$k\in[1,q]$, we choose a countable subset $\{a_{k,i}\}_{i}$ of $E_n(c,\bold i)$ with the following properties:

-   the intervals $\{I_{t_k}(a_{k,i})\}_{i}$ are pairwise disjoint and $E_n(c,\bold i)\subset \bigcup _{i}L^{-r_k/3}\cdot
    {I}_{t_k}(a_{k,i});$

-   for each $k\in[2,q]$ and $a_{k,i}$ there exists $a_{k-1,j}$ such that ${I}_{t_k}(a_{k,i})\subset
    2L^{-r_{k-1}/3}\cdot{I}_{{t_{k-1}}}(a_{k-1,j})$.

Observe that the desired estimate follows from this.

For the definition of such a subset we need two combinatorial statements. The following elementary fact from Lemma \[exp\] is used in the proofs of these two sublemmas: if $a\in E_n(c,\bold i)$, then $(\gamma_{t_k}^{(c)}|I_{t_k}(a))^{-1}(c^{(k)})$ consists of a single point and is contained in $L^{-r_k/3}\cdot I_{t_k}(a)$.

\[lem1\] If $a$, $a'\in E_n(c,\bold i)$ and $a'\notin I_{t_k}(a)$, then $I_{t_k}(a)\cap I_{t_k}(a')=\emptyset$ .

Suppose $I_{t_k}(a)\cap I_{t_k}(a')\neq\emptyset$. Lemma \[dist\] gives $|I_{t_k}(a)|\approx |I_{t_k}(a')|$. This and $a'\in I_{t_k}(a)$ imply $(\gamma_{t_k}^{(c)}|I_{t_k}(a))^{-1}(c^{(k)})\neq(\gamma_{t_k}^{(c)}|I_{t_k}(a'))^{-1}(c^{(k)}).$ On the other hand, by the definition of the intervals $I_{t_k}(\cdot)$, $\gamma_{t_k}^{(c)}$ is injective on $I_{t_k}(a)\cup I_{t_k}(a')$. A contradiction arises.

\[lem2\] If $a$, $a'\in E_n
(c,\bold i)$ and $a'\in L^{-r_k/3}\cdot I_{k}(a)$, then $I_{t_{k+1}}(a')\subset
2L^{-r_k/3}\cdot I_{t_k}(a)$.

We have $(\gamma_{t_k}^{(c)}|I_{t_k}(a))^{-1}(c^{(k)})
\notin I_{t_{k+1}}(a')$, for otherwise the distortion of $\gamma_{\nu_{k+1}}^{(c)}$ on $I_{t_{k+1}}(a')$ is unbounded. This and the assumption together imply that one of the connected components of $I_{t_{k+1}}(a')-\{a'\}$ is contained in $L^{-r_k/3}\cdot I_{t_k}(a)$. This implies the inclusion.

We are in position to choose subsets 