**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
from collections import defaultdict,deque
T=int(input())
for t in range(T):
    N=int(input())
    S=[[int(s=="A") for s in input()] for i in range(2)]
    dp=defaultdict(int)
    dp[(0,0)]=0
    queue=deque([(0,0)])
    set_queue=set()
    set_queue.add((0,0))
    while queue:
        i,j=queue.popleft()
        for di in range(4):
            dj=3-di
            ii=i+di
            jj=j+dj
            if abs(ii-jj)<3 and ii<=N and jj<=N:
                c=int(sum(S[0][i:ii]+S[1][j:jj])>=2)
                if dp[ii,jj]<dp[i,j]+c:
                    dp[ii,jj]=dp[i,j]+c
                if not (ii,jj) in set_queue:
                    set_queue.add((ii,jj))
                    queue.append((ii,jj))
    ans=max(dp.values())
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
9
AJAJAJAJA
AJAJAJAJA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
from collections import defaultdict as ddic, deque
from itertools import *
import random
import math
from bisect import *
import string
from functools import cache

# MOD = 998244353
MOD = 10**9 + 7
inf = float("inf")

fmax = lambda x, y: x if x > y else y

def count(a, b, c):
    if a == b:
        return a
    if a == c:
        return a
    return b

def solve():
    N = rint()
    A = [[+(c == "A") for c in input()] for _ in range(2)]
    dp = [[-inf] * 3 for _ in range(N + 3)]
    dp[N][0] = 0
    for i in range(N)[::-1]:
        if i + 3 <= N:
            a, b, c = A[0][i : i + 3]
            d, e, f = A[1][i : i + 3]
            score = max(
                count(a, b, c) + count(d, e, f),
                count(a, b, d) + count(c, e, f),
                count(a, d, e) + count(b, c, f),
            )
            score2 = count(a, b, d)
            score3 = count(a, d, e)

            dp[i][0b00] = fmax(
                dp[i][0b00],
                max(
                    dp[i + 3][0b00] + score,
                    dp[i + 1][0b01] + score2,
                    dp[i + 1][0b10] + score3,
                ),
            )

            dp[i][0b01] = fmax(dp[i][0b01], dp[i + 2][0b00] + count(b, d, e))
            if i + 4 <= N:
                x = A[0][i + 3]
                dp[i][0b01] = fmax(
                    dp[i][0b01], dp[i + 3][0b01] + count(b, c, x) + count(d, e, f)
                )

            dp[i][0b10] = fmax(dp[i][0b10], dp[i + 2][0b00] + count(a, b, e))
            if i + 4 <= N:
                y = A[1][i + 3]
                dp[i][0b10] = fmax(
                    dp[i][0b10], dp[i + 3][0b10] + count(a, b, c) + count(e, f, y)
                )

    return dp[0][0]

def main():
    T = 1
    T = rint()

    for tc in range(1, 1 + T):
        ans = solve()

        print(ans)
        # print(*ans)

        # print("Yes" if ans else "No")
        # print("YES" if ans else "NO")
        # print("Alice" if ans else "Bob")
        # print("First" if ans else "Second")
        # print("Case #{}: {}".format(tc, ans))
        # print(len(ans))
        # for row in ans: print(*row)

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
rint = lambda: int(input())

def rlist():
    return list(map(int, input().split()))

def rgrid(n):
    return [rlist() for _ in range(n)]

# endregion

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 generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability in faulty solutions. Here are the steps to create such a test input:

1. **Understand the Bug**: The provided code uses dynamic programming to solve the problem, but there might be issues with how it handles certain configurations of houses and districts. Specifically, the code might not handle cases where the optimal districting requires a more complex pattern of connections.

2. **Edge Cases**:
   - **Small Grids**: Test with the smallest possible grid (3 columns) to see if the code handles the simplest case correctly.
   - **Uniform Grids**: Test with grids where all houses vote for the same candidate to ensure the code can handle uniform cases.
   - **Alternating Patterns**: Test with grids where the houses vote in alternating patterns to check if the code can handle more complex configurations.
   - **Mixed Patterns**: Test with grids where the houses vote for different candidates in a mixed pattern to ensure the code can handle mixed cases.

3. **Generate the Test Input**:
   - **Small Grids**: A 3x3 grid where all houses vote for Álvaro.
   - **Uniform Grids**: A 6x6 grid where all houses vote for Álvaro.
   - **Alternating Patterns**: A 6x6 grid where the houses vote in an alternating pattern.
   - **Mixed Patterns**: A 9x9 grid where the houses vote in a mixed pattern.

Given the constraints, we will focus on a small-scale test input that is complete and valid. Here is the test input:

```plaintext
4
3
AAA
AAA
6
AAAAAA
AAAAAA
6
AJAJAJ
AJAJAJ
9
AJJAJJAJJ
JAJJAJJAJ
```

This test input includes:
1. A 3x3 grid where all houses vote for Álvaro.
2. A 6x6 grid where all houses vote for Álvaro.
3. A 6x6 grid where houses vote in an alternating pattern.
4. A 9x9 grid where houses vote in a mixed pattern.

These cases should help expose any bugs in the provided code by testing a variety of scenarios.

4
3
AAA
AAA
6
AAAAAA
AAAAAA
6
AJAJAJ
AJAJAJ
9
AJJAJJAJJ
JAJJAJJAJ


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    r1 = input()
    r2 = input()
    av_win = 0
    for i in range(0, n, 3):
        a_cnt = 0
        for j in range(3):
            if r1[i+j] == 'A':
                a_cnt += 1
            if r2[i+j] == 'A':
                a_cnt += 1
        if a_cnt >= 4:
            av_win += 2
        elif a_cnt >= 2:
            av_win += 1
    print(av_win)

**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 given buggy code, we need to consider the following:

1. **Edge Cases**: The code might fail when the distribution of 'A' and 'J' is not uniform or when there are specific patterns that the code does not handle correctly.
2. **Optimal Districts**: The code should correctly count the number of districts Álvaro can win, but it might miss some optimal configurations.

Let's consider a test case where the distribution of 'A' and 'J' is such that the code might incorrectly count the number of districts Álvaro can win.

