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 = 200005, MaxC = 1000000000, NA = -1;
const long double eps = 1E-9;
int a[MaxN];
int n;
int main(void) {
  int i, j;
  long double lo, me, hi;
  long double p1, p2, p3;
  while (scanf(" %d", &n) != EOF) {
    for (i = 0; i < n; i++) scanf(" %d", &a[i]);
    sort(a, a + n);
    lo = 0.0;
    hi = MaxC;
    a[n] = MaxC;
    for (i = 0; i < 100; i++) {
      me = (lo + hi) * 0.5;
      j = 0;
      p1 = a[j] + me;
      while (j < n && a[j] < p1 + me) j++;
      p2 = a[j] + me;
      while (j < n && a[j] < p2 + me) j++;
      p3 = a[j] + me;
      while (j < n && a[j] < p3 + me) j++;
      assert(j <= n);
      if (j == n)
        hi = me;
      else
        lo = me;
    }
    lo = hi;
    lo += eps;
    printf("%.6lf\n", (double)lo);
    j = 0;
    p1 = a[j] + me;
    while (j < n && a[j] <= p1 + me + eps) j++;
    p2 = a[j] + me;
    while (j < n && a[j] <= p2 + me + eps) j++;
    p3 = a[j] + me;
    while (j < n && a[j] <= p3 + me + eps) j++;
    assert(j <= n);
    printf("%.6lf %.6lf %.6lf\n", (double)p1, (double)p2, (double)p3);
  }
  return 0;
}