**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
import sys
input = sys.stdin.readline
from collections import deque

class Node:
    def __init__(self, val):
        self.val = val
        self.prev = None
        self.next = None
        self.valid = True

def solve():
    n = int(input())
    s = input().strip()
    r = input().strip()
    flag = True

    nodes = [Node(ch) for ch in s]
    for i in range(len(nodes)):
        if i > 0:
            nodes[i].prev = nodes[i-1]
        if i < len(nodes)-1:
            nodes[i].next = nodes[i+1]

    diff = deque()
    diffs = set()

    for i in range(len(nodes)-1):
        if nodes[i].val != nodes[i+1].val:
            diff.append(nodes[i])
            diffs.add(nodes[i])

    for i in range(n - 1):
        while diff and not diff[0].valid:
            node = diff.popleft()
            diffs.discard(node)
        if not diff:
            flag = False
            break

        node = diff.popleft()
        diffs.discard(node)

        if not node.valid:
            continue

        nx = node.next
        if not nx or not nx.valid:
            continue

        prv = node.prev
        nx_node = nx.next

        node.valid = False
        nx.valid = False

        new = Node(r[i])

        if prv:
            prv.next = new
            new.prev = prv
        else:
            new.prev = None

        if nx_node:
            nx_node.prev = new
            new.next = nx_node
        else:
            new.next = None

        if prv and prv.valid:
            if prv.val != new.val and prv not in diffs:
                diff.append(prv)
                diffs.add(prv)
            elif prv.val == new.val and prv in diffs:
                diffs.discard(prv)
        if nx_node and nx_node.valid:
            if new.val != nx_node.val and new not in diffs:
                diff.append(new)
                diffs.add(new)
            elif new.val == nx_node.val and new in diffs:
                diffs.discard(new)

    if flag:
        print("YES")
    else:
        print("NO")

t = int(input())
for _ in range(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:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. Specifically, we need to consider scenarios where the string `s` and the string `r` create a situation where the algorithm might incorrectly determine the outcome.

### Key Points to Consider:
1. **Initial Condition**: Ensure that the initial string `s` has alternating characters to maximize the number of valid operations.
2. **String `r`**: Design the string `r` such that it creates a situation where the algorithm might miss a valid operation or incorrectly determine the outcome.
3. **Edge Cases**: Test cases where the length of `s` is small (e.g., 2, 3) and where the length of `s` is large (e.g., 100) to ensure the algorithm handles all edge cases correctly.

### Test Input:
```plaintext
4
2
01
0
3
101
00
4
1101
001
5
10101
1000
```

### Explanation:
1. **Test Case 1**:
   - `n = 2`
   - `s = "01"`
   - `r = "0"`
   - **Expected Output**: "YES"
   - **Reason**: The only operation is to replace `01` with `0`, resulting in `s = "0"`.

2. **Test Case 2**:
   - `n = 3`
   - `s = "101"`
   - `r = "00"`
   - **Expected Output**: "YES"
   - **Reason**: The operations are:
     - Replace `10` with `0` (resulting in `s = "01"`)
     - Replace `01` with `0` (resulting in `s = "0"`)

3. **Test Case 3**:
   - `n = 4`
   - `s = "1101"`
   - `r = "001"`
   - **Expected Output**: "YES"
   - **Reason**: The operations are:
     - Replace `11` with `0` (resulting in `s = "001"`)
     - Replace `00` with `0` (resulting in `s = "01"`)
     - Replace `01` with `0` (resulting in `s = "0"`)

4. **Test Case 4**:
   - `n = 5`
   - `s = "10101"`
   - `r = "1000"`
   - **Expected Output**: "YES"
   - **Reason**: The operations are:
     - Replace `10` with `1` (resulting in `s = "1101"`)
     - Replace `11` with `0` (resulting in `s = "001"`)
     - Replace `00` with `0` (resulting in `s = "01"`)
     - Replace `01` with `0` (resulting in `s = "0"`)

