**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
import sys
import sys

def main():
    import sys
    import sys
    from sys import stdin
    import sys

    def input_gen():
        for line in sys.stdin:
            for num in line.strip().split():
                yield num

    tokens = input_gen()
    try:
        t = int(next(tokens))
    except StopIteration:
        t = 0

    for _ in range(t):
        try:
            l = int(next(tokens))
            n = int(next(tokens))
            m = int(next(tokens))
        except StopIteration:
            l, n, m = 0, 0, 0

        a = []
        for _ in range(l):
            try:
                a.append(int(next(tokens)))
            except StopIteration:
                a.append(0)

        # Read matrix b
        b = []
        list_x = [[] for _ in range(8)]  # 1 to 7
        for r in range(n):
            for c in range(m):
                try:
                    val = int(next(tokens))
                except StopIteration:
                    val = 0
                if 1 <= val <=7:
                    list_x[val].append(r * m + c)  # Flattened index
                # No need to store the entire matrix

        # Initialize dp_next as a list of False
        size = (n + 2) * (m + 2)
        dp_next = [False] * size

        for k in range(l-1, -1, -1):
            a_k = a[k]
            current_positions = list_x[a_k]
            grid = [0] * size  # Using integers 0/1 instead of booleans

            for pos in current_positions:
                r = pos // m +1  # 1-based
                c = pos % m +1
                r_p1 = r +1
                c_p1 = c +1
                if r_p1 > n or c_p1 > m:
                    grid[r * m + c] =1
                else:
                    idx = (r_p1) * m + (c_p1)
                    if not dp_next[idx]:
                        grid[r * m + c] =1

            # Compute suffix OR in reverse order
            suffix = [0] * size
            for r in range(n, 0, -1):
                row_or = 0
                for c in range(m, 0, -1):
                    pos = r * m + c
                    row_or |= grid[pos]
                    next_pos = (r +1) * m + c
                    suffix[pos] = suffix[next_pos] | row_or

            # Update dp_current based on suffix
            dp_current = [False] * size
            for r in range(1, n+1):
                for c in range(1, m+1):
                    pos = r * m + c
                    if suffix[pos]:
                        dp_current[pos] = True

            # Prepare for next iteration
            dp_next = dp_current

        # After processing all moves, check if starting position is winning
        start_pos = m +1  # (1,1) in flattened index
        if n >=1 and m >=1 and dp_next[1 * m +1]:
            print("T")
        else:
            print("N")

if __name__ == "__main__":
    main()

**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 buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The key aspects to consider are:

1. **Empty Submatrix**: A scenario where a player is forced to choose from an empty submatrix.
2. **Multiple Occurrences of the Same Element**: A scenario where the same element appears multiple times in the matrix.
3. **Boundary Conditions**: A scenario where the matrix or array has minimal or maximal sizes.

Let's construct a small-scale test input that covers these aspects:

### Test Input
```plaintext
3
2 2 2
1 2
1 2
3 4
2 2 2
1 1
1 1
1 1
1 2 3
1 2
1 2 3
4 5 6
```

### Explanation
1. **Test Case 1**:
   - Array: `1 2`
   - Matrix:
     ```
     1 2
     3 4
     ```
   - Tsovak starts and finds `1` at `(1,1)`.
   - Narek then needs to find `2` in the submatrix starting from `(2,2)`, which contains `4`.
   - Narek loses because he cannot find `2` in the submatrix.
   - **Expected Output**: `T`

2. **Test Case 2**:
   - Array: `1 1`
   - Matrix:
     ```
     1 1
     1 1
     ```
   - Tsovak starts and finds `1` at `(1,1)`.
   - Narek then needs to find `1` in the submatrix starting from `(2,2)`, which contains `1`.
   - Narek finds `1` at `(2,2)`.
   - Tsovak then needs to find `1` in the submatrix starting from `(3,3)`, which is empty.
   - Tsovak loses because the submatrix is empty.
   - **Expected Output**: `N`

3. **Test Case 3**:
   - Array: `1 2`
   - Matrix:
     ```
     1 2 3
     4 5 6
     ```
   - Tsovak starts and finds `1` at `(1,1)`.
   - Narek then needs to find `2` in the submatrix starting from `(2,2)`, which contains `5`.
   - Narek loses because he cannot find `2` in the submatrix.
   - **Expected Output**: `T`

### Final Test Input
```plaintext
3
2 2 2
1 2
1 2
3 4
2 2 2
1 1
1 1
1 1
1 2 3
1 2
1 2 3
4 5 6
```

This test input should help expose any issues in the buggy code by covering edge cases and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
def solve():
    import sys
    input_data = sys.stdin.read().strip().split()
    t = int(input_data[0])
    pos = 1

    # We'll solve each test case independently.
    # Approach:
    # 1. We have an array a of length l and a matrix b of size n x m.
    # 2. Two players (Tsovak starts first) alternate turns, each time choosing a cell that contains a[i].
    #    The chosen cell for the i-th element must be strictly "down-right" from the previously chosen cell.
    #
    # If at some turn a player cannot find the required element in the allowed submatrix, that player loses.
    # If a player manages to choose the last element a[l-1], the next player loses because the array ended.
    #
    # We need to determine the winner if both play optimally.
    #
    # Observations:
    # - Values in a and b are at most min(7, n*m). So values are from 1 to at most 7.
    # - We must play a game of perfect information. This suggests we can use DP or memoized recursion.
    # - Let's define a recursive DP:
    #   dp(i, r, c) = whether the current player (who must place a[i]) has a winning strategy,
    #   given that the previously chosen cell is (r, c). For the first call, we can use (r, c) = (0, 0)
    #   to mean "no previous cell chosen" and we must place a[0] anywhere in the matrix.
    #
    #   Transition:
    #   - On turn i, we look for a[i]. We must pick a cell (r', c') where:
    #       r' > r and c' > c, and b[r'][c'] == a[i].
    #   - If we cannot find such a cell, dp(i, r, c) = False (losing state for the current player).
    #   - Otherwise, for each possible choice (r', c'):
    #       we compute dp(i+1, r', c'), which represents the next state for the opponent.
    #       If at least one leads to a losing state for the opponent, dp(i, r, c) = True (current can force a win).
    #   If all possible moves lead to winning states for the opponent, dp(i, r, c) = False.
    #
    # The array and matrix sizes can be up to 300x300 = 90000 cells (per test), sum over tests <= 10^5.
    # We must be efficient.
    #
    # To quickly find the next cell containing a[i] below and to the right of (r, c):
    # - We know values are small (<=7), so we can store occurrences of each value in a sorted list of (r, c).
    # - To find a cell with r' > r and c' > c, we can binary search for r and then among those with r' > r, find one with c' > c.
    #
    # We'll implement a memoized recursion and use binary search to find next moves.
    # This might still be large, but we rely on the fact that many states won't be reached.
    #
    # Implementation details:
    # - Store positions of each value (1 to 7) in separate lists sorted by (r, c).
    # - For each dp(i, r, c), perform a binary search on occ[a[i]] to find positions that satisfy r' > r, c' > c.
    # - Use memo to avoid repeated computations.
    #
    # Let's go ahead.

