the normalization map. Then

-   either $\Delta(\bar{X},\pi^*(K_X+\Lambda))=1$ and $(X,\Lambda)$ is as in Thm \[delta1,1\],

-   or $\Delta(\bar{X},\pi^*(K_X+\Lambda))=0$ and $(\bar{X},\bar{\Lambda}+\bar{D})$ is as in Thm \[delta1,0\].

If $X$ is reducible, write $X=\bigcup X_i$, where $X_i$ is an irreducible component. Then $\Delta(X_i,(K_X+\Lambda)|_{X_i})=1$ or $0$ for each component $X_i$, $X$ has a unique minimal connected component $U$ such that $\Delta(U,(K_X+\Lambda)|_{U})=1$, $X\setminus U$ is composed with several trees $T_j$ with $\Delta(T_j,(K_X+\Lambda)|_{T_j})=0$ and $X$ is glued by $U$ and $T_j$ along lines.

Moreover, $U$ is one of the followings:

-   $X=U$ is a string of log surfaces $X_i$ with $\Delta(X_i,(K_X+\Lambda)|_{X_i})=1$ and $|(K_{X}+\Lambda)|_{X_i}|$ composed with a pencil of elliptic curves. The connecting curves are all fibers. Moreover, in this case $|K_X+\Lambda|$ is composed with a pencil of elliptic curves.

-   $U$ is a string of surfaces glued along lines. The end surfaces $X_i$ of the string $U$ are non-normal with $\Delta(X_i,(K_X+\Lambda)|_{X_i})=1$.

-   $U$ is composed with a single irreducible log surface $X_k$ with $\Delta(X_k,(K_X+\Lambda)|_{X_k})=1$ and $(K_{X}+\Lambda)|_{X_k}$ very ample. Moreover, in this case $K_X+\Lambda$ is very ample.

-   $U=X_j\cup X_k $ where $X_j$, $X_k$ are two Gorenstein log surfaces with $\Delta(X_j,(K_X+\Lambda)|_{X_j})=\Delta(X_k,(K_X+\Lambda)|_{X_k})=0$. All the connecting curves of $X$ are lines except $X_j\cap X_k$.

-   $U$ is a cycle of 