The answer is $$$\left \lfloor \log_{10} (2^m - 1) \right \rfloor$$$. Apparently, there does not exist $$$10^k = 2^m$$$ for positive integers $$$k$$$, $$$m$$$, which results in $$$\left \lfloor \log_{10} (2^m - 1) \right \rfloor = \left \lfloor \log_{10} 2^m \right \rfloor = \left \lfloor m \log_{10} 2 \right \rfloor$$$.

Each letter's contribution to the answer can be treated as a number in the base of $$$26$$$, so the problem is equivalent to multiply these contributions with weights ranged from $$$0$$$ to $$$25$$$ and thus make the answer maximum. Obviously, the largest contribution matches $$$25$$$, the second largest matches $$$24$$$, and so on. Doing this greedy algorithm by sorting could be enough, except for the only case where it might cause leading zeros, which is not permitted.

Time complexity is $$$\mathcal{O}\left(\sum_{i = 1}^{n}{|s_i|} + \max(|s_1|, |s_2|, \ldots, |s_n|) \log C\right)$$$, where $$$C = 26$$$.

A permutation can be decomposed into one or more disjoint cycles, which are found by repeatedly tracing the application of the permutation on some elements.

For each element $$$i$$$ in a $$$l$$$-cycle of permutation $$$a$$$, we have

which implies $$$f(i)$$$ must be some element in a cycle of permutation $$$b$$$, whose length is a factor of $$$l$$$.

Besides, if $$$f(i)$$$ is fixed, the other $$$(l - 1)$$$ numbers in this cycle can be determined by $$$f(i) = b_{f(a_i)}$$$.

Consequently, the answer is $$$\prod_{i = 1}^{k} \sum_{j | l_i} {j \cdot c_j}$$$, where $$$k$$$ is the number of cycles obtained from permutation $$$a$$$, $$$l_i$$$ represents the length of the $$$i$$$-th cycle, and $$$c_j$$$ indicates the number of cycles in permutation $$$b$$$ whose length are equal to $$$j$$$.

Due to $$$\sum_{i = 1}^{k} \sum_{j | l_i}{1} \leq \sum_{i = 1}^{k}{2 \sqrt{l_i}} \leq 2 \sqrt{k \sum_{i = 1}^{k}{l_i}} \leq 2 n$$$, the time complexity is $$$\mathcal{O}(n + m)$$$.

The answers for fixed $$$n$$$ form an almost periodic pattern. It is not difficult to find out that is $$$\underbrace{1, 2, \ldots, n}_{n \text{ numbers}}$$$, $$$\underbrace{1, 2, \ldots, n - 1}_{(n - 1) \text{ numbers}}$$$, $$$\underbrace{1, 2, \ldots, n - 2, n}_{(n - 1) \text{ numbers}}$$$, $$$\underbrace{1, 2, \ldots, n - 1}_{(n - 1) \text{ numbers}}$$$, $$$\underbrace{1, 2, \ldots, n - 2, n}_{(n - 1) \text{ numbers}}$$$, $$$\ldots$$$