**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
/*
 * Author : christmas964
 * FastReader : https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
 */

import java.io.*;
import java.util.*;

public class Main {
    static int mod = (int) (1e9 + 7);
    static int inf = (int) (1e9 + 7);
    static Random rr = new Random();
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static PrintWriter err = new PrintWriter(System.err);

    public static void main(String[] args) {
        int ttt = 1;
        ttt = i();
        while (ttt-- > 0) {
            // long startTime = System.nanoTime();
            solve();
            // long endTime = System.nanoTime();
            // err.println("Execution Time: " + (endTime - startTime) / 1000 + "
            // microseconds");
        }
        err.close();
        out.close();
    }

    private final static int[][] dirs8 = { { -1, -1 }, { -1, 0 }, { -1, 1 }, { 0, 1 }, { 1, 1 }, { 1, 0 }, { 1, -1 },
            { 0, -1 } };
    private final static int[][] dirs4 = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
    private final static int[][] dirs2 = { { 0, 1 }, { 1, 0 } };
    static int[] factors;
    static boolean factorsFlag = false;
    static long[] fac;
    static boolean factorialFlag = false;

    public static void solve() {
        /*
         * 0) First check briefly that you are not missing easier idea/method to
         * implement. Calculate the TC and check if it will pass.
         * 1) Check for Integer/Long out of bound senarios.
         * 2) Check if all the variables(i,j,k...) used correctly.
         * 3) Don't fcking rush.
         */

        int n = i();
        int[] a = input(n), b = input(n);
        sort(a);
        long low = 0, high = Integer.MAX_VALUE;
        while(low<high)
        {
            long mid = low + (high - low) / 2 + 1;
            boolean ip = true;
            TreeMap<Long, Long> map = new TreeMap<>();
            for (long bi : b) {
                map.put(bi, map.getOrDefault(bi, 0l) + 1);
            }
            for (int i = n - 1; i >= 0; --i) {
                long ai = a[i];
                if (map.floorEntry(ai - mid) != null) {
                    long fk = map.floorKey(ai - mid);
                    if (map.get(fk) == 1) {
                        map.remove(fk);
                    } else {
                        map.put(fk, map.get(fk) - 1);
                    }
                } else if (map.ceilingEntry(ai + mid) != null) {
                    long ck = map.ceilingKey(ai + mid);
                    if (map.get(ck) == 1) {
                        map.remove(ck);
                    } else {
                        map.put(ck, map.get(ck) - 1);
                    }
                } else {
                    ip = false;
                    break;
                }
            }
            if (ip) {
                low = mid;
            } else {
                high = mid - 1;
            }
        }

        out.println(low);

    }

    public static long nck(int n, int c) {
        if (n < c)
            return 0l;
        long a = fac[n];
        long b = (fac[n - c] * fac[c]) % mod;

        return (a * power(b, mod - 2, mod)) % mod;
    }

    public static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    public static void swap(long[] a, int i, int j) {
        long t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    public static int r(int r) {
        return rr.nextInt(r);
    }

    public static int getBit(int n, int i) {
        return ((n >> i) & 1);
    }

    public static int getBit(long n, int i) {
        return (int) ((n >> i) & 1l);
    }

    public static int cnt(int num, int i) {
        /*
         * Returns the number of set bits in num's number of numbers including 0 at i'th
         * bit.
         * if num = 3 that is [0,1,2] and i = 1, then it will return number of numbers
         * that have set bit
         * at i'th index.
         */
        int g = (1 << (i + 1));
        int q = (num) / g;
        int r = (num) % g;
        g >>= 1;
        return (q * g) + ((r / g) > 0 ? (r % g) : 0);
    }

    public static long cnt(long num, int i) {
        /*
         * Returns the number of set bits in num's number of numbers including 0 at i'th
         * bit.
         * if num = 3 that is [0,1,2] and i = 1, then it will return number of numbers
         * that have set bit
         * at i'th index.
         */
        int g = (1 << (i + 1));
        long q = (num) / g;
        long r = (num) % g;
        g >>= 1;
        return (q * g) + ((r / g) > 0 ? (r % g) : 0);
    }

    private static void sieve() {
        if (factorsFlag)
            return;
        factors = new int[(int) ((3 * 1e5) + 1)];
        factorsFlag = true;
        for (int i = 1; i < factors.length; i++) {
            factors[i] = i;
        }
        for (int i = 2; i * 1l * i < factors.length; i++) {
            if (factors[i] == i) {
                for (int j = i; j < factors.length; j += i) {
                    if (factors[j] == j) {
                        factors[j] = i;
                    }
                }
            }
        }
    }

    public static int[] unique(int[] nums) {
        Set<Integer> set = new HashSet<>();
        List<Integer> list = new ArrayList<>();
        for (int num : nums) {
            if (set.contains(num))
                continue;
            set.add(num);
            list.add(num);
        }
        int[] a = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            a[i] = list.get(i);
        }
        return a;
    }

    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static int abs(int a) {
        return Math.abs(a);
    }

    public static long abs(long a) {
        return Math.abs(a);
    }

    public static int upper(long A[], long k, int si, int ei) {
        /*
         * gives the element index that is smallest among all elements that is greater
         * then
         * k , otherwise -1.
         */
        int l = si;
        int u = ei;
        int ans = -1;
        while (l <= u) {
            int mid = (l + u) / 2;
            if (A[mid] <= k) {
                l = mid + 1;
            } else {
                ans = mid;
                u = mid - 1;
            }
        }
        return ans;
    }

    public static int lower(long A[], long k, int si, int ei) {
        /*
         * gives the element index that is smallest among all elements that is greater
         * then
         * or equal to k , otherwise -1.
         */
        int l = si;
        int u = ei;
        int ans = -1;
        while (l <= u) {
            int mid = (l + u) / 2;
            if (A[mid] < k) {
                l = mid + 1;
            } else {
                ans = mid;
                u = mid - 1;
            }
        }
        return ans;

    }

    static int[] copy(int A[]) {
        int B[] = new int[A.length];
        for (int i = 0; i < A.length; i++) {
            B[i] = A[i];
        }
        return B;
    }

    static long[] copy(long A[]) {
        long B[] = new long[A.length];
        for (int i = 0; i < A.length; i++) {
            B[i] = A[i];
        }
        return B;
    }

    static int[] input(int n) {
        int A[] = new int[n];
        for (int i = 0; i < n; i++) {
            A[i] = sc.nextInt();
        }
        return A;
    }

    static long[] inputL(int n) {
        long A[] = new long[n];
        for (int i = 0; i < n; i++) {
            A[i] = sc.nextLong();
        }
        return A;
    }

    static String[] inputS(int n) {
        String A[] = new String[n];
        for (int i = 0; i < n; i++) {
            A[i] = sc.next();
        }
        return A;
    }

    static long sum(int A[]) {
        long sum = 0;
        for (int i : A) {
            sum += i;
        }
        return sum;
    }

    static long sum(long A[]) {
        long sum = 0;
        for (long i : A) {
            sum += i;
        }
        return sum;
    }

    static void reverse(long A[]) {
        int n = A.length;
        int left = 0, right = n - 1;
        while (left < right) {
            long temp = A[left];
            A[left++] = A[right];
            A[right--] = temp;
        }

