I have canonical divisor $K$ of curve $\mathbb{P}^1$, and I would like to find $l(2K)$, $l(3K)$ and $l(-K)$ using Riemann-Roch theorem. I know that $g=0$ in this case, so $deg(K)=2\cdot 0-2=-2$.
My only idea is trying something like this $$l(2K)=deg(2K)+1-g+l(K-2K)$$ I know that $deg(2K)=2deg(K)=-4$ so I get $$l(2K)=-3+l(-K)$$ but I can't think of a way to get just one of them? After that it is easy to calculate other two.