CF1106F.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 $i \gt k$, $f_i$ can be obtained by the following recursive equation:

$$f\_i = \\left(f\_{i - 1} ^ {b\_1} \\cdot f\_{i - 2} ^ {b\_2} \\cdot \\cdots \\cdot f\_{i - k} ^ {b\_k}\\right) \\bmod p,$$

which in short is

$$f\_i = \\left(\\prod\_{j = 1}^{k} f\_{i - j}^{b\_j}\\right) \\bmod p,$$

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 $y$.

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.

农历新年即将来临，Bob最近收到了朋友送的一份礼物——一个递归序列！他非常喜爱这个序列，想要好好玩一玩。

设 $f_1, f_2, \ldots, f_i, \ldots$ 是一个无限的正整数序列。Bob 知道：当 $i > k$ 时，$f_i$ 可由如下递推公式得到：

$$f\_i = \\left(f\_{i - 1} ^ {b\_1} \\cdot f\_{i - 2} ^ {b\_2} \\cdot \\cdots \\cdot f\_{i - k} ^ {b\_k}\\right) \\bmod p,$$

简写为：

$$f\_i = \\left(\\prod\_{j = 1}^{k} f\_{i - j}^{b\_j}\\right) \\bmod p,$$

其中 $p = 998\,244\,353$（一个被广泛使用的素数），$b_1, b_2, \ldots, b_k$ 是已知的整数常数，且 $x \bmod y$ 表示 $x$ 除以 $y$ 的余数。

Bob 不慎遗失了序列的前 $k$ 项 $f_1, f_2, \ldots, f_k$，这极为棘手——它们可是整个序列的基础！幸运的是，Bob 还记得序列的前 $k-1$ 项：$f_1 = f_2 = \ldots = f_{k - 1} = 1$，以及第 $n$ 项：$f_n = m$。请找出任意一个可能的 $f_k$ 值。若不存在解，则只需告诉 Bob：无论他多么伤心，都无法恢复他最钟爱的这个序列。

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 $b_1, b_2, \ldots, b_k$ ($1 \leq b_i \lt p$).

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

第一行包含一个正整数 $k$（$1 \leq k \leq 100$），表示序列 $b_1, b_2, \ldots, b_k$ 的长度。

第二行包含 $k$ 个正整数 $b_1, b_2, \ldots, b_k$（$1 \leq b_i \lt p$）。

第三行包含两个正整数 $n$ 和 $m$（$k \lt n \leq 10^9$，$1 \leq m \lt p$），表示 $f_n = m$。

Output a possible value of $f_k$, where $f_k$ is a positive integer satisfying $1 \leq f_k \lt p$. 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 $1 \leq f_k \lt p$.

输出 $f_k$ 的一个可能取值，其中 $f_k$ 是满足 $1 \leq f_k \lt p$ 的正整数。若存在多个答案，输出任意一个即可。若不存在这样的 $f_k$ 使得 $f_n = m$，则输出 $-1$。

容易证明：若存在满足条件的 $f_k$，则其中必至少有一个满足 $1 \leq f_k \lt p$。

In the first sample, we have $f_4 = f_3^2 \cdot f_2^3 \cdot f_1^5$. Therefore, applying $f_3 = 4$, we have $f_4 = 16$. Note that there can be multiple answers.

In the third sample, applying $f_7 = 1$ makes $f_{23333} = 1$.

In the fourth sample, no such $f_1$ makes $f_{88888} = 66666$. Therefore, we output $-1$ instead.

在第一个样例中，我们有 $f_4 = f_3^2 \cdot f_2^3 \cdot f_1^5$。因此，代入 $f_3 = 4$，可得 $f_4 = 16$。注意，可能存在多个答案。

在第三个样例中，令 $f_7 = 1$ 可使得 $f_{23333} = 1$。

在第四个样例中，不存在满足 $f_{88888} = 66666$ 的 $f_1$。因此，我们输出 $-1$。