Understanding __call__ with metaclasses

From my understanding the __call__ method inside a class implements the function call operator, for example:

class Foo:
    def __init__(self):
        print("I'm inside the __init__ method")

    def __call__(self):
        print("I'm inside the __call__ method")

x = Foo() #outputs "I'm inside the __init__ method"
x() #outputs "I'm inside the __call__ method"

However, I'm going through the Python Cookbook and the writer defined a metaclass to control instance creation so that you can't instantiate an object directly. This is how he did it:

class NoInstance(type):
    def __call__(self, *args, **kwargs):
        raise TypeError("Can't instantaite class directly")


class Spam(metaclass=NoInstance):
    @staticmethod
    def grok(x):
        print("Spam.grok")

Spam.grok(42) #outputs "Spam.grok"

s = Spam() #outputs TypeError: Can't instantaite class directly

However, what I don't get is how s() wasn't called, yet it's __call__ method was called. How does this work?