Lunar New Year and a Recursive Sequence

有一串 $n$ 个数的数列，给你 $b_1\sim b_k$。当 $i>k$ 时： $$f_i=(\prod\limits_{j=1}^kf_{i-j}^{b_j})\bmod{998244353}$$ 已知 $f_1=f_2=\cdots=f_{k-1}=1,f_n=m$，问 $f_k$ 可能是多少。 $k \leq 100$，$n \leq 10^9$。

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 $ .

3
2 3 5
4 16

4

5
4 7 1 5 6
7 14187219

6

8
2 3 5 6 1 7 9 10
23333 1

1

1
2
88888 66666

-1

3
998244352 998244352 998244352
4 2

-1

10
283 463 213 777 346 201 463 283 102 999
2333333 6263423

382480067

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.