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;
bool sePuede(int N, int casas[], int D);
void cosa(int N, int casas[], int D);
int main() {
  int N;
  int casas[200000];
  int ini, fin, piv;
  int i;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%d", &casas[i]);
  }
  sort(casas, casas + N);
  ini = 0;
  fin = (1 << 30);
  while (ini != fin) {
    piv = (ini + fin) / 2;
    if (sePuede(N, casas, piv)) {
      fin = piv;
    } else {
      ini = piv + 1;
    }
  }
  printf("%lf\n", ini / 2.0);
  cosa(N, casas, ini);
  return 0;
}
bool sePuede(int N, int casas[], int D) {
  int pos, usd, posUlt;
  usd = 0;
  pos = 0;
  while (usd < 3 && pos < N) {
    posUlt = casas[pos];
    while (pos < N && casas[pos] <= posUlt + D) {
      pos++;
    }
    usd++;
  }
  return pos >= N;
}
void cosa(int N, int casas[], int D) {
  int pos, usd, posUlt;
  usd = 0;
  pos = 0;
  while (usd < 3 && pos < N) {
    posUlt = casas[pos];
    if (usd) printf(" ");
    printf("%lf", 1.0 * posUlt + (D / 2.0));
    while (pos < N && casas[pos] <= posUlt + D) {
      pos++;
    }
    usd++;
  }
  while (usd < 3) {
    if (usd) printf(" ");
    printf("1.000000");
    usd++;
  }
  printf("\n");
  return;
}