A python solution

I will just explain my algorithm below using the example a='1011'=11 and b='11'=3 with n=5

the problem ask us to compute

'1011'xor'11' =(1000) = bx2^0+a+err1

+'1011'xor'110' =(1101) = bx2^1+a+err2

+'1011'xor'1100' =(111) = bx2^2+a+err3

+'1011'xor'11000' =(10011) = bx2^3+a+err4

+'1011'xor'110000' =(111011) = bx2^4+a (no err from here)

+'1011'xor'1100000' =(1101011) = bx2^5+a

with err1=4+2, err2=4, err3=16, err4=16 with total error=sum=42.

so we can reduce the problem into two parts

1) compute the trivial sum bx(2^0+2^1+...+2^n)+ax(n+1) %1e9+7

2) compute the error terms

here i will explain how to compute the total errors by this example, the detailed algorithm can be found below

these errors come from when shifting b to the left, each '1' in b will only hit the digits '1' of a on the left.

for example,

'1 1 ' will hit ' 1 0 11 '

leading to an error 2x(1+2+8)=2+4+16 (times 2 here since both of those 1 become 0 after we take xor)

also ' 1 1' will only hit ' 1 0 1 1'

leading to an error 2x(2+8)=4+16

you can see that 2,4,16,4,16 are exactly the terms in err1--err4.

from collections import defaultdict

M=1000000007
a = str(input())
b = str(input())

# need to change n into 314159
n=314159

# pre-compute powers of 2 up to n=314159
pof2=[1]
for i in range(n+1):
    pof2.append((pof2[-1]*2)%M)

# use a default list to count the number of 1's in b in the right k digits
len_a=len(a)
len_b=len(b)
#print(len_a,len_b)
cnt_b=defaultdict(int)
cnt_b[0]=int(b[-1])
for i in range(1,len_b):
    cnt_b[i]=cnt_b[i-1]+int(b[-i-1])
# extend to the length of a
for i in range(len_b,len_a):
    cnt_b[i]=cnt_b[len_b-1]

# using pof2 to compute num_a%M,num_b%M
num_a=0
num_b=0
for i in range(len_a):
    if int(a[-1-i])!=0:
        num_a+=pof2[i]
for i in range(len_b):
    if int(b[-1-i])!=0:
        num_b+=pof2[i]
#print(num_a,num_b)

#compute the offset
offset=0
for i in range(len_a):
    if int(a[-i-1])!=0:
        offset+=pof2[i+1]*cnt_b[i]
#print(offset)

# n=314159, which is very large
answer=(num_b*(pof2[n+1]-1)+num_a*(n+1)-offset)%M
print(answer)