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 int INF = 1e9 + 10;
map<int, bool> vis;
vector<int> raw;
int ans[114514], M, s;
inline int idxLB(int x) {
  return lower_bound(raw.begin(), raw.end(), x) - raw.begin();
}
inline int idxUB(int x) {
  return upper_bound(raw.begin(), raw.end(), x) - raw.begin();
}
bool judge(int d) {
  int idx = 0;
  for (int i = 1; i <= 3; i++) {
    idx = idxUB(raw[idx] + d);
    ans[i] = idx;
    if (idx == s) return 1;
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    if (!vis[tmp]) {
      vis[tmp] = 1;
      raw.push_back(tmp);
    }
  }
  sort(raw.begin(), raw.end());
  s = raw.size();
  int L = 0, R = INF;
  while (L <= R) {
    M = L + (R - L) / 2;
    if (judge(M))
      R = M - 1;
    else
      L = M + 1;
  }
  if (judge(M) ? 0 : M++) judge(M);
  printf("%.6lf\n", M / 2.0);
  for (int i = 0; i <= 2; i++)
    printf("%.6lf ", raw[ans[i] >= s ? 0 : ans[i]] + (M / 2.0));
  return 0;
}