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;
int a[2 * int(1e5 + 5)];
double d1, d2, d3;
bool check(double d) {
  d1 = 1.0 * a[0] + d;
  d3 = 1.0 * a[n - 1] - d;
  bool flg = 0;
  for (int i = 0; i < n; ++i) {
    if (fabs(d1 - 1.0 * a[i]) - d < 0.0000000001 ||
        fabs(d3 - 1.0 * a[i]) - d < 0.0000000001)
      continue;
    if (flg) {
      if (fabs(d2 - 1.0 * a[i]) - d > 0.0000000001) return 0;
      continue;
    }
    flg = 1;
    d2 = 1.0 * a[i] + d;
  }
  return 1;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  sort(a, a + n);
  double st = 0, en = a[n - 1];
  double res;
  for (int i = 0; i < 62; ++i) {
    double md = (st + en) / 2;
    if (check(md)) {
      en = md;
      res = d2;
    } else
      st = md;
  }
  cout << setprecision(6) << fixed << st << endl;
  cout << d1 << ' ' << res << ' ' << d3 << endl;
}