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 spf[101];
long long fac[101];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 101; i++) spf[i] = i;
  for (long long i = 4; i < 101; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 101; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 101; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> getfactor(long long a) {
  map<long long, long long> m;
  while (a > 1) {
    m[spf[a]]++;
    a /= spf[a];
  }
  return m;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long inverse(long long a, long long p) { return power(a, p - 2, p); }
long long ncr(long long n, long long r, long long p) {
  if (r == 0) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
long long n;
long double a[200009];
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long double l = 0, r = 2e9;
  long double ans = 0;
  vector<long long> v, v1;
  while (r - l >= .0000001) {
    long double mid = (l + r) / 2;
    long long count = 1, ind = 0;
    v.push_back(ind);
    for (long long i = 0; i < n; i++) {
      if (a[i] <= 2 * mid + a[ind])
        continue;
      else {
        count++;
        ind = i;
        v.push_back(ind);
      }
    }
    if (count <= 3) {
      v1 = v;
      ans = mid;
      r = mid - .0000001;
    } else
      l = mid + .0000001;
    v.clear();
  }
  cout << fixed << setprecision(6) << ans << '\n';
  cout << fixed << setprecision(6) << a[v1[0]] + ans << " "
       << a[v1[1 % v1.size()]] + ans << " " << a[v1[2 % v1.size()]] + ans
       << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}