These test cases are designed to cover a range of scenarios, including small and medium-sized inputs, and to ensure that the algorithm correctly handles alternating and non-alternating patterns in the strings `s` and `r`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
def ii(): return int(input())
def mii(): return map(int, input().split())
def lii(): return list(map(int, input().split()))

def solve():
    n = ii()
    s = input()
    t = input()
    o = s.count('1')
    z = s.count('0')
    if s.count('1') == n or s.count('0') == n: print("NO")
    elif min(o, z) < max(o, z) // 2: print("NO")
    else:
        f = 1
        c1 = t.count('1')
        c2 = t.count('0')
        if c1 and c1 > z: f = 0
        if c2 and c2 > o: f = 0
        print("YES" if f else "NO")
    pass

for _ in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
2
10
1
4
1010
010
6
101010
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

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

public class B {
    final static int MOD = 1000000007;
    final static int N = 1;
    private static long[] fact = new long[N];
    static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            String s = sc.next();
            String r = sc.next();

            StringBuilder sb  = new StringBuilder(s);
            boolean valid = false;
            int index = 1;
            for(int i = 0 ; i < n-1 ; i++){
                valid = false;
                char ri = r.charAt(i);
                for(int j = index ; j < sb.length() ; j++){
                    if(sb.charAt(j) != sb.charAt(j-1)){
                        sb.replace(j-1,j+1,ri+"");
                        valid = true;
                        break;
                    }
                    else{
                        index = j;
                    }
                }
                if(!valid) {
                    out.println("NO");
                    break;
                }
            }
            if(valid){
                out.println("YES");
            }

        }
        out.close();

    }

    /* template methods */
    public static void dfs(Vertix<Integer> v){
        Stack<Vertix<Integer>> s = new Stack<>();
        s.push(v);
        while(!s.isEmpty()){
            Vertix<Integer> curr  = s.pop();
            if(!curr.visited) {
                curr.setVisited(true);

                out.println(curr.getData()); // do operation here

                curr.getNeighbors().forEach(e -> {
                    s.push(e);
                });
            }
        }
    }

    public static void bfs(Vertix<Integer> v){
        Queue<Vertix<Integer>> s = new LinkedList<>();
        s.offer(v);
        while(!s.isEmpty()){
            Vertix<Integer> curr  = s.poll();
            if(!curr.visited) {
                curr.setVisited(true);

                out.println(curr.getData()); // do operation here

                curr.getNeighbors().forEach(e -> {
                    s.offer(e);
                });
            }
        }
    }
    public static double distance2Points(int x1, int y1,int x2,int y2){
        double distance = Math.sqrt(Math.pow(Math.abs(x1-x2),2) + Math.pow(Math.abs(y1-y2),2));
        return distance;
    }

    private static void precomputeFactorial() {
        fact[0] = 1;
        for(int i = 1 ; i < N ; i++){
            fact[i] = fact[i-1] * i;
        }
    }

    // The Euclidean Algorithm is an efficient method to compute GCD of two numbers
    static int gcd(int a, int b) {
        while (b != 0) {
            int t = a;
            a = b;
            b = t % b;
        }
        return a;
    }

    static int lcm(int a, int b){
        return Math.abs(a*b)/gcd(a,b);
    }

    // Find all primes in range 2 to n quickly
    static boolean [] sieveOfEratosthenes(int n)
    {
        boolean prime[] = new boolean[n+1];
        for(int i=0;i<=n;i++)
            prime[i] = true;

        for(int p = 2; p*p <=n; p++)
        {
            if(prime[p] == true)
            {
                for(int i = p*p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;
    }

    static long pw(long a, long b) {
        long r = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                r = (r * a) % MOD;
            }
            b /= 2;
            a = (a * a) % MOD;
        }
        return r;
    }

    static long nCr(long n, long k) {
        if (k > n) return 0;
        long result = 1;

        for (int i = 0; i < k; i++) {
            result = result * (n - i) % MOD;
            result = result * pw(i + 1, MOD - 2) % MOD;
        }

        return result;
    }

    static long C(long n, long k) {
        if (n < k) return 0;
        return (fact[(int) n] * pw((fact[(int) (n - k)] * fact[(int) k]) % MOD, MOD - 2)) % MOD;
    }
    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long[] readArrayL(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

