AT_abc452_e [ABC452E] You WILL Like Sigma Problem

> Your lucky Greek letter for today is sigma. Solve this problem that uses sigma twice, and good fortune will surely come your way. You are given a sequence of positive integers $ A = (A_1, \cdots, A_N) $ of length $ N $ and a sequence of positive integers $ B = (B_1, \cdots, B_M) $ of length $ M $ . Find the value, modulo $ 998244353 $ , of $ \displaystyle \sum_{i=1}^{N} \sum_{j=1}^{M} A_i \cdot B_j \cdot (i \bmod j) $ .

The input is given from Standard Input in the following format: > $ N $ $ M $ $ A_1 $ $ A_2 $ $ \cdots $ $ A_N $ $ B_1 $ $ B_2 $ $ \cdots $ $ B_M $

Output the answer on a single line.

### Sample Explanation 1 The sum of the following $ 24 $ values is $ 508 $ . - $ A_1 \cdot B_1 \cdot (1 \bmod 1) = 1 \cdot 1 \cdot 0 = 0 $ - $ A_1 \cdot B_2 \cdot (1 \bmod 2) = 1 \cdot 10 \cdot 1 = 10 $ - $ A_1 \cdot B_3 \cdot (1 \bmod 3) = 1 \cdot 3 \cdot 1 = 3 $ - $ A_1 \cdot B_4 \cdot (1 \bmod 4) = 1 \cdot 7 \cdot 1 = 7 $ - $ A_2 \cdot B_1 \cdot (2 \bmod 1) = 6 \cdot 1 \cdot 0 = 0 $ - $ A_2 \cdot B_2 \cdot (2 \bmod 2) = 6 \cdot 10 \cdot 0 = 0 $ - $ A_2 \cdot B_3 \cdot (2 \bmod 3) = 6 \cdot 3 \cdot 2 = 36 $ - $ A_2 \cdot B_4 \cdot (2 \bmod 4) = 6 \cdot 7 \cdot 2 = 84 $ - $ A_3 \cdot B_1 \cdot (3 \bmod 1) = 9 \cdot 1 \cdot 0 = 0 $ - $ A_3 \cdot B_2 \cdot (3 \bmod 2) = 9 \cdot 10 \cdot 1 = 90 $ - $ A_3 \cdot B_3 \cdot (3 \bmod 3) = 9 \cdot 3 \cdot 0 = 0 $ - $ A_3 \cdot B_4 \cdot (3 \bmod 4) = 9 \cdot 7 \cdot 3 = 189 $ - $ A_4 \cdot B_1 \cdot (4 \bmod 1) = 2 \cdot 1 \cdot 0 = 0 $ - $ A_4 \cdot B_2 \cdot (4 \bmod 2) = 2 \cdot 10 \cdot 0 = 0 $ - $ A_4 \cdot B_3 \cdot (4 \bmod 3) = 2 \cdot 3 \cdot 1 = 6 $ - $ A_4 \cdot B_4 \cdot (4 \bmod 4) = 2 \cdot 7 \cdot 0 = 0 $ - $ A_5 \cdot B_1 \cdot (5 \bmod 1) = 3 \cdot 1 \cdot 0 = 0 $ - $ A_5 \cdot B_2 \cdot (5 \bmod 2) = 3 \cdot 10 \cdot 1 = 30 $ - $ A_5 \cdot B_3 \cdot (5 \bmod 3) = 3 \cdot 3 \cdot 2 = 18 $ - $ A_5 \cdot B_4 \cdot (5 \bmod 4) = 3 \cdot 7 \cdot 1 = 21 $ - $ A_6 \cdot B_1 \cdot (6 \bmod 1) = 1 \cdot 1 \cdot 0 = 0 $ - $ A_6 \cdot B_2 \cdot (6 \bmod 2) = 1 \cdot 10 \cdot 0 = 0 $ - $ A_6 \cdot B_3 \cdot (6 \bmod 3) = 1 \cdot 3 \cdot 0 = 0 $ - $ A_6 \cdot B_4 \cdot (6 \bmod 4) = 1 \cdot 7 \cdot 2 = 14 $ ### Constraints - $ 1 \leq N,M \leq 5 \times 10^5 $ - $ 1 \leq A_i, B_j \leq 5 \times 10^5 $ - All input values are integers.