    sys.setrecursionlimit(10**7)
    from bisect import bisect_left

    results = []
    for _ in range(t):
        l, n, m = map(int, input_data[pos:pos+3])
        pos += 3
        a_arr = list(map(int, input_data[pos:pos+l]))
        pos += l
        b_mat = []
        for __ in range(n):
            row = list(map(int, input_data[pos:pos+m]))
            pos += m
            b_mat.append(row)

        # Value occurrences: For each v in [1..7], store a list of (r, c) (1-indexed)
        # Actually index from 1 to n and 1 to m as given.
        occ = [[] for __ in range(8)]  # 1-based indexing for values
        for i_row in range(n):
            for j_col in range(m):
                v = b_mat[i_row][j_col]
                occ[v].append((i_row+1, j_col+1))

        # Sort occurrences by row, and then by column
        for v in range(1, 8):
            occ[v].sort()

        # Memo dictionary: key: (i, r, c) -> bool
        # (i: index in a_arr, r: last chosen row, c: last chosen col)
        memo = {}

        def can_win(i, r, c):
            # If we've placed all elements, then the previous player placed the last element.
            # That means the current player has no next element to place and thus loses.
            # So if i == l, current player cannot move => losing state
            if i == l:
                return False

            key = (i, r, c)
            if key in memo:
                return memo[key]

            val = a_arr[i]
            candidates = occ[val]

            # We must find a candidate (r', c') with r' > r and c' > c.
            # Since candidates are sorted by (r', c'), we can:
            # 1) Binary search by row: find first occurrence with r' > r.
            # After that, we must only consider those with c' > c.

            # Find the position in candidates where r' > r using binary search by r dimension:
            # We'll do a binary search on r first:
            # But we have sorted by (r, c). If we find the first element with r' > r,
            # then among those with the same r', we must find c' > c.
            #
            # A strategy:
            #   Use bisect to find the insertion point for (r+1, 0) to get the first with r' >= r+1.
            #   Then from that index forward, we check if c' > c and try moves.

            start_idx = bisect_left(candidates, (r+1, 0))
            # Now we iterate from start_idx until out of range or until no suitable found:
            # We must find at least one candidate (r', c') with c' > c. We can skip those with c' <= c.
            # Once we find a suitable candidate that leads the opponent to a losing state (dp(i+1, r', c') = False),
            # we return True.

            # To speed up checks, we can just linearly scan from start_idx.
            # Because maximum candidates might be large, but we rely on partial pruning.
            # This should be okay given constraints.

            found_win = False
            for (nr, nc) in candidates[start_idx:]:
                if nr <= r:
                    continue
                if nc <= c:
                    continue
                # Try this move:
                nxt = can_win(i+1, nr, nc)
                # nxt is whether the opponent can win from that state.
                # If the opponent cannot win => dp for current = True
                if not nxt:
                    found_win = True
                    break

            memo[key] = found_win
            return found_win

        # Initial call: Tsovak starts with i=0 and no previously chosen cell.
        # Use (r, c) = (0, 0) as a base (so next chosen cell must have r'>0 and c'>0, i.e. any cell)
        first_player_win = can_win(0, 0, 0)
        results.append("T" if first_player_win else "N")

    print("\n".join(results))

if __name__ == "__main__":
    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
3
2 2 2
1 2
1 2
3 4
2 2 2
1 2
1 1
2 2
2 2 2
2 1
3 4
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
/* package codechef; // don't place package name! */

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

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

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;
    }

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

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

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

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n;
    Boolean[][][] dp;
    int m , l;
    int[] val;
    int[][] a;
    List<List<int[]>> list = new ArrayList<>();
    boolean help(int i, int j, int index){
        if(i>=n||j>=m||index>=l) return false;
        if(dp[i][j][index]!= null) return dp[i][j][index];
        boolean win  = false;
        for(int[] e: list.get(val[index])){
            int x= e[0];
            int y = e[1];
            if(a[x][y]==val[index]&&x>=i&&y>=j){
                    win|= !help(x+1, y+1, index+1);
            }
        }
        return dp[i][j][index]=win;
    }
    public void solve() throws IOException {
        t = sc.nextInt();
        while (t-- > 0) {
            l = i();
            n = i();
            m = i();
            list = new ArrayList<>(7);
            for(int i = 0;i<=7;i++) list.add(new ArrayList<>());
            val = sc.readIntArray(l);
            a = new int[n][m];
            dp = new Boolean[n][m][l];
            for(int i = 0;i<n;i++) for(int j = 0;j<m;j++) a[i][j] = i();
            for(int i = 0;i<n;i++) for(int j =0;j<m;j++) list.get(a[i][j]).add(new int[]{i,j});
            out.print(help(0, 0, 0)?'T':'N');
            out.print("\n");
        }
    }
}

**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 where the faulty logic in the code might fail. Here are some considerations:

1. **Matrix and Array Size**: Use the maximum allowed values for \( l \), \( n \), and \( m \) to stress the algorithm.
2. **Matrix Values**: Use values that are within the constraints but can create complex submatrix scenarios.
3. **Array Values**: Ensure the array values can be found in the matrix but in positions that make the game challenging to play optimally.

Let's create a test case that fits these criteria:

- **Test Case 1**: A small but complex case where the matrix and array values are such that the submatrix constraints are tight and the players have to make strategic choices.

### Test Case 1
```plaintext
1
3 3 3
1 2 3
1 2 3
4 5 6
7 8 9
```

### Explanation:
- **Array \( a \)**: [1, 2, 3]
- **Matrix \( b \)**:
  ```
  1 2 3
  4 5 6
  7 8 9
  ```

### Analysis:
- **Tsovak's First Move**: Tsovak can choose (1,1) for the value 1.
- **Narek's First Move**: Narek needs to choose (2,2) for the value 2.
- **Tsovak's Second Move**: Tsovak needs to choose (3,3) for the value 3.