    }

    static void reverse(int A[]) {
        int n = A.length;
        int left = 0, right = n - 1;
        while (left < right) {
            int temp = A[left];
            A[left++] = A[right];
            A[right--] = temp;
        }

    }

    static int[][] input(int n, int m) {
        int A[][] = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                A[i][j] = i();
            }
        }
        return A;
    }

    static char[][] inputc(int n, int m) {
        char A[][] = new char[n][m];
        for (int i = 0; i < n; i++) {
            String s = s();
            for (int j = 0; j < m; j++) {
                A[i][j] = s.charAt(j);
            }
        }
        return A;
    }

    static int nextPowerOf2(int n) {
        /*
         * Returns the power of two greater then equal to n
         */
        if (n <= 0)
            return 1;
        if ((n & (n - 1)) == 0)
            return n;

        n--;
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        n++;

        return n;
    }

    static long nextPowerOf2(long n) {
        /*
         * Returns the power of two greater then equal to n
         */
        if (n <= 0)
            return 1;
        if ((n & (n - 1)) == 0)
            return n;

        n--;
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        n++;

        return n;
    }

    static int max(int... A) {
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < A.length; i++) {
            max = Math.max(max, A[i]);
        }
        return max;
    }

    static int min(int... A) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < A.length; i++) {
            min = Math.min(min, A[i]);
        }
        return min;
    }

    static long max(long... A) {
        long max = Long.MIN_VALUE;
        for (int i = 0; i < A.length; i++) {
            max = Math.max(max, A[i]);
        }
        return max;
    }

    static long min(long... A) {
        long min = Long.MAX_VALUE;
        for (int i = 0; i < A.length; i++) {
            min = Math.min(min, A[i]);
        }
        return min;
    }

    static void fill(int dp[], int val) {
        Arrays.fill(dp, val);
    }

    static void fill(int dp[][], int val) {
        for (int i = 0; i < dp.length; i++)
            fill(dp[i], val);
    }

    static void fill(int dp[][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                fill(dp[i][j], val);
            }
        }
    }

    static void fill(int dp[][][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                for (int k = 0; k < dp[0][0].length; k++) {
                    fill(dp[i][j][k], val);
                }
            }
        }
    }

    static void fill(long dp[], int val) {
        Arrays.fill(dp, val);
    }

    static void fill(long dp[][], int val) {
        for (int i = 0; i < dp.length; i++)
            fill(dp[i], val);
    }

    static void fill(long dp[][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                fill(dp[i][j], val);
            }
        }
    }

    static void fill(long dp[][][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                for (int k = 0; k < dp[0][0].length; k++) {
                    fill(dp[i][j][k], val);
                }
            }
        }
    }

    static long power(long x, long y, long p) {

        if (y == 0)
            return 1;
        if (x == 0)
            return 0;
        long res = 1;
        x = x % p;

        while (y > 0) {

            if (y % 2 == 1)
                res = (res * x) % p;

            y = y >> 1;
            x = (x * x) % p;
        }

        return res;
    }

    static long power(long x, long y) {

        if (y == 0)
            return 1;
        if (x == 0)
            return 0;
        long res = 1;

        while (y > 0) {

            if (y % 2 == 1)
                res = (res * x);

            y = y >> 1;
            x = (x * x);
        }

        return res;
    }

    static void print(int A[]) {
        for (int i : A) {
            out.print(i + " ");
        }
        out.println();
    }

    static void printI(List<Integer> A) {
        for (int i : A) {
            out.print(i + " ");
        }
        out.println();
    }

    static void print(long A[]) {
        for (long i : A) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    static void printL(List<Long> A) {
        for (long i : A) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    static long mod(long x) {
        return ((x % mod + mod) % mod);
    }

    static String reverse(String s) {
        StringBuilder p = new StringBuilder(s);
        p.reverse();
        return p.toString();
    }

    static int i() {
        return sc.nextInt();
    }

    static String s() {
        return sc.next();
    }

    static String sl() {
        return sc.nextLine();
    }

    static long l() {
        return sc.nextLong();
    }

    static void sort(int[] A) {
        int n = A.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            int tmp = A[i];
            int randomPos = i + rnd.nextInt(n - i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    static void sort(long[] A) {
        int n = A.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            long tmp = A[i];
            int randomPos = i + rnd.nextInt(n - i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    static void sort(int[] A, Comparator<Integer> com) {
        int n = A.length;
        List<Integer> l = new ArrayList<>();
        for (int ai : A) {
            l.add(ai);
        }
        Collections.sort(l, com);
        for (int i = 0; i < n; ++i) {
            A[i] = l.get(i);
        }
    }

    static void sort(long[] A, Comparator<Long> com) {
        int n = A.length;
        List<Long> l = new ArrayList<>();
        for (long ai : A) {
            l.add(ai);
        }
        Collections.sort(l, com);
        for (int i = 0; i < n; ++i) {
            A[i] = l.get(i);
        }
    }

    static String sort(String s) {
        int n = s.length();
        char A[] = s.toCharArray();
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            char tmp = A[i];
            int randomPos = i + rnd.nextInt(n - i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
        StringBuilder st = new StringBuilder("");
        for (int i = 0; i < s.length(); i++) {
            st.append(A[i]);
        }
        return st.toString();
    }

    static boolean isPrime(long n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        double sq = Math.sqrt(n);

        for (int i = 5; i <= sq; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }

    static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through the problem:

1. **Understand the Problem:**
   - We need to maximize the minimum charm of the meals.
   - The charm of a meal is the absolute difference between the spiciness of the appetizer and the main dish.

2. **Identify Potential Bugs:**
   - The binary search logic might have issues with the mid calculation.
   - The TreeMap usage might have edge cases where it fails to find the correct floor or ceiling entries.
   - The sorting of the arrays might not handle duplicate values correctly.

