\\
\mbox{(iii)} & 3m^2 +mnd + n^2(g-1) \geq 0
\end{array}$$ We claim that there exist no divisors $D\subset S$ with $D^2 > 0$, satisfying (i)-(iii).

[*Case $n<0$.* ]{} From (iii), we have that either $m < -n$ or $m >-\frac{2a}{3} n$. In the first case, by using inequality (i) we get $$2< -6n +d n =n(2a - 3),$$ which is a contradiction since $n<0$ and $a\geq5$. Suppose $m > - 2an/3>0$. Inequality (ii) implies that $n(2-d/3)<1$, that is, $ (-n) (2a-3) < 3 $. Hence $2a-3 <3 $, which contradicts the hypothesis $a \geq 5$.

[*Case $n>0$.* ]{} Again, from condition (iii), we have either $m<-\frac{2a}{3}n$ or $m> -n$. In the first case, using (i) we obtain $2 < n(d-4a)$, which is impossible since $d=2a+3<4a$. Suppose now that $-n<m<0$. From (ii) we have that $2a(2n-1) \leq  -md <  nd$, which implies $n < \frac{2a}{ 4a-d} = \frac{2a}{2a-3} < 2.$ Then $n=1 > -m >0$, therefore the case $n>0$ does not occur.

[*Case $n=0$.* ]{} From (ii), $m \leq \frac{g-1}{d} = \frac{2a}{2a+3} <1$, but this yields a contradiction since, $m>0$. This completes the proof of the claim.

We are left with checking that $\mbox{Cliff}(\OO_C(D))\geq a$, for all primitive effective classes $D\in 