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 cpt = 10000000;
long long x[200003];
long long nc[5];
int n;
bool out;
bool can(long long r) {
  int now = 1;
  nc[1] = x[0] + r;
  for (int i = 0; i < n; ++i) {
    if (x[i] > nc[now] + r) nc[++now] = x[i] + r;
    if (now > 3) return false;
  }
  while (now < 3) nc[++now] = nc[1];
  return true;
}
int main() {
  out = false;
  cin >> n;
  for (int i = 0; i < n; ++i) scanf("%lld", &x[i]), x[i] *= cpt;
  sort(x, x + n);
  long long l = 0, r = 20000000000000000ll, m;
  while (l < r) {
    m = (l + r) >> 1;
    if (can(m))
      r = m;
    else
      l = m + 1;
  }
  long double res = l;
  res /= cpt;
  printf("%.7Lf\n", res);
  out = true;
  can(l);
  for (int i = 1; i <= 3; ++i) {
    res = nc[i];
    res /= cpt;
    printf("%.7Lf ", res);
  }
}