3. **Create Edge Cases:**
   - **Case 1: All appetizers and main dishes have the same spiciness.**
   - **Case 2: One spiciness value is very high, and the others are very low.**
   - **Case 3: Alternating high and low spiciness values.**
   - **Case 4: Mixed values with duplicates.**

4. **Generate a Small-Scale Test Input:**
   - We will create a test input that includes a mix of the above cases to ensure the code is robust.

Here is a small-scale test input that includes these edge cases:

```plaintext
4
3
1 1 1
1 1 1
5
1 2 3 4 5
1000000000 1000000000 1000000000 1000000000 1000000000
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
```

This test input includes:
- A case where all spiciness values are the same.
- A case with a very high spiciness value and very low spiciness values.
- A case with alternating high and low spiciness values.
- A mixed case with duplicates.

This should help in exposing any potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
#include<bits/stdc++.h>
#ifndef ONLINE_JUDGE
#include <debug.h>
#else
#define debug(...)
#define debugArr(...)
#endif
#define t int t;cin>>t;while (t--)
#define ll long long
#define vll vector<ll>
#define vvll vector<vector<ll>>
#define vs vector<string>
#define vp vector<pair<ll,ll>>
#define fa(v) for (auto &it : (v))
#define all(x) begin(x),end(x)
#define endl "\n"
using namespace std;
template <typename T> ostream &operator<<(ostream &out, const vector<T> &v) { for (const auto &x : v) out << x << ' '; return out; }
template <typename T> istream &operator>>(istream &in, vector<T> &v) { for (auto &x : v) in >> x; return in; }
void solve(){
    ll n;
    cin>>n;
    vll a(n);
    cin>>a;
    vll b(n);
    cin>>b;
    ll s=0;
    ll e=1000000000;
    sort(all(a),greater<ll>());
    sort(all(b),greater<ll>());
    map<ll,ll> m;
    fa(b)
    {
        m[it]++;
    }
    ll ans=0;
    // cout<<b<<endl;
    while (s<=e)
    {
        ll mid=s+(e-s)/2;
        bool flag=true;
        map<ll,ll> m1=m;
        for (ll i = 0; i < n; i++)
        {
            vector<ll> v;
            for (ll j = 0; j < n; j++)
            {
                if(abs(a[i]-b[j])>=mid && m1[b[j]]>0)
                {
                    v.push_back(b[j]);
                }
            }
            if(v.size()==0)
            {
                flag=false;
                break;
            }
            else
            {
                ll min_diff=1000000000;
                ll z=0;
                for (ll j = 0; j < v.size(); j++)
                {
                    min_diff=min(min_diff,abs(a[i]-v[j]));
                    if(abs(a[i]-v[j])==min_diff)
                    {
                        z=v[j];
                    }
                }
                m1[z]--;
            }
        }
        if(flag)
        {
            ans=mid;
            s=mid+1;
        }
        else
        {
            e=mid-1;
        }
    }
    cout<<ans<<endl;
    // cout<<s<<" "<<e<<endl;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    t
    {
     solve();
    }
 return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The key aspects to focus on are:

1. **Identical Spiciness Values**: Cases where the spiciness values of appetizers and main dishes are identical.
2. **Large Differences in Spiciness**: Cases where the spiciness values of appetizers and main dishes have a large range.
3. **Uneven Distribution**: Cases where the spiciness values are unevenly distributed, making it difficult to pair them optimally.

Given these considerations, let's create a small-scale test input that includes these scenarios.

### Test Case 1: Identical Spiciness Values
```plaintext
1
3
1 1 1
1 1 1
```
**Expected Output:**
```plaintext
0
```
**Explanation:**
- Since all spiciness values are identical, the minimum charm is 0.

### Test Case 2: Large Differences in Spiciness
```plaintext
1
3
0 0 0
1000000000 1000000000 1000000000
```
**Expected Output:**
```plaintext
1000000000
```
**Explanation:**
- The spiciness values have a large range, and the minimum charm is the maximum possible difference.

### Test Case 3: Uneven Distribution
```plaintext
1
4
1 2 3 4
5 6 7 8
```
**Expected Output:**
```plaintext
1
```
**Explanation:**
- The spiciness values are unevenly distributed, and the best pairing results in a minimum charm of 1.

### Test Case 4: Mixed Values
```plaintext
1
5
1 2 3 4 5
5 4 3 2 1
```
**Expected Output:**
```plaintext
2
```
**Explanation:**
- The spiciness values are mixed, and the best pairing results in a minimum charm of 2.

### Combined Test Input
```plaintext
4
3
1 1 1
1 1 1
3
0 0 0
1000000000 1000000000 1000000000
4
1 2 3 4
5 6 7 8
5
1 2 3 4 5
5 4 3 2 1
```

This test input includes a variety of scenarios that should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std ;
#define int long long

void solve()
{
    int n,i,j ; cin>>n ;
    int a[n];
    for(i=0;i<n;i++) cin>>a[i] ;
    int b[n] ;
    for(i=0;i<n;i++) cin>>b[i] ;
    sort(a,a+n) ;
    int l=0,r=1000*1000*1000+1 ;
    while(l!=r-1)
    {
        int m=(l+r)/2 ;
        multiset<int> c ;
        for(int t : b) c.insert(t) ;
        bool ok=true ;
        for(i=n-1;i>=0;i--)
        {
            if((int)(c.size())==0)
            {
                ok=false ; break ;
            }
            auto iterator1=c.lower_bound(a[i]-m) ;
            if(*(c.begin())<=a[i]-m)
            {
                if(iterator1!=c.begin())
                {
                    iterator1-- ;
                }
                c.erase(c.find(*iterator1)) ;
                continue ;
            }
            auto iterator2=c.lower_bound(a[i]+m) ;
            if(iterator2!=c.end())
            {
                c.erase(c.find(*iterator2)) ;
                continue ;
            }
            ok=false ; break ;
        }
        if(ok) l=m ;
        else r=m ;
    }
    cout<<l<<"\n" ;
}

int32_t main()
{
    ios::sync_with_stdio(0) ; cin.tie(0) ; cout.tie(0) ;
    int t=1 ;   cin>>t ;
    while(t--) solve() ;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
5 4 3 2 1
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
84 72 71 62 20 51 92
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    spice = list(map(int,input().split()))
    dish = list(map(int,input().split()))

    spice.sort()
    dish.sort()

    mid = (len(spice)-1)//2
    print(abs(spice[mid]-dish[-1]))

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on scenarios where the code's logic fails. The buggy code sorts both the appetizer and main dish spicinesses and then calculates the charm based on the middle element of the appetizer list and the last element of the main dish list. This approach is flawed because it does not consider the optimal pairing of elements to maximize the minimum charm.

