Tinkoff Internship Warmup Round 2018 and Codeforces Round 475 (Div. 1)
A
Alternating Sum
Codeforces
1800
原题
←B →
翻译中…

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 $s_{i} = s_{i - k}$.

翻译中…

Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.

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Note that the modulo is unusual!

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Input

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The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.

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The second line contains a sequence of length $k$ consisting of characters '+' and '-'.

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If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.

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Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.

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Output

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Output a single integer — value of given expression modulo $10^{9} + 9$.

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Note

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In the first example:

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$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7

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In the second example:

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$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$.

翻译中…

样例

Input #1
2 2 3 3
+-+
Output #1
7
InputCopy OutputCopy
Input #2
4 1 5 1
-
Output #2
999999228
InputCopy OutputCopy
题目信息
题号
963A
时间限制
1 s
内存限制
256 MB
标签
数学
数论

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。