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 double eps = 1e-8;
int n, tot;
double l, r, mid, x[200000], posnow[200000];
int cmp(double a, double b) {
  if (fabs(a - b) < eps) return 0;
  if (a < b) return -1;
  return 1;
}
bool check(double dis) {
  tot = 0;
  posnow[tot++] = x[0] + dis;
  for (int i = 0; i < n; i++)
    if (cmp(fabs(x[i] - posnow[tot - 1]), dis) > 0) posnow[tot++] = x[i] + dis;
  return tot <= 3;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) scanf("%lf", &x[i]);
    sort(x, x + n);
    l = 0.0;
    r = (double)(x[n - 1] - x[0]) / 6.0;
    for (int tims = 0; tims < 100; tims++) {
      mid = (l + r) / 2;
      if (check(mid) == true)
        r = mid;
      else
        l = mid;
    }
    printf("%.6f\n", l);
    tot = 0;
    posnow[tot++] = x[0] + r;
    for (int i = 0; i < n; i++)
      if (cmp(fabs(x[i] - posnow[tot - 1]), r) > 0) {
        posnow[tot++] = x[i] + r;
      }
    if (tot < 3)
      for (int i = tot; i < 3; i++) posnow[i] = posnow[tot - 1];
    for (int i = 0; i < 3; i++) printf("%.6f ", posnow[i]);
    printf("\n");
  }
  return 0;
}