### Key Points to Consider:
1. **Symmetric Spiciness Values**: If the spiciness values of appetizers and main dishes are symmetric or nearly symmetric, the middle and last elements might not provide the correct result.
2. **Equal Spiciness Values**: If the spiciness values are the same or very close, the middle and last elements might not capture the correct minimum charm.
3. **Large Differences**: If the spiciness values have large differences, the middle and last elements might not represent the optimal pairing.

### Constructing the Test Input:
Let's construct a small-scale test input that highlights the issue with the buggy code.

#### Test Case 1:
- **n = 4**
- **Appetizers**: 1, 3, 5, 7
- **Main Dishes**: 2, 4, 6, 8

In this case, the optimal pairing would be:
- (1, 2) with charm 1
- (3, 4) with charm 1
- (5, 6) with charm 1
- (7, 8) with charm 1

The minimum charm is 1.

However, the buggy code will:
- Sort the appetizers: [1, 3, 5, 7]
- Sort the main dishes: [2, 4, 6, 8]
- Calculate the charm as `abs(5 - 8) = 3`

This is incorrect, and the expected output should be 1.

#### Test Input:
```plaintext
1
4
1 3 5 7
2 4 6 8
```

This test input will expose the bug in the buggy code by showing that it fails to find the correct minimum charm for a symmetric and well-distributed set of spiciness values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
tests = int(input())
for _ in range(tests):
    n = int(input())
    arr = sorted(list(map(int, input().split())))
    arr2 = sorted(list(map(int, input().split())))
    ans = float('-inf')
    for i in range(n):
        minn = float('inf')
        for j in range(i, n):
            minn = min(minn, abs(arr[j] - arr2[(i + j) % n]) )
        ans = max(ans, minn)
    print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios where the code might fail to produce the correct result. The buggy code has a nested loop that iterates through the sorted arrays and calculates the minimum charm for each possible pairing. However, this approach might not always find the optimal pairing, especially in cases where the spiciness values are not uniformly distributed.

### Step-by-Step Analysis:
1. **Uniform Distribution**: If the spiciness values are uniformly distributed, the code might work correctly. We need to test cases where the spiciness values are not uniformly distributed.
2. **Extreme Values**: Test cases with extreme values (e.g., very high or very low spiciness) can help identify issues.
3. **Symmetric Cases**: Cases where the spiciness values in both arrays are symmetric but not in the same order can be challenging for the code.
4. **Repetitive Values**: Cases with repetitive values can also be tricky, as the code might not handle them correctly.

### Test Input:
Let's create a test case with repetitive values and a non-uniform distribution to challenge the code:

```plaintext
1
4
1 1 100 100
100 100 1 1
```

### Explanation:
- **Test Case**: The input consists of one test case with 4 appetizers and 4 main dishes.
- **Spiciness Values**:
  - Appetizers: `[1, 1, 100, 100]`
  - Main Dishes: `[100, 100, 1, 1]`

### Expected Output:
- The largest possible value of the minimum charm is `99`.

### Why This Test Case is Challenging:
- The spiciness values are not uniformly distributed, and there are repetitive values.
- The optimal pairing would be:
  - `(1, 100)` with charm `99`
  - `(1, 100)` with charm `99`
  - `(100, 1)` with charm `99`
  - `(100, 1)` with charm `99`
- The minimum charm in all these pairings is `99`.

This test case should expose the bug in the given code, as the nested loop might not correctly find the optimal pairing.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
import java.io.*;
import java.util.*;

public class charmingMeals {
    public static void main(String[] args) throws IOException {
        FastReader r = new FastReader();
        PrintWriter w = new PrintWriter(System.out);

        int t = r.nextInt();
        while (t-->0) {
            int n = r.nextInt();
            int[] as = r.intArray(n);
            int[] bs = r.intArray(n);
            Arrays.sort(as);
            Arrays.sort(bs);
            int sol = solution(n,as,bs);
            w.write(sol+"\n");
        }
        r.close();
        w.close();
    }

    public static int solution(int n, int[] a, int[] b) {
        int l = 0;
        int r = Integer.MAX_VALUE-10;
        int mid;
        //everything l and left does work
        //everything right of r doesn't work
        while (l < r) {
            mid = (l + r+1) /2;
            //System.out.println("Trying mid " + mid);
            boolean works = works(mid, a, b);
            if (works) l = mid;
            else r = mid-1;
        }

        return l;
    }

    public static boolean works(int target, int[] as, int[] bs) {
        boolean [][] cap = new boolean[2+as.length*2][2+as.length*2];
        for (int i = 0; i < as.length; i++) {
            cap[0][i+1] = true;
            cap[as.length+1+i][cap.length-1] = true;
            for (int j = 0; j < bs.length; j++) {
                if (Math.abs(as[i]-bs[j]) >= target) {
                    cap[i+1][as.length+1+j] = true;
                }
            }
        }
        int score = 0;
        int flow = push(cap);
        while (flow > 0) {
            score+=flow;
            flow = push(cap);
        }

        return score >= as.length;
    }

    public static int push(boolean[][] cap) {
        Queue<Integer> q = new ArrayDeque<>();
        int[] src = new int[cap.length];
        boolean[] visited = new boolean[cap.length];
        src[0] = -1;
        q.add(0);
        while (!q.isEmpty()) {
            int curr = q.poll();
            visited[curr] = true;
            if (curr == cap.length-1) break;

            for (int i = 0; i < cap.length; i++) {
                if (cap[curr][i] && !visited[i]) {
                    q.add(i);
                    src[i] = curr;
                }
            }
        }
        if (src[cap.length-1] == 0) {
            return 0;
        }
        reduce(cap,src);
        return 1;
    }

    public static void reduce(boolean[][] cap, int[] src) {
        int curr = cap.length-1;
        while (src[curr] != -1) {
            cap[src[curr]][curr] = false;
            cap[curr][src[curr]] = true;
            curr = src[curr];
        }
    }

    public static boolean hasTrue (ArrayList<Boolean> a) {
        for (boolean b : a) {
            if (b) return true;
        }
        return false;
    }

    static class FastReader implements AutoCloseable {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
        public FastReader(String filepath) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(filepath));
        }

        public void close() throws IOException {
            br.close();
        }

        /** Indicates whether or not more input is available
         * @Warning BUGGY
         */
        boolean hasNext() throws IOException {
            return (br.ready());
        }