### Test Case Analysis
- **Test Case 1**: A simple case where the distribution of 'A' and 'J' is such that the code might miss an optimal configuration.
- **Test Case 2**: A more complex case where the distribution of 'A' and 'J' is more varied and the code might fail to count the correct number of districts.

### Challenging Test Input
```plaintext
2
6
AAJAAJ
AAJAAJ
9
AJJAJJAJJ
AJJAJJAJJ
```

### Explanation
1. **Test Case 1**:
   - Input: `6\nAAJAAJ\nAAJAAJ`
   - Expected Output: `4`
   - Explanation: The grid can be divided into the following districts:
     - District 1: (1,1), (1,2), (2,1) -> 3 'A's, so Álvaro wins.
     - District 2: (1,3), (2,2), (2,3) -> 2 'A's, so Álvaro wins.
     - District 3: (1,4), (1,5), (2,4) -> 3 'A's, so Álvaro wins.
     - District 4: (1,6), (2,5), (2,6) -> 2 'A's, so Álvaro wins.
   - The code might fail to count the correct number of districts because it might not consider the optimal configuration.

2. **Test Case 2**:
   - Input: `9\nAJJAJJAJJ\nAJJAJJAJJ`
   - Expected Output: `3`
   - Explanation: The grid can be divided into the following districts:
     - District 1: (1,1), (1,2), (2,1) -> 2 'A's, so Álvaro wins.
     - District 2: (1,3), (1,4), (2,2) -> 2 'A's, so Álvaro wins.
     - District 3: (1,5), (1,6), (2,3) -> 2 'A's, so Álvaro wins.
     - The remaining houses form districts where José wins.
   - The code might fail to count the correct number of districts because it might not consider the optimal configuration.

This test input should help expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            String a = sc.next();
            String b = sc.next();
            ArrayList<Integer> l = new ArrayList<Integer>();
            for(int i=0;i<n;i++){
                if(a.charAt(i)=='A'){
                    l.add(i);
                }
            }
            for(int i=0;i<n;i++){
                if(b.charAt(i)=='A'){
                    l.add(i);
                }
            }
            Collections.sort(l);
            int k = 0;
            int i = 0;
            while(i<l.size()-1){
                if(l.get(i+1)-l.get(i)<=1){
                    k++;
                    i += 2;
                }else{
                    i++;
                }
            }
            System.out.println(k);
        }
    }
}

**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 given code, we need to focus on edge cases where the code's logic might fail. The code attempts to count the number of adjacent 'A' characters in the two strings, but it does not correctly handle the formation of districts that can span both rows.

### Step-by-Step Analysis:
1. **Understanding the Bug**: The code only considers pairs of 'A' characters that are adjacent in the same row. It does not consider the possibility of forming districts that span both rows, which is crucial for maximizing the number of districts Álvaro can win.

2. **Edge Cases**:
   - **Single Row Districts**: The code should handle cases where districts can be formed within a single row.
   - **Cross-Row Districts**: The code should handle cases where districts are formed by combining houses from both rows.

3. **Test Input Design**:
   - **Test Case 1**: A simple case where all 'A' characters are in one row.
   - **Test Case 2**: A case where 'A' characters are distributed across both rows in a way that maximizes the number of districts Álvaro can win.
   - **Test Case 3**: A case where the optimal solution involves forming districts that span both rows.

### Challenging Test Input:
```plaintext
3
6
AAAAAA
JJJJJJ
6
AABBBB
BBBAAA
9
AABBBBAAA
BBBAAABBB
```

### Explanation:
1. **Test Case 1**:
   - Input: `6\nAAAAAA\nJJJJJJ`
   - Output: `2`
   - Explanation: All 'A' characters are in the first row. The code should correctly identify that Álvaro can win 2 districts by forming districts within the first row.

2. **Test Case 2**:
   - Input: `6\nAABBBB\nBBBAAA`
   - Output: `2`
   - Explanation: 'A' characters are distributed across both rows. The code should correctly identify that Álvaro can win 2 districts by forming districts that span both rows.

3. **Test Case 3**:
   - Input: `9\nAABBBBAAA\nBBBAAABBB`
   - Output: `3`
   - Explanation: 'A' characters are distributed across both rows in a way that maximizes the number of districts Álvaro can win. The code should correctly identify that Álvaro can win 3 districts by forming districts that span both rows.

This test input should expose the bug in the given code by demonstrating scenarios where the code's logic fails to consider cross-row district formation.

3
6
AAAAAA
JJJJJJ
6
AABBBB
BBBAAA
9
AABBBBAAA
BBBAAABBB


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
import java.util.Scanner;

public class C{
    static int l,r;
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt(),k=0;
            String s=sc.next(),p=sc.next();
            while(l<n&&r<n){
                int w=func(s,p,2);
                if(w==0){
                    w=func(s,p,3);
                    if(w==0){
                        if(l==r){
                            if(p.charAt(r+1)=='J')r++;
                            else l++;
                            l++;
                            r++;
                        }else l=r=Math.max(l,r)+1;
                    }
                    else k+=w;
                }else k+=w;
            }
            System.out.println(k);
            l=r=0;
        }
    }
    static int func(String s,String p,int q){
        if(l<=r&&(s.substring(l,l+2)+p.charAt(r)).replace("J","").length()==q){
            l+=2;
            r++;
            return 1;
        }else if(r<=l&&(s.charAt(l)+p.substring(r,r+2)).replace("J","").length()==q){
            l++;
            r+=2;
            return 1;
        }else if(l+3<=s.length()&&s.substring(l,l+3).replace("J","").length()==q){
            if(r+3<=s.length()&&p.substring(r,r+3).replace("J","").length()==q)return 2;
            r+=3;
            l+=3;
            return 1;
        }else if(r+3<=s.length()&&p.substring(r,r+3).replace("J","").length()==q){
            r+=3;
            l+=3;
            return 1;
        }
        return 0;
    }
}

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

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

Think step by step.

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

1. **Initial Conditions**: Ensure the grid is set up in a way that the algorithm might not handle correctly.
2. **Edge Cases**: Test with grids that have specific patterns that could trip up the logic.
3. **Optimal District Formation**: Create a grid where the optimal district formation is not straightforward and requires careful consideration.

