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 int MOD = 1e9 + 7;
inline long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
inline long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
inline long long int add(long long int a, long long int b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
inline long long int multi(long long int a, long long int b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
inline long long int sub(long long int a, long long int b) {
  a %= MOD;
  b %= MOD;
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline long long int power(long long int a, long long int b) {
  a %= MOD;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = multi(res, a);
    }
    a = multi(a, a);
    b >>= 1;
  }
  return res;
}
bool isPrime(long long int x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) {
    return x == 2 || x == 3;
  }
  for (long long int i = 5; i * i <= x; i += 6) {
    if (x % i == 0 || x % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
long double ep = 1e-15;
long double x, y, z;
long double get(long double m, long double arr[], long long int n) {
  x = arr[0] + m;
  long long int p = 0;
  while (p < n && x + m >= arr[p]) p++;
  if (p < n)
    y = arr[p] + m;
  else
    return arr[n - 1] + 1;
  while (p < n && y + m >= arr[p]) p++;
  if (p < n)
    z = arr[p] + m;
  else
    return arr[n - 1] + 1;
  return z + m;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  set<long long int> st;
  long long int xx, j = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> xx;
    st.insert(xx);
  }
  n = st.size();
  long double arr[n];
  for (long long int i : st) {
    arr[j++] = i;
  }
  long double low = 0, high = arr[n - 1] - arr[0];
  long double ans = 9223372036854775807;
  long long int first = 300;
  while (high - low > ep && first--) {
    long double mid = (low + high) / 2.0;
    long double res = get(mid, arr, n);
    if (res - arr[n - 1] > ep) {
      ans = min(mid, ans);
      high = mid;
    } else
      low = mid;
  }
  get(ans, arr, n);
  if (n <= 3) {
    ans = 0;
    x = arr[0];
    y = arr[min(1LL, n - 1)];
    z = arr[min(2LL, n - 1)];
  }
  cout << fixed << setprecision(12) << ans << endl;
  cout << fixed << setprecision(12) << x << " " << y << " " << z;
}