F. Lunar New Year and a Recursive Sequence
math; matrices; number theory
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.
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 \(1 \leq f_k < p\).
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.