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 long double eps = 1e-8;
int n, x[200010];
bool check(long double m, bool flag) {
  long double fx = x[0] + 2 * m;
  int i = 0;
  while (i < n && x[i] < fx + eps) i++;
  if (flag) printf("%.6f ", x[0] + (double)m);
  if (i == n) {
    if (flag) printf("%.6f %.6f\n", 1.0 * x[n - 1], 1.0 * x[n - 1]);
    return true;
  }
  fx = x[i] + 2 * m;
  if (flag) printf("%.6f ", x[i] + (double)m);
  while (i < n && x[i] < fx + eps) i++;
  if (i == n) {
    if (flag) printf("%.6f\n", 1.0 * x[n - 1]);
    return true;
  }
  fx = x[i] + 2 * m;
  if (flag) printf("%.6f\n", x[i] + (double)m);
  while (i < n && x[i] < fx + eps) i++;
  return i == n;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) scanf("%d", &x[i]);
  sort(x, x + n);
  long double l = 0, r = 1e10, m;
  while (r - l > 1e-7) {
    m = (l + r) * 0.5;
    if (check(m, false))
      r = m;
    else
      l = m;
  }
  printf("%.6f\n", (double)r);
  check(r, true);
}