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 main() {
  int n;
  cin >> n;
  vector<long long> v;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    v.push_back(x);
  }
  sort(v.begin(), v.end());
  long long lo = 0, hi = 1000000000;
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    int pos = 0;
    for (int i = 0; i < 3; i++) {
      int cur = pos;
      while (cur < v.size() && v[pos] + mid >= v[cur]) cur++;
      pos = cur;
    }
    if (pos == v.size())
      hi = mid;
    else
      lo = mid + 1;
  }
  printf("%.6lf\n", lo / 2.0);
  int pos = 0;
  for (int i = 0; i < 3; i++) {
    if (i > 0) printf(" ");
    if (pos < v.size())
      printf("%.6lf", (double)v[pos] + lo / 2.0);
    else
      printf("%.6lf", 1.0);
    int cur = pos;
    while (cur < v.size() && v[pos] + lo >= v[cur]) cur++;
    pos = cur;
  }
  assert(pos == v.size());
  printf("\n");
}