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 int INF = 1000 * 1000 * 1000;
const long long LINF = 1LL * INF * INF;
const int MAX = 100010;
const long double PI = acos(-1.);
const double EPS = 1e-6;
long long mod = 1000LL * 1000 * 1000 + 7;
int x[2 * MAX];
double l1, l2, l3;
int n;
bool iscan(double d) {
  l1 = l2 = l3 = 1e9;
  int ind = 0;
  l1 = x[ind] + d;
  while (ind < n && x[ind] <= l1 + d) {
    ind++;
  }
  if (ind == n) return 1;
  l2 = x[ind] + d;
  while (ind < n && x[ind] <= l2 + d) {
    ind++;
  }
  if (ind == n) return 1;
  l3 = x[ind] + d;
  while (ind < n && x[ind] <= l3 + d) {
    ind++;
  }
  if (ind == n) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> x[i];
  sort(x, x + n);
  double l = 0.0, r = 2e9;
  for (int i = 0; i < 60; ++i) {
    double m = (l + r) / 2;
    if (iscan(m))
      r = m;
    else
      l = m;
  }
  iscan(r);
  cout << fixed << setprecision(6) << r << endl << l1 << " " << l2 << " " << l3;
  return 0;
}