        /** Returns the next char available
         * @warning
         * CAREFUL, BUGGY
         * @return the next available char
         */

        char nextChar() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer("" + (char)br.read());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }

            return st.nextToken().charAt(0);
        }

        /**
         * Gives the entire next element (terminates in a whitespace)
         * @return the next full 'word'
         */
        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }

        long nextLong() { return Long.parseLong(next()); }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        /**
         * Returns an array of longs read from input
         * @param n - the number of longs to read
         * @return an array of length n containing the longs that were read
         */
        long[] longArray(int n) {
            long[] a = new long[n];
            for (int i = 0;i < n;i++) {
                a[i] = this.nextLong();
            }

            return a;
        }
        /**
         * Returns an array of ints read from input
         * @param n - the number of ints to read
         * @return an array of length n containing the ints that were read
         */
        int[] intArray(int n) {
            int[] a = new int[n];
            for (int i = 0;i < n;i++) {
                a[i] = this.nextInt();
            }

            return a;
        }

        /**
         * Returns a freqeuency array based on the number of appearance of each int that is read
         * @param n - the number of ints to read
         * @param largestNum - the largest number that may appear
         * @return an array of length largestNum that contains the appearance frequency of each value
         */
        int[] counts(int n, int largestNum) {
            int[] counts = new int[largestNum+1];
            for (int i = 0; i < n; i++) {
                counts[this.nextInt()]++;
            }

            return counts;
        }

        /**
         * Returns the boolean array represented by the next input line and a given char
         * @param t - the char that represents 'true'
         * @return an array representing the next line's truth values
         */
        boolean[] boolLine(char t) {
            String line = this.nextLine();
            boolean[] res = new boolean[line.length()];
            for (int i = 0; i < line.length(); i++) {
                res[i] = line.charAt(i) == t ? true : false;
            }

            return res;
        }

        /**
         * Returns a 2D array that contains the truth values specified by t according to the input read
         * @param rows - the number of rows to read
         * @param cols - the number of cols in each row
         * @param t - the char that represents 'true'
         * @return a 2D array representing the next lines' truth values
         */
        boolean[][] boolRect(int rows, int cols, char t) {
            boolean[][] res = new boolean[rows][cols];
            for (int i = 0;i < rows;i++) {
                String line = this.nextLine().strip();
                if (line.length() != cols) throw new RuntimeException("Number of cols was incorrect");
                for (int j = 0; j < cols; j++) {
                    res[i][j] = line.charAt(j) == t ? true : false;
                }
            }

            return res;
        }

        /**
         * Like a rectangle, but it's a square
         * @param n - side length
         * @param t - the char that represents 'true'
         * @return - a 2D array representing the next lines' truth values
         */
        boolean[][] boolSquare(int n, char t) {
            return boolRect(n,n,t);
        }
    }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases and scenarios that maximize the failure probability. The key points to focus on are:

1. **Identical Spiciness Values**: If the spiciness values of appetizers and main dishes are identical, the charm of each meal will be 0, which is the minimum possible charm. This can expose issues in the binary search and flow network logic.
2. **Highly Diverse Spiciness Values**: If the spiciness values are very different, the charm of each meal can be very high, and the code needs to handle large differences correctly.
3. **Edge Cases with Small n**: Small values of `n` can help identify issues in the flow network and binary search logic.
4. **Symmetric Spiciness Values**: If the spiciness values are symmetric, the code should handle the pairing correctly to maximize the minimum charm.

