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 $$$\mathcal{O}(n + m)$$$.

As the graph is a cactus graph, it is no doubt that one edge of each cycle needs to be removed in order to make a spanning tree. Hence, the problem can be rewritten as that we have $$$M$$$ ($$$M \leq \frac{m}{3}$$$) arrays consisting of integers and we want to pick a number from each array so that we can get their sum, however, we want you to find possible ways with the largest $$$K$$$ sums and report the sum of these values.

It is a classic problem that can be solved by sequence merge algorithms, for example, we can maintain a set of $$$K$$$ largest values obtained from the first $$$x$$$ arrays, and then merge it with the $$$(x + 1)$$$-th array and find $$$K$$$ largest new values by using a heap or priority queue. More specifically, let's assume that we are going to merge two non-increasing arrays $$$A$$$, $$$B$$$ and pick $$$K$$$ largest values obtained from $$$(A_i + B_j)$$$. As we know $$$B_j \geq B_{j + 1}$$$, so if we pick $$$(A_i + B_{j + 1})$$$, we must pick $$$(A_i + B_j)$$$ first. That inspires us to set a counter $$$c_i$$$ for each $$$A_i$$$, which represents if we will pick $$$A_i$$$ with some value, we are going to pick $$$(A_i + B_{c_i})$$$. Then, we can use a data structure to maintain the set of all possible $$$(A_i + B_{c_i})$$$ and then query and erase the largest value, which we can just repeat at most $$$K$$$ times to get all the merged values we need.

It seems the above method runs in time complexity $$$\mathcal{O}(MK \log K)$$$, but actually, it can run faster. Let the sizes of these arrays are $$$m_1$$$, $$$m_2$$$, $$$\ldots$$$, $$$m_M$$$ ($$$m_i \geq 3$$$ for each $$$i$$$) respectively. If we instead to maintain $$$(A_{c_j} + B_j)$$$ in the data structure, where $$$B$$$ is the next array to be merged, the complexity will be $$$\mathcal{O}\left(\sum_{i = 1}^{M}{K \log{m_i}}\right) = \mathcal{O}\left(K \log{\prod_{i = 1}^{M}{m_i}}\right) = \mathcal{O}\left(M K \log{\frac{\sum_{i = 1}^{M}{m_i}}{M}}\right) = \mathcal{O}\left(M K \log{\frac{m}{M}}\right)$$$. As $$$M \leq \frac{m}{3}$$$, we can conclude the worst complexity is $$$\mathcal{O}(m K)$$$, when $$$M = \frac{m}{3}$$$.

By the way, there exist solutions in time complexity $$$\mathcal{O}(K \log K)$$$.

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