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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 equivalent_release_for_product(self, product):
 """ Returns the release for a specified product with the same channel and major version with the highest minor version, or None if no such releases exist """

        releases = self._default_manager.filter(
            version__startswith=self.major_version() + '.',
            channel=self.channel, product=product).order_by('-version')
        if not getattr(settings, 'DEV', False):
            releases = releases.filter(is_public=True)
        if releases:
            return sorted(
                sorted(releases, reverse=True,
                       key=lambda r: len(r.version.split('.'))),
                reverse=True, key=lambda r: r.version.split('.')[1])[0]