This test case is designed to be small and manageable but still challenging. It ensures that the players have to make optimal choices, and the submatrix constraints are tight, which can expose any issues in the logic of the code.

### Output:
```plaintext
T
```

This test case should help in identifying any issues with the logic in the provided code, especially in handling the submatrix constraints and optimal play.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class c {
    static long mod = 1000000007;

    static long GCD(long u, long v) {
        if (u == 0)
            return v;
        return GCD(v % u, u);
    }

    // LCM of two numbers
    static long LCM(long u, long v) {
        return (u / GCD(u, v)) * v;
    }

    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 {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    // static FastReader sc = new FastReader();

    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static int[][][] dp;

    private static void solve() {
        int l = sc.nextInt();
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[l];
        for (int i = 0; i < l; i++) {
            a[i] = sc.nextInt();
        }
        int[][] b = new int[n][m];

        for (int i = 0; i < b.length; i++) {
            for (int j = 0; j < b[0].length; j++) {
                b[i][j] = sc.nextInt();
            }
        }

        dp = new int[n][m][l];
        for (int[][] z : dp) {
            for (int[] y : z) {
                    Arrays.fill(y, -1);

            }
        }
        boolean ans = f1(a, b, l, 0, 0, 0);
        System.out.println(!ans ? "N" : "T");
    }

    private static boolean f1(int[] a, int[][] b, int l, int i, int j, int ind) {
        if (i >= b.length || j >= b[0].length) {
            return false;
        }
        if (ind >= a.length) {
            return false;
        }
        if (dp[i][j][ind] != -1) {
            int ans = dp[i][j][ind];
            return ans == 1 ? true : false;
        }
        boolean ans = false;

        for (int x = i; x < b.length; x++) {
            for (int y = j; y < b[0].length; y++) {
                if (a[ind] == b[x][y]) {

                    boolean temp = f1(a, b, l, x + 1, y + 1, ind + 1);
                    if (!temp)
                        ans = true;

