Maze generating one liner The famous C64 basic one liner

10 PRINT CHR$(205.5+RND(1)); : GOTO 10


prints a maze of slashes and backslashes.

\\/\\\//\/\////\\/\/
\/\///\\///////\//\/
/\\\//\//\////\\//\\
\////\//\//\/\\\\\\/
/\/\\///\\\\/\\\\/\\
\/\//\\\\\\//\/\////
/\//\\///\/\///\////
\/\\\//\\/\\\//\\/\/
//////\\/\\/\/\/\///
\\/\/\\////\/\/\\/\/


Read in such maze made of diagonal walls from stdin and print out the same maze with horizontal and vertical walls consisting of the wall character "#"

For example the small maze

/\\
\\/
///


translates to 

     #####
     #   #
     # # # #
     # # # #
 ##### # # #
       #   #
   #########

     #####    


To be precise, each isolated wall segment has the length of five characters, adjacent wall segments share a corner. Moving a character to the right/left/top/down in the matrix of slashes and backslashes corresponds to a diagonal translation by 2 characters in vertical and 2 characters in horizontal direction in the #-matrix.
Python 3, 226 224 bytes

My first Python golf, so probably very sub-optimal.
It produces a whole lot of trailing whitespace, but there are no preceding newlines, and at most two preceding spaces.
The input needs to be given by hand from command line (maybe someone knows a shorter way to get multiline input in Python...).

e,z,s=enumerate,'0',list(iter(input,""))
p=''.join(s)*5
r=[len(p)*[' ']for _ in p]
for y,l in e(s):
 for x,c in e(l):
  for i in range(-2,3):r[2*(x+y+(s&gt;[z]))+i*(c&gt;z)][2*(x+len(s)-y)+i*(c&lt;z)]='#'
for l in r:print(''.join(l))


The idea is to initialize a huge array of spaces r, then iterate through the input and replace the spaces with # as needed, and finally print the whole array.
A trick I used is to compare characters to z = '0' instead of testing equality to '/' or '\', which saves a bunch of bytes.