it << " ";
    }
  } else if (v.size() == 2) {
    for (auto it : v) {
      cout << it << " ";
    }
    cout << a[0] << "\n";
  } else if (v.size() == 1) {
    for (auto it : v) {
      cout << it << " ";
    }
    cout << a[0] << "\n";
    cout << a[1] << "\n";
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int t = 1, u = 0;
  while (t--) {
    solve();
  }
}