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 + 10;
int n, a[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int lo = 0, hi = 1e9 + 10;
  for (int i = 0; i < 40; i++) {
    int mid = (lo + hi) / 2;
    int pos1 = upper_bound(a, a + n, a[0] + mid) - a;
    int pos2 = upper_bound(a, a + n, a[pos1] + mid) - a;
    int pos3 = upper_bound(a, a + n, a[pos2] + mid) - a;
    if (pos1 == n or pos2 == n or pos3 == n)
      hi = mid;
    else
      lo = mid;
  }
  setprecision(6);
  cout << fixed;
  cout << double(hi) / 2 << endl;
  int pos1 = upper_bound(a, a + n, a[0] + hi) - a;
  cout << double(a[pos1 - 1] + a[0]) / 2 << " ";
  int pos2 = upper_bound(a, a + n, a[pos1] + hi) - a;
  cout << double(a[pos2 - 1] + a[pos1]) / 2 << " ";
  cout << double(a[n - 1] + a[pos2]) / 2 << endl;
  return 0;
}