Time cost of security reduction

Question

The following theorem is the description of one-more unforgeability in the paper lattice based blind signatures. I want to ask if the time $t^{'}$ is an exponential time since the $q_{H}$ and $q_{Sign}$ are about $2^{60}$. If the time $t^{'}$ is an exponential time then the adversary acturally cannot solve the hard problem, namely there is no contradiction exist.

Theorem 3.8 (One-more unforgeability. Let $\text{Sign}$ be the signature oracle. Let $T_{\text{Sign}}$ and $T_\text{H}$ be the cost functions for simulating the oracles $\text{Sign}$ and $\text{H}$, and let $c \lt 1$ be the probability for a restart in the protocol. $\text{BS}$ is $(t, q_\text{Sign}, q_\text{H}, \delta)$-one-more unforgeable if $\text{com}$ is $(t', \delta / 2)$-binding and $Col(\mathcal{H}(\mathcal{R}, m), D)$ is $(t', \delta / 2)$-hard with $t' = t + q_{\text{H}}^{q_\text{Sign}}(q_\text{Sign}T_\text{Sign} + q_\text{H}T_\text{H})$ and non-negligible $\delta'$ if $\delta$ is non-negligible.

Answer 1

Even if $q_H$ and $q_{Sign}$ are $\mathsf{poly}(n)$, the security guarantees turn out meaningless since the loss is exponential in $q_{Sign}$. As the author points out in the paragraph (Page 18) following the proof of the theorem, $q_{Sign}$ has to be restricted to $o(n)$ so that security can be based on sub-exponential hardness of the underlying lattice problem.