Let's think through a few scenarios:

1. **All Houses Voting for Álvaro**: This should be a simple case where Álvaro wins all districts.
2. **All Houses Voting for José**: This should be a simple case where José wins all districts.
3. **Mixed Voting Patterns**: This is where the algorithm might struggle. We can create a grid with a mix of votes that requires careful district formation.

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

```plaintext
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
```

This test input includes:
- A simple 3x2 grid where Álvaro can win 2 districts.
- A 6x2 grid with a mixed pattern where Álvaro can win 2 districts.
- Another 6x2 grid with a different mixed pattern where Álvaro can win 3 districts.
- A 9x2 grid with a complex pattern where Álvaro can win 2 districts.

These test cases should help expose any bugs in the algorithm, especially in how it handles mixed voting patterns and optimal district formation.

4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        string row1, row2;
        cin >> row1 >> row2;
        int alvaro_votes = 0;
        for (int i = 0; i < n; i += 3) {
            int count_A = 0;
            if (row1[i] == 'A') count_A++;
            if (row1[i + 1] == 'A') count_A++;
            if (row1[i + 2] == 'A') count_A++;
            if (row2[i] == 'A') count_A++;
            if (row2[i + 1] == 'A') count_A++;
            if (row2[i + 2] == 'A') count_A++;
            if (count_A >= 4) {
                alvaro_votes += 2;
            } else if (count_A >= 2) {
                alvaro_votes += 1;
            }
        }
        cout << alvaro_votes << "\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:
```plaintext
1
9
AJJJJAJAJ
JAAJJJJJA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
#include <bits/stdc++.h>
#define int long long
using namespace std;
int mod = 1000000007;  // Define mod if needed

signed main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    int t;
    cin >> t;  // Number of test cases
    while (t--) {
        int n ;
        cin>>n;
        vector<int> a,b,r1,r2;
        a.resize(n),b.resize(n),r1.resize(n),r2.resize(n);
        for(int i=0;i<n;i++)
        {
            char c;
            cin>>c;
            if(c=='A')
            a[i]=1;
           // cout<<a[i]<<" ";
        }
        //cout<<endl;
        for(int i=0;i<n;i++)
        {
            char c;
            cin>>c;
            if(c=='A')
            b[i]=1;
          //  cout<<b[i]<<" ";
        }
        //cout<<endl;
        int count =0;
        for(int i=0;i<n;i++)
        {
            if(a[i]==1&&b[i]==1)
            {
                a[i] = -1;
                b[i] = -1;
                if(i<n-1){
                if(a[i+1]==0)
                {a[i+1] = -1;
                count++;

                }
                else if(b[i+1]==0)
                {
                    b[i+1] = -1;
                    count++;

                }

                else
                {
                    a[i+1]=-1;
                    count++;
                }
                }
                if(i>0){
                a[i] = -1;
                b[i] = -1;
                if(a[i-1]==0)
                {a[i-1] = -1;
                count++;

                }
                else if(b[i-1]==0)
                {
                    b[i-1] = -1;
                    count++;

                }
                else
                {
                    b[i-1]=-1;
                    count++;
                }
                }

            }
            if((a[i]==0 && b[i]==1)|| (a[i]==1 && b[i]==0))
            {

                if(i<n-1){
                if(a[i+1]==1)
                {
                a[i] = -1;
                b[i] = -1;
                a[i+1] = -1;
                count++;

                }
                else if(b[i+1]==1)
                {
                    a[i] = -1;
                    b[i] = -1;
                    b[i+1] = -1;
                    count++;

                }
                }
                if(i>0){

                if(a[i-1]==1)
                {
                a[i] = -1;
                b[i] = -1;
                a[i-1] = -1;
                count++;

                }
                else if(b[i-1]==1)
                {
                    a[i] = -1;
                    b[i] = -1;
                    b[i-1] = -1;
                    count++;

                }
                }
            }
        if(i>0 && i<n-1)
        {
            if((a[i] + a[i+1] + a[i-1] >= 2)&&(a[i]*a[i+1]*a[i-1]>=0))
            {
                a[i]=-1;
                a[i+1]=-1;
                a[i-1] =-1;
                count++;
            }
            if((b[i] + b[i+1] + b[i-1] >= 2)&&(a[i]*a[i+1]*a[i-1]>=0))
            {
                b[i]=-1;
                b[i+1]=-1;
                b[i-1] =-1;
                count++;
            }
        }
        }
        cout<<count<<endl;

    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The code appears to have issues with how it handles the formation of districts, especially when it tries to optimize the number of districts Álvaro can win.

### Analysis of the Buggy Code
1. **District Formation Logic**: The code attempts to form districts by marking houses as `-1` when they are used in a district. It tries to form districts by checking the current house and its neighbors.
2. **Overwriting Issues**: The code might overwrite houses that are already part of a district, leading to incorrect district formation.
3. **Edge Cases**: The code might not handle edge cases where the grid is very small or where the distribution of `A` and `J` is very uneven.

### Challenging Test Input
To expose the bug, we can create a test case where the distribution of `A` and `J` is such that the code's logic fails to form the optimal number of districts. Here is a small-scale test input that can challenge the code:

```plaintext
1
9
AJAJAJAJA
AJAJAJAJA
```

### Explanation
- **Grid Size**: The grid size is 9, which is a multiple of 3.
- **Distribution**: The distribution of `A` and `J` is alternating, which can create a challenging scenario for the code to form districts optimally.

### Expected Output
- The optimal number of districts Álvaro can win is 3.

### Why This Test Case is Challenging
- **Alternating Pattern**: The alternating pattern of `A` and `J` can cause the code to struggle with forming districts that maximize the number of votes for Álvaro.
- **Small Size**: The small size of the grid ensures that the test case is valid and fits within the constraints.

By using this test case, we can observe whether the buggy code correctly forms the optimal number of districts or fails to do so, thus exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
from sys import *
input = stdin.readline
inputlist = lambda: map(int, stdin.readline().split())

def solve():
    n = int(input())
    row1 = " " + input().strip()
    row2 = " " + input().strip()

    def majority(top, bottom):
        acount = 0
        for index in top:
            if row1[index] == "A": acount += 1
        for index in bottom:
            if row2[index] == "A": acount += 1
        return acount >= 2

    dp = [[-1, -1, -1] for _ in range(len(row1))]
    dp[0] = [0, 0, 0]
    for i in range(1, len(dp)):
        if i >= 3:
            if dp[i-3][0] != -1: dp[i][0] = max(dp[i][0], dp[i-3][0] + majority([i, i-1, i-2], []) + majority([], [i, i-1, i-2]))
            if dp[i-2][2] != -1: dp[i][0] = max(dp[i][0], dp[i-2][2] + majority([i-1, i], [i]))
            if dp[i-2][1] != -1: dp[i][0] = max(dp[i][0], dp[i-2][1] + majority([i], [i, i-1]))
        if i >= 1 and i < len(dp) - 1:
            if dp[i-1][0] != -1: dp[i][1] = max(dp[i][1], dp[i-1][0] + majority([i, i+1], [i]))
            if i >= 4:
                if dp[i-3][1] != -1: dp[i][1] = max(dp[i][1], dp[i-3][1] + majority([i-1, i, i+1], [i-2, i-1, i]))
        if i >= 1 and i < len(dp) - 1:
            if dp[i-1][0] != -1: dp[i][2] = max(dp[i][2], dp[i-1][0] + majority([i], [i, i+1]))
            if i >= 4:
                if dp[i-3][2] != -1: dp[i][2] = max(dp[i][2], dp[i-3][2] + majority([i-1, i, i-2], [i+1, i-1, i]))

    return dp[-1][0]

for _ in range(int(input())):
    print(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
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
9
AJJJJAJAJ
JAAJJJJJA
```

This test input includes three test cases that are designed to challenge the buggy code:

1. **Test Case 1 (n=3):**
   - This is a simple case where the grid is small and the optimal solution is straightforward. It should return `2` as the answer, which is the maximum number of districts Álvaro can win.

2. **Test Case 2 (n=6):**
   - This case includes a mix of both candidates, and the optimal solution requires careful districting to maximize Álvaro's votes. The expected output is `2`.

3. **Test Case 3 (n=9):**
   - This case is more complex, with a larger grid and a mix of both candidates. The optimal solution is `2` districts for Álvaro, but the complexity of the grid may expose issues in the buggy code, especially in the dynamic programming transitions and the majority function.

These test cases are designed to cover a range of scenarios, including small and large grids, and different distributions of candidate selections, to ensure that the buggy code is thoroughly tested.

3
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
9
AJJJJAJAJ
JAAJJJJJA


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
t = int(input())

for _ in range(t):
    n = int(input())
    s1 = input()
    s2 = input()
    arr1 = [1 if s1[i]=='A' else 0 for i in range(n)]
    arr2 = [1 if s2[i]=='A' else 0 for i in range(n)]

    cache = {}

    def f(i,j):
        if i>=n or j>=n:
            return 0
        if (i,j) in cache:
            return cache[(i,j)]
        ans = 0

        if i<=j+1 and j+1<n:
            ans = max(ans, int(arr1[i]+arr2[j]+arr2[j+1]>=2) + f(i+1, j+2))
        if i+1>=j and i+1<n:
            ans = max(ans, int(arr1[i]+arr1[i+1]+arr2[j]>=2) + f(i+2, j+1))

        if i+2<n and j+2<n:
            ans = max(ans, int(arr1[i]+arr1[i+1]+arr1[i+2]>=2) + int(arr2[j]+arr2[j+1]+arr2[j+2]>=2) + f(i+3, j+3))

        cache[(i,j)] =  ans
        return ans

    print(f(0,0))

**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 challenging test input that exposes the bug in the provided buggy code, let's consider the following points:

1. **Edge Cases**: Test cases where the grid has the minimum number of columns (3) or a small number of columns (6, 9) to ensure the input is small-scale.
2. **Optimal District Formation**: Create scenarios where the optimal district formation is not straightforward, such as when the grid has a mix of 'A' and 'J' in a way that requires careful consideration of the district boundaries.
3. **Symmetry and Asymmetry**: Test cases with both symmetric and asymmetric distributions of 'A' and 'J' to ensure the code handles different patterns.

Given these considerations, let's construct a test input:

```plaintext
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
```

This test input includes:
- A 3x3 grid with all 'A's in the first row and a mix of 'A' and 'J' in the second row.
- A 6x2 grid with alternating 'A' and 'J' in both rows.
- A 6x2 grid with a more complex distribution of 'A' and 'J'.
- A 9x2 grid with a more complex distribution of 'A' and 'J', including a mix of long sequences and alternating patterns.

These test cases should expose any issues in the buggy code, especially in how it handles different patterns and the optimal formation of districts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;

public class c {
    static FastReader fr = new FastReader();
    static FastWriter fw = new FastWriter();
    public static void solve(){
        int n = fr.nextInt();
        String a = fr.next();
        String b = fr.next();

        int ans = 0;
        for(int i=0;i<n/3;i++){
            int cnt = 0;
            boolean tog = false;
            boolean prev = false;
            for(int j=0;j<3;j++){
                if(a.charAt(3*i + j)=='A'){
                    cnt++;
                    if(prev) tog = true;
                    prev = true;
                }
                if(b.charAt(3*i + j)=='A'){
                    cnt++;
                    if(prev) tog = true;
                    prev = true;
                }
                if(a.charAt(3*i + j)=='A' && b.charAt(3*i + j)=='j')
                prev = false;
            }
            if(cnt>=4) ans +=2;
            else if(cnt>=3) ans++;
            else if(cnt==2 && tog) ans++;
        }
        fw.println(ans);
    }
    public static void main(String[] args) throws NumberFormatException, IOException{
        //long start = System.nanoTime();
        int t = fr.nextInt();
        //System.out.println(solve());
        //System.out.println(solve()?"YES":"NO");
        while(t-->0){
            solve();
            //System.out.println(solve());
            //System.out.println(solve()?"YES":"NO");
        }
        //System.out.println(((long)System.nanoTime()-start)/1000000);
    }
}