                }
            }
        }

        dp[i][j][ind] = ans ? 1 : 0;
        return 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:
```plaintext
1
3 3 3
1 2 3
1 4 5
6 2 7
8 9 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fi first
#define se second
#define endl "\n"
#define rep(a,b,c) for(ll a=b; a<c; a++)
#define rep2(a,b,c,d) for(ll a=b; a<c; a+=d)
#define repr(a,b,c) for(ll a=b-1; a>c-1; a--)
#define repa(a,b) for(const auto &a: b)
#define multicase() int t; cin>>t; while(t--)
#define fastIO() ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define valid(c) cout<<(c ? "YES" : "NO")<<endl;
#define valid2(c,a,b) cout<<(c ? a : b)<<endl;
#define pq_min(a) priority_queue<a, vector<a>, greater<a>>
#define pq_max(a) priority_queue<a>
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define mid ((l+r)>>1)
#define flush endl<<flush

using namespace std;
using namespace __gnu_pbds;

using vi = vector<int>;
using ll = long long;
template <typename T>
using vec = vector<T>;
template <typename T>
using ordered_set = tree<T, null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update>;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());

const ll inf=1e18;

int main(){
	fastIO();
	multicase(){
		int n, x, y, k;
		cin>>n;
		ll sz=2*n+1, a[sz], b[sz][4]{}, dp[2], dp2[2];
		bool vis[sz]{};
		set<ll> nums;
		map<ll,ll> mn;
		vec<pll> adj[sz];
		vi cycle;
		k=n/2;
		rep(i,1,sz) cin>>a[i];
		if(n&1){
			rep(i,1,n+1){
				x=(i+k-1)%n+1;
				y=(i+k)%n+1;
				adj[i].pb({x,a[2*i-1]-a[2*x]});
				adj[i].pb({y,a[2*i]-a[2*y-1]});
			}
			y=1;
			do{
				x=y;
				y=1;
				vis[x]=true;
				if(vis[adj[x][0].fi]) swap(adj[x][0],adj[x][1]);
				y=adj[x][0].fi;
				if(vis[y] && y!=1) swap(adj[x][0],adj[x][1]);
				rep(j,0,4){
					b[x][j]=a[2*x]+a[2*x-1];
					if(j&1) b[x][j]-=adj[x][0].se;
					if(j&2) b[x][j]-=adj[x][1].se;
					nums.insert({b[x][j]});
					//cout<<b[x][j]<<endl;
				}
				cycle.pb(x);
			}while(!vis[y]);
			//repa(e,cycle) cout<<e<<" ";
			//cout<<endl;
			rep(i,0,4){
				repa(e,nums){
					dp[0]=dp[1]=inf;
					if(b[1][i]>=e) dp[i&1]=b[1][i];
					rep(i,1,n){
						dp2[0]=dp[0];
						dp2[1]=dp[1];
						dp[0]=inf;
						dp[1]=inf;
						x=cycle[i];
						if(b[x][0]>=e) dp[0]=min(dp[0],max(dp2[0],b[x][0]));
						if(b[x][2]>=e) dp[0]=min(dp[0],max(dp2[1],b[x][2]));
						if(b[x][1]>=e) dp[1]=min(dp[1],max(dp2[0],b[x][1]));
						if(b[x][3]>=e) dp[1]=min(dp[1],max(dp2[1],b[x][3]));
						//cout<<x<<" "<<e<<" "<<dp[0]<<" "<<dp[1]<<endl;
					}
					if(!mn[e]) mn[e]=dp[(i&2)>>1];
					mn[e]=min(mn[e],dp[(i&2)>>1]);
					//cout<<i<<" "<<e<<" "<<mn[e]<<endl;
				}
			}
		}else{
			rep2(i,1,k+1,2){
				nums.insert(a[2*i-1]+a[2*i]);
				nums.insert(a[2*i+n-1]+a[2*i]);
				nums.insert(a[2*i-1]+a[2*i+n]);
				nums.insert(a[2*i+n-1]+a[2*i+n]);
			}
			rep2(i,1,k+1,2){
				repa(e,nums){
					mn[e]=inf;
					if(min(a[2*i-1]+a[2*i],a[2*i+n-1]+a[2*i+n])>=e) mn[e]=(a[2*i-1]+a[2*i],a[2*i+n-1]+a[2*i+n]);
					if(min(a[2*i+n-1]+a[2*i],a[2*i-1]+a[2*i+n])>=e) mn[e]=(a[2*i+n-1]+a[2*i],a[2*i-1]+a[2*i+n]);
					if(min(a[2*i-1]+a[2*i+n],a[2*i+n-1]+a[2*i])>=e) mn[e]=(a[2*i-1]+a[2*i+n],a[2*i+n-1]+a[2*i]);
				}
			}
		}
		ll ans=inf;
		repa(e,nums) ans=min(ans,mn[e]-e);
		cout<<ans<<endl;
	}
}

**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
5
1
1 1000000000
2
1 1000000000 1 1000000000
3
1 1000000000 1 1000000000 1 1000000000
4
1 1000000000 1 1000000000 1 1000000000 1 1000000000
5
1 1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> process(vector<pair<int, int>> v) {
  vector<pair<int, int>> ret;
  sort(v.begin(), v.end(), [&](auto i, auto j) {
    if (i.second != j.second) return i.second < j.second;
    return i.first > i.second;
  });
  int mx = -2e9 - 5;
  for (auto [l, r] : v) {
    if (mx < l) {
      ret.push_back({l, r});
      mx = l;
    }
  }
  return ret;
}
void testCase() {
  int n;
  cin >> n;
  vector<int> a(2 * n);
  for (int &i : a) cin >> i;
  auto solve_even = [&]() {
    vector<pair<int, int>> dp = {{2e9, -2e9}};
    for (int i = 0; i < n; i += 2) {
      vector<pair<int, int>> ndp;
      for (int mask = 0; mask < 4; mask++) {
        int s1 = a[i + (mask & 1) * n] + a[i + 1 + (mask & 2) * n];
        int s2 = (long long) a[i] + a[i + 1] + a[n + i] + a[n + i + 1] - s1;
        for (auto [l, r] : dp) {
          ndp.push_back({min({l, s1, s2}), max({r, s1, s2})});
        }
      }
      dp = ndp;
      dp = process(dp);
    }
    int ans = 2e9 + 5;
    for (auto [l, r] : dp) {
      ans = min(ans, r - l);
    }
    return ans;
  };
  auto solve_odd = [&]() {
    int ans = 2e9 + 5;
    for (int it : {0, 1}) {
      vector<pair<int, int>> dp[2];
      dp[0] = dp[1] = {{2e9, -2e9}};
      for (int i = 1; i < n; i++) {
        vector<pair<int, int>> ndp[2];
        for (int p : {0, 1}) {
          if (i == 1 && p != it) continue;
          for (int nw : {0, 1}) {
            int s;
            if (i & 1) s = a[i - 1 + p * n] + a[i + nw * n];
            else s = a[i - 1 + n - p * n] + a[i + n - nw * n];
            for (auto [l, r] : dp[p]) {
              ndp[nw].push_back({min(l, s), max(r, s)});
            }
          }
        }
        for (int j : {0, 1}) {
          dp[j] = process(ndp[j]);
        }
      }
      for (int nw : {0, 1}) {
        int s = a[n - it * n] + a[n - 1 + nw * n];
        for (auto [l, r] : dp[nw]) {
          ans = min(ans, max(s, r) - min(s, l));
        }
      }
    }
    return ans;
  };
  if (n % 2 == 0) {
    cout << solve_even() << "\n";
  } else {
    cout << solve_odd() << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tc;
  cin >> tc;
  while (tc--) {
    testCase();
  }
  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
5
1
1 1000000000
2
1 2 3 4
3
1 1000000000 1 1000000000 1 1000000000
4
1 1000000000 1 1000000000 1 1000000000 1 1000000000
5
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

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

using ll  = long long;
using ld  = long double;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using ull = unsigned long long;

#define X               first
#define Y               second
#define SZ(x)           int(x.size())
#define all(x)          x.begin(), x.end()
#define mins(a,b)       (a = min(a,b))
#define maxs(a,b)       (a = max(a,b))
#define pb              push_back
#define Mp              make_pair
#define lc              id<<1
#define rc              lc|1
#define mid             ((l+r)/2)
mt19937_64              rng(chrono::steady_clock::now().time_since_epoch().count());

const ll  INF = 3e9 + 23;
const ll  MOD = 1e9 + 7;
const int MXN = 4e3 + 5;
const int LOG = 23;

int n, a[MXN];

void Main() {
    cin >> n;
    for(int i=1; i<=(n<<1); i++) cin >> a[i];
    if(n%2 == 0) {
        int mn = INF, mx = -INF;
        for(int i=1; i<=n/2; i++) {
            int j = i+n/2;
            vector<int> vec = {a[2*i-1], a[2*i], a[2*j-1], a[2*j]};
            pii best = {INF, 0};
            sort(all(vec));
            do {
                mins(best, Mp(abs((vec[0]+vec[1])-(vec[2]+vec[3])), vec[0]+vec[1]));
            } while(next_permutation(all(vec)));
            best.X = vec[0]+vec[1]+vec[2]+vec[3]-best.Y;
            if(best.X>best.Y) swap(best.X, best.Y);
            mins(mn, best.X);
            maxs(mx, best.Y);
        }
        cout << mx-mn << '\n';
    }
    else {
        int ans=INF;
        for(int t=0; t<=100000; t++) {
            int mn=INF, mx=-INF;
            for(int i=1; i<=n; i++) {
                mins(mn, a[2*i-1]+a[2*i]);
                maxs(mx, a[2*i-1]+a[2*i]);
            }
            mins(ans, mx-mn);
            pii msum = {INF, -1};
            for(int i=1; i<=n; i++)
                mins(msum, Mp(a[2*i-1]+a[2*i], i));
            int i = msum.Y;
            vector<pii> vec;
            if(2*i-1<=n) {
                if(a[2*i-1+n]>a[2*i-1]) vec.pb(Mp(2*i-1, 2*i-1+n));
            }
            else {
                if(a[2*i-1-n]>a[2*i-1]) vec.pb(Mp(2*i-1, 2*i-1-n));
            }

            if(2*i<=n) {
                if(a[2*i+n]>a[2*i]) vec.pb(Mp(2*i, 2*i+n));
            }
            else {
                if(a[2*i-n]>a[2*i]) vec.pb(Mp(2*i, 2*i-n));
            }

            if(SZ(vec)) {
                int j = rng()%SZ(vec);
                swap(a[vec[j].X], a[vec[j].Y]);
            }
        }
        cout << ans << '\n';
    }
}

int32_t main() {
    cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while(T--) Main();
    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 create a test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code has different logic for even and odd values of `n`, so we should consider both cases.

### Steps to Identify a Challenging Test Input:

1. **Even `n` Case:**
   - The code uses a permutation approach to minimize the difference between the sums of pairs.
   - A challenging test input could be one where the sums of pairs are very close, but the permutation logic might fail to find the optimal solution.

2. **Odd `n` Case:**
   - The code uses a random swapping approach to minimize the difference.
   - A challenging test input could be one where the initial configuration is already optimal or very close to optimal, making it hard for the random swaps to improve the result.

### Challenging Test Input:

Let's create a test input with both even and odd `n` values to cover different scenarios.

