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;
int n;
double num[200004], a1, a2, a3, ans;
int cmp(double x) { return x > 1e-10 ? 1 : x < -1e-10 ? -1 : 0; }
int test(double x) {
  int re1, re2, i;
  a1 = num[1] + x;
  a3 = num[n] - x;
  for (i = 2; i <= n; i++) {
    if (cmp(num[i] - num[1] - 2 * x) > 0) break;
  }
  re1 = i;
  for (i = n - 1; i >= 1; i--) {
    if (cmp(num[n] - num[i] - 2 * x) > 0) break;
  }
  re2 = i;
  if (cmp(num[re2] - num[re1] - 2 * x) <= 0) {
    a2 = num[re1] + x;
    return 1;
  }
  return 0;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%lf", &num[i]);
    sort(num + 1, num + n + 1);
    double l, r, mid;
    l = 0;
    r = 1e10;
    int time = 200;
    while (time--) {
      mid = (l + r) / 2;
      if (test(mid)) {
        ans = mid;
        r = mid;
      } else {
        l = mid;
      }
    }
    printf("%.6lf\n", ans);
    printf("%.6lf %.6lf %.6lf\n", a1, a2, a3);
  }
  return 0;
}