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;
    }
    int[] nextIntArray() {
        String line = nextLine();
        String[] strNums = line.split("\\s+");
        int[] arr = new int[strNums.length];
        for (int i = 0; i < strNums.length; i++) {
            arr[i] = Integer.parseInt(strNums[i]);
        }
        return arr;
    }
    long[] nextLongArray() {
        String line = nextLine();
        String[] strNums = line.split("\\s+");
        long[] arr = new long[strNums.length];
        for (int i = 0; i < strNums.length; i++) {
            arr[i] = Long.parseLong(strNums[i]);
        }
        return arr;
    }
    char[] nextCharArray() {
        String line = nextLine();
        String[] strNums = line.split("\\s+");
        char[] arr = new char[strNums.length];
        for (int i = 0; i < strNums.length; i++) {
            arr[i] = strNums[i].charAt(0);
        }
        return arr;
    }
    char[] StringtoChar() {
        String line = nextLine();
        return line.toCharArray();
    }
}

class FastWriter {
    private final PrintWriter writer;

    public FastWriter() {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), true);
    }

    public void print(Object object) {
        writer.print(object);
    }

    public void println(Object object) {
        writer.println(object);
    }

    public void flush() {
        writer.flush();
    }
}

**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 given buggy code, let's consider the following:

1. **Identify the Bug**: The buggy code has a logical error in the way it counts the number of 'A' (Álvaro) and 'J' (José) votes within each district. Specifically, the `prev` and `tog` variables are not correctly tracking the connectivity of 'A' votes within a district.

2. **Edge Cases**:
   - **All 'A' in a district**: This should always be counted as 2 votes for Álvaro.
   - **Two 'A' and one 'J' in a district**: This should be counted as 1 vote for Álvaro if the 'A' votes are connected.
   - **One 'A' and two 'J' in a district**: This should not count as a vote for Álvaro.

3. **Test Case Design**:
   - Create a small grid where the connectivity of 'A' votes is crucial to determining the number of votes for Álvaro.
   - Ensure that the grid has a mix of connected and non-connected 'A' votes to test the logic.

Let's design a small-scale test input with 3 columns (n=3) and 2 rows, where the connectivity of 'A' votes is important:

```plaintext
1
3
AJA
JAJ
```

### Explanation:
- **Test Case**:
  - **n = 3**: The grid has 3 columns.
  - **First Row**: "AJA"
  - **Second Row**: "JAJ"

### Expected Output:
- **District 1**:
  - Houses: (1,1) = A, (1,2) = J, (1,3) = A
  - Houses: (2,1) = J, (2,2) = A, (2,3) = J

- **Counting Votes**:
  - In the district, there are 2 'A' votes (at positions (1,1) and (1,3)).
  - The 'A' votes are connected (both in the same row).
  - Therefore, the district should count as 1 vote for Álvaro.

### Final Test Input:
```plaintext
1
3
AJA
JAJ
```

This test input should expose the bug in the buggy code by requiring the correct handling of connected 'A' votes within a district.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCases = sc.nextInt();
        StringBuilder result = new StringBuilder();

        while (testCases-- > 0) {
            int columns = sc.nextInt();
            String firstRow = sc.next();
            String secondRow = sc.next();
            int alvaroVotes = calculateVotes(firstRow, secondRow, columns);
            result.append(alvaroVotes).append("\n");
        }

        System.out.print(result);
    }

    public static int calculateVotes(String firstRow, String secondRow, int columns) {
        int votes = 0;

        for (int index = 0; index < columns; index += 3) {
            int countA = 0;
            if (firstRow.charAt(index) == 'A') countA++;
            if (firstRow.charAt(index + 1) == 'A') countA++;
            if (firstRow.charAt(index + 2) == 'A') countA++;
            if (secondRow.charAt(index) == 'A') countA++;
            if (secondRow.charAt(index + 1) == 'A') countA++;
            if (secondRow.charAt(index + 2) == 'A') countA++;

            if (countA >= 4) votes += 2;
            else if (countA >= 2) votes += 1;
        }

        return votes;
    }
}

**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
3
AAA
AAA
6
AJA
JAJ
9
AJAJAJ
AJAJAJ
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
//   Bismillahir Rahmanir Rahim      //
                                                 //     After hardship comes ease     //
                                                //         AUTHOR : iamarman         //

// pragmas
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization ("unroll-loops")
// #pragma GCC optimization ("strict-overflow")

#include<bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

                                                    ////       TEMPLATE       ////

//---------------------------------------------------------------------------------------------------------------------------------|

# define    el '\n'
# define    sp " "
# define    ff first
# define    ss second
# define    ll long long
# define    pb push_back
# define    mp make_pair
# define    yess1 cout<<1<<el
# define    noo cout<<"NO"<<el
# define    yess cout<<"YES"<<el
# define    siz(x) (int)x.size()
# define    ull unsigned long long
# define    all(v) v.begin(),v.end()
# define    allr(v) v.rbegin(),v.rend()
# define    torad(x) ((x) * ((2*acos(0))/180.0))
# define    todeg(x) ((x) * (180.0/(2*acos(0))))

constexpr ll mod=998244353 ;
constexpr ll INF=LLONG_MAX;
constexpr double PI= acos(-1);
constexpr double eps=1e-9;

# define mem(a,b) memset(a,b,sizeof(a))
# define sqr(a) ((a)*(a))
# define lcm(a,b) (a*b)/__gcd(a,b)

# define optimise   { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); }
# define fraction(a) cout.unsetf(ios::floatfield); cout.precision(a); cout.setf(ios::fixed,ios::floatfield);
# define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>

// find_by_order() - Returns an iterator to the k-th largest element (counting from zero)
// order_of_key()  - The number of items in a set that are strictly smaller than our item
// greater instead of less for descending order
// less_equal works as ordered multiset
// we can use pair<int,int> instead of int for pair of orderd set
// for multiset st.lower_bound(x) works as upper bound and st.upper_bound(x) works as lower bound

inline void file() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif // ONLINE_JUDGE
}
//----------------------------------------------------------------------------------------------------------------------------------|

