Game with Marbles (Hard Version)
Contest: Codeforces Round 916 (Div. 3)
Difficulty rating: 1400
Tags: games, greedy, sortings

The easy and hard versions of this problem differ only in the constraints on the number of test cases and $$$n$$$. In the hard version, the number of test cases does not exceed $$$10^4$$$, and the sum of values of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Furthermore, there are no additional constraints on $$$n$$$ in a single test case.

Recently, Alice and Bob were given marbles of $$$n$$$ different colors by their parents. Alice has received $$$a_1$$$ marbles of color $$$1$$$, $$$a_2$$$ marbles of color $$$2$$$,..., $$$a_n$$$ marbles of color $$$n$$$. Bob has received $$$b_1$$$ marbles of color $$$1$$$, $$$b_2$$$ marbles of color $$$2$$$, ..., $$$b_n$$$ marbles of color $$$n$$$. All $$$a_i$$$ and $$$b_i$$$ are between $$$1$$$ and $$$10^9$$$.

After some discussion, Alice and Bob came up with the following game: players take turns, starting with Alice. On their turn, a player chooses a color $$$i$$$ such that both players have at least one marble of that color. The player then discards one marble of color $$$i$$$, and their opponent discards all marbles of color $$$i$$$. The game ends when there is no color $$$i$$$ such that both players have at least one marble of that color.

The score in the game is the difference between the number of remaining marbles that Alice has and the number of remaining marbles that Bob has at the end of the game. In other words, the score in the game is equal to $$$(A-B)$$$, where $$$A$$$ is the number of marbles Alice has and $$$B$$$ is the number of marbles Bob has at the end of the game. Alice wants to maximize the score, while Bob wants to minimize it.

Calculate the score at the end of the game if both players play optimally.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of three lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the number of colors;
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of marbles of the $$$i$$$-th color that Alice has;
- the third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$), where $$$b_i$$$ is the number of marbles of the $$$i$$$-th color that Bob has.

Additional constraint on the input: the sum of $$$n$$$ for all test cases does not exceed $$$2 \cdot 10^5$$$.

Output format:
For each test case, output a single integer — the score at the end of the game if both Alice and Bob act optimally.

Constraints:
Time limit: 3.5 s
Memory limit: 256.0 MB

Note:
In the first example, one way to achieve a score of $$$1$$$ is as follows:

1. Alice chooses color $$$1$$$, discards $$$1$$$ marble. Bob also discards $$$1$$$ marble;
2. Bob chooses color $$$3$$$, discards $$$1$$$ marble. Alice also discards $$$1$$$ marble;
3. Alice chooses color $$$2$$$, discards $$$1$$$ marble, and Bob discards $$$2$$$ marble.

As a result, Alice has $$$a = [3, 1, 0]$$$ remaining, and Bob has $$$b = [0, 0, 3]$$$ remaining. The score is $$$3 + 1 - 3 = 1$$$.

It can be shown that neither Alice nor Bob can achieve a better score if both play optimally.

In the second example, Alice can first choose color $$$1$$$, then Bob will choose color $$$4$$$, after which Alice will choose color $$$2$$$, and Bob will choose color $$$3$$$. It can be shown that this is the optimal game.

Example input:
5
3
4 2 1
1 2 4
4
1 20 1 20
100 15 10 20
5
1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1
3
5 6 5
2 1 7
6
3 2 4 2 5 5
9 4 7 9 2 5

Example output:
1
-9
2999999997
8
-6

Editorial:
Let's change the game in the following way:
• Firstly, we'll let Bob make all moves. It means that for each color $$$i$$$ Bob discarded one marble, while Alice discarded all her marbles. So the score $$$S$$$ will be equal to $$$0 - \sum_{i}{(b_i - 1)}$$$.
• Alice makes the first move by choosing some color $$$i$$$ and "takes back color $$$i$$$". In means that we cancel Bob's move on color $$$i$$$ and Alice makes that move instead. How score $$$S$$$ will change? Initially, we had $$$-(b_i - 1)$$$ contribution to the score, but now contribution becomes $$$+(a_i - 1)$$$. In other words, choosing color $$$i$$$ gives score $$$S' = S + (a_i + b_i - 2)$$$. Note that, the greater $$$(a_i + b_i)$$$ the greater $$$S'$$$ becomes.
• Bob makes the second move by "saving some color $$$i$$$", i. e. he forbids Alice to choose color $$$i$$$. It doesn't change score $$$S$$$ but now Alice can't choose color $$$i$$$ on her turns.
• Alice and Bob continue playing, by "taking" and "saving" colors, until Alice will take all non-forbidden colors.
As a result, we showed that optimum strategy in the initial game is the same: sort colors by $$$a_i + b_i$$$ in decreasing order. Alice chooses $$$1$$$-st, $$$3$$$-rd, $$$5$$$-th and so on colors, while Bob chooses $$$2$$$-nd, $$$4$$$-th, $$$6$$$-th and so on colors.
The resulting complexity is $$$O(n \log{n})$$$.