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

    static class Vertix<T>{
        T data;
        boolean visited;
        List<Vertix<T>> neighbors = new LinkedList<>();

        public Vertix(T data){
            this.data = data;
        }
        public Vertix(T data, List<Vertix<T>>  neighbors){
            this.data = data;
            this.neighbors = neighbors;
        }

        public T getData() {
            return data;
        }

        public void setData(T data) {
            this.data = data;
        }

        public boolean isVisited() {
            return visited;
        }

        public void setVisited(boolean visited) {
            this.visited = visited;
        }

        public List<Vertix<T>> getNeighbors() {
            return neighbors;
        }

        public void setNeighbors(List<Vertix<T>> neighbors) {
            this.neighbors = neighbors;
        }

    }
}

**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
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
import java.util.*;

public class Main {
    public static void sol(int n, String s, String r) {
        StringBuilder string1 = new StringBuilder(s);
        // StringBuilder string2 = new StringBuilder(r);
        for (int i = 0; i < n - 2; i++) {
            boolean found = false;
            if (r.charAt(i) == '1') {

                for (int j = 0; j < string1.length() - 2; j++) {
                    if ((string1.charAt(j) == '1' && string1.charAt(j + 1) == '0' && string1.charAt(j + 2) == '0') ||
                            (string1.charAt(j) == '0' && string1.charAt(j + 1) == '1'
                                    && string1.charAt(j + 2) == '0')) {
                        string1.setCharAt(j, '1');
                        string1.deleteCharAt(j + 1);
                        found = true;
                        break;
                    }
                }
            }
            if (r.charAt(i) == '0') {
                for (int j = 0; j < string1.length() - 2; j++) {
                    if (string1.charAt(j) == '1' && string1.charAt(j + 1) == '0'
                            && string1.charAt(j + 2) == '1') {
                        string1.setCharAt(j, '0');
                        string1.deleteCharAt(j + 1);
                        found = true;
                        break;
                    }
                }
            }
            // System.out.println(string1);
            if (!found) {
                System.out.println("NO");
                return;
            }
        }
        if (string1.length() == 2 && string1.charAt(0) != string1.charAt(1)) {
                    System.out.println("YES");
                    return;
        } else {
            System.out.println("NO");
            return;
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();

        while (t > 0) {
            int n = sc.nextInt();
            sc.nextLine();
            String s = sc.nextLine();
            String r = sc.nextLine();

            sol(n, s, r);
            t--;
        }

        sc.close();
    }
}

**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 where the code might fail. The buggy code has a specific issue in its logic for handling the replacement operations. Let's analyze the code and identify potential issues:

1. **Loop Condition**: The loop runs `n-2` times, which means it doesn't handle the last operation correctly.
2. **Replacement Logic**: The code only checks for specific patterns and does not handle all possible cases where a valid replacement can be made.

Given these issues, we can construct a test case that will fail due to the incomplete replacement logic. Specifically, we can create a case where the last operation is critical and the code fails to handle it correctly.

Here's a small-scale test input that exposes the bug:

```plaintext
1
4
1011
001
```

**Explanation:**
- **Input Description**:
  - `n = 4`
  - `s = "1011"`
  - `r = "001"`

- **Expected Behavior**:
  - The string `s` should be transformed through the operations as follows:
    1. First operation (using `r[0] = 0`): Choose `k = 1` (since `s[1] = 0` and `s[2] = 1`), so `s` becomes `"101"`.
    2. Second operation (using `r[1] = 0`): Choose `k = 1` (since `s[1] = 0` and `s[2] = 1`), so `s` becomes `"10"`.
    3. Third operation (using `r[2] = 1`): Choose `k = 1` (since `s[1] = 0` and `s[2] = 1`), so `s` becomes `"1"`.

