Create a mapped type from an interface with overloaded functions where only the properties that match a specific type is returned?

Question

Consider

interface MagicMonkey {
   goCrazy(): Magic
   goCrazy(): Magic & string
   dance(): Magic
   dance(): Magic & string
   state: "Full" | "Hungry" | "Tired"
}

I want to create a mapped type when applied to the interface above results in something that strictly enforces the below interface

interface MagicMonkey {
    goCrazy: Magic & string
    dance: Magic & string
}

I've managed to map all the return types by isolating the methods using the trick that was suggested in this question. Like so:

type AnyMethod = (...args: any[]) => any;

type MatchingPropertyKeys<T, U> = ({
    [P in keyof T]: T[P] extends U ? P : never
})[keyof T];

type JustMethodKeys<T> = MatchingPropertyKeys<T, AnyMethod>;
type JustMethods<T> = Pick<T, JustMethodKeys<T>>;

type MagicMonkeyMethods = JustMethods<MagicMonkey>

type MagicMonkeyMethodReturnTypes = {
   [P in keyof MagicMonkeyMethods]: ReturnType<MagicMonkeyMethods[P]>
}

But it seems to be happy with either implementation of the overloaded methods return types, so the interface below also adheres to the type.

interface MagicMoneyMethods {
   goCrazy: Magic & string;
   dance: Magic;
}

Which makes sense. But nothing I've tried to apply to filter it seems to work. Because I always end up with a set of keys I can use to Pick from the original type, which just gets me back to the same place.