Resolve invalid names when scalar! is applied to a type with a type parameter

[IvanUkhov]

Hello,

I understand that scalar! has an argument for controlling the name, but by default, it generates an invalid name when applied to a type with a type parameters, such as scalar!(Foo<Bar>) resulting in "name": String("Foo<Bar>") in the schema introspection.

[sunli829]

rename to FooBar

scalar!(Foo<Bar>, "FooBar");