Print a flag in as little lines as possible!

Question

Just for fun, I wanted to see how little lines I could solve this problem in, using python. I managed to condense my code into 9 lines.

size = int(input("Size: "))
a = [x for x in range(size-2, -1, -1)] if size % 2 == 0 else [x for x 
in range(size-2, 0, -1)]
b, counter = a[::-1], 0
print("\n"+"\\"+"#"*size+"/") #top
for i in a[::2]: print("#"+" "*counter+"\\"+" "*i +"/"+ " "*counter+ 
"#"); counter += 1 
if size % 2 != 0: print("#"+" "*counter+"X"+ " "*counter+ "#")
counter = int((size-2)/2)
for i in b[::2]: print("#"+" "*counter+"/"+" "*i +"\\"+ " "*counter+ 
"#"); counter -= 1
print("/"+"#"*size+"\\"+"\n") #bottom
as you can probably tell by the output the code is meant to produce 
flags based on an integer from the user

with even number flags looking like this

\####/
#\  /#
# \/ #
# /\ #
#/  \#
/####\

and odd number flags looking like this, with the cross in the middle

\#####/
#\   /#
# \ / #
#  X  #
# / \ #
#/   \#
/#####\

I just wanted to see if anyone else could beat me, one of my classmates did it in four lines. Good luck :)

code with the least number of lines and bytes wins!