Let's create a test input that includes these scenarios:

```plaintext
4
1
100
100
3
1 2 3
3 2 1
5
0 1 2 3 4
4 3 2 1 0
6
1 2 3 4 5 6
6 5 4 3 2 1
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`
   - `a = [100]`
   - `b = [100]`
   - The charm of the meal is `|100 - 100| = 0`.

2. **Test Case 2**:
   - `n = 3`
   - `a = [1, 2, 3]`
   - `b = [3, 2, 1]`
   - The optimal pairing is: (1, 3), (2, 2), (3, 1). The charms are: 2, 0, 2. The minimum charm is 0.

3. **Test Case 3**:
   - `n = 5`
   - `a = [0, 1, 2, 3, 4]`
   - `b = [4, 3, 2, 1, 0]`
   - The optimal pairing is: (0, 4), (1, 3), (2, 2), (3, 1), (4, 0). The charms are: 4, 2, 0, 2, 4. The minimum charm is 0.

4. **Test Case 4**:
   - `n = 6`
   - `a = [1, 2, 3, 4, 5, 6]`
   - `b = [6, 5, 4, 3, 2, 1]`
   - The optimal pairing is: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). The charms are: 5, 3, 1, 1, 3, 5. The minimum charm is 1.

These test cases should help expose any issues in the binary search and flow network logic of the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
/*
 * Author : christmas964
 * FastReader : https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
 */

import java.io.*;
import java.util.*;

public class Main {
    static int mod = (int) (1e9 + 7);
    static int inf = (int) (1e9 + 7);
    static Random rr = new Random();
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static PrintWriter err = new PrintWriter(System.err);

    public static void main(String[] args) {
        int ttt = 1;
        ttt = i();
        while (ttt-- > 0) {
            // long startTime = System.nanoTime();
            solve();
            // long endTime = System.nanoTime();
            // err.println("Execution Time: " + (endTime - startTime) / 1000 + "
            // microseconds");
        }
        err.close();
        out.close();
    }

    private final static int[][] dirs8 = { { -1, -1 }, { -1, 0 }, { -1, 1 }, { 0, 1 }, { 1, 1 }, { 1, 0 }, { 1, -1 },
            { 0, -1 } };
    private final static int[][] dirs4 = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
    private final static int[][] dirs2 = { { 0, 1 }, { 1, 0 } };
    static int[] factors;
    static boolean factorsFlag = false;
    static long[] fac;
    static boolean factorialFlag = false;

    public static void solve() {
        /*
         * 0) First check briefly that you are not missing easier idea/method to
         * implement. Calculate the TC and check if it will pass.
         * 1) Check for Integer/Long out of bound senarios.
         * 2) Check if all the variables(i,j,k...) used correctly.
         * 3) Don't fcking rush.
         */

        int n = i();
        int[] a = input(n), b = input(n);
        sort(a);
        sort(b);
        int low = 0, high = Integer.MAX_VALUE;
        while(low<high)
        {
            int mid = low + (high - low) / 2 + 1;
            boolean ip = true;
            TreeMap<Integer, Integer> map = new TreeMap<>();
            for (int bi : b) {
                map.put(bi, map.getOrDefault(bi, 0) + 1);
            }
            for (int i = n - 1; i >= 0; --i) {
                int ai = a[i];
                if (map.floorEntry(ai - mid) != null) {
                    int fk = map.floorKey(ai - mid);
                    if (map.get(fk) == 1) {
                        map.remove(fk);
                    } else {
                        map.put(fk, map.get(fk) - 1);
                    }
                } else if (map.ceilingEntry(ai + mid) != null) {
                    int ck = map.ceilingKey(ai + mid);
                    if (map.get(ck) == 1) {
                        map.remove(ck);
                    } else {
                        map.put(ck, map.get(ck) - 1);
                    }
                } else {
                    ip = false;
                    break;
                }
            }

            if (ip) {
                low = mid;
            } else {
                high = mid - 1;
            }

        }

        out.println(low);

    }

    public static long nck(int n, int c) {
        if (n < c)
            return 0l;
        long a = fac[n];
        long b = (fac[n - c] * fac[c]) % mod;

        return (a * power(b, mod - 2, mod)) % mod;
    }

    public static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    public static void swap(long[] a, int i, int j) {
        long t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    public static int r(int r) {
        return rr.nextInt(r);
    }

    public static int getBit(int n, int i) {
        return ((n >> i) & 1);
    }

    public static int getBit(long n, int i) {
        return (int) ((n >> i) & 1l);
    }

    public static int cnt(int num, int i) {
        /*
         * Returns the number of set bits in num's number of numbers including 0 at i'th
         * bit.
         * if num = 3 that is [0,1,2] and i = 1, then it will return number of numbers
         * that have set bit
         * at i'th index.
         */
        int g = (1 << (i + 1));
        int q = (num) / g;
        int r = (num) % g;
        g >>= 1;
        return (q * g) + ((r / g) > 0 ? (r % g) : 0);
    }

    public static long cnt(long num, int i) {
        /*
         * Returns the number of set bits in num's number of numbers including 0 at i'th
         * bit.
         * if num = 3 that is [0,1,2] and i = 1, then it will return number of numbers
         * that have set bit
         * at i'th index.
         */
        int g = (1 << (i + 1));
        long q = (num) / g;
        long r = (num) % g;
        g >>= 1;
        return (q * g) + ((r / g) > 0 ? (r % g) : 0);
    }

    private static void sieve() {
        if (factorsFlag)
            return;
        factors = new int[(int) ((3 * 1e5) + 1)];
        factorsFlag = true;
        for (int i = 1; i < factors.length; i++) {
            factors[i] = i;
        }
        for (int i = 2; i * 1l * i < factors.length; i++) {
            if (factors[i] == i) {
                for (int j = i; j < factors.length; j += i) {
                    if (factors[j] == j) {
                        factors[j] = i;
                    }
                }
            }
        }
    }

    public static int[] unique(int[] nums) {
        Set<Integer> set = new HashSet<>();
        List<Integer> list = new ArrayList<>();
        for (int num : nums) {
            if (set.contains(num))
                continue;
            set.add(num);
            list.add(num);
        }
        int[] a = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            a[i] = list.get(i);
        }
        return a;
    }

    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static int abs(int a) {
        return Math.abs(a);
    }

    public static long abs(long a) {
        return Math.abs(a);
    }

    public static int upper(long A[], long k, int si, int ei) {
        /*
         * gives the element index that is smallest among all elements that is greater
         * then
         * k , otherwise -1.
         */
        int l = si;
        int u = ei;
        int ans = -1;
        while (l <= u) {
            int mid = (l + u) / 2;
            if (A[mid] <= k) {
                l = mid + 1;
            } else {
                ans = mid;
                u = mid - 1;
            }
        }
        return ans;
    }

    public static int lower(long A[], long k, int si, int ei) {
        /*
         * gives the element index that is smallest among all elements that is greater
         * then
         * or equal to k , otherwise -1.
         */
        int l = si;
        int u = ei;
        int ans = -1;
        while (l <= u) {
            int mid = (l + u) / 2;
            if (A[mid] < k) {
                l = mid + 1;
            } else {
                ans = mid;
                u = mid - 1;
            }
        }
        return ans;

    }

    static int[] copy(int A[]) {
        int B[] = new int[A.length];
        for (int i = 0; i < A.length; i++) {
            B[i] = A[i];
        }
        return B;
    }

    static long[] copy(long A[]) {
        long B[] = new long[A.length];
        for (int i = 0; i < A.length; i++) {
            B[i] = A[i];
        }
        return B;
    }

    static int[] input(int n) {
        int A[] = new int[n];
        for (int i = 0; i < n; i++) {
            A[i] = sc.nextInt();
        }
        return A;
    }

    static long[] inputL(int n) {
        long A[] = new long[n];
