the hierarchies between $v_{\alpha\gamma}^{+1}$ and $v_{\beta\gamma}^{+1}$, and $v_{\alpha\beta}^{+1}$ and $v_{\beta\gamma}^{-1}$ have length less than $8M^2$.

Now, make a hierarchy triangle $v_{\alpha\beta}^{+1}v_{\alpha\gamma}^{-1}v_{\beta\gamma}^{+1}$ in ${\mathcal{P}}(S \backslash v)$, see Figure \[links are thin\] for how this fits in with above. By Theorem \[hierarchy k-centered\], $v_{\alpha\beta}^{+1}v_{\alpha\gamma}^{-1}v_{\beta\gamma}^{+1}$ in ${\mathcal{P}}(S \backslash v)$ is $8,900$ centered, call the point at the center $z$. Then by Theorem \[hierarchy k-centered\] and Lemma \[centered to thin\], the hierarchy triangles $v_{\alpha\beta}^{+1}v_{\alpha\beta}^{-1}v_{\alpha\gamma}^{-1}$, $v_{\beta\gamma}^{-1}v_{\beta\gamma}^{+1}v_{\alpha\gamma}^{+1}$, and $v_{\alpha\gamma}^{-1}v_{\alpha\gamma}^{+1}v_{\beta\gamma}^{+1}$ are $35,600$ thin. Therefore $z$ is at most $124,500$ away from each $[v_*^{+1}, v_{*}^{-1}]_{S \backslash v}$. This implies that $\{z, v\}$ is at most $124,500$-centered in the relative 3-archy triangle $\alpha\beta\gamma$.

**Case 2:** For the same reasons as in Theorem \[hierarchy k-centered\] case 2, this case can be reduced to case 3.

**Case 3:** This proceeds with the same strategy as in case 3 of Theorem \[hierarchy k-centered\]. By Theorem \[curve hyp\], we know the triangle of main geodesics, $g_{\alpha\beta}g_{\beta\gamma}g_{\alpha\gamma}$ in ${\mathcal{C}}(S)$ is $17$-centered. Let $c$ be the curve that is at the center of this triangle. Connect $c$ to $g_{\alpha\beta}$, $g_{\beta\gamma}$, and $g_{\alpha\gamma}$ with a geodesic in ${\mathcal{C}}(S)$. Define $v_{\alpha \beta}$ to be the vertex in $g_{\alpha \beta}$ which is the least distance to $c$, and 