```plaintext
3
2
1 1 1 1
3
1 2 3 4 5 6
4
1 2 3 4 5 6 7 8
```

### Explanation:

1. **Test Case 1:**
   - `n = 2`
   - Input: `1 1 1 1`
   - Expected Output: `0`
   - Explanation: All pairs have the same sum, so the difference is `0`.

2. **Test Case 2:**
   - `n = 3`
   - Input: `1 2 3 4 5 6`
   - Expected Output: `4`
   - Explanation: The optimal pairs are `(1, 6)`, `(2, 5)`, and `(3, 4)`, resulting in sums `7`, `7`, and `7` respectively, with a difference of `0`. However, the random swapping might not find this optimal solution, leading to a larger difference.

3. **Test Case 3:**
   - `n = 4`
   - Input: `1 2 3 4 5 6 7 8`
   - Expected Output: `4`
   - Explanation: The optimal pairs are `(1, 8)`, `(2, 7)`, `(3, 6)`, and `(4, 5)`, resulting in sums `9`, `9`, `9`, and `9` respectively, with a difference of `0`. The permutation logic should handle this correctly, but it's a good test to ensure the logic works as expected.

This test input is small-scale and complete, covering both even and odd values of `n` to expose 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: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
def minn_diff(t,test_case):
    res=[]
    for n,a in test_case:
        sorted_a=sorted(a)
        paired_sums=[]
        for i in range(n):
            paired_sum=sorted_a[i] + sorted_a[i+n]
            paired_sums.append(paired_sum)
        max_sum=max(paired_sums)
        min_sum=min(paired_sums)
        min_diff=max_sum-min_sum
        res.append(min_diff)
    return res
t=int(input())
test_case=[]
for _ in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    test_case.append((n,a))
res=minn_diff(t,test_case)
for re in res:
    if(re==8):
        print(re//2)
    else:
        print(re)

**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
1
4
1 1000000000 1 1000000000 1 1000000000 1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
from sys import stdin
input=lambda :stdin.readline()[:-1]

inf=1<<60
def calc_fix_min(n,a,mn):
  if n%2==0:
    ans=0
    for i in range(0,n,2):
      ai,aj=a[i],a[i+1]
      bi,bj=a[i+n],a[i+1+n]
      res=inf
      if ai+bi>=mn and aj+bj>=mn:
        res=min(res,max(ai+bi,aj+bj))
      if ai+bj>=mn and aj+bi>=mn:
        res=min(res,max(ai+bj,aj+bi))
      ans=max(ans,res)
    return ans-mn

  ans=inf
  # a[n-1] fix
  for _ in range(2):
    prev1=[a[n-1],0]
    prev2=[-inf,inf]
    for i in range(n):
      # a[i]
      nprev1=[a[i+n],inf]
      if prev1[0]+a[i]>=mn:
        nprev1[1]=min(nprev1[1],max(prev1[1],a[i]+prev1[0]))
      if prev2[0]+a[i]>=mn:
        nprev1[1]=min(nprev1[1],max(prev2[1],a[i]+prev2[0]))
      nprev2=[a[i],inf]
      if prev1[0]+a[i+n]>=mn:
        nprev2[1]=min(nprev2[1],max(prev1[1],a[i+n]+prev1[0]))
      if prev2[0]+a[i+n]>=mn:
        nprev2[1]=min(nprev2[1],max(prev2[1],a[i+n]+prev2[0]))
      prev1,prev2=nprev1,nprev2
    ans=min(ans,prev2[1])
    a[n-1],a[2*n-1]=a[2*n-1],a[n-1]
  return ans-mn

def solve():
  n=int(input())
  a=list(map(int,input().split()))
  if n%2==0:
    ans=inf
    for i in range(0,n,2):
      ai,aj=a[i],a[i+1]
      bi,bj=a[i+n],a[i+1+n]
      ans=min(ans,calc_fix_min(n,a,ai+aj))
      ans=min(ans,calc_fix_min(n,a,ai+bj))
      ans=min(ans,calc_fix_min(n,a,bi+aj))
      ans=min(ans,calc_fix_min(n,a,bi+bj))
    print(ans)
    return
  ans=inf
  for i in range(0,2*n,2):
    ai,aj=a[i],a[i+1]
    bi,bj=a[(i+n)%(2*n)],a[(i+1+n)%(2*n)]
    ans=min(ans,calc_fix_min(n,a,ai+aj))
    ans=min(ans,calc_fix_min(n,a,ai+bj))
    ans=min(ans,calc_fix_min(n,a,bi+aj))
    ans=min(ans,calc_fix_min(n,a,bi+bj))
  print(ans)

for _ in range(int(input())):
  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
3
1
1 1000000000
2
1 1000000000 1000000000 1
3
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

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

using ll  = long long;
using ld  = long double;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using ull = unsigned long long;

#define X               first
#define Y               second
#define SZ(x)           int(x.size())
#define all(x)          x.begin(), x.end()
#define mins(a,b)       (a = min(a,b))
#define maxs(a,b)       (a = max(a,b))
#define pb              push_back
#define Mp              make_pair
#define lc              id<<1
#define rc              lc|1
#define mid             ((l+r)/2)
mt19937_64              rng(chrono::steady_clock::now().time_since_epoch().count());

const ll  INF = 3e9 + 23;
const ll  MOD = 1e9 + 7;
const int MXN = 4e3 + 5;
const int LOG = 23;

int n, a[MXN];

void Main() {
    cin >> n;
    for(int i=1; i<=(n<<1); i++) cin >> a[i];
    if(n%2 == 0) {
        int mn = INF, mx = -INF;
        for(int i=1; i<=n/2; i++) {
            int j = i+n/2;
            vector<int> vec = {a[2*i-1], a[2*i], a[2*j-1], a[2*j]};
            pii best = {INF, 0};
            sort(all(vec));
            do {
                mins(best, Mp(abs((vec[0]+vec[1])-(vec[2]+vec[3])), vec[0]+vec[1]));
            } while(next_permutation(all(vec)));
            best.X = vec[0]+vec[1]+vec[2]+vec[3]-best.Y;
            if(best.X>best.Y) swap(best.X, best.Y);
            mins(mn, best.X);
            maxs(mx, best.Y);
        }
        cout << mx-mn << '\n';
    }
    else {
        int ans=INF;
        for(int t=0; t<=20000; t++) {
            int mn=INF, mx=-INF;
            for(int i=1; i<=n; i++) {
                mins(mn, a[2*i-1]+a[2*i]);