                                                               // DEBUGGER //

//----------------------------------------------------------------------------------------------------------------------------------|

template < typename F, typename S > ostream& operator << ( ostream& os, const pair< F, S > & p ) { return os << "(" << p.first << ", " << p.second << ")"; }
template < typename T > ostream &operator << ( ostream & os, const vector< T > &v ) { os << "{"; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << ", ";  os << *it; }  return os << "}";  }
template < typename T >  ostream &operator << ( ostream & os, const set< T > &v ) { os << "["; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << ", "; os << *it;  } return os << "]"; }
template < typename T > ostream &operator << ( ostream & os, const multiset< T > &v ) { os << "["; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << ", "; os << *it; } return os << "]"; }
template < typename F, typename S > ostream &operator << ( ostream & os, const map< F, S > &v ) { os << "["; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << ", "; os << it -> first << " = " << it -> second ; } return os << "]";  }
#define dbg(args...) do {cerr << #args << " : "; iamarman(args); } while(0)
void iamarman () { cerr << endl; }
template <typename T> void iamarman( T a[], int n ) {   for(int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl;  }
template <typename T, typename ... hello>  void iamarman( T arg, const hello &... rest) {   cerr << arg << ' ';  iamarman(rest...);  }

//--------------------------------------------------------------------------------------------------------------------------------------|

                                                           /////    FUNCTIONS     /////

ll bigmod(ll base,ll power){ ll res=1; ll p=base%mod; while(power>0) { if(power%2==1) {  res=((res%mod)*(p%mod))%mod; }  power/=2; p=((p%mod)*(p%mod))%mod; } return res; }

ll inversemod(ll base) { return bigmod(base,mod-2); }

ll poww(ull a,ull b) { ull ans=1; if(!b) return 1; while(b>1) {  if(b&1) { ans=ans*a%mod; } a=a*a%mod; b/=2; }return a*ans%mod; }

ll gcd(ll a,ll b) { ll rem; while(b%a!=0)  {  rem=b%a;  b=a;  a=rem; } return a; }

ll sqrtt(ll a){ long long x = sqrt(a) + 2; while (x * x > a) x--; return x;}

ll sqrt(ll n) {ll low=0,high=1e10; while(high-low>1){ ll mid=low+(high-low)/2; if(mid*mid<=n) low=mid; else high=mid; }return low; }

long double sqrtd(long double n){ long double low=0,high=n,mid; for(int i=0;i<100;i++) { mid=(low+high)/2; if(mid*mid<=n) low=mid; else high=mid;} return low;}

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());

inline ll getrandom(ll a,ll b) { return uniform_int_distribution<ll>(a,b)(rng); }

int dx[]={-1, 1 , 0 , 0 , -1 ,-1, 1, 1};
int dy[]={ 0, 0 ,-1 , 1 , -1 , 1,-1, 1};

// up = { -1,0 } , down = { 1,0 } , right = { 0,1 } , left = { 0,-1 }
// up-right = { -1,1 } , up-left = { -1,-1 } , down-right = { 1,1 } , down-left = { 1,-1 }

                                                   ///  ____________CODE STARTS FROM HERE____________    ///

void solve()
{
   int n;
   cin>>n;

   string a,b;
   cin>>a>>b;

   int ans=0;
   for(int i=0;i<n;i+=3)
   {
        int cnt=0;
        cnt+=(a[i]=='A');
        cnt+=(a[i+1]=='A');
        cnt+=(a[i+2]=='A');
        cnt+=(b[i]=='A');
        cnt+=(b[i+1]=='A');
        cnt+=(b[i+2]=='A');

        cnt/=2;
        cnt=min(2,cnt);
        ans+=cnt;
   }

    cout<<ans<<el;

}

int main()
{
    optimise;
    file();
    clock_t start= clock();
    int t=1;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
       // cout<<"Case #"<<i<<": ";
        solve();

    }
    cerr << "Run Time : " <<((double)(clock() - start) / CLOCKS_PER_SEC)<<el;

}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given code, we need to focus on edge cases where the algorithm might fail to correctly count the number of districts Álvaro can win. The provided code seems to count the number of 'A' votes in each block of 3 houses and then divides by 2 to determine the number of districts Álvaro can win. However, this approach might not always be optimal.

### Step-by-Step Analysis

1. **Basic Case**: The code should handle simple cases correctly, where the number of 'A' votes is clearly more than 2 in a block.
2. **Edge Case 1**: When the number of 'A' votes is exactly 2 in a block, the code should count this as one district.
3. **Edge Case 2**: When the number of 'A' votes is less than 2 in a block, the code should not count this as a district.
4. **Complex Case**: When the grid is such that the optimal way to form districts is not straightforward, the code might fail to correctly count the districts.

### Test Input Generation

Let's create a test input where the grid is arranged in a way that the optimal division of districts is not immediately obvious. Specifically, we will create a grid where the number of 'A' votes is not uniformly distributed across the blocks.

### Test Input

```plaintext
1
9
AJJJAJAJA
AJAJAJAJA
```

### Explanation

- **Grid Layout**:
  - The grid has 2 rows and 9 columns.
  - The first row is "AJJJAJAJA".
  - The second row is "AJAJAJAJA".

- **Expected Output**:
  - The optimal division of districts should maximize the number of districts where Álvaro wins.
  - For the first block (columns 1-3): "A J J" and "A J A" -> 2 'A' votes, so 1 district for Álvaro.
  - For the second block (columns 4-6): "J J A" and "J A J" -> 2 'A' votes, so 1 district for Álvaro.
  - For the third block (columns 7-9): "J A J" and "J A A" -> 2 'A' votes, so 1 district for Álvaro.
  - Total: 3 districts for Álvaro.

