Fault-tolerant Network
Contest: Educational Codeforces Round 124 (Rated for Div. 2)
Difficulty rating: 1500
Tags: brute force, data structures, implementation

There is a classroom with two rows of computers. There are $$$n$$$ computers in each row and each computer has its own grade. Computers in the first row has grades $$$a_1, a_2, \dots, a_n$$$ and in the second row — $$$b_1, b_2, \dots, b_n$$$.

Initially, all pairs of neighboring computers in each row are connected by wire (pairs $$$(i, i + 1)$$$ for all $$$1 \le i < n$$$), so two rows form two independent computer networks.

Your task is to combine them in one common network by connecting one or more pairs of computers from different rows. Connecting the $$$i$$$-th computer from the first row and the $$$j$$$-th computer from the second row costs $$$|a_i - b_j|$$$.

You can connect one computer to several other computers, but you need to provide at least a basic fault tolerance: you need to connect computers in such a way that the network stays connected, despite one of its computer failing. In other words, if one computer is broken (no matter which one), the network won't split in two or more parts.

That is the minimum total cost to make a fault-tolerant network?

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

The first line of each test case contains the single integer $$$n$$$ ($$$3 \le n \le 2 \cdot 10^5$$$) — the number of computers in each row.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the grades of computers in the first row.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$) — the grades of computers in the second row.

It's guaranteed that the total sum of $$$n$$$ doesn't exceed $$$2 \cdot 10^5$$$.

Output format:
For each test case, print a single integer — the minimum total cost to make a fault-tolerant network.

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

Note:
In the first test case, it's optimal to connect four pairs of computers:

1. computer $$$1$$$ from the first row with computer $$$2$$$ from the second row: cost $$$|1 - 4| = 3$$$;
2. computer $$$3$$$ from the first row with computer $$$2$$$ from the second row: cost $$$|1 - 4| = 3$$$;
3. computer $$$2$$$ from the first row with computer $$$1$$$ from the second row: cost $$$|10 - 20| = 10$$$;
4. computer $$$2$$$ from the first row with computer $$$3$$$ from the second row: cost $$$|10 - 25| = 15$$$;

In the second test case, it's optimal to connect $$$1$$$ from the first row with $$$1$$$ from the second row, and $$$4$$$ from the first row with $$$4$$$ from the second row.

Example input:
2
3
1 10 1
20 4 25
4
1 1 1 1
1000000000 1000000000 1000000000 1000000000

Example output:
31
1999999998

Editorial:
There is a criterion when the given network becomes fault-tolerant: the network becomes fault-tolerant if and only if each of corner computers (let's name them $$$A_1$$$, $$$A_n$$$, $$$B_1$$$ and $$$B_n$$$) is connected to the other row.
From the one side: if, WLOG, $$$A_1$$$ is not connected to other row then if $$$A_2$$$ is broken — $$$A_1$$$ loses connection to the other network (since $$$A_1$$$ is connected only with $$$A_2$$$).
From the other side: suppose, WLOG, $$$A_i$$$ is broken, then the row $$$A$$$ is falling in at most two parts: $$$A_1 - \dots - A_{i - 1}$$$ and $$$A_{i + 1} - \dots - A_n$$$. But since both $$$A_1$$$ and $$$A_n$$$ are connected to row $$$B$$$ and $$$B$$$ is still connected, then the resulting network is still connected.
Now the question is: how to connect all corner computers? Because sometimes it's optimal not to connect corners directly. One of the approaches is described below.
Let's look at $$$A_1$$$. Essentially, there are three ways to connect it to row $$$B$$$: to $$$B_1$$$, $$$B_n$$$ or $$$\mathrm{best}_B(A_1)$$$ (where $$$\mathrm{best}_B(A_1)$$$ is $$$B_j$$$ with minimum possible $$$|a_i - b_j|$$$). The same applies to $$$A_n$$$.
So, let's just iterate over all these $$$3 \times 3$$$ variants. For each of these variants,
• if we didn't cover $$$B_1$$$ then we should also add one more connection between $$$B_1$$$ and $$$\mathrm{best}_A(B_1)$$$;
• if we didn't cover $$$B_n$$$ then we should also add one more connection between $$$B_n$$$ and $$$\mathrm{best}_A(B_n)$$$;
As a result, we choose the best variant.