        for (int i = 0; i < n; i++) {
            A[i] = sc.nextLong();
        }
        return A;
    }

    static String[] inputS(int n) {
        String A[] = new String[n];
        for (int i = 0; i < n; i++) {
            A[i] = sc.next();
        }
        return A;
    }

    static long sum(int A[]) {
        long sum = 0;
        for (int i : A) {
            sum += i;
        }
        return sum;
    }

    static long sum(long A[]) {
        long sum = 0;
        for (long i : A) {
            sum += i;
        }
        return sum;
    }

    static void reverse(long A[]) {
        int n = A.length;
        int left = 0, right = n - 1;
        while (left < right) {
            long temp = A[left];
            A[left++] = A[right];
            A[right--] = temp;
        }

    }

    static void reverse(int A[]) {
        int n = A.length;
        int left = 0, right = n - 1;
        while (left < right) {
            int temp = A[left];
            A[left++] = A[right];
            A[right--] = temp;
        }

    }

    static int[][] input(int n, int m) {
        int A[][] = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                A[i][j] = i();
            }
        }
        return A;
    }

    static char[][] inputc(int n, int m) {
        char A[][] = new char[n][m];
        for (int i = 0; i < n; i++) {
            String s = s();
            for (int j = 0; j < m; j++) {
                A[i][j] = s.charAt(j);
            }
        }
        return A;
    }

    static int nextPowerOf2(int n) {
        /*
         * Returns the power of two greater then equal to n
         */
        if (n <= 0)
            return 1;
        if ((n & (n - 1)) == 0)
            return n;

        n--;
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        n++;

        return n;
    }

    static long nextPowerOf2(long n) {
        /*
         * Returns the power of two greater then equal to n
         */
        if (n <= 0)
            return 1;
        if ((n & (n - 1)) == 0)
            return n;

        n--;
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        n++;

        return n;
    }

    static int max(int... A) {
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < A.length; i++) {
            max = Math.max(max, A[i]);
        }
        return max;
    }

    static int min(int... A) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < A.length; i++) {
            min = Math.min(min, A[i]);
        }
        return min;
    }

    static long max(long... A) {
        long max = Long.MIN_VALUE;
        for (int i = 0; i < A.length; i++) {
            max = Math.max(max, A[i]);
        }
        return max;
    }

    static long min(long... A) {
        long min = Long.MAX_VALUE;
        for (int i = 0; i < A.length; i++) {
            min = Math.min(min, A[i]);
        }
        return min;
    }

    static void fill(int dp[], int val) {
        Arrays.fill(dp, val);
    }

    static void fill(int dp[][], int val) {
        for (int i = 0; i < dp.length; i++)
            fill(dp[i], val);
    }

    static void fill(int dp[][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                fill(dp[i][j], val);
            }
        }
    }

    static void fill(int dp[][][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                for (int k = 0; k < dp[0][0].length; k++) {
                    fill(dp[i][j][k], val);
                }
            }
        }
    }

    static void fill(long dp[], int val) {
        Arrays.fill(dp, val);
    }

    static void fill(long dp[][], int val) {
        for (int i = 0; i < dp.length; i++)
            fill(dp[i], val);
    }

    static void fill(long dp[][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                fill(dp[i][j], val);
            }
        }
    }

    static void fill(long dp[][][][], int val) {
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                for (int k = 0; k < dp[0][0].length; k++) {
                    fill(dp[i][j][k], val);
                }
            }
        }
    }

    static long power(long x, long y, long p) {

        if (y == 0)
            return 1;
        if (x == 0)
            return 0;
        long res = 1;
        x = x % p;

        while (y > 0) {

            if (y % 2 == 1)
                res = (res * x) % p;

            y = y >> 1;
            x = (x * x) % p;
        }

        return res;
    }

    static long power(long x, long y) {

        if (y == 0)
            return 1;
        if (x == 0)
            return 0;
        long res = 1;

        while (y > 0) {

            if (y % 2 == 1)
                res = (res * x);

            y = y >> 1;
            x = (x * x);
        }

        return res;
    }

    static void print(int A[]) {
        for (int i : A) {
            out.print(i + " ");
        }
        out.println();
    }

    static void printI(List<Integer> A) {
        for (int i : A) {
            out.print(i + " ");
        }
        out.println();
    }

    static void print(long A[]) {
        for (long i : A) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    static void printL(List<Long> A) {
        for (long i : A) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    static long mod(long x) {
        return ((x % mod + mod) % mod);
    }

    static String reverse(String s) {
        StringBuilder p = new StringBuilder(s);
        p.reverse();
        return p.toString();
    }

    static int i() {
        return sc.nextInt();
    }

    static String s() {
        return sc.next();
    }

    static String sl() {
        return sc.nextLine();
    }

    static long l() {
        return sc.nextLong();
    }

    static void sort(int[] A) {
        int n = A.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            int tmp = A[i];
            int randomPos = i + rnd.nextInt(n - i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    static void sort(long[] A) {
        int n = A.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            long tmp = A[i];
            int randomPos = i + rnd.nextInt(n - i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    static void sort(int[] A, Comparator<Integer> com) {
        int n = A.length;
        List<Integer> l = new ArrayList<>();
        for (int ai : A) {
            l.add(ai);
        }
        Collections.sort(l, com);
        for (int i = 0; i < n; ++i) {
            A[i] = l.get(i);
        }
    }

    static void sort(long[] A, Comparator<Long> com) {
        int n = A.length;
        List<Long> l = new ArrayList<>();
        for (long ai : A) {
            l.add(ai);
        }
        Collections.sort(l, com);
        for (int i = 0; i < n; ++i) {
            A[i] = l.get(i);
        }
    }

    static String sort(String s) {
        int n = s.length();
        char A[] = s.toCharArray();
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            char tmp = A[i];
            int randomPos = i + rnd.nextInt(n - i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
        StringBuilder st = new StringBuilder("");
        for (int i = 0; i < s.length(); i++) {
            st.append(A[i]);
        }
        return st.toString();
    }

    static boolean isPrime(long n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        double sq = Math.sqrt(n);

        for (int i = 5; i <= sq; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }

    static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that might cause the algorithm to fail. The algorithm uses a binary search to find the maximum possible minimum charm, and it uses a `TreeMap` to manage the spiciness levels of the main dishes.

Here are some potential edge cases to consider:
1. **All spiciness levels are the same**: This can cause issues if the algorithm doesn't handle duplicate values correctly.
2. **Spiciness levels are very close**: This can expose issues with the binary search and the floor/ceiling entry checks.
3. **Spiciness levels are spread out**: This can test the algorithm's ability to handle large differences in spiciness levels.
4. **Small and large values mixed**: This can test the algorithm's ability to handle a wide range of values.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of these scenarios.

### Test Input
```plaintext
1
5
1 2 3 4 5
5 4 3 2 1
```

### Explanation
- **Number of test cases**: 1
- **Number of appetizers and main dishes**: 5
