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;
vector<long long> p;
long long M, mn, mx, h1, h2;
set<long long> st;
long long n, b;
bool check(long long d) {
  if (d >= M) {
    h1 = mn;
    h2 = mx;
    return 1;
  }
  h1 = *(upper_bound(p.begin(), p.end(), mn + d));
  h2 = *(--(lower_bound(p.begin(), p.end(), mx - d)));
  return (h2 - h1 <= d);
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d", &n);
  st.insert(-100);
  st.insert(200000000001LL);
  mn = 2000000000000LL;
  mx = -1;
  for (int i = 0; i <= (int)(n)-1; i++) {
    scanf("%d", &b);
    st.insert(b);
    mn = ((mn) < (b) ? mn : b);
    mx = ((mx) < (b) ? b : mx);
  }
  for (set<long long>::iterator it = st.begin(); it != st.end(); it++)
    p.push_back(*it);
  long long lo = 0, hi = 200000000000LL, mid;
  M = mx - mn;
  while (hi - lo > 1) {
    mid = (hi + lo) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid;
  }
  if (check(lo))
    mid = lo;
  else
    mid = hi;
  check(mid);
  double D = 0.5 * mid;
  printf("%.6f\n%.6f %.6f %.6f", D, double(mn) + D, 0.5 * (h1 + h2),
         double(mx) - D);
  return 0;
}