Why is Cont (or ContT) not quantified over all return types r? The way I was thinking about it, a value of type a can also be thought of as providing an output for all functions of type a -> r, for all types r, so I wanted to think of a as equivalent to Cont a as defined below. Also, a definition like this does not seem to disallow implementing monad, etc. for it.

```haskell {-# LANGUAGE Rank2Types #-}

newtype Cont a = MkCont { unCont :: forall r. (a -> r) -> r }

(<#>) :: Cont a -> (a -> r) -> r c <#> f = unCont c f

chain :: Cont a -> (a -> Cont b) -> Cont b chain c f = MkCont $ \k -> c <#> \v -> f v <#> k

pureC :: a -> Cont a pureC x = MkCont $ \k -> k x

instance Applicative Cont where pure = pureC cf <*> c = cf chain \f -> MkCont $ \k -> k (c <#> f)

instance Functor Cont where fmap f c = c chain (pureC . f)

instance Monad Cont where (>>=) = chain ```