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 a[200009];
long long n;
bool check(long double d, bool code) {
  long long j = 0;
  long double last = -1;
  for (long long i = 0; i < n; i++) {
    if (last == -1 || (last + d < a[i])) {
      last = a[i] + d;
      if (code == 1) {
        cout << setprecision(29) << last << ' ';
      }
      j++;
    }
  }
  if (j > 3) return 0;
  if (code == 1) {
    for (long long i = 0; i < 3 - j; i++) {
      cout << 2000000000 << ' ';
    }
  }
  return 1;
}
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  set<long long> ss;
  for (long long i = 0; i < n; i++) {
    long long xx;
    cin >> xx;
    ss.insert(xx);
  }
  n = ss.size();
  for (long long i = 0; i < n; i++) {
    a[i] = *ss.begin();
    ss.erase(ss.begin());
  }
  if (n <= 3) {
    cout << 0 << endl;
    for (long long i = 0; i < n; i++) {
      cout << a[i] << ' ';
    }
    for (long long i = 0; i < 3 - n; i++) {
      cout << 2000000000 << ' ';
    }
    return 0;
  }
  long double l = 0, r = 1000000000;
  while (r - l > 0.000001) {
    long double m = (r + l) / 2;
    if (check(m, 0)) {
      r = m;
    } else {
      l = m;
    }
  }
  cout << setprecision(29) << r << endl;
  check(r, 1);
  return 0;
}