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 N = 2e5 + 7;
const int INF = 1e9 + 7;
int x[N], n;
vector<double> ans;
bool gao(int lim, bool sav = false) {
  int ned = 0;
  for (int l = 0, r; l < n; l = r) {
    for (r = l; r < n && x[r] - x[l] <= lim; r++)
      ;
    if (sav) ans.push_back(x[l] + (x[r - 1] - x[l]) / 2.0);
    ned++;
  }
  return ned <= 3;
}
int main() {
  cin >> n;
  for (int i = (0); i < (n); i++) scanf("%d", &x[i]);
  sort(x, x + n);
  int ret = -1;
  for (int l = 0, r = INF, mid = (l + r) >> 1; l <= mid; mid = (l + r) >> 1) {
    if (gao(mid))
      r = (ret = mid) - 1;
    else
      l = mid + 1;
  }
  gao(ret, true);
  while ((int)((ans).size()) < 3) ans.push_back(ans.back());
  printf("%.10f\n", ret / 2.0);
  for (int i = (0); i < ((int)((ans).size())); i++) printf("%.10f ", ans[i]);
  return 0;
}