### Final Test Input

```plaintext
1
9
AJJJAJAJA
AJAJAJAJA
```

This test input should expose any issues in the given code by requiring a more nuanced approach to counting the districts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+5,inf=0x3f3f3f3f;
string s[2];
int dp[2][N];
bool check(char a,char b,char c)
{
    int num=(a=='A')+(b=='A')+(c=='A');
    return num>=2;
}
void solve()
{
    int n;
    cin>>n;
    cin>>s[0]>>s[1];
    //cout<<s[0]<<" "<<s[1]<<endl;
    memset(dp,-0x3f,sizeof dp);
    dp[0][0]=0;
    for(int i=0;i<n;i++)
    {
        if(i%3==0)
        {
            dp[0][i+3]=max(dp[0][i+3],dp[0][i]+check(s[0][i],s[0][i+1],s[0][i+2])+check(s[1][i],s[1][i+1],s[1][i+2]));
            dp[0][i+1]=max(dp[0][i+1],dp[0][i]+check(s[0][i],s[0][i+1],s[1][i]));
            dp[1][i+1]=max(dp[1][i+1],dp[0][i]+check(s[1][i],s[1][i+1],s[1][i]));
        }
        else if(i%3==1)
        {
            if(i<n-3)
            {
                dp[0][i+3]=max(dp[0][i+3],dp[0][i]+check(s[0][i+1],s[0][i+2],s[0][i+3])+check(s[1][i],s[1][i+1],s[1][i+2]));
                dp[1][i+3]=max(dp[1][i+3],dp[1][i]+check(s[0][i],s[0][i+1],s[0][i+2])+check(s[1][i+1],s[1][i+2],s[1][i+3]));
            }
            dp[0][i+2]=max(dp[0][i+2],dp[0][i]+check(s[0][i+1],s[1][i],s[1][i+1]));
            dp[0][i+2]=max(dp[0][i+2],dp[1][i]+check(s[0][i],s[0][i+1],s[1][i+1]));
        }
    }
    cout<<dp[0][n]<<endl;
}
signed main()
{
    int t;
    cin>>t;
    while(t--)
    {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
6
AAAJJJ
JJJAAA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
import sys
t = int(sys.stdin.readline())
total_n = 0

neighbors = {
    0: [1, 3],
    1: [0, 2, 4],
    2: [1, 5],
    3: [0, 4],
    4: [1, 3, 5],
    5: [2, 4],
}

from itertools import combinations
cells = range(6)
all_triplets = list(combinations(cells, 3))

def is_connected(triplet):
    visited = set()
    stack = [triplet[0]]
    while stack:
        node = stack.pop()
        if node not in visited and node in triplet:
            visited.add(node)
            for neighbor in neighbors[node]:
                if neighbor in triplet:
                    stack.append(neighbor)
    return len(visited) == 3

connected_triplets = []
for triplet in all_triplets:
    if is_connected(triplet):
        connected_triplets.append(set(triplet))

partitions = []
for triplet1 in connected_triplets:
    triplet2 = set(cells) - triplet1
    if is_connected(tuple(triplet2)):
        partitions.append((triplet1, triplet2))

precompute = [0] * 64
for mask in range(64):
    max_wins = 0
    for p in partitions:
        wins = 0
        count = 0
        for idx in p[0]:
            if mask & (1 << idx):
                count += 1
        if count >= 2:
            wins += 1
        count = 0
        for idx in p[1]:
            if mask & (1 << idx):
                count += 1
        if count >= 2:
            wins += 1
        if wins > max_wins:
            max_wins = wins
    precompute[mask] = max_wins

for _ in range(t):
    n = int(sys.stdin.readline())
    total_n += n
    s1 = sys.stdin.readline().strip()
    s2 = sys.stdin.readline().strip()
    total_wins = 0

    for i in range(0, n, 3):
        mask = 0
        idx = 0
        for row in [s1, s2]:
            for j in range(i, i + 3):
                if row[j] == 'A':
                    mask |= (1 << idx)
                idx += 1
        total_wins += precompute[mask]
    print(total_wins)

**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
9
AAAAA AAA
JJJJJ JJJ
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
#author: sushmanth
from sys import stdin , stdout
input = stdin.readline

meta = lambda : print("Case #%d: "%(T + 1) , end = '')
inp = lambda : list(map(int , input().split()))

"""
dp[i][2][2][2][2] ?
"""
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

sz = (10 ** 5)
dp = [[[[[-1 , -1] for i2 in range(2)] for i1 in range(2)] for i in range(2)] for j in range(sz)]
def answer():

    @bootstrap
    def solve(i , t1 , t2 , nt1 , nt2):

        if(i == n - 1):
            if(t1 == 1 and t2 == 1):
                yield 0
            else:
                yield -Inf

        if(dp[i][t1][t2][nt1][nt2] != -1):
            yield dp[i][t1][t2][nt1][nt2]

        ans = -Inf
        if(t1 == 1 and t2 == 1):
            ans = max(ans , (yield solve(i + 1 , nt1 , nt2 , 0 , 0)))
        elif(t1 == 1 and t2 == 0):
            if(i + 1 < n and nt1 == 0 and nt2 == 0):
                count = (a[i][1] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))

            if(i + 2 < n):
                count = (a[i][1] == 'A') + (a[i + 1][1] == 'A') + (a[i + 2][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 0 , 1 , 0 , 1)) + (count >= 2))
        elif(t1 == 0 and t2 == 1):
            if(i + 1 < n and nt1 == 0 and nt2 == 0):
                count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))

            if(i + 2 < n):
                count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 2][0] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 0 , 1 , 0)) + (count >= 2))
        else:
            if(i + 1 < n):
                count = (a[i][1] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))

                count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))

                count = (a[i][0] == 'A') + (a[i][1] == 'A') + (a[i + 1][0] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 0 , 0 , 0)) + (count >= 2))

                count = (a[i][0] == 'A') + (a[i][1] == 'A') + (a[i + 1][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 0 , 1 , 0 , 0)) + (count >= 2))

            if(i + 2 < n):
                count = (a[i][1] == 'A') + (a[i + 1][1] == 'A') + (a[i + 2][1] == 'A')
                ans = max(ans , (yield solve(i + 1 , 0 , 1 , 0 , 1)) + (count >= 2))

                count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 2][0] == 'A')
                ans = max(ans , (yield solve(i + 1 , 1 , 0 , 1 , 0)) + (count >= 2))

        dp[i][t1][t2][nt1][nt2] = ans
        yield ans

    for i in range(n):
        for i1 in range(2):
            for i2 in range(2):
                for i3 in range(2):
                    for i4 in range(2):
                        dp[i][i1][i2][i3][i4] = -1

    ans = solve(0 , 0 , 0 , 0 , 0)

    return ans

for T in range(int(input())):

    n = int(input())
    b = [input().strip() for i in range(2)]
    a = [['?' , '?'] for i in range(n)]
    for i in range(2):
        for j in range(n):
            a[j][i] = b[i][j]

