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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _etextno_to_uri_subdirectory(etextno):
 """Returns the subdirectory that an etextno will be found in a gutenberg mirror. Generally, one finds the subdirectory by separating out each digit of the etext number, and uses it for a directory. The exception here is for etext numbers less than 10, which are prepended with a 0 for the directory traversal. '0/1' '1/19' '1/5/4/5/15453' """

    str_etextno = str(etextno).zfill(2)
    all_but_last_digit = list(str_etextno[:-1])
    subdir_part = "/".join(all_but_last_digit)
    subdir = "{}/{}".format(subdir_part, etextno)  # etextno not zfilled
    return subdir