If $D$ is nef, then $h^1(D)=h^2(D)=0$

Question

Let $k$ be a number field and $S$ a smooth algebraic surface over $k$ which is birational to $\Bbb{P}^2_k$.

I'm trying to understand why the following is true:

If $D$ is nef divisor on $S$, then $h^1(S,\mathcal{O}_S(D))=h^2(S,\mathcal{O}_S(D))=0$.

[I've read this claim in the the lecutre notes of a course on arithmetic geometry]

Apparently this is supposed to be a trivial statement, but I can't prove it.

The only tools I know how to use here is Riemann-Roch for surfaces or Serre duality. But I don't what to do about $\chi(S,\mathcal{O}_S)$ and $\mathcal{O}_S(K_S-D)$.

I also keep wondering whether or not if $k$ being a number field and $S$ being birational to $\Bbb{P}^2_k$ are necessary in order to conclude that $h^1(D)=h^2(D)=0$.

Any help will be appreciated, thank you!