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 = 2 * 1000 * 100 + 10;
double d[maxn];
double a[maxn];
int n;
bool check(double x) {
  int y = 0, k = 0, xa;
  while (y < n && k < 3) {
    xa = y;
    y = lower_bound(a, a + n, a[y] + 2 * x) - a;
    if (a[xa] + 2 * x == a[y]) y++;
    k++;
  }
  if (a[xa] + 2 * x >= a[n - 1]) return 1;
  return 0;
}
double ans1, ans2, ans3;
int as, bs, cs;
void check1(double x) {
  ans1 = a[0] + x;
  as = lower_bound(a, a + n, ans1 + x) - a;
  ans2 = a[as] + x;
  bs = lower_bound(a, a + n, ans2 + x) - a;
  if (bs == n)
    ans3 = a[n - 1] + x;
  else
    ans3 = a[bs] + x;
}
double help(double s, double b) {
  double u = (b + s) / 2;
  if (b - s < 0.000001) {
    if (check(s) == 1) return s;
    return b;
  }
  if (check(u) == 0) return help(u, b);
  if (check(u) == 1) return help(s, u);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  double ans = help(0, 1000 * 1000 * 1000);
  if (ans == 0) {
    cout << 0 << endl;
    cout << fixed;
    cout << setprecision(6) << a[0] << " " << a[1] << " " << a[2] << endl;
  } else {
    cout << fixed << setprecision(6) << ans << endl;
    check1(ans);
    cout << ans1 << " " << ans2 << " " << ans3 << endl;
  }
  return 0;
}