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 maxn = 5e5 + 5;
const double eps = 1e-6;
int n;
double a[maxn];
double x[5], sx[5];
bool check(double mid) {
  int s = 1;
  for (int i = 1; i <= 3; ++i) {
    sx[i] = a[s];
    s = upper_bound(a + 1, a + n + 1, a[s] + mid * 2) - a;
    if (s > n) return 1;
  }
  return s > n;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lf", &a[i]);
  sort(a + 1, a + n + 1);
  if (check(0.0)) {
    for (int i = 1; i <= 3; ++i) x[i] = sx[i];
    double as = 0;
    printf("%.6f\n", as);
    for (int i = 1; i <= 3; ++i) printf("%.6f ", x[i] + as);
    return 0;
  }
  double l = 0, r = a[n], ans;
  while (r - l > eps) {
    double mid = (l + r) / 2;
    if (check(mid)) {
      r = mid, ans = mid;
      for (int i = 1; i <= 3; ++i) x[i] = sx[i];
    } else
      l = mid;
  }
  printf("%.6f\n", ans);
  for (int i = 1; i <= 3; ++i) printf("%.6f ", x[i] + ans);
  return 0;
}