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 PYTHON):

Solution:
from bisect import bisect_left as L,bisect_right as R
F=lambda x:'%.6lf'%x
n=input()
a=sorted(map(lambda x:int(x)*2,raw_input().split()))
l,h=0,1<<31
while l<h:
  d=(l+h)//2
  s=d*2
  x,y=R(a,a[0]+s),L(a,a[-1]-s)
  if x<y and a[y-1]-a[x]>s:
    l=d+1
  else:
    h=d
print F(l/2.)
x,y=R(a,a[0]+l*2),L(a,a[-1]-l*2)
if x>y:x=y
print ' '.join(map(F,((a[0]+a[x-1])/4.,(a[x]+a[y-1])/4.,(a[y]+a[-1])/4.)))