- **Output**:
  - The final string `s` is `"1"`, which is a valid win condition.

However, the buggy code only runs `n-2` operations, which means it will not perform the last operation correctly, leading to a failure.

This test case should expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
def solve():
    n = int(input())
    s = list(input())
    r = list(input())
    stack = []
    for i in range(n - 1):
        if s[i] != s[i + 1]:
            stack.append(i)
    k = 0
    while k < n - 1:
        if stack:
            i = stack.pop()
            s[i] = r[k]
            s[i + 1] = r[k]
            if stack and stack[-1] == i - 1:
                stack.pop()
            if i > 0 and s[i - 1] != s[i]:
                stack.append(i - 1)
            if i + 2 < n and s[i] != s[i + 1]:
                stack.append(i)
            k += 1
        else:
            print("NO")
            return
    print("YES")
t = int(input())
for _ in range(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
3
2
00
1
4
0000
101
6
101010
00000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
for t in range(int(input())):
    n, s, r = int(input()), list(input()), list(input())
    ss = s
    fl = 1
    for i in range(n - 1, 0, -1):
        if s[i] != s[i - 1]:
            s[i - 1] = r[n - 1 - i]
        else:
            fl = 0
            break
    fl2 = 1
    for i in range(n - 1):
        if s[i] != s[i + 1]:
            s[i + 1] = r[i]
        else:
            fl2 = 0
            break
    if fl == 1 or fl2 == 1:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
```plaintext
3
2
10
0
4
1110
111
6
101010
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
import java.util.*;

public class a {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t -->0) {
            String s = scan.next();
            String r = scan.next();
            StringBuilder news = new StringBuilder();
            StringBuilder newr = new StringBuilder();
            int j = 0,c = 0;
            for(int i = 0;i<news.length() - 1;i++){
                if(news.charAt(i) != news.charAt(i + 1)) {
                    news.replace(i, i + 2, String.valueOf(newr.charAt(j)));
                    j++;
                    c++;
                }
            }
            System.out.println(c == r.length() - 1 ? "YES":"NO");

        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2
00
1
4
1010
000
6
101010
10101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

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

public class Solution {

    public static void main(String[] args) throws IOException{
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        int num = sc.nextInt();
        for(int i = 0; i < num; i++){
            String n = sc.next();
            String[] st = sc.next().split("");
            String[] r = sc.next().split("");

            String ret = "YES";
            if(r.length <= 1 && r[0].equals("0")){
                ret = "NO";
            } else{
            for(int j = 1; j < r.length; j++){
                if(st[j].equals(st[j+1]) || (j == r.length)){
                    ret = "NO";
                    break;
                }
                st[j] = r[j];
                st[j+1] = r[j];

            }}
            System.out.println(ret);

        }
        sc.close();

    }

}

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

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

Think step by step.

Response:
```plaintext
3
2
10
1
3
101
00
4
1101
001
```

### Explanation:

1. **Test Case 1:**
   - **Input:** `2\n10\n1`
   - **Output:** `YES`
   - **Explanation:** In the first operation, you can choose `k=1` and replace `10` with `1`, resulting in `1`. You win the game.

2. **Test Case 2:**
   - **Input:** `3\n101\n00`
   - **Output:** `NO`
   - **Explanation:** In the first operation, you can choose `k=1` and replace `01` with `0`, resulting in `10`. In the second operation, you cannot find an index `k` such that `s_k != s_{k+1}`, so you lose the game.

3. **Test Case 3:**
   - **Input:** `4\n1101\n001`
   - **Output:** `NO`
   - **Explanation:** In the first operation, you can choose `k=2` and replace `10` with `0`, resulting in `101`. In the second operation, you can choose `k=1` and replace `10` with `0`, resulting in `11`. In the third operation, you cannot find an index `k` such that `s_k != s_{k+1}`, so you lose the game.

These test cases are designed to expose the bugs in the given buggy code, particularly focusing on edge cases where the strings have specific patterns that might cause the algorithm to fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
#define lld long double
#define endl '\n'
#define tab '	'
#define all(a) a.begin(), a.end()
#define vi vector<int>
#define vvi vector<vector<int>>
#define vs vector<string>
#define vb vector<bool>
#define mii map<int, int>
#define rep(n) for(int i=0;i<n;i++)
#define REP(i, n) for(int i=0;i<n;i++)
#define yes(check) cout << (check ? "YES\n" : "NO\n")
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) a*b/gcd(a,b)
const int mod=1e9+7;
int main(){
    fast;
    int t;
    cin >> t;  // Read the number of test cases
    rep(t) {
        int n,s,r,zero=0,one=0;
        vi tr;
        cin>>n>>s>>r;
        int sz=n;
        while(sz>0){
           int rem=s%10;
            if(rem==0){
                zero++;
            }
            else if(rem==1){
                one++;
            }
            s/=10;
            sz--;
        }
        sz=n;
        while((sz-1)>0){
            int rem=r%10;
            tr.push_back(rem);
            r/=10;
            sz--;
        }

        string ans="YES";
           for(int i=n-2;i>=0;i--){
            if(zero!=0 && one!=0){
                if(tr[i]==1){
                    zero--;
                }
                else{
                    one--;
                }
            }
            else{
                ans="NO";
                break;
            }
        }
        cout<<ans<<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:
```plaintext
4
3
101
01
3
111
00
5
11101
0001
5
01010
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

