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;
long long n, x;
long long a[N];
bool Can(long long dis) {
  long long last = 0;
  int used = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] > last) {
      last = a[i] + (2 * dis);
      used++;
    }
    if (used > 3) return 0;
  }
  return 1;
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    a[i] = 2 * x;
  }
  sort(a, a + n);
  long long high, mid, low;
  low = 0;
  high = 2e9;
  while (high - low > 0) {
    mid = (high + low) / 2;
    if (Can(mid))
      high = mid;
    else
      low = mid + 1;
  }
  printf("%.6lf \n", double(low) / 2);
  double u;
  int g = 0;
  long long l = 0;
  long long o = 0;
  while (o < n) {
    if (a[o] > l) {
      l = a[o] + (2 * low);
      u = (a[o] + low);
      printf("%.6lf ", double(u) / 2);
      g++;
    }
    o++;
  }
  while (g < 3) {
    cout << 1 << endl;
    g++;
  }
  return 0;
}