Counting DFS Trees

You are studying the Depth First Search (DFS) tree traversal algorithm. Consider a rooted tree of $n$ nodes, numbered from $1$ to $n$, with depth $K$ (levels numbered from $1$ to $K$). The root, the node at depth $1$, is node $1$. All leaves are located at the same depth, that is, depth $K$. Node $i$ has an array of children $c_i$, which may be empty if $i$ is a leaf. The algorithm is written in pseudocode below.

DFS(u, depth):
  let res be an empty array
  append depth to res

  for each v in c[u]:
    let D be an array initialized with DFS(v, depth + 1)
    for each x in D:
      append x to res

return res

In the trees shown in the illustration, the return values of DFS(1, 1) are $[1, 2, 3, 3, 3]$ for the left tree and $[1, 2, 3, 2, 3]$ for the right tree.

Let $f_K(n)$ be the number of distinct return values of DFS(1, 1) over all trees with $n$ nodes in which all leaves are at depth $K$. You are given $M$ integers $A_1, A_2, \dots, A_M$. Find the value of $f_K(A_1) \times f_K(A_2) \times \cdots \times f_K(A_M)$ modulo $998\,244\,353$.

The first line contains two integers $K$ and $M$ ($1 \le K, M \le 100\,000$). The second line contains $M$ integers $A_1, A_2, \dots, A_M$ ($K \le A_i \le 200\,000$).

Print the value of $f_K(A_1) \times f_K(A_2) \times \cdots \times f_K(A_M)$ modulo $998\,244\,353$.