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 OO = (int)1e8;
double arr[200001];
int t;
bool isok(double x) {
  double pos = 0;
  int tot = 0;
  while (1) {
    int idx = upper_bound(arr, arr + t, pos) - arr;
    if (idx >= t) break;
    tot++;
    pos = arr[idx] + 2 * x;
  }
  return tot <= 3;
}
void build(double x) {
  double pos = 0;
  int tot = 0;
  while (1) {
    int idx = upper_bound(arr, arr + t, pos) - arr;
    if (idx >= t) break;
    if (tot) printf(" ");
    printf("%.6lf", arr[idx] + x);
    tot++;
    pos = arr[idx] + 2 * x;
  }
  while (tot < 3) {
    tot++;
    printf(" %.6lf", pos);
  }
}
int main() {
  scanf("%d", &t);
  double end = 0;
  for (int i = 0; i < t; ++i) {
    scanf("%lf", arr + i);
    end += arr[i];
  }
  sort(arr, arr + t);
  double st = 0, sz = end / 2, ret;
  while (sz > 1e-7) {
    double mid = st + sz;
    if (isok(mid)) {
      ret = mid;
    } else
      st = mid;
    sz *= 0.5;
  }
  ret *= 1000000;
  ret = round(ret);
  ret /= 1000000;
  if (ret < 1e-7)
    printf("0\n");
  else
    printf("%.6lf\n", ret);
  build(ret);
  return 0;
}