$D$ a divisor numerically 1-connected implies that $H^0(\mathcal{O}_D)$=constants

Question

I'm trying to understand the following:

Let $X$ be a nonsingular complex projective surface and $D$ an effective divisor. Then $D$ numerically 1-connected implies that $H^0(\mathcal{O}_D)=constants$.

Suppose that $D_1 \subset D$ is a maximal divisor for which $h^0(\mathcal{O}_{D_1})=1$. I can assume this since for every irreducible component $\Gamma\subset D$ it holds $h^0(\mathcal{O}_{\Gamma})=1$. I claim that $D_1=D$. If not, then $D>D_1>0$ and so, by 1-connectedness, $D_1(D-D_1)\geq1$ and i can find a component $\Gamma \subset D-D_1$ such that $D_1\Gamma\geq 1$. Now i want to show that $h^0(\mathcal{O}_{D_1+\Gamma})=1$ so to get a contradiction with maximality.

I know that there exists a short exact sequence $0 \to \mathcal{O}_{\Gamma}(-D_1) \to \mathcal{O}_{D_1+\Gamma} \to \mathcal{O}_{D_1} \to 0$, where $\mathcal{O}_{\Gamma}(-D_1):=\mathcal{O}_X(-D_1)\otimes \mathcal{O}_{\Gamma}$.

This sequence induces the exact equence

$H^0(\mathcal{O}_{\Gamma}(-D_1))\to H^0(\mathcal{O}_{D_1+\Gamma})\to H^0(\mathcal{O}_{D_1})$.

I know that the last term is one-dimensional but i cannot see why the first term is zero, so to conclude. I think that this should come by the fact that $h^0(\mathcal{O}_X(-D_1))=0$ but even using the definition of sheafification (for $\mathcal{O}_{\Gamma}(-D_1))$) i cannot see it.

Thanks in advance.