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 long long INF = 1000000000 + 7;
const double esp = 1e-13;
int n, a[200000 + 10], dem;
double ans, fi, mid, la, d;
void solve(double mid, int k) {
  dem = 0;
  d = 0;
  for (long long i = 1; i <= n; i++)
    if (dem > 3)
      break;
    else {
      if (i == 1) {
        dem++;
        d = mid + a[i];
        if (k == 1) printf("%0.9lf ", d);
      } else if (d + mid - a[i] > -0.0000000001)
        ;
      else {
        dem++;
        d = a[i] + mid;
        if (k == 1) printf("%0.9lf ", d);
      }
    }
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  fi = 0;
  la = INF;
  ans = INF;
  for (long long i = 1; i <= 100; i++) {
    mid = (fi + la) / 2;
    solve(mid, 2);
    if (dem <= 3) {
      la = mid;
      ans = min(ans, mid);
    } else
      fi = mid;
  }
  printf("%0.9lf\n", ans);
  solve(ans, 1);
  if (dem <= 1) printf("%0.9lf ", a[n] + 1);
  if (dem <= 2) printf("%0.9lf ", a[n] + 2);
}