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;
using namespace std;
vector<long long> x;
double f(const vector<long long>::const_iterator& i, long long m) {
  if (i == x.end()) {
    return 0.0;
  }
  return min((double)(*i / 2) + (double)m / 2.0, 2e9);
}
int main() {
  cout << fixed;
  cout.precision(6);
  int n;
  cin >> n;
  x.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
    x[i] *= 2;
  }
  sort(x.begin(), x.end());
  vector<long long>::iterator e = unique(x.begin(), x.end());
  x.erase(e, x.end());
  e = x.end();
  if (x.size() <= 3) {
    cout << 0.0 << "\n";
    int r = 0;
    for (vector<long long>::iterator i = x.begin(); i != e; ++i) {
      cout << (double)(*i) / 2 << " ";
      ++r;
    }
    for (; r < 3; ++r) {
      cout << 0.0 << " ";
    }
    return 0;
  }
  long long l = 0, r = x.back() - x.front();
  vector<long long>::iterator i, j, k;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    i = lower_bound(x.begin(), e, x.front() + 2 * m + 1);
    if (i == x.end()) {
      j = k = i;
    } else {
      j = lower_bound(i, e, *i + 2 * m + 1);
      if (j == x.end()) {
        k = j;
      } else {
        k = lower_bound(j, e, *j + 2 * m + 1);
      }
    }
    if (k == e) {
      r = m;
    } else {
      l = m;
    }
  }
  long long m = r;
  i = lower_bound(x.begin(), e, x.front() + 2 * m + 1);
  if (i == e) {
    j = k = i;
  } else {
    j = lower_bound(i, e, *i + 2 * m + 1);
    if (j == x.end()) {
      k = j;
    } else {
      k = lower_bound(j, e, *j + 2 * m + 1);
    }
  }
  cout << (double)m / 2.0 << endl;
  cout << f(x.begin(), m) << " " << f(i, m) << " " << f(j, m) << "\n";
  return 0;
}