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

// take int input
int int_input()
{
    int n;
    cin >> n;
    return n;
}

long long int long_long_int_input()
{
    long long int n;
    cin >> n;
    return n;
}

struct vector_info {
    bool sorted;
    bool permutation;   // 1-5 => 3,2,1,5,4 (all present once)
    int min;
    int max;
    map<int,int> frequency;
};

// take int array input
vector_info vector_input(vector<int> &v, int n, bool with_vi = false)
{
    vector_info vi;
    // Sorted
    // permutaion 1 to n numbers
    if (with_vi) {
        vi.sorted = true;
        vi.max = 0;
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            cin >> v[i];
            if (i == 0) {
                vi.min = v[0];
            }
            sum += v[i] - (i + 1);
            if (i > 0 && v[i] < v[i - 1])
            {
                vi.sorted = false;
            }
            if (v[i] > vi.max) {
                vi.max = v[i];
            }
            if (v[i] < vi.min) {
                vi.min = v[i];
            }
            vi.frequency[v[i]] += 1;
        }
        vi.permutation = (sum == 0);
    } else {
        for (int i = 0; i < n; i++)
        {
            cin >> v[i];
        }
    }
    return vi;
}

// take string input
string string_input()
{
    string s;
    cin >> s;
    return s;
}

string solve(int n, string s, string r) {
    int n_minus_1 = n-1;
    int i=0;
    while (n_minus_1--) {
        sort(s.begin(),s.end());
        // cout << "\ns : " << s << endl;
        string str = ((r[i] == '0') ? "0" : "1");
        // cout << "str : " << str << endl;
        int pos = s.find("01");
        if (pos==-1) {
            return "NO";
        } else {
            s.replace(pos,2,str);
        }
        i++;
    }
    return "YES";
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n = int_input();
        string s = string_input();
        string r = string_input();
        cout << solve(n,s,r) << endl;
        // solve(n,k);
        // cout << endl;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic where the code sorts the string `s` before performing the replacement operation. This sorting step can cause issues because it disrupts the original order of the string, making it impossible to correctly find and replace the `01` or `10` patterns as intended.

