Bit Game (Hard Version)
Contest: Codeforces Round 982 (Div. 2)
Difficulty rating: 3100
Tags: bitmasks, dp, math

This is the hard version of this problem. The only difference is that you need to output the number of choices of games where Bob wins in this version, where the number of stones in each pile are not fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $$$n$$$ piles. Initially, there are $$$x_i$$$ stones in the $$$i$$$-th pile, and it has an associated value $$$a_i$$$. A player can take $$$d$$$ stones away from the $$$i$$$-th pile if and only if both of the following conditions are met:

- $$$1 \le d \le a_i$$$, and
- $$$x \, \& \, d = d$$$, where $$$x$$$ is the current number of stones in the $$$i$$$-th pile and $$$\&$$$ denotes the bitwise AND operation.

The player who cannot make a move loses, and Alice goes first.

You're given the $$$a_i$$$ values of each pile, but the number of stones in the $$$i$$$-th pile has not been determined yet. For the $$$i$$$-th pile, $$$x_i$$$ can be any integer between $$$1$$$ and $$$b_i$$$, inclusive. That is, you can choose an array $$$x_1, x_2, \ldots, x_n$$$ such that the condition $$$1 \le x_i \le b_i$$$ is satisfied for all piles.

Your task is to count the number of games where Bob wins if both players play optimally. Two games are considered different if the number of stones in any pile is different, i.e., the arrays of $$$x$$$ differ in at least one position.

Since the answer can be very large, please output the result modulo $$$10^9 + 7$$$.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 1000$$$). The description of the test cases follows.

The first line of each test case contains $$$n$$$ ($$$1 \le n \le 10^4$$$) — the number of piles.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i < 2^{30}$$$).

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i < 2^{30}$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output format:
Output a single integer, the number of games where Bob wins, modulo $$$10^9 + 7$$$.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first test case, no matter which values of $$$x_2$$$ and $$$x_3$$$ we choose, the second and third piles will always be chosen exactly once before no more stones can be taken from them. If $$$x_1 = 2$$$, then no stones can be taken from it, so Bob will make the last move. If $$$x_1 = 1$$$ or $$$x_1 = 3$$$, then exactly one move can be made on that pile, so Alice will make the last move. So Bob wins when $$$x = [2, 1, 1]$$$ or $$$x = [2, 1, 2]$$$ or $$$x = [2, 2, 1]$$$ or $$$x = [2, 2, 2]$$$.

In the second test case, Bob wins when $$$x_1 = 14$$$ or $$$x_1 = 30$$$ by removing $$$14 - k$$$ stones, where $$$k$$$ is the number of stones Alice removes on her turn. Bob also wins when $$$x_1 = 16$$$ or $$$x_1 = 32$$$ since Alice does not have any moves to begin with.

Example input:
7
3
1 2 3
3 2 2
1
13
45
5
5 4 7 8 6
4 4 5 5 5
4
6 4 8 8
12 13 14 12
3
92856133 46637598 12345678
29384774 73775896 87654321
2
65 12
110 31
4
677810235 275091182 428565855 720629731
74522416 889934149 3394714 230851724

Example output:
4
4
0
6552
722019507
541
665443265

Editorial:
Let's continue on from the editorial for E1. We figured out that the nimbers lie in one of these four cases:
1. $$$g(2^k - 2) = 0$$$, for all $$$k \ge 1$$$.
2. $$$g(2^k - 1) = k$$$, for all $$$k \ge 0$$$.
3. $$$g(2^k) = k + (-1)^k$$$, for all $$$k \ge 0$$$.
4. $$$g(2^k+1) = g(2^k+2) = \ldots = g(2^{k+1} - 3) = k + 1$$$, for all $$$k \ge 2$$$.
From these, it's clear to see that the nimber of each pile is always $$$\le 31$$$ given the constraints. So, let's say we can count the amount of values of $$$x_i \le b_i$$$ exist such that $$$f(x_i, a_i) = p$$$ for all $$$p$$$, then we can easily maintain a knapsack to count the number of games where the nimbers of piles have a bitwise XOR of $$$0$$$.
All that's left is to count the number of values of $$$x_i$$$ which yield some $$$a'$$$ in each of the cases above, which is probably the hardest part of the problem.
We can use digit DP for this. The state should store the position of the most significant one in $$$a'$$$, and whether $$$a'$$$ is $$$0$$$ or in the form $$$1$$$, $$$11 \cdots 1$$$, $$$11 \cdots 0$$$, $$$100 \cdots 0$$$ or anything else. To find whether a $$$1$$$ added to $$$x$$$ becomes $$$0$$$ or $$$1$$$ in $$$a'$$$, have another flag to check whether a good bit has occurred yet. Also, to ensure that $$$x \le b$$$, have one more flag to check whether the current prefix has fallen strictly below $$$b$$$ yet. Please see the model code's comments for a better understanding of how the transitions are computed.
Time complexity: $$$\mathcal{O}(n \log^2 C)$$$, where $$$C$$$ is the upper limit on $$$a_i$$$ and $$$b_i$$$.