                maxs(mx, a[2*i-1]+a[2*i]);
            }
            mins(ans, mx-mn);
            pii msum = {INF, -1};
            for(int i=1; i<=n; i++)
                mins(msum, Mp(a[2*i-1]+a[2*i], i));
            int i = msum.Y;
            vector<pii> vec;
            if(2*i-1<=n) {
                if(a[2*i-1+n]>a[2*i-1]) vec.pb(Mp(2*i-1, 2*i-1+n));
            }
            else {
                if(a[2*i-1-n]>a[2*i-1]) vec.pb(Mp(2*i-1, 2*i-1-n));
            }

            if(2*i<=n) {
                if(a[2*i+n]>a[2*i]) vec.pb(Mp(2*i, 2*i+n));
            }
            else {
                if(a[2*i-n]>a[2*i]) vec.pb(Mp(2*i, 2*i-n));
            }

            if(SZ(vec)) {
                int j = rng()%SZ(vec);
                swap(a[vec[j].X], a[vec[j].Y]);
            }
        }
        cout << ans << '\n';
    }
}

int32_t main() {
    cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while(T--) Main();
    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
10 1
2
1 10 10 1
3
1 10 1 10 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
n = int(input())
for _ in range(n):
    x = int(input())
    m = map(int, input().split())
    M = list(m)
    diff = []

    def swap(x, y):
        p = x
        x = y
        y = p

    def pair(x, p):
        global diff
        sum = []
        for i in range(len(M) // 2):
            sum.append(M[2 * i] + M[2 * i + 1])
        sum.sort()
        diff.append(sum[-1] - sum[0])
        for i in range(p, x):
            z = M[i]
            M[i] = M[i + x]
            M[i + x] = z
            pair(x, p + 1)

    pair(x, 0)
    diff.sort()
    print(diff[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
10 1
2
1 10 10 1
3
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class Solution{

    public static void caseEven(int[] intellgence,int n){

        int maxPair = Integer.MIN_VALUE;
        int minPair = Integer.MAX_VALUE;
        for (int left = 1,right = 2; right <= n; left+=2,right+=2) {
            int pair1 = Math.max(intellgence[left],intellgence[left+n])
                + Math.min(intellgence[right],intellgence[right+n]);
            int pair2 = intellgence[left] + intellgence[left+n]
                + intellgence[right] + intellgence[right+n] - pair1;
            if (pair1 > maxPair) maxPair = pair1;
            if (pair1 < minPair) minPair = pair1;
            if (pair2 > maxPair) maxPair = pair2;
            if (pair2 < minPair) minPair = pair2;
        }

        System.out.println(maxPair - minPair);
    }

    //数组很适合
    // record PairInfo(int zeroZero,int zeroOne,int oneZero,int oneOne){};
    static final int zeroZero = 0;
    static final int zeroOne = 1;
    static final int oneZero = 2;
    static final int oneOne = 3;
    static int myMax(int x,int y){
        if (x == 0
            || y == 0)
        {
            return 0;
        }
        return Math.max(x, y);
    }
    static int myMin(int x,int y){
        if (x == 0
            || y == 0)
        {
            return Math.max(x, y);
        }
        return Math.min(x, y);
    }
    record SumInfo(int pairIndex,int sumIndex,int value){}

    static int[] merge(int[] p1,int[] p2){
        return new int[]{
            Math.min(Math.max(p1[zeroZero],p2[zeroZero]),Math.max(p1[zeroOne],p2[oneZero])),
            Math.min(Math.max(p1[zeroZero],p2[zeroOne]),Math.max(p1[zeroOne],p2[oneOne])),
            Math.min(Math.max(p1[oneZero],p2[zeroZero]),Math.max(p1[oneOne],p2[oneZero])),
            Math.min(Math.max(p1[oneZero],p2[zeroOne]), Math.max(p1[oneOne], p2[oneOne]))};
    }

    static void caseOdd(int[] knight,int n){

        int[][] pairArr = new int[n][0];
        ArrayList<Solution.SumInfo> sumList = new ArrayList<SumInfo>();
        for (int i = 1; i <= n; i++) {
            int left,right,anotherLeft,anotherRight;
            if (i % 2 == 1) {
                left = i;
                right = i+1;
                anotherLeft = i+n;
                anotherRight = (i+1+n)%(2*n);
            }
            else {
                anotherLeft = i;
                anotherRight = i+1;
                left = i+n;
                right = i+1+n;
            }

            int[] thisPair = new int[]{
                knight[left]+knight[right],
                knight[left]+knight[anotherRight],
                knight[anotherLeft]+knight[right],
                knight[anotherLeft]+knight[anotherRight]};
            for (int j = 0; j < 4;j++){
                sumList.add(new SumInfo(i-1, j, thisPair[j]));
            }
            pairArr[i-1] = thisPair;
        }
        sumList.sort((s1,s2)->(s1.value - s2.value));

        int minDiff = Integer.MAX_VALUE;
        int prevMinSum = -1;
        for (SumInfo info : sumList) {
            if (info.value == prevMinSum) {
                pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
                continue;
            }

            //检测prevMinSum
            int minSum = info.value;
            int[] dp = pairArr[0];
            int[] dpWithMin = new int[]{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE};
            for (int i = 0; i < 4; i++) {
                if (dp[i] == minSum) {
                    dpWithMin[i] = minSum;
                }
            }

            for (int i = 1; i < pairArr.length; i++) {
                int[] thisPair = pairArr[i];

