human
I found an interesting problem on https://atcoder.jp/contests/arc081/tasks/arc081_b:
We have a board with a 2 \times N grid.
Snuke covered the board with N dominoes without overlaps.
Here, a domino can cover a 1 \times 2 or 2 \times 1 square.
Then, Snuke decided to paint these dominoes using three colors: red, cyan and green.
Two dominoes that are adjacent by side should be painted by different colors.
Here, it is not always necessary to use all three colors.
Find the number of such ways to paint the dominoes, modulo 1000000007.
The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner:
 - Each domino is represented by a different English letter (lowercase or uppercase).
 - The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

-----Constraints-----
 - 1 \leq N \leq 52
 - |S_1| = |S_2| = N
 - S_1 and S_2 consist of lowercase and uppercase English letters.
 - S_1 and S_2 represent a valid arrangement of dominoes.

-----Input-----
Input is given from Standard Input in the following format:
N
S_1
S_2

-----Output-----
Print the number of such ways to paint the dominoes, modulo 1000000007.

-----Sample Input-----
3
aab
ccb

-----Sample Output-----
6

There are six ways as shown below:

I tried it in Python, but could not do it. Can you solve it?

 ... Write me a Python code to find the number of such ways to paint the dominoes, modulo 1000000007?
assistant
#Here is Python code to find the number of such ways to paint the dominoes, modulo 1000000007.

import sys
from collections import deque
from copy import deepcopy
from bisect import bisect_left, bisect_right, insort_left, insort_right
from heapq import heapify, heappop, heappush
from itertools import product, permutations, combinations, combinations_with_replacement
from functools import reduce
from math import gcd, sin, cos, tan, asin, acos, atan, degrees, radians

sys.setrecursionlimit(10**6)
INF = 10**20
eps = 1.0e-20
MOD = 10**9+7

def lcm(x,y):
    return x*y//gcd(x,y)
def lgcd(l):
    return reduce(gcd,l)
def llcm(l):
    return reduce(lcm,l)
def powmod(n,i,mod=MOD):
    return pow(n,mod-1+i,mod) if i<0 else pow(n,i,mod)
def div2(x):
    return x.bit_length()
def div10(x):
    return len(str(x))-(x==0)
def intput():
    return int(input())
def mint():
    return map(int,input().split())
def lint():
    return list(map(int,input().split()))
def ilint():
    return int(input()), list(map(int,input().split()))
def judge(x, l=['Yes', 'No']):
    print(l[0] if x else l[1])
def lprint(l, sep='\n'):
    for x in l:
        print(x, end=sep)
def ston(c, c0='a'):
    return ord(c)-ord(c0)
def ntos(x, c0='a'):
    return chr(x+ord(c0))
class counter(dict):
    def __init__(self, *args):
        super().__init__(args)
    def add(self,x,d=1):
        self.setdefault(x,0)
        self[x] += d
    def list(self):
        l = []
        for k in self:
            l.extend([k]*self[k])
        return l
class comb():
    def __init__(self, n, mod=None):
        self.l = [1]
        self.n = n
        self.mod = mod
    def get(self,k):
        l,n,mod = self.l, self.n, self.mod
        k = n-k if k>n//2 else k
        while len(l)<=k:
            i = len(l)
            l.append(l[i-1]*(n+1-i)//i if mod==None else (l[i-1]*(n+1-i)*powmod(i,-1,mod))%mod)
        return l[k]
def pf(x,mode='counter'):
    C = counter()
    p = 2
    while x>1:
        k = 0
        while x%p==0:
            x //= p
            k += 1
        if k>0:
            C.add(p,k)
        p = p+2-(p==2) if p*p<x else x
    if mode=='counter':
...