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;
using ll = long long;
using pii = pair<int, int>;
using pib = pair<int, bool>;
using vi = vector<int>;
const int INF = (int)1e9;
const ll INFLL = (long long)2e18;
const int MOD = (int)1e9 + 7;
const int N = (int)1e6 + 5;
const double EPS = 1E-9;
const double EPS_ANG = 1E-16;
const static double PI = 3.141592653589793;
const int dx[8] = {+0, +0, +1, -1, +1, +1, -1, -1};
const int dy[8] = {+1, -1, +0, +0, +1, -1, +1, -1};
int main() {
  int t = 1;
  for (int cas = 1; cas <= t; ++cas) {
    int n;
    scanf("%d", &n);
    int a[n];
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    sort(a, a + n);
    double lo = 0, hi = (a[0] + a[n - 1]) / 2.0;
    double x, y, z, d;
    while (hi - lo > EPS) {
      double mid = lo + (hi - lo) / 2;
      double curBase = a[0] + mid;
      int i = 0;
      while (i < n) {
        if (a[i] > curBase + mid) break;
        ++i;
      }
      double curBase2 = curBase;
      if (i < n) {
        curBase2 = a[i] + mid;
        while (i < n) {
          if (a[i] > curBase2 + mid) break;
          ++i;
        }
      }
      double curBase3 = curBase2;
      if (i < n) {
        curBase3 = a[i] + mid;
        while (i < n) {
          if (a[i] > curBase3 + mid) break;
          ++i;
        }
      }
      if (i == n) {
        d = mid;
        x = curBase, y = curBase2, z = curBase3;
        hi = mid - 0.000000000001;
      } else
        lo = mid + 0.000000000001;
    }
    printf("%.6lf\n", d);
    printf("%.6lf %.6lf %.6lf\n", x, y, z);
  }
  return 0;
}