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 mm = 2e5 + 9;
const double oo = 1e39;
const double ex = 1e-7;
double locat[mm];
int n, ans[mm];
int blook(int x) {
  int l = 0, r = n, mid = (l + r) / 2;
  while (l < r) {
    if (locat[mid] > x)
      r = mid;
    else
      l = mid + 1;
    mid = (l + r) / 2;
  }
  return mid;
}
bool ok(int x) {
  int z = 0;
  for (int i = 0; i < 3; i++) {
    z = blook(locat[z] + x);
    ans[i] = z;
    if (z == n) return 1;
  }
  return 0;
}
int main() {
  while (cin >> n) {
    for (int i = 0; i < n; i++) cin >> locat[i];
    sort(locat, locat + n);
    int r = 2e9 + 9, l = 0, mid;
    while (l < r) {
      mid = (l + r) / 2;
      if (ok(mid))
        r = mid;
      else
        l = mid + 1;
    }
    ok(l);
    double len = l;
    len /= 2.0;
    double a, b, c;
    a = locat[ans[0] - 1] + locat[0];
    b = locat[ans[1] - 1] + locat[ans[0]];
    c = locat[ans[2] - 1] + locat[ans[1]];
    cout.setf(ios::fixed);
    cout << setprecision(6) << len << "\n"
         << a / 2.0 << " " << b / 2.0 << " " << c / 2.0 << endl;
  }
}