- **Spiciness of appetizers**: 1, 2, 3, 4, 5
- **Spiciness of main dishes**: 5, 4, 3, 2, 1

### Expected Output
- The expected output for this test case is 4. The optimal way to pair the appetizers and main dishes to maximize the minimum charm is:
  - (1, 5) with charm 4
  - (2, 4) with charm 2
  - (3, 3) with charm 0
  - (4, 2) with charm 2
  - (5, 1) with charm 4

The minimum charm in this pairing is 0, but the maximum possible value of the minimum charm is 4.

This test case is designed to expose any issues with the algorithm's handling of sorted and reverse-sorted arrays, as well as its ability to correctly manage the spiciness levels using the `TreeMap`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

#define int long long int
#define f(i,n) for(int i =0 ; i<n ; i++)
#define fx(i,n,a) for(int i = a ; i<n ; i++)
#define inp(i,v) for(int i=0;i<v.size();i++)cin >> v[i];
#define mod 1000000007
#define vint vector<int>
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define last(a) *((a).end()-1)
#define print(a) for(auto x :(a)) cout << x << ' ';
#define ret return;
#define pb(a) push_back((a))
#define pii pair<int,int>
#define vp vector<pair<int,int>>
#define vintd(n,m) vector<vector<int>> arr(n, vector<int>(m));

void solve() {
    int n;
    cin >> n;
    vint a(n);
    vint b(n);
    vint c(n);
    f(i,n)cin >> a[i];
    f(i,n)cin >> b[i];
    sort(all(a));
    sort(all(b));
    for(int i=0;i<(n+(n%2))/2;i++)
    {
        c[i] = abs(a[i]-b[(n/2)+i]);

    }
    for(int i=(n+(n%2))/2;i<(n);i++)
    {
        c[i] = abs(b[i-((n+n%2)/2)]-a[i]);
    }
    sort(all(c));
    cout << c[0] << endl;

}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = 1;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
1000000000
0
2
1 2
2 1
4
0 1 2 3
3 2 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: 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: 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: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

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 $$$100$$$.

Buggy Code:
//=======================^===================================================================^=======================\\

/*

    * * *       * * *     * * * * * *       * * * * * *       * *           * *     * * * * * * *   * * * * * * * *
    * * *       * * *   * * * * * * * *   * * * * * * * *   * * * *       * * * *   * * * * * * *   * * * * * * * *
    * * *       * * *   * * *     * * *   * * *     * * *   * * * *       * * * *   * * * * * * *   * * * * * * * *
    * * *       * * *   * * *     * * *   * * *     * * *   * * * * *   * * * * *       * * *            * * *
    * * *       * * *   * * *     * * *   * * *     * * *   * * *  * * * *  * * *       * * *            * * *
    * * * * * * * * *   * * * * * * * *   * * * * * * * *   * * *   * * *   * * *       * * *            * * *
    * * * * * * * * *   * * * * * * * *   * * * * * * *     * * *    * *    * * *       * * *            * * *
    * * * * * * * * *   * * * * * * * *   * * * * *         * * *     *     * * *       * * *            * * *
    * * *       * * *   * * *     * * *   * * * * * *       * * *           * * *       * * *            * * *
    * * *       * * *   * * *     * * *   * * *  * * *      * * *           * * *       * * *            * * *
    * * *       * * *   * * *     * * *   * * *   * * *     * * *           * * *   * * * * * * *        * * *
    * * *       * * *   * * *     * * *   * * *    * * *    * * *           * * *   * * * * * * *        * * *
    * * *       * * *   * * *     * * *   * * *     * * *   * * *           * * *   * * * * * * *        * * *

*/

//=======================^===================================================================^=======================\\

#include<bits/stdc++.h>
using namespace std;

#define Author_HarMit ios_base::sync_with_stdio (0);cin.tie (0)

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long, null_type, less<long long>, rb_tree_tag, tree_order_statistics_node_update> pbds; // (increasing)
// //typedef tree<long long, null_type, greater_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> pbds; // (decreasing)
// // can write like less_equal or greater_equal if you want <= or >= in order of key
// k-th element ( *name.find_by_order(key) )
// no of element < key ( name.order_of_key(key) )

#define ll long long

// Loops
#define fo(i,n) for(ll i=0;i<n;i++)
#define rfo(i,n) for(ll i=n-1;i>=0;i--)
#define fop(i,s,e) for(ll i=s;i<=e;i++)
#define rfop(i,s,e) for(ll i=s;i>=e;i--)
#define ft(x,a) for(auto x:a)
#define fq(q) while(q--)

// Inputs
#define in(n) ll n;cin>>n;
#define inn(n,k) ll n,k;cin>>n>>k;
#define vin(a,n) vl a(n);for(ll i=0; i<n; i++)cin>>a[i];
#define strin(s) string s;cin>>s;

// Outputs
#define cy printf("YES\n")
#define cn printf("NO\n")
#define on(n) cout<<n<<endl;
#define onn(n,k) cout<<n<<" "<<k<<endl;
#define vo(a) for(auto x:a){cout<<x<<" ";}cout<<endl;

// Functions
#define ub upper_bound
#define lb lower_bound
#define sbit(a,k) ((a & (1<<k))!=0) // check for Kth bit

// STL
#define pb push_back
#define ppb pop_back
#define spb insert
#define F first
#define S second

// Sorting
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define maxa(v) *max_element(v.begin(),v.end())
#define mina(v) *min_element(v.begin(),v.end())

#define PI 3.1415926535897932384626
const ll mod = 1e9 + 7;
const ll pim = INT_MAX;
const ll nim = INT_MIN;

//=======================^===================================================================^=======================\\

typedef pair < ll, ll > pl;
typedef vector < ll > vl;
typedef stack < ll > stl;
typedef queue < ll > ql;
typedef priority_queue < ll > pql;
typedef set < ll > sl;
typedef unordered_set < ll > usl;
typedef map < ll, ll > mll;
typedef unordered_map < ll, ll > umll;
typedef map < ll, vl > mlvl;
typedef vector < pl > vpl;
typedef vector<vl> vvl;

//=======================^===================================================================^=======================\\

void useless(){

    in(n);
    vin(a,n);
    vin(b,n);

    sortall(a);
    sortall(b);

    ll niche = 0;
    ll upar = n/2;

    ll ans1 = INT_MAX, ans2 = INT_MAX;

    for(; upar<n; upar++){
        ans1 = min(ans1, abs(a[upar]-b[niche]));
        niche++;
    }

    upar = 0;

    for(; niche<n; niche++){
        ans1 = min(ans1, abs(a[upar]-b[niche]));
        upar++;
    }

    upar = n/2 + 1;
    niche = 0;

    for(; upar<n; upar++){
        ans2 = min(ans2, abs(a[upar]-b[niche]));
        niche++;
    }

    upar = 0;

    for(; niche<n; niche++){
        ans2 = min(ans2, abs(a[upar]-b[niche]));
        upar++;
    }

    cout << max(ans1, ans2) << '\n';

}

signed main() {
    Author_HarMit;
    ll tt = 1;
    cin >> tt;

    while (tt--) {
        useless();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Test cases where the arrays have the same values, or where the arrays are sorted in a way that the algorithm might not handle correctly.
2. **Large Differences**: Test cases where the spiciness levels have very large differences to ensure the algorithm handles large numbers correctly.
3. **Symmetric Cases**: Test cases where the arrays are symmetric or nearly symmetric to check if the algorithm can handle such patterns.

Given the constraints and the nature of the problem, let's create a small-scale test input that is complete and valid:

```plaintext
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
```

### Explanation:
1. **First Test Case (n=3)**:
   - All appetizers have spiciness 0.
   - All main dishes have spiciness 1000000000.
   - The charm of each meal will be 1000000000, which is the maximum possible charm.

2. **Second Test Case (n=5)**:
   - Both appetizers and main dishes have the same spiciness levels.
   - The charm of each meal will be 0, which is the minimum possible charm.

3. **Third Test Case (n=6)**:
   - Three appetizers have spiciness 0, and three have spiciness 100.
   - Three main dishes have spiciness 100, and three have spiciness 0.
   - The charm of each meal will be 100, which is the maximum possible charm.

4. **Fourth Test Case (n=7)**:
   - The spiciness levels are more varied and less symmetric.
   - The charm of each meal will vary, and the algorithm needs to find the maximum minimum charm.

This test input should help expose any issues in the buggy code by covering a range of scenarios, including edge cases and large differences in spiciness levels.