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 A[MAXN];
int N;
double ans_d, x, y, z;
bool check(double d) {
  ans_d = d;
  int i = 0;
  x = A[i] + d;
  i = upper_bound(A, A + N, x + d) - A;
  if (i == N) return true;
  y = A[i] + d;
  i = upper_bound(A, A + N, y + d) - A;
  if (i == N) return true;
  z = A[i] + d;
  if (z + d >= A[N - 1]) return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];
  sort(A, A + N);
  double L = 0, R = 1e9 + 1;
  while (R - L > 1e-7) {
    double m = (L + R) / 2;
    if (check(m)) {
      R = m;
    } else
      L = m + 1e-7;
  }
  check(R);
  printf("%.6lf\n", ans_d);
  printf("%.6lf %.6lf %.6lf\n", x, y, z);
  return 0;
}