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 maxn = 2e5 + 100;
int v[maxn];
int n;
vector<double> ans1;
bool check(double d) {
  double last = -3e9;
  int cnt = 0;
  vector<double> a;
  for (int i = 0; i < n; i++) {
    if (a.empty() || (v[i] - a.back() > d)) {
      a.push_back(v[i] + d);
    }
  }
  if (a.size() > 3) return false;
  while (a.size() < 3) a.push_back(a.back());
  ans1 = a;
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  sort(v, v + n);
  double lo = 0;
  double hi = 2e9;
  double ans = 0;
  for (int it = 0; it < 300; it++) {
    double mid = (lo + hi) / 2;
    if (check(mid)) {
      ans = mid;
      hi = mid;
    } else
      lo = mid;
  }
  printf("%.6lf\n", ans);
  sort(ans1.begin(), ans1.end());
  for (int i = 0; i < 3; i++) {
    printf("%.6lf ", ans1[i]);
  }
  puts("");
  return 0;
}