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 double EPS = (1e-5);
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int findXOR(int n) {
  switch (n % 4) {
    case 0:
      return n;
    case 1:
      return 1;
    case 2:
      return n + 1;
    case 3:
      return 0;
  }
}
int rangeXOR(int l, int r) { return (findXOR(l - 1) ^ findXOR(r)); }
int getbit(int mask, int bit) { return (mask & (1 << bit)); }
void setbit(int& mask, int bit, int val) {
  if (val)
    mask |= (1 << bit);
  else
    mask &= ~(1 << bit);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const int N = 2e5 + 10;
const int M = 1e2 + 1;
const long long INF = (long long)1e10;
long long a[N];
int n;
bool can(long long mid) {
  int pos = a[0] + mid;
  int cnt = 1;
  for (int i = 0; i < n; ++i) {
    int diff = abs(a[i] - pos);
    if (diff > mid) {
      pos = a[i] + mid;
      cnt++;
      if (cnt > 3) return false;
    }
  }
  return true;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i] *= 2ll;
  }
  sort(a, a + n);
  long long lo = -1, hi = INF;
  while (hi - lo > 1) {
    long long mid = (lo + hi + 1ll) / 2ll;
    if (can(mid)) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  vector<long long> output(3);
  output[0] = a[0] + hi;
  for (int i = 0, j = 1; i < n; ++i) {
    if (output[j - 1] + hi < a[i] && j <= 2) {
      output[j] = a[i] + hi;
      j++;
    }
  }
  printf("%.6f\n%.6f %.6f %.6f\n", (double)hi / 2.0, (double)output[0] / 2.0,
         (double)output[1] / 2.0, (double)output[2] / 2.0);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tc = 1;
  while (tc--) {
    solve();
  }
}