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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 make_name(text, delim=u'-', maxlength=50, checkused=None, counter=2):
 u""" Generate an ASCII name slug. If a checkused filter is provided, it will be called with the candidate. If it returns True, make_name will add counter numbers starting from 2 until a suitable candidate is found. :param string delim: Delimiter between words, default '-' :param int maxlength: Maximum length of name, default 50 :param checkused: Function to check if a generated name is available for use :param int counter: Starting position for name counter 'this-is-a-title' 'invalid-url-slug-here' 'this-that' 'this-that' 'how-bout-this' 'hows-that' 'k-d' 'billion-pageviews' 'hindii-slug' u'your-webapps-should-talk-not-just-in-english-but-in-espanol-kiswahili-guang-zhou-hua-and-asmiiyaa-too' 'name' 'how_about_this' 'and_that' 'umlauts-in-motorhead' 'candidate2' 'candidate1' 'candidate3' 'long-title-but-snipp' 20 'long-cand2' 'lankaran' 'example-example-com' 'trailing-d' 'trailing' 'test-this-newline' u'testing-an-emoji' 'testing-more-slashes' 'what-if-a-html-tag' 'these-are-equivalent-to-through' """

    name = text.replace('@', delim)
    name = unidecode(name).replace('@', 'a')  # We don't know why unidecode uses '@' for 'a'-like chars
    name = six.text_type(delim.join([_strip_re.sub('', x) for x in _punctuation_re.split(name.lower()) if x != '']))
    if isinstance(text, six.text_type):
        # Unidecode returns str. Restore to a unicode string if original was unicode
        name = six.text_type(name)
    candidate = name[:maxlength]
    if candidate.endswith(delim):
        candidate = candidate[:-1]
    if checkused is None:
        return candidate
    existing = checkused(candidate)
    while existing:
        candidate = name[:maxlength - len(str(counter))] + str(counter)
        counter += 1
        existing = checkused(candidate)
    return candidate