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 = 222222;
int x[maxn];
double ans[3];
int n;
bool yes(double d) {
  double temp[3];
  int m = 0;
  double sum = 0;
  temp[0] = temp[1] = temp[2] = x[n - 1] - d;
  for (int i = 1; i < n; i++) {
    sum += x[i] - x[i - 1];
    if (sum > d * 2) {
      if (m == 2) return false;
      sum -= x[i] - x[i - 1];
      temp[m++] = max(0.0, x[i - 1] - d);
      sum = 0;
    }
  }
  for (int i = 0; i < 3; i++) {
    ans[i] = temp[i];
  }
  return true;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &x[i]);
    }
    sort(x, x + n);
    double lo = 0;
    double hi = 1e9;
    int T = 100;
    while (T--) {
      double mid = (lo + hi) / 2;
      if (yes(mid))
        hi = mid;
      else
        lo = mid;
    }
    printf("%.6lf\n", lo);
    for (int i = 0; i < 3; i++) {
      printf("%lf%c", ans[i], i == 2 ? '\n' : ' ');
    }
  }
  return 0;
}