First, we consider the case that $t=1$, which is to say $\#\{1\}=0$ and $\#\{-1\}=2k-s$. The way we choose $2k-s$ positions from the total $2k$ (the total number of length is $4k$, only the odd ones are allowed for “-1”) is $$\begin{pmatrix}
     2k \\
     2k-s \\
   \end{pmatrix}.$$ Then since we have $s$ noncoincident $J$-vertices on the $J$-line, the noncoincident vertical edges is at most $2s$ (since we have each edge repeated at least twice). Therefore, the expectation $$\begin{aligned}
{\mathbb{E}}[\varepsilon_{j_1\,i_1}\varepsilon_{j_1\,i_2}\varepsilon_{j_2\,s+i_2}\varepsilon_{j_2\,s+i_3}\varepsilon_{j_3\,i_3}\varepsilon_{j_3\,i_4}\varepsilon_{j_4\,s+i_4}\varepsilon_{j_4\,s+i_5}\cdots \varepsilon_{j_{2k}\,s+i_{2k}}\varepsilon_{j_{2k}\,s+i_1}]\end{aligned}$$ is bounded by $M^{4k-2s}$. Therefore, we have the contribution to : $$\begin{aligned}
\label{11}
&\quad \quad\frac{1}{pT^{2k}}\sum_{s=1}^{2k-1}Tp^sM^{4k-2s}\begin{pmatrix}
                                                 2k \\
                                                 2k-s \\
                                               \end{pmatrix}\nonumber\\
&=\left(\frac{\sqrt p}{\delta^2}\right)^{2k}p^{k-1}T^{1-2k}\delta^{4k}\sum_{s=1}^{2k-1}\left(\frac{\sqrt p}{\delta^2}\right)^{s-2k}\begin{pmatrix}
                                                                           2k \\
                                                                           s \\
                                                                         \end{pmatrix}\nonumber\\
&=\left(\frac{\sqrt p}{\delta^2}\right)^{2k}p^{k-1}T^{1-2k}\delta^{4k}\sum_{i=1}^{2k-1}\left(\frac{\delta^2}{\sqrt p}\right)^{i}\begin{pmatrix}
                                                                           2k \\
                                                                           i \\
                                                                         \end{pmatrix}\nonumber\\
&\leq\left(\frac{\sqrt p}{\delta^2}\right)^{2k}p^{k-1}T^{1-2k}\delta^{4k}\sum_{i=1}^{2k-1}\left(\frac{2k\delta^2}{\sqrt p}\right)^{i}\nonumber\\
&=O\left(\left(\frac{\sqrt p}{\delta^2}\right)^{2k}p^{k-1}T^{1-2k}\delta^{4k}\frac{2k\delta^2}{\sqrt p}\right)\nonumber\\
&=O\left(\frac{2k\delta^2}{\sqrt p}y^{2k-1}\right)~,\end{aligned}$$ which is due to the fact that $k=(\log p)^{1.01}$.

Then consider the case that $t>1$. Since $\#\{1\}=t-1<\#\{-1\}=2k-s$, we can first construct a characteristic sequence that satisfies (1)(2)(3)(4)(5) in . Therefore, we have the degree of each $I$-vertex at least four and each edge in the $Q$-graph repeated exactly twice, which ensures that the $Q$-graph will not fall in Category 2. And the possible ways for constructing such a characteristic sequence is $f_{t-1}(k)$ by definition. Since in the 