According to Python 2.7.12 documentation, 3.4.3. Customizing class creation:

__metaclass__ This variable can be any callable accepting arguments for name, bases, and dict. Upon class creation, the callable is used instead of the built-in type().

New in version 2.2.

However, this article argues:

Q: Wow! Can I use any type object as the __metaclass__?

A: No. It must be a subclass of the type of the base object. ...

So I did an experiment on my own:

class metacls(list):    # <--- subclassing list, rather than type
    def __new__(mcs, name, bases, dict):
        dict['foo'] = 'metacls was here'
        return type.__new__(mcs, name, bases, dict)

class cls(object):
    __metaclass__ = metacls
    pass

This gives me:

Traceback (most recent call last):
  File "test.py", line 6, in 
    class cls(object):
  File "test.py", line 4, in __new__
    return type.__new__(mcs, name, bases, dict)
TypeError: Error when calling the metaclass bases
    type.__new__(metacls): metacls is not a subtype of type

So is the document really wrong?