For convenience, we assume that for \(0 \le i \le N-1\), the \(i\)-th (\(0\)-indexed) lowest digits of \(A\) and \(B\) are different.

Let \(a_i\) and \(b_i\) be the \(i\)-th lowest digits of \(A\) and \(B\), respectively. Then, we have:

\(\begin{aligned} A \times B &= \sum_{i=0}^{N-1} \sum_{j=0}^{N-1} a_ib_j10^{i+j} \\ &= \sum_{i=0}^{N-1} a_ib_i10^{2i} + \sum_{i=0}^{N-1} \sum_{j=i+1}^{N-1} (a_ib_j+a_jb_i) 10^{i+j}. \end{aligned}\)

The value \(\sum_{i=0}^{N-1} a_ib_i10^{2i}\) does not change no matter what we do. Our objective is to minimize \(\sum_{i=0}^{N-1} \sum_{j=i+1}^{N-1} (a_ib_j+a_jb_i) 10^{i+j}\).

Here, which of \(a_ib_j+a_jb_i\) and \(a_ib_i + a_jb_j\) is the smaller? If \(a_i-a_j\) and \(b_i-b_j\) have the same sign, the former; otherwise, the letter.

Therefore, we can minimize \(A \times B\) by specifying the smaller of \(a_i\) and \(b_i\) as \(a_i\) and the larger as \(b_i\).

Now, we can find \(\sum_{i=0}^{N-1} a_i10^i\) and \(\sum_{i=0}^{N-1} b_i10^i\) modulo \(998244353\) and compute their product to solve the problem in \(\mathrm{O}(N)\) time.