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 A[200010];
int starts[200010];
int N;
bool check(int x) {
  int idx = 0, cnt = 0;
  while (idx < N) {
    int st = A[idx];
    starts[cnt] = st;
    while (idx < N && A[idx] <= st + x) idx += 1;
    cnt += 1;
  }
  if (cnt <= 3) return true;
  return false;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &A[i]);
  sort(A, A + N);
  int lo = 0, hi = 2000000000;
  while (lo < hi) {
    int mid = (lo + hi) >> 1;
    if (check(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  if (check(lo)) hi = lo;
  cout << (hi >> 1) << (hi & 1 ? ".500000" : ".000000") << endl;
  for (int i = 0; i < 3; i++) {
    int st = starts[i];
    if (st == -1) st = starts[0];
    int ed = st + hi;
    cout << ((st + ed) >> 1) << ((st + ed) & 1 ? ".500000" : ".000000") << " ";
  }
  cout << endl;
}