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 A[300000], S[3];
bool B[300000], fl;
int n;
int main() {
  int i, j;
  long long l, r, m;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%I64d", &A[i]), A[i] *= 2ll;
  sort(A, A + n);
  l = -1;
  r = ((long long)1e10);
  while (r - l > 1ll) {
    m = (l + r + 1ll) / (2ll);
    fl = 1;
    memset(B, 0, sizeof(B));
    S[0] = A[0] + m;
    for (i = 0, j = 1; i < n; i++) {
      if (S[j - 1] + m < A[i] && j <= 2) S[j] = A[i] + m, j++;
      if (j == 3 && S[j - 1] + m < A[i]) {
        fl = 0;
        break;
      }
    }
    if (fl)
      r = m;
    else
      l = m;
  }
  S[0] = A[0] + r;
  for (i = 0, j = 1; i < n; i++) {
    if (S[j - 1] + r < A[i] && j <= 2) S[j] = A[i] + r, j++;
  }
  printf("%.2lf\n%.2lf %.2lf %.2lf\n", ((double)r) / 2.0, ((double)S[0]) / 2.0,
         ((double)S[1]) / 2.0, ((double)S[2]) / 2.0);
  return 0;
}