F. Lunar New Year and a Recursive Sequencetime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLunar 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 isf_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.InputThe 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.OutputOutput 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.ExamplesInput
3
2 3 5
4 16
Output
4
Input
5
4 7 1 5 6
7 14187219
Output
6
Input
8
2 3 5 6 1 7 9 10
23333 1
Output
1
Input
1
2
88888 66666
Output
-1
Input
3
998244352 998244352 998244352
4 2
Output
-1
Input
10
283 463 213 777 346 201 463 283 102 999
2333333 6263423
Output
382480067
NoteIn 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.
3
2 3 5
4 16

4 
['math', 'matrices', 'number theory', '*2400']