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;
int n, k;
vector<int> v;
bool check(double mid) {
  double last = v[0] + 2 * mid;
  int cnt = 1;
  for (int i = 1; i < v.size(); i++) {
    if (v[i] > last) {
      cnt++;
      last = v[i] + 2 * mid;
    }
  }
  if (cnt <= 3) return 1;
  return 0;
}
void print(double mid) {
  double last = v[0] + 2 * mid;
  printf("%.6lf", v[0] + mid);
  int cnt = 1;
  for (int i = 1; i < v.size(); i++) {
    if (v[i] > last) {
      cnt++;
      last = v[i] + 2 * mid;
      printf(" %.6lf", v[i] + mid);
    }
  }
  while (cnt < 3) {
    printf(" %.6lf", (double)v[0]);
    cnt++;
  }
}
int main() {
  cin >> n;
  set<int> s;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (s.find(a) == s.end()) {
      s.insert(a);
      v.push_back(a);
    }
  }
  sort(v.begin(), v.end());
  if (v.size() <= 3) {
    cout << 0 << endl;
    print(0);
  } else {
    int l = 0, r = 2 * 1e9;
    double ans = 0;
    while (l < r) {
      double mid = (l + r) / 2.0;
      if (check(mid)) {
        r = mid;
        ans = mid;
      } else {
        l = mid + 1;
      }
    }
    ans = max(0.0, ans + 1);
    while (check(ans)) ans -= .5;
    ans += .5;
    printf("%.6lf\n", ans);
    print(ans);
  }
}