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 house[200005];
int N;
long long p1, p2, p3;
int is_enough(long long d) {
  p1 = house[0] + d;
  int k = 0;
  while (k < N && house[k] <= p1 + d) k++;
  p2 = house[k] + d;
  while (k < N && house[k] <= p2 + d) k++;
  p3 = house[k] + d;
  while (k < N && house[k] <= p3 + d) k++;
  if (k >= N) return 1;
  return 0;
}
int main() {
  int i, j;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%I64d", &house[i]);
    house[i] *= 1e7;
  }
  sort(house, house + N);
  long long High = (house[N - 1] + house[0]) / 6 + 1;
  long long Low = 0, mid;
  while (Low < High) {
    mid = Low + (High - Low) / 2;
    if (is_enough(mid))
      High = mid;
    else
      Low = mid + 1;
  }
  is_enough(Low);
  printf("%.6f\n", ((double)Low / 1e7));
  printf("%.6f %.6f %.6f\n", (double)p1 / 1e7, (double)p2 / 1e7,
         (double)p3 / 1e7);
  return 0;
}