                dpWithMin = merge(dpWithMin,thisPair);
                for (int j = 0; j < 4; j++) {
                    if (thisPair[j] != minSum) continue;
                    int[] auxPair = new int[]{
                        thisPair[0] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[1] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[2] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[3] == minSum ? minSum : Integer.MAX_VALUE};
                    int[] auxDp = merge(dp,auxPair);
                    for (int k = 0; k < 4; k++) {
                        if (auxDp[k] < dpWithMin[k])
                            dpWithMin[k] = auxDp[k];
                    }
                }
                dp = merge(dp,thisPair);
            }
            if (dpWithMin[0] == Integer.MAX_VALUE
                && dpWithMin[1] == Integer.MAX_VALUE
                && dpWithMin[2] == Integer.MAX_VALUE
                && dpWithMin[3] == Integer.MAX_VALUE)
            {
                break ;
            }
            minDiff = Math.min(minDiff,dpWithMin[0] - minSum);
            minDiff = Math.min(minDiff,dpWithMin[3] - minSum);

            prevMinSum = info.value;
            pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
        }

        System.out.println(minDiff);
    }

    public static void slove(Scanner scanner){

        int t = scanner.nextInt();
        while (t > 0 ) {t--;
            int n = scanner.nextInt();
            int[] knight = new int[2*n+1];
            for (int i = 1; i < knight.length; i++) {
                knight[i] = scanner.nextInt();
            }
            if (n % 2 == 0) {
                caseEven(knight,n);
                continue;
            }
            if (n == 1){
                System.out.println(0);
                continue;
            }
            caseOdd(knight,n);
        }
    }

    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in);) {
            slove(scanner);
        }
    }
}

**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
2
1 1000000000 1000000000 1
3
1 2 3 4 5 6
5
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

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

const int NM = 2000, inf = 2e9+11;

int ntest, n, a[2*NM+5];
vector <int> arr, v;
int pos[2*NM+5], dp[NM+5][2];

int pre(int x){
	return (x+2*n-1)%(2*n);
}

int nxt(int x){
	return (x+1)%(2*n);
}

int solve(int x){
	int mn = a[arr[x]]+a[arr[x+1]];
	v.clear();
	for (int i = nxt(x+1), cnt = 1; cnt <= 2*n-2; i = nxt(i), cnt++){
		v.push_back(arr[i]);
	}
	dp[0][0] = (a[v[0]]+a[v[1]] >= mn ? a[v[0]]+a[v[1]] : -1);
	dp[0][1] = (a[v[0]]+a[v[2]] >= mn ? a[v[0]]+a[v[2]] : -1);
	for (int i = 1; i <= n-2; i++){
		int x = ((dp[i-1][0] == -1 || a[v[2*i]]+a[v[2*i+1]] < mn) ? -1 : max(dp[i-1][0], a[v[2*i]]+a[v[2*i+1]])),
		    y = ((dp[i-1][1] == -1 || a[v[2*i-1]]+a[v[2*i+1]] < mn) ? -1 : max(dp[i-1][1], a[v[2*i-1]]+a[v[2*i+1]]));
		if (x == -1 && y == -1) dp[i][0] = -1; else if (x == -1) dp[i][0] = y;
		else if (y == -1) dp[i][0] = x; else dp[i][0] = min(x, y);
		if (i == n-2) break;

		x = ((dp[i-1][0] == -1 || a[v[2*i]]+a[v[2*i+2]] < mn) ? -1 : max(dp[i-1][0], a[v[2*i]]+a[v[2*i+2]])),
		y = ((dp[i-1][1] == -1 || a[v[2*i-1]]+a[v[2*i+2]] < mn) ? -1 : max(dp[i-1][1], a[v[2*i-1]]+a[v[2*i+2]]));
		if (x == -1 && y == -1) dp[i][1] = -1; else if (x == -1) dp[i][1] = y;
		else if (y == -1) dp[i][1] = x; else dp[i][1] = min(x, y);
	}
	if (dp[n-2][0] == -1) return +inf;
	return dp[n-2][0]-mn;
}

void solve(){
	cin >> n;
	for (int i = 1; i <= 2*n; i++){
		cin >> a[i];
		pos[i] = -1;
	}
	if (n == 1){
		cout << "0\n";
		return;
	}
	if (n%2 == 0){
		int ans = 0;
		for (int i = 1; i <= n; i += 2){
			int x = a[i]+a[i+1], y = a[i+n]+a[i+n+1], z = a[i]+a[i+n+1], t = a[i+1]+a[i+n];
			ans = max(ans, min(abs(x-y), abs(z-t)));
		}
		cout << ans << '\n';
		return;
	}
	arr.clear();
	arr.push_back(1);
	pos[1] = 0;
	for (int i = 1; i < 2*n; i++){
		int x = (arr[i-1]%2 == 0 ? arr[i-1]-1 : arr[i-1]+1), y = (arr[i-1] <= n ? arr[i-1]+n : arr[i-1]-n);
		if (pos[x] == -1){
			arr.push_back(x);
			pos[x] = i;
		}
		else{
			arr.push_back(y);
			pos[y] = i;
		}
	}
	int ans = +inf;
	for (int i = 0; i < 2*n; i += 2)
		for (int t = 0; t < 2; t++){
			ans = min(ans, solve(i));
			swap(a[arr[i]], a[arr[pre(i)]]);
			ans = min(ans, solve(i));
			swap(a[arr[i+1]], a[arr[nxt(i+1)]]);
		}
	cout << ans << '\n';
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	cin >> ntest;
	while (ntest--){
		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
2
1
10 17
3
1 10 1 10 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
namespace Le0{
#define int long long
#define pii pair<int,int>
#define fi first
#define se second
const int N=2e3+5,inf=1e18;
int T,n;
int a[N];
namespace sub1{
int aim(int x){
    if(x<=n) return x+n;
    return x-n;
}
int oth(int x){return (x&1)?x+1:x-1;}
int maxn;
pii c[N];
int bel[N],vis[N],f[N][2];
int cal2(int op){
    for(int i=2;i<=n;i++){
        int val=c[i].fi+c[i-1].se;
        if(!vis[i]||vis[i]==1){
            if(val<=maxn){
                // cout<<i<<" "<<f[i-1][0]<<" "<<val<<"??\n";
                f[i][0]=max(f[i][0],min(val,f[i-1][0]));
            }
            val=c[i].fi+c[i-1].fi;
            if(val<=maxn){
                f[i][0]=max(f[i][0],min(val,f[i-1][1]));
            }
        }
        if(!vis[i]||vis[i]==2){
            val=c[i].se+c[i-1].se;
            if(val<=maxn){
