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 long long MOD = 1000 * 1000 * 1000 + 7;
const long long MAXN = 223456 + 100;
const long long MAXM = 200 + 50;
const long long INF = 1e9 + 7;
const long long dx[4] = {0, 0, -1, +1};
const long long dy[4] = {+1, -1, 0, 0};
ifstream fin("input.txt");
ofstream fout("output.txt");
long long a[MAXN];
int n;
bool check(long long t) {
  long long last = a[0];
  int cnt = 1;
  for (int i = 1; i < n; i++)
    if (last + t < a[i]) cnt++, last = a[i];
  return (cnt <= 3);
}
int main() {
  ;
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cout << fixed << setprecision(1);
  int mohammadjavadakbari = 1379;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  n = unique(a, a + n) - a;
  if (check(0)) {
    cout << 0 << '\n';
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
    for (int i = 0; i < 3 - n; i++) cout << mohammadjavadakbari++ << ' ';
    exit(0);
  }
  long long MAX = 1e15, MIN = 0;
  while (MAX - MIN > 1) {
    long long MID = (MIN + MAX) / 2;
    if (!check(MID))
      MIN = MID;
    else
      MAX = MID;
  }
  long double res = MAX / 2.00;
  cout << res << '\n' << a[0] + res << ' ';
  long long last = a[0];
  int cnt = 2;
  for (int i = 1; i < n; i++)
    if (last + MAX < a[i]) cnt--, last = a[i], cout << a[i] + res << ' ';
  while (cnt--) cout << mohammadjavadakbari++ << ' ';
}