Think simple.

Real simple.

$$$1 + 2 \equiv 0 \pmod{3}$$$.

Can you notice something that does not change, an invariant?

Can we convert this complicated operation to something simpler?

We can convert the operation to basically swapping adjacent elements and it still works. Try to prove why?

Swapping adjacent elements in a distinct array is basically trying to equate two permutations using adjacent swaps. When is it possible?

Inversions in a permutation

Can we fix an upper bound to count the number of subarrays having xor-pair less than that?

We can use binary search to fix an upper bound and calculate all subarrays with xor-pair less than the given value.

In order to count the number of subarrays we can use a popular data-structure used for a lot of xor-operations.

We can binary search on the upper bound and use tries to count the number of subarrays with xor-pair less than given value. This way we can find the kth-smallest value.

Try looking at the problem bit by bit.

What does the contribution of a particular bit over a path look like?

Can you think of some simpler queries you can easily solve the problem for?

Root the tree.

For a query from a node $$$u$$$ to its ancestor $$$v$$$, the countribution from the counting term $$$(0,1,2,3, \ldots)$$$ changes for the first time at an ancestor at a jump of a power of two.

Let $$$dp[u][k]$$$ be the contribution of bit $$$k$$$ to the query from $$$u$$$ to the root.

Keep track of a few additional nodes for each query. Read hint 5 again.