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;
template <typename T>
inline T abs(T a) {
  return ((a < 0) ? -a : a);
}
template <typename T>
inline T sqr(T a) {
  return a * a;
}
const int INF = (int)1E9 + 7;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
int n, x[300000];
long double ans[50];
bool can(long double d) {
  long double rg = -1E100;
  int it = 0;
  for (int i = 0; i < int(n); ++i) {
    if (fabs(x[i] - rg) < EPS) continue;
    if (x[i] <= rg) continue;
    ans[it++] = x[i] + d;
    rg = max(x[i] + 2 * d, rg);
    if (it > 3) break;
  }
  return it <= 3;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < int(n); ++i) scanf("%d", &x[i]);
  sort(x, x + n);
  n = unique(x, x + n) - x;
  long double lf = 0.0, rg = 1E9 + 7;
  for (int i = 0; i < int(100); ++i) {
    long double mid = (lf + rg) / 2;
    if (can(mid))
      rg = mid;
    else
      lf = mid;
  }
  can(rg);
  cout.precision(6);
  cout << fixed;
  cout << (lf + rg) / 2 << endl;
  for (int i = 0; i < int(3); ++i)
    cout << min(max(0.0L, ans[i]), 1E9L + 100) << " ";
  cout << endl;
  return 0;
}