A. Alternating Sum
math; number theory
You are given two integers \(a\) and \(b\). Moreover, you are given a sequence \(s_0, s_1, \dots, s_{n}\). All values in \(s\) are integers \(1\) or \(-1\). It's known that sequence is \(k\)-periodic and \(k\) divides \(n+1\). In other words, for each \(k \leq i \leq n\) it's satisfied that \(s_{i} = s_{i - k}\).Find out the non-negative remainder of division of \(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}\) by \(10^{9} + 9\).Note that the modulo is unusual!
The first line contains four integers \(n, a, b\) and \(k\) \((1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})\).The second line contains a sequence of length \(k\) consisting of characters '+' and '-'. If the \(i\)-th character (0-indexed) is '+', then \(s_{i} = 1\), otherwise \(s_{i} = -1\).Note that only the first \(k\) members of the sequence are given, the rest can be obtained using the periodicity property.
Output a single integer — value of given expression modulo \(10^{9} + 9\).
In the first example:\((\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})\) = \(2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}\) = 7In the second example:\((\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}\).