Hints: The 67th XOR Problem

Try the process by hand for $n = 2$ and $n = 3$. After each operation, write the remaining numbers in terms of the original $a_1, \ldots, a_n$.

What simple expression do you get for the last number when only two values are left?

Answer to Hint 1: For $n = 2$, you XOR both entries by $a_1$ and delete $a_1$, so the survivor is $a_1 \oplus a_2$. For $n = 3$, whichever element you delete last, a short case check shows the final value is always the XOR of two of the original three numbers (never a single $a_i$ alone).

Does the final answer always depend on only two original indices?

You want to prove that for any $n \ge 2$, the remaining value is always $a_i \oplus a_j$ for some distinct $i, j$.

Each operation XORs the whole array with the chosen $a_i$, then removes one index. If you track how each surviving entry is built from the originals, what algebraic structure appears?

Answer to Hint 2: Think in terms of which original $a_k$ still “contributes” to each position. XORing everything by the chosen value and deleting an index toggles contributions in a structured way; equivalently, the final number is a XOR of some subset of the original values, but with a parity constraint that forces exactly two distinct indices to remain “active” in the XOR.

Focus on showing the final value is always of the form $a_i \oplus a_j$.

Fix $n \ge 2$. Show that for every pair $i < j$, there is some sequence of operations whose final number equals $a_i \oplus a_j$.

If you can do that, what does “maximize the final value” reduce to?

Answer to Hint 3: Once you know every value $a_i \oplus a_j$ is achievable and no other values are possible, the answer is simply

So the problem becomes purely combinatorial: compute the best pair under XOR.

How heavy can this be, given the constraints on $n$ and the sum of $n$ over test cases?

You do not need a fancy trie or linear basis for this contest version: brute force over pairs is enough.

What is the time complexity of trying all unordered pairs $(i, j)$ per test case?

Answer to Hint 4: A double loop over $i < j$ costs $O(n^2)$ per test case. Here $\sum n \le 3105$, so $\sum n^2$ is also small in practice—well within limits.

Implementation: initialize ans = 0, then for each $i < j$ set ans = max(ans, a[i] ^ a[j]) and print ans.