Can a surjective continuous function from the reals to the reals assume each value an even number of times?

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and onto. Is it possible for $f$ to assume each of its values an even number of times?

To clarify, some values might be taken 2 times, some 4, some 6, etc., but always an even (and therefore finite) number. I don't require that there be a value that is assumed any particular number of times. For example, the function might never take on any value exactly twice.

Here is a closely related question with an excellent answer.

This is impossible. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is such that the preimage of every point contains an even number of points. There are only countably many points in $\mathbb{R}$ that are either local minima or local maxima of $f$ (since for each one, we can choose an interval with rational endpoints on which it is the unique global maximum/minimum). It follows that there exists some $a\in \mathbb{R}$ such that none of the preimages $x_1<x_2<\dots<x_n$ of $a$ are local minima or local maxima. That is, $f(x)-a$ changes sign at each $x_i$. Since $n$ must be even, it follows that $f(x)-a$ has the same sign on all of $\mathbb{R}\setminus [x_1,x_n]$. Since $f$ is bounded on $[x_1,x_n]$, it follows that $f$ is either bounded below or bounded above (depending on which sign it has on $\mathbb{R}\setminus [x_1,x_n]$).

(This argument is borrowed from this nice partial answer to a near-duplicate of your previous question.)