XOR(Exclusive OR) problem Given 2 multisets A and B, |A| = |B|. The task is to find such X that A ^ X = B, where A ^ X means foreach element of A xor it with X; or say that theres no such X.  |A| <= 10^5, elements of A, B are decimal number from [0; 10^18].
For example -
If A = {0, 2} and B = {1, 3} then 1 is the answer: A ^ 1 = {1, 3}.
I mean I am looking for algorithm with asymptotic << O(|A|^3 * Log(|A|)) which is straightforward brute force.
Edit: well its seems that intended solution is some 'smart' brute force. FML

We will exploit property of xor operation which says that:
even $\oplus$ even= even $\tag {1}$
 similiarly odd $\oplus$ odd = even $\tag{2}$
 even $\oplus$ odd= odd $\oplus$ even=odd $\tag{3}$. We have two sets A and B according to question containing integers ($\ge0$).
The basic concept is that there are only two kinds of numbers either even or odd which is also the key fact in our solution to the problem. Xor operation is like a function here because we want a single X which can map A to B that is 
$$B=f(X)=A\oplus X$$ For all even A's the B's will be same and similarly in the case for odd A's also. So basically if in the set A we have m even numbers then m numbers in B would be either even or odd. Suppose m numbers are odd in B. then it is something like this:-
$$A(evens) \oplus odd=B(odds) $$ i.e. m evens of A are mapped to m odds of B, the remainings in A which are odd has to be mapped to evens which is true. if at any point the corresponding match is not found then X doesn't exist.
Algorithm:
The algorithm is recursive ,you can convert it into iterative but recursion is more obvious here.
Hint:
XXOR(A,B)
(1)map A to B according to the equality mentioned above.
initially we will get two sets of equivalent mappings from A to B, let it be $f_1$(A(even) to B(odd)) and so $f_2$(A(odd) to B(even)) i.e. $X=odd+X$.
if a division is not possible that generates same X in both $f_1$ and $f_2$, then no X exists then exit.
else if we have reached the m.s.b. of the numbers, return X;
(2) right shift each of the integer in both the sets A and B by one position.
(3) According to the example 
XXOR(A(even)>>1 ,B(odd)>>1)
              XXOR(A(odd)>>1, B(even)>>1)
Complexity: The worst case will be when we will have 2 classes of divisions.In that case:
$$T(|A|,10^{18})=4T\left(\frac{|A|}{2},10^{18}/2\right)+4|A|$$
if $$T\left(\frac{|A|}{k},1\right)$$ is reached then terminate.
The complexity of this algorithm is $O(|A|^2\times 10^{18})$ in the worst case.
