$ {\widetilde}{N}_0(D) $ such that for any $ N {\geqslant}{\widetilde}{N}_0(D) $, we have $$\label{e.ProbabilityTau}
{\mathbb{P}}(\tau=1) > 1-N^{-D}.$$

*Step 1.* To see the claim would be enough, we first show the following sets inclusion $$\label{e.SetsInclusion}
\{ \tau=1 \} \bigcap_{\substack{1{\leqslant}l,m {\leqslant}N^{10} \\ -N {\leqslant}k {\leqslant}N}} A_{l,m,k} {\subseteq}\bigcap_{z \in {\mathscr{S}}, 0<t<1} \left\{ {\text{\rm Im}\hspace{0.1cm}}f_t(z) \lesssim \varphi \dfrac{\kappa(E)^{1/2}}{\max(\kappa(E)^{1/2},t)} \right\},$$ where $$A_{l,m,k} := \left\{ \sup_{t_l {\leqslant}u {\leqslant}t_{l+1}} \left| \int_{t_l}^u \dfrac{e^{-\frac{s}{2\xi}} {\mathfrak{v}}_k(s) {\mathrm{d}}B_k(s)}{(z^{(m)} - x_k(s))^2} \right| <N^{-3} \right\}.$$ To prove this, for any given $ z $ and $ t $, choose $ t_l $ and $ z^{(m)} $ such that $ t_l {\leqslant}t <t_{l+1} $ and $ |z-z^{(m)}|<N^{-5} $. Note that by rigidity and the maximum principle (Lemma \[l.MaxPrin\]) we have $ |{\mathfrak{v}}_k(t)| \lesssim \varphi N^{-2/3} $. Combining this with the definition of $ f_t $, we have $ |f_t(z) - f_t(z^{(m)})| < N^{-2} $. Moreover, note that we have the following estimates $$\begin{gathered}
\sup_{t_l {\leqslant}u {\leqslant}t_{l+1}} \left| \int_{t_l}^u \dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{-N {\leqslant}k {\leqslant}N} \dfrac{{\mathfrak{v}}_k}{(x_k-z)^2(x_k+z)} {\mathrm{d}}t \right|\\
\lesssim N^{-10}  \dfrac{1}{2N} \varphi N^{-2/3} \max_{t_l {\leqslant}t {\leqslant}t_{l+1}} \sum_{-N {\leqslant}k {\leqslant}N} \left| \dfrac{1}{(x_k-z)^2(x_k+z)} \right| < N^{-5},\end{gathered}$$ and $$\begin{gathered}
\sup_{t_l {\leqslant}u {\leqslant}t_{l+1}} \left| \int_{t_l}^u \left( 1 - \dfrac{1}{\xi} \right) e^{-\frac{t}{2\xi}} \sum_{-N {\leqslant}k {\leqslant}N} \dfrac{z {\mathfrak{v}}_k}{x_k^2(x_k -z) (x_k+z)} {\mathrm{d}}t \right|\\
\lesssim N^{-10} 