Let's solve the problem for the whole sequence instead of a given range.
We can store each number in a trie according to its binary representation from the most significant bit to the least significant bit. Each trie node represents a prefix in binary form and stores the number of integers of that prefix. Let this number be of the node. Each node can have at most two child nodes (left & right) where the left node represents the bit of 0 after the current prefix and the right node represents the bit of 1 after the current prefix. Now let's iterate from the most significant bit to the least significant bit of and calculate the maximum sum of .
We will start from the root node of the trie. If the current bit of is , then it's optimal to take the numbers where the current bit is . If the count of such numbers ( of the left node of the current node) is greater than or equal to , we can move to the left node and we have to take all the integers from that node. Otherwise, we can take all numbers of the left node and move to the right node and we have to take the remaining integers from the right node where is the number of integers () of the left node. While taking all the numbers from the left node, we have to calculate their contribution to the sum of XOR. To do this, in each node, we can also store the number of integers among the subtree of the current node having 1 at the -th bit for each from to (as the maximum value of the input is less than ).
If the current bit of is , we can process it similarly. In this case, we try to take as many numbers as possible from the right node and the remaining from the left node.
This way we can get the maximum sum of XOR. How can we get the count of subsequences? The count of subsequences depends on the last node we are picking the numbers from. If the count of integers to the last node is and we need to pick integers from that node, the count of subsequences is as there are ways to pick integers from .
To solve the problem for a given range, we have to use Persistent Trie instead of a simple trie. The complexity is ) for building the trie with necessary information and ) for each query where is the maximum value. Overall complexity ).