A. Alternating Sumtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou 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!InputThe 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.OutputOutput a single integer — value of given expression modulo 10^{9} + 9.ExamplesInput2 2 3 3+-+Output7Input4 1 5 1-Output999999228NoteIn 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}.
Input2 2 3 3+-+
Output7
['math', 'number theory', '*1800']