A12281 | Lunar New Year and a Recursive Sequence

Lunar New Year is approaching, and Bob received a gift from his friend recently — a recursive sequence! He loves this sequence very much and wants to play with it.

Let $f_1, f_2, \ldots, f_i, \ldots$ be an infinite sequence of positive integers. Bob knows that for $i > k$ , $f_i$ can be obtained by the following recursive equation:

$$$$f_i = \left(f_{i - 1} ^ {b_1} \cdot f_{i - 2} ^ {b_2} \cdot \cdots \cdot f_{i - k} ^ {b_k}\right) \bmod p, $$ </p><p>which in short is</p><p> $$ f_i = \left(\prod_{j = 1}^{k} f_{i - j}^{b_j}\right) \bmod p, $$ </p><p>where $p = 998\\,244\\,353$ (a widely-used prime), $b\_1, b\_2, \\ldots, b\_k$ are known integer constants, and $x \\bmod y$ denotes the remainder of $x$ divided by $y$ .</p><p>Bob lost the values of $f\_1, f\_2, \\ldots, f\_k$ , which is extremely troublesome – these are the basis of the sequence! Luckily, Bob remembers the first $k - 1$ elements of the sequence: $f\_1 = f\_2 = \\ldots = f\_{k - 1} = 1$ and the $n$ -th element: $f\_n = m$ . Please find any possible value of $f\_k$$$. If no solution exists, just tell Bob that it is impossible to recover his favorite sequence, regardless of Bob's sadness.

The first line contains a positive integer $k$ ( $1 \leq k \leq 100$ ), denoting the length of the sequence $b_1, b_2, \ldots, b_k$ .

The second line contains $k$ positive integers $b_1, b_2, \ldots, b_k$ ( $1 \leq b_i < p$ ).

The third line contains two positive integers $n$ and $m$ ( $k < n \leq 10^9$ , $1 \leq m < p$ ), which implies $f_n = m$ .

Output a possible value of $f_k$ , where $f_k$ is a positive integer satisfying $1 \leq f_k < p$ . If there are multiple answers, print any of them. If no such $f_k$ makes $f_n = m$ , output $-1$ instead.

It is easy to show that if there are some possible values of $f_k$ , there must be at least one satisfying $1 \leq f_k < p$ .