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 double eps = 1e-9;
inline bool lt(double a, double b) { return a < b; }
inline bool eq(double a, double b) { return !lt(a, b) && !lt(b, a); }
inline bool lteq(double a, double b) { return lt(a, b) || eq(a, b); }
inline bool gteq(double a, double b) { return lt(b, a) || eq(a, b); }
int x[200002], n;
double pos1, pos2, pos3;
int prvog(double p, double d) {
  int lo = 0, hi = n - 1, mid = 0, ret = 0;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    if (lteq(x[mid], p + d)) {
      lo = mid + 1;
      ret = mid;
    } else
      hi = mid - 1;
  }
  return ret + 1;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  sort(x, x + n);
  double lo = 0, hi = 1000000000, d = 0;
  for (int it = 0; it < 1000; ++it) {
    d = (lo + hi) / 2;
    int i = 0;
    double pos = x[0] + d;
    pos1 = pos2 = pos3 = pos;
    i = prvog(pos, d);
    if (i < n) {
      double pos = x[i] + d;
      pos2 = pos3 = pos;
      i = prvog(pos, d);
    }
    if (i < n) {
      double pos = x[i] + d;
      pos3 = pos;
      i = prvog(pos, d);
    }
    if (i < n)
      lo = d;
    else
      hi = d;
  }
  printf("%lf\n", d);
  printf("%.6lf %.6lf %.6lf\n", pos1, pos2, pos3);
  return 0;
}