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 lim = 2000000000;
int n, t[200010];
bool spr(double d) {
  double l1 = t[0] + d;
  if (l1 + d >= t[n - 1]) return 1;
  int pos = upper_bound(t, t + n, (int)(l1 + d)) - t;
  double l2 = t[pos] + d;
  if (l2 + d >= t[n - 1]) return 1;
  pos = upper_bound(t, t + n, (int)(l2 + d)) - t;
  double l3 = t[pos] + d;
  if (l3 + d >= t[n - 1]) return 1;
  return 0;
}
double BS(double p, double k) {
  double prev = p;
  while (true) {
    double s = (p + k) / 2.0;
    if (spr(s))
      k = s;
    else
      p = s;
    if (fabs(prev - s) < 0.0000000001) break;
    prev = s;
  }
  return k;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &t[i]);
  sort(t, t + n);
  double d = BS(0, t[n - 1] - t[0]);
  double l1 = t[0] + d, l2, l3;
  if (l1 + d >= t[n - 1]) {
    l2 = l3 = l1;
    printf("%.6lf\n%.6lf %.6lf %.6lf\n", d, l1, l2, l3);
    return 0;
  }
  int pos = upper_bound(t, t + n, (int)(l1 + d)) - t;
  l2 = t[pos] + d;
  if (l2 + d >= t[n - 1]) {
    l3 = l2;
    printf("%.6lf\n%.6lf %.6lf %.6lf\n", d, l1, l2, l3);
    return 0;
  }
  pos = upper_bound(t, t + n, (int)(l2 + d)) - t;
  l3 = t[pos] + d;
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", d, l1, l2, l3);
  return 0;
}