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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 get_pages(url):
 """ Return the 'pages' from the starting url Technically, look for the 'next 50' link, yield and download it, repeat """

    while True:
        yield url
        doc = html.parse(url).find("body")
        links = [a for a in doc.findall(".//a") if a.text and a.text.startswith("next ")]
        if not links:
            break
        url = urljoin(url, links[0].get('href'))