Igor the analyst is at work. He learned about a feature in his text editor called "Replace All". Igor is too bored at work and thus he came up with the following problem:
Given two strings x and y which consist of the English letters 'A' and 'B' only, a pair of strings (s, t) is called good if:
For example, if x = AAB, y = BB and n = 4, then (01, 0101) is one of good pairs of strings, because both obtained after replacing strings are "01010101".
The flexibility of a pair of strings x and y is the number of pairs of good strings (s, t). The pairs are ordered, for example the pairs ( 0, 1 ) and ( 1, 0 ) are different.
You're given two strings c and d. They consist of characters 'A', 'B' and '?' only. Find the sum of flexibilities of all possible pairs of strings (c', d') such that c' and d' can be obtained from c and d respectively by replacing the question marks with either 'A' or 'B', modulo 10^{9} + 7.
The first line contains the string c (1 ≤ |c| ≤ 3·10^{5}).
The second line contains the string d (1 ≤ |d| ≤ 3·10^{5}).
The last line contains a single integer n (1 ≤ n ≤ 3·10^{5}).
Output a single integer: the answer to the problem, modulo 10^{9} + 7.
A?
?
3
2
A
B
10
2046
For the first sample, there are four possible pairs of (c', d').
If (c', d') = ( AA , A ), then the flexibility is 0.
If (c', d') = ( AB , A ), then the flexibility is 0.
If (c', d') = ( AA , B ), then the flexibility is 2, as the pairs of binary strings ( 1 , 11 ), ( 0 , 00 ) are the only good pairs.
If (c', d') = ( AB , B ), then the flexibility is 0.
Thus, the total flexibility is 2.
For the second sample, there are 2^{1} + 2^{2} + ... + 2^{10} = 2046 possible binary strings of length not greater 10, and the set of pairs of good strings is precisely the set of pairs (s, s), where s is a binary string of length not greater than 10.
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