2 - c_2$ and $f_1 \ge 2 + (c_1 - 1) d$. If $c_2 > 1$ then $\lceil L \rceil \ge C_1 + 2 C_2 + F_1 > -K_X$, whence $H^0(X, K_X + \lceil L \rceil) \ne 0$. So we may assume that $c_2 \le 1$. Then $c_1 \ge 1$ and $f_1 \ge 2$. Thus $\lceil L \rceil \ge C_1 + C_2 + 2F_1 \sim -K_X$, whence $H^0(X, K_X + \lceil L \rceil) \ne 0$.

Consider Case (iii). Since $L$ is a connected tree, we may assume that $C_1 . C_2 = 1$. So $C_2 \sim n(C_1 + d F) + F$ for some integer $n \ge 1$. Since $C_i$ ($i \ge 2$) is irreducible, we have $C_i \ge C_1 + d F$ by [@Ha]. If $k \ge 3$ or $n \ge 2$, then we see that $\lceil L \rceil \ge \sum_{i=1}^k C_i >  - K_X$. So assume that $k = 2$ and $n = 1$. By the inequalities displayed above, we have $c_1 \ge 2 - c_2$ and $c_2 \ge 2 + (c_1-1) d$. If $c_2 > 1$ then $\lceil L \rceil \ge C_1 + 2C_2 > -K_X$. So assume that $c_2 \le 1$. Then $c_1 \ge 2 - 