B"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 = <= ft(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 = <= ft( 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 xe2 x80 x93 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 <= q k <= q 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 <= q b_i < p ). The third line contains two positive integers n and m ( k < n <= q 10^9 , 1 <= q m < p ), which implies f_n = m . Output a possible value of f_k , where f_k is a positive integer satisfying 1 <= q 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 <= q f_k < p . In the first sample, we have f_4 = f_3^2 cdot f_2^3 cd"...