    Inf = n + 10

    print(answer())

**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
9
AJJJJAJAJ
JAAJJJJJA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

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

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Gerrymandering {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int tc = Integer.parseInt(br.readLine());
        while(tc-- > 0) {
            int cols = Integer.parseInt(br.readLine());
            char[] top = br.readLine().toCharArray();
            char[] bottom = br.readLine().toCharArray();
            boolean[] combined1 = new boolean[cols/3];
            boolean[] combined2 = new boolean[cols/3];
            int[] scores = new int[cols/3];
            int curMaxScore = calculateMaxOfSegment(top, bottom, 0);
            scores[0] = curMaxScore;
            for(int i = 1; i < cols/3; i++) {
                // keep note of += vs = if alr combined
                int optimalVotes = calculateMaxOfSegment(top, bottom, i);
                scores[i] = optimalVotes;
                int combined1Score = -1;
                int combined2Score = -1;
                boolean wasCombined1 = combined1[i-1];
                boolean wasCombined2 = combined2[i-1];
                if(combined1[i-1]) {
                    combined1Score = getCombined1(top, bottom, i-1, i);
                } else if(combined2[i-1]) {
                    combined2Score = getCombined2(top,  bottom, i-1, i);
                } else {
                    int temp1 = getCombined1Start(top, bottom, i) + getCombined1(top, bottom, i-1, i) - scores[i-1];
                    int temp2 = getCombined2Start(top, bottom, i) + getCombined2(top, bottom, i-1, i) - scores[i-1];
                    if(temp1 > temp2) {
                        combined1Score = temp1;
                    } else {
                        combined2Score = temp2;
                    }
                }

                if(combined1Score > optimalVotes) {
                    combined1[i] = true;
                    curMaxScore += combined1Score;
                    if (!wasCombined1) {
                        combined1[i - 1] = true;
                    }
                } else if(combined2Score > optimalVotes) {
                    combined2[i] = true;
                    curMaxScore += combined2Score;
                    if (!wasCombined2) {
                        combined2[i - 1] = true;
                    }
                } else {
                    curMaxScore += optimalVotes;
                }
            }
            System.out.println(curMaxScore);
        }
    }

    public static int calculateMaxOfSegment(char[] top, char[] bot, int segNum) {
        int col = segNum * 3;
        int aCount = 0;
        for(int i = col; i < col + 3; i++) {
            if(top[i] == 'A')
                aCount++;
            if(bot[i] == 'A')
                aCount++;
        }
        if(aCount == 1)
            return 0;
        if(aCount == 2) {
            if(top[col] == 'A' && bot[col+2] == 'A' || top[col+2] == 'A' && bot[col] == 'A')
                return 0;
            return 1;
        }
        if(aCount == 3)
            return 1;

        return 2;
    }

    public static int getCombined1Start(char[] top, char[] bot, int segNumToAdd) {
        int prevSegment = segNumToAdd - 1;
        int prevCol = prevSegment*3;

        int beginCount1 = (top[prevCol] == 'A') ? 1 : 0;
        beginCount1 = (top[prevCol+1] == 'A') ? beginCount1+1 : beginCount1;
        beginCount1 = (bot[prevCol] == 'A') ? beginCount1+1 : beginCount1;

        return (beginCount1 >= 2) ? 1 : 0;
    }

    public static int getCombined2Start(char[] top, char[] bot, int segNumToAdd) {
        int prevSegment = segNumToAdd - 1;
        int prevCol = prevSegment*3;

        int beginCount2 = (top[prevCol] == 'A') ? 1 : 0;
        beginCount2 = (bot[prevCol] == 'A') ? beginCount2+1 : beginCount2;
        beginCount2 = (bot[prevCol+1] == 'A') ? beginCount2+1 : beginCount2;
        return (beginCount2 >= 2) ? 1 : 0;
    }

    public static int getCombined1(char[] top, char[] bot, int prevSeg, int seg) {
        // * * - - - *
        // * - - - * *
        int prevCol = prevSeg * 3;
        int col = seg * 3;

        int topCount = 0;
        int botCount = 0;
        for(int i = prevCol + 2; i < prevCol + 5; i++) {
            int topCol = i;
            int botCol = i-1;
            if(top[topCol] == 'A')
                topCount++;
            if(bot[botCol] == 'A')
                botCount++;
        }
        int total = (topCount >= 2) ? 1 : 0;
        total = (botCount >= 2) ? total+1 : total;

        int endCount = (top[col+2] == 'A') ? 1 : 0;
        endCount = (bot[col+1] == 'A') ? endCount + 1 : endCount;
        endCount = (bot[col+2] == 'A') ? endCount + 1 : endCount;
        total = (endCount >= 2) ? total+1 : total;

        return total;
    }

    public static int getCombined2(char[] top, char[] bot, int prevSeg, int seg) {
        int prevCol = prevSeg * 3;
        int col = seg * 3;
        int topCount = 0;
        int botCount = 0;
        for(int i = prevCol + 2; i < prevCol + 5; i++) {
            int botCol = i;
            int topCol = i-1;
            if(top[topCol] == 'A')
                topCount++;
            if(bot[botCol] == 'A')
                botCount++;
        }
        int total = (topCount >= 2) ? 1 : 0;
        total = (botCount >= 2) ? total+1 : total;

        int endCount = (bot[col+2] == 'A') ? 1 : 0;
        endCount = (top[col+1] == 'A') ? endCount + 1 : endCount;
        endCount = (top[col+2] == 'A') ? endCount + 1 : endCount;
        total = (endCount >= 2) ? total+1 : total;

        return total;
    }
}

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

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

Think step by step.

Response:
```plaintext
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
```