function dfs(int cur, int parent):
print(`(`)
for all nxt that cur is adjacent to:
dfs(nxt, cur)
print(`)`)
You might notice that Cuber QQ print a "(" when entering a node, and a ")" when leaving a node. So when he finishes this DFS, in his console, he will see a bracket sequence of length 2n2n, where nn is the number of vertices in the tree.
Obviously, if the tree is undirected and the nodes are unlabelled (meaning that all the nodes are treated equally), you can get a lot of different bracket sequences when you do the DFS. There are two reasons accounting for this. Firstly, when you are at cur, you can follow any permutation of the nodes that cur is adjacent to when you visit nxt. Secondly, the entrance to the tree, that is the root, is undeterministic when you start your DFS.
So Cuber QQ couldn't help wondering how many distinct bracket sequences he can get possibly. As the answer can be very large, output it modulo 998 244 353998 244 353.