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>
static const int MAXN = 2e5 + 3;
int n;
long long x[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &x[i]);
    x[i] <<= 1;
  }
  std::sort(x, x + n);
  int lo = -1, hi = 2e9 + 15, mid;
  int p;
  while (lo < hi - 1) {
    mid = (lo + hi) >> 1;
    p = std::upper_bound(x, x + n, x[0] + mid * 2) - &x[0];
    if (p < n) p = std::upper_bound(x + p, x + n, x[p] + mid * 2) - &x[0];
    if (p < n) p = std::upper_bound(x + p, x + n, x[p] + mid * 2) - &x[0];
    if (p < n)
      lo = mid;
    else
      hi = mid;
  }
  printf("%.6lf\n", (double)hi / 2);
  printf("%.6lf ", (double)(x[0] + hi) / 2);
  p = std::upper_bound(x, x + n, x[0] + hi * 2) - &x[0];
  printf("%.6lf ", (double)(x[p] + hi) / 2);
  if (p < n) p = std::upper_bound(x + p, x + n, x[p] + hi * 2) - &x[0];
  if (p >= n) p = n / 2;
  printf("%.6lf\n", (double)(x[p] + hi) / 2);
  return 0;
}