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;
long long ar[200010];
void solve() {
  long long a = 1, b = 0, n, i, j, k, z, q, o, c;
  long long p = 0, sm = 0, cnt = 0, mx = -1000000007, mn = 1000000007;
  double x, y;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) scanf("%lld", &ar[i]);
  sort(ar, ar + n);
  if (n == 1)
    cout << 0 << "\n" << ar[0] << ' ' << ar[0] << ' ' << ar[0] << "\n";
  else if (n == 2)
    cout << 0 << "\n" << ar[0] << ' ' << ar[0] << ' ' << ar[1] << "\n";
  else if (n == 3)
    cout << 0 << "\n" << ar[0] << ' ' << ar[1] << ' ' << ar[2] << "\n";
  else {
    long long l = 0, r = ar[n - 1] - ar[0];
    while (l < r) {
      long long mid = (l + r) / 2;
      c = 0, mx = 0;
      for (i = 0; i < n; i++) {
        if (ar[i] > mx) {
          mx = ar[i] + mid;
          c++;
        }
      }
      if (c > 3)
        l = mid + 1;
      else
        r = mid;
    }
    x = (l * 1.0) / 2.0;
    printf("%.6f\n", x);
    c = 0, mx = 0;
    for (i = 0; i < n; i++) {
      if (ar[i] > mx) {
        mx = ar[i] + l;
        c++;
        printf("%.6f ", ar[i] * 1.0 + x);
      }
    }
    while (c < 3) {
      cout << 0 << ' ';
      c++;
    }
  }
}
int main() {
  solve();
  return 0;
}