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;
long long int power(long long int x, long long int y, long long int m) {
  if (y == 0) return 1;
  long long int a = power(x, y / 2, m);
  if (y % 2) {
    return (a * ((a * x) % m)) % m;
  } else {
    return (a * a) % m;
  }
}
long long int mod_inverse(long long int x, long long int m) {
  return power(x, m - 2, m);
}
long long int fact(long long int n, long long int m) {
  if (n <= 1) return 1;
  return (fact(n - 1, m) * n) % m;
}
long long int ncr(long long int n, long long int r, long long int m) {
  if (r > n) return 0;
  long long int n1 = 1, d1 = 1, d2 = 1;
  n1 = fact(n, m);
  d1 = fact(r, m);
  d2 = fact(n - r, m);
  long long int ans = mod_inverse((d1 * d2) % m, m);
  ans = (ans * n1) % m;
  return ans;
}
int ispal(string s) {
  int len = s.size();
  int flag = 1;
  for (int i = 0; i < len; ++i) {
    if (s[i] != s[len - i - 1]) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long int sroot(long long int n, long long int low = 1,
                    long long int high = 1e9 + 1) {
  if (low == high) return low;
  if (low == high - 1) {
    if (high * high <= n)
      return high;
    else
      return low;
  }
  long long int mid = (low + high) / 2;
  long long int a = mid * mid;
  if (a > n)
    return sroot(n, low, mid - 1);
  else
    return sroot(n, mid, high);
}
long long int croot(long long int n, long long int low = 1,
                    long long int high = 1e6 + 1) {
  if (low == high) return low;
  if (low == high - 1) {
    if (high * high * high <= n)
      return high;
    else
      return low;
  }
  long long int mid = (low + high) / 2;
  long long int a = mid * mid * mid;
  if (a > n)
    return croot(n, low, mid - 1);
  else
    return croot(n, mid, high);
}
class solver {
 public:
  int n;
  double x[200010];
  set<double> st;
  set<double>::iterator itr, it;
  double eb = 0.0000001;
  bool check(double pw) {
    double val;
    it = st.begin();
    val = (*it);
    it = st.upper_bound(val + 2 * pw);
    if (it == st.end()) return true;
    val = (*it);
    it = st.upper_bound(val + 2 * pw);
    if (it == st.end()) return true;
    val = (*it);
    it = st.upper_bound(val + 2 * pw);
    if (it == st.end()) return true;
    return false;
  }
  double bs(double low, double high) {
    if ((high - low) < eb) {
      return low;
    }
    double mid = (low + high) / 2;
    bool fg = check(mid);
    if (fg) {
      return bs(low, mid);
    } else {
      return bs(mid, high);
    }
  }
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> x[i];
    for (int i = 1; i <= n; ++i) st.insert(x[i]);
    double d = bs(0, 2000000000);
    d += eb;
    double p1 = 0, p2 = 0, p3 = 0;
    it = st.begin();
    double val = (*it);
    p1 = val + d;
    it = st.upper_bound(val + 2 * d);
    if (it == st.end()) {
      p2 = p3 = p1;
    } else {
      val = (*it);
      p2 = val + d;
      it = st.upper_bound(val + 2 * d);
      if (it == st.end()) {
        p3 = p2;
      } else {
        val = (*it);
        p3 = val + d;
      }
    }
    p1 += eb;
    p2 += eb;
    p3 += eb;
    cout << fixed << setprecision(6) << d << '\n'
         << p1 << ' ' << p2 << ' ' << p3 << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test = 1;
  while (test--) {
    solver o;
    o.solve();
  }
  return 0;
}