k f(k-1)\\
&\geq (2\alpha -1.5)(\beta \log n)f(k)n + 2.
\end{split}$$

For the first inequality, it seems we have overcounted the weight. we explain the reason. When $(x, x^{'})$ is rooted at $\gamma_i$ and $(y,y^{'})$ lies in some $\gamma_j$ if we extends $(x, x^{'})$ to $(y,y^{'})$ to complete a $\overline{C_4}$-type cycle $\gamma=(x,
x^{'},y,y^{'})$, the contribution to the weight is only $n_i^{-1}f(k-1)$ in the above inequality. Similarly, when $(x,
x^{'})$ is rooted at the same position as $(y,y^{'})$ in $\gamma_j$ and $(y,y^{'})$ lies in the same position as $(x, x^{'})$ in $\gamma_i$, the contribution to the weight is $n_j^{-1}f(k-1)$. Plus these two weight, $(n_j^{-1}+n_i^{-1})f(k-1)$ is exactly $w(E,E^{'})$ needed to be considered by the definition of $w$, where $E\oplus E^{'}=\gamma$. Hence, though each $\overline{C}_4$ cycle is counted twice, the weight not. The last inequality follows immediately from $k\geq 2$ and $f(k-1)\geq 1$.    $\Box$\
\
**Proof of Claim 2.** For each $E\in F_k$, we now proceed to bound $w^{-}(E)$. Let $(E^{'},E)$ be an arc in $K$. It is straightforward to verify that the $\overline{C_4}$-type $\gamma=(x,
x^{'},y,y^{'})= E\oplus E^{'}$ must contain two edges $(x, x^{'})$ and $(y,y^{'})$ from a single $\gamma_i$ of $E$, and $(y,x^{'}),
(x,y^{'})\in E^{'}$. Removing these two edges from $\gamma_i$ leaves a 