1949B · Charming Meals

Problem: The Czech cuisine features $n$ appetizers and $n$ main dishes. The $i$-th appetizer has spiciness $a_i$, and the $i$-th main dish has spiciness $b_i$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $n$ appetizers and $n$ main dishes into $n$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $x$ and a main dish with spiciness $y$ has charm equal to $|x-y|$.

You want to maximize the minimum charm of the resulting $n$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input Format: Each test contains multiple test cases. The first line contains an integer $t$ ($1\le t\le 1\,000$) — the number of test cases. The descriptions of the $t$ test cases follow.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 5\,000$) —the number of appetizers and main dishes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^{9}$) — the spicinesses of the $n$ appetizers.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($0 \leq b_i \leq 10^{9}$) — the spicinesses of the $n$ main dishes.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25\cdot 10^6$.

Output Format: For each test case, print the largest possible value of the minimum charm you can achieve.

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $0$ and a main dish with spiciness $1000000000$, so the charm of each meal will be $1000000000$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $(1, 5)$, $(2, 4)$, $(3, 1)$, $(4, 2)$, $(5, 3)$. The corresponding meals have charms: $4$, $2$, $2$, $2$, $2$. The resulting minimum charm is $2$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $0$ with the three main dishes with spiciness $100$, and the three appetizers with spiciness $100$ with the three main dishes with spiciness $0$. Doing so, the charm of each meal will be exactly $100$.