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 double pi = 3.1415926535897932384626433;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
vector<long double> q(3);
long long n;
vector<long double> a;
bool check(long double d) {
  long long i = 0;
  long long c = 3;
  long long k = 0;
  q.assign(3, 0);
  while (i < n and c--) {
    long double tc = a[i] + d;
    q[c] = tc;
    k++;
    long long ind = upper_bound(a.begin(), a.end(), a[i] + 2 * d) - a.begin();
    i = ind;
  }
  if (i >= n) {
    return true;
  } else
    return false;
}
void solve() {
  cin >> n;
  a.resize(n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  long double lo = 0;
  long double hi = 2e9;
  vector<long double> ansv;
  long double ans = -1;
  long long cnt = 80;
  while (lo <= hi or cnt--) {
    long double mid = lo + (hi - lo) / 2;
    if (check(mid)) {
      ansv = {q[0], q[1], q[2]};
      ans = mid;
      hi = mid - 0.000001;
    } else {
      lo = mid + 0.000001;
    }
  }
  cout << fixed << setprecision(10);
  cout << abs(ans) << "\n";
  cout << ansv[0] << " " << ansv[1] << " " << ansv[2] << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}