Lunar New Year and a Recursive Sequence

3000ms

262144K

Description:

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:

$$$$$$f_i = \left(f_{i - 1} ^ {b_1} \cdot f_{i - 2} ^ {b_2} \cdot \cdots \cdot f_{i - k} ^ {b_k}\right) \bmod p,$$$$$$

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.

Input:

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:

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 $$$1 \leq f_k < p$$$.

Sample Input:

3
2 3 5
4 16

Sample Output:

Sample Input:

5
4 7 1 5 6
7 14187219

Sample Output:

Sample Input:

8
2 3 5 6 1 7 9 10
23333 1

Sample Output:

Sample Input:

1
2
88888 66666

Sample Output:

-1

Sample Input:

3
998244352 998244352 998244352
4 2

Sample Output:

-1

Sample Input:

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

Sample Output:

382480067

Note:

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.

Informação

Provedor Codeforces

Código CF1106F