                f[i][1]=max(f[i][1],min(val,f[i-1][0]));
            }
            val=c[i].se+c[i-1].fi;
            if(val<=maxn){
                f[i][1]=max(f[i][1],min(val,f[i-1][1]));
            }
        }
        // cout<<f[i][0]<<" "<<f[i][1]<<"bif\n";
    }
    if(op==1){
        int val=c[n].se+c[1].fi;
        if(val<=maxn) f[n][0]=min(val,f[n][0]);
        else f[n][0]=-inf;
        val=c[n].fi+c[1].fi;
        if(val<=maxn) f[n][1]=min(val,f[n][1]);
        else f[n][1]=-inf;
    }
    else{
        int val=c[n].se+c[1].se;
        if(val<=maxn) f[n][0]=min(val,f[n][0]);
        else f[n][0]=-inf;
        val=c[n].fi+c[1].se;
        if(val<=maxn) f[n][1]=min(val,f[n][1]);
        else f[n][1]=-inf;
    }
    return max(f[n][0],f[n][1]);
}
int cal(){
    int mx=-inf;
    if(!vis[1]||vis[1]==1){
        for(int j=1;j<=n;j++) f[j][0]=f[j][1]=-inf;
        f[1][0]=inf;
        int rs=cal2(1);
        if(rs!=-inf) mx=max(mx,rs);
    }
    if(!vis[1]||vis[1]==2){
        for(int j=1;j<=n;j++) f[j][0]=f[j][1]=-inf;
        f[1][1]=inf;
        int rs=cal2(2);
        if(rs!=-inf) mx=max(mx,rs);
    }
    return mx;
}
void work(){
    int x=1;
    for(int i=1;i<=2*n;i++){
        bel[i]=(x-1)/2+1;
        int y=aim(x);
        c[i]={a[x],a[y]};
        x=oth(y);
    }
    // for(int i=1;i<=n;i++) cout<<bel[i]<<" " ;cout<<'\n';
    // for(int i=1;i<=n;i++) cout<<c[i].fi<<" "<<c[i].se<<"\n";
    int ans=inf;
    for(int i=1;i<=n;i++){
        pii A=c[i],B=(i==1?c[n]:c[i-1]);
        maxn=A.fi+B.se;
        // cout<<maxn<<"asda\n";
        for(int j=1;j<=n;j++) vis[j]=0;
        vis[i]=1;
        vis[i==1?n:i-1]=1;
        int rs=cal();
        if(rs!=-1) ans=min(ans,maxn-rs);
        // cout<<ans<<"\n";
        maxn=A.fi+B.fi;
        for(int j=1;j<=n;j++) vis[j]=0;
        vis[i]=1;
        vis[i==1?n:i-1]=2;
        rs=cal();
        if(rs!=-1) ans=min(ans,maxn-rs);
        // cout<<ans<<"\n";
        maxn=A.se+B.se;
        for(int j=1;j<=n;j++) vis[j]=0;
        vis[i]=2;
        vis[i==1?n:i-1]=1;
        rs=cal();
        if(rs!=-1) ans=min(ans,maxn-rs);
        // cout<<ans<<"\n";
        maxn=A.se+B.fi;
        for(int j=1;j<=n;j++) vis[j]=0;
        vis[i]=2;
        vis[i==1?n:i-1]=2;
        rs=cal();
        if(rs!=-1) ans=min(ans,maxn-rs);
        // cout<<ans<<"\n";
    }
    cout<<ans<<"\n";
}
}
namespace sub2{
int pos;
pii c[N],d[N];
int cal(int maxn){
    int mn=0;
    for(int i=1;i<=n;i++){
        if(pos==i) continue;
        int mx=-1;
        if(c[i].fi<=maxn){
            mx=max(mn,c[i].se);
        }
        if(d[i].fi<=maxn){
            mx=max(mn,d[i].se);
        }
        if(mn==-1) return -1;
        mn=min(mn,mx);
    }
    return mn;
}
void work(){
    int ans=inf;
    for(int i=1;i<=n;i++){
        int x=2*i-1,y=2*i;
        int x2=x+n,y2=y+n;
        int val=a[x]+a[y];
        int val2=a[x2]+a[y2];
        if(val<val2) swap(val,val2);
        c[i]={val,val2};
        val=a[x]+a[y2];
        val2=a[y]+a[x2];
        if(val<val2) swap(val,val2);
        d[i]={val,val2};
    }
    for(int i=1;i<=n;i++){
        pos=i;
        int rs=cal(c[i].fi);
        if(rs!=-1){
            ans=min(ans,c[i].fi-max(rs,c[i].se));
        }
        pos=i;
        rs=cal(d[i].fi);
        if(rs!=-1){
            ans=min(ans,d[i].fi-max(rs,d[i].se));
        }
    }
    cout<<ans<<"\n";
}
}
int main(){
    cin>>T;
    while(T--){
        cin>>n;
        for(int i=1;i<=2*n;i++) cin>>a[i];
        if(n==1){cout<<0<<'\n';continue;}
        if(n&1) sub1::work();
        else sub2::work();
    }
    return 0;
}
}
signed main(){
    #ifdef LOCAL
    freopen("1.in","r",stdin);
    #endif
    ios::sync_with_stdio(0),cin.tie(0);
    return Le0::main();
}

**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 challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The key points to consider are:

1. **Small Scale**: The test input should be small enough to fit within the constraints but still be challenging.
2. **Edge Cases**: Consider cases where the algorithm might struggle, such as:
   - All pairs having the same intelligence.
   - Pairs with very different intelligence values.
   - Specific patterns in the intelligence values that might cause issues with the swapping logic.

Given these considerations, let's create a test input that includes these edge cases:

```plaintext
3
2
1 1 1000000000 1000000000
3
1 2 3 4 5 6
4
1 10 100 1000 10000 100000 1000000 10000000
```

### Explanation:
1. **Test Case 1**:
   - \( n = 2 \)
   - Intelligence values: \( 1, 1, 1000000000, 1000000000 \)
   - This case tests the scenario where the intelligence values are extremely different, making it challenging for the algorithm to balance the pairs.

2. **Test Case 2**:
   - \( n = 3 \)
   - Intelligence values: \( 1, 2, 3, 4, 5, 6 \)
   - This case tests a simple increasing sequence of intelligence values, which can help identify issues with the algorithm's logic for balancing pairs.

3. **Test Case 3**:
   - \( n = 4 \)
   - Intelligence values: \( 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000 \)
   - This case tests a sequence with a wide range of intelligence values, which can help identify issues with the algorithm's ability to handle large differences in intelligence values.

This test input should help expose any bugs in the provided code by presenting it with challenging and varied scenarios.