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 double EPS(1.0e-10);
long double delta(1.0e-12);
int idx_upperthan(int begin, long double threshold, vector<int>& dat,
                  long double eps) {
  for (int i = (int)(begin); i < (int)(dat.size()); i++)
    if (dat[i] > threshold + eps) return i;
  return (int)dat.size();
}
bool coverable(long double d, vector<int>& dat) {
  long double sep = dat[0] + d;
  int idx = idx_upperthan(0, sep, dat, delta);
  if (idx == dat.size()) return true;
  sep = dat[idx] + d;
  idx = idx_upperthan(idx, sep, dat, delta);
  if (idx == dat.size()) return true;
  long double end = dat[idx] + d;
  if (end + delta >= dat.back())
    return true;
  else
    return false;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> dat(n);
  for (int i = (int)(0); i < (int)(n); i++) {
    scanf("%d", &dat[i]);
  }
  sort((dat).begin(), (dat).end());
  long double minv(0), maxv(1e10);
  for (int i = (int)(0); i < (int)(100); i++) {
    long double d = (minv + maxv);
    long double midv(d * 0.5);
    if (coverable(d, dat)) {
      maxv = midv;
    } else {
      minv = midv;
    }
  }
  printf("%.6f\n", (double)minv);
  long double p1 = dat[0] + minv;
  int idx = idx_upperthan(0, p1 + minv, dat, EPS);
  long double p2 = min(idx != (int)dat.size() ? dat[idx] + minv : p1 + minv,
                       2.0L * 1000000000);
  long double p3 = max(0.0L, dat.back() - minv);
  printf("%.6f %.6f %.6f\n", (double)p1, (double)p2, (double)p3);
  return 0;
}