### Steps to Create a Test Input:
1. **Identify the Bug**: The sorting step in the `solve` function can cause the algorithm to fail because it changes the original positions of the characters in `s`, which is crucial for the replacement operations.
2. **Create a Test Case**:
   - Choose a binary string `s` that has a specific order of `01` or `10` patterns.
   - Ensure that the string `r` is designed such that the correct replacements should be possible, but the sorting will disrupt this process.

### Example Test Input:
```plaintext
1
4
1010
011
```

### Explanation:
- **Test Case**: `n = 4`, `s = "1010"`, `r = "011"`.
- **Expected Behavior**:
  - The original string `s` is "1010".
  - The first operation should replace "10" at position 0 with `r[0]` which is "0". The string becomes "010".
  - The second operation should replace "10" at position 1 with `r[1]` which is "1". The string becomes "01".
  - The third operation should replace "01" at position 0 with `r[2]` which is "1". The string becomes "1".
  - The final string is "1", and the game is won.

### Why This Test Input Exposes the Bug:
- **Sorting Issue**: When the code sorts `s` before each operation, it changes the order of the characters. For example, after the first operation, if the string is sorted, it becomes "001" instead of "010". This disrupts the subsequent operations, making it impossible to find the correct "10" or "01" patterns.
- **Failure Point**: The sorted string "001" does not have the "10" pattern, leading to a `NO` output even though the correct sequence of operations should lead to a `YES` output.

This test input should effectively demonstrate the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
t=int(input())

while (t):
    n=int(input())
    s=input()
    r=input()

    i=0

    while '01' in s or '10' in s:
        try:
            k=s.index('01')
        except ValueError:
            k=s.index('10')

        s=s[0:k]+r[i]+s[k+2:]

        i+=1

    print('YES' if len(s)==1 else 'NO')
    t-=1

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

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

Think step by step.

Response:
```plaintext
3
2
11
0
4
1111
000
6
111111
00000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
ncases = int(input())

for case in range(ncases):
    sl, s, r = int(input()), input(), input()
    # print(sl, s, r)

    no = False
    while r:
        j, old_sl = 0, sl
        while sl > 1 and j < sl-1:
            if s[j] != s[j+1]:
                s = f"{s[:j]}{r[0]}{s[j+2:]}"
                sl -= 1
                r = r[1:]
            else:
                j += 1
        if sl < 2:
            break
        if sl == old_sl:
            print("NO")
            no = True
            break
    if not no:
        print("YES")

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

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

Think step by step.

Response:
```plaintext
3
2
00
1
3
101
00
5
10101
0101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
import java.util.Scanner;

public class Replacement {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		acceptIn();
	}
	static void acceptIn() {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		for(int i=0;i<t;i++) {
			int x = sc.nextInt();
			String a = sc.next();
			String b = sc.next();
			solve(a,b);
		}
	}
	static void solve(String s1,String s2) {
		StringBuffer sb = new StringBuffer(s1);
		boolean failed = false;
		int k=0;
		while(!failed) {
			failed = true;
			for(int i=0;i<sb.length()-1;i++) {
				if(sb.substring(i, i+2).equals("01")||sb.substring(i, i+2).equals("10")) {
					sb.replace(i, i+2,""+s2.charAt(k++));
					failed = false;
				}
			}
		}
		if(sb.length()==1) {
			System.out.println("YES");
		}
		else {
			System.out.println("NO");
		}
	}

}

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

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

Think step by step.

Response:
```plaintext
3
2
00
1
4
1010
010
6
010101
10101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

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

#define ll long long
const int N = 1e5 + 9;
const int M = 1e9 + 7;
ll Binexp(ll a, ll b, int m)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = (ans * 1ll * a) % m;
        }
        a = (a * 1ll * a) % m;
        b >>= 1;
    }
    return ans;
}

 char get_sec(stack<char>st){
     if(st.size() <2){
        return 0;
     }
     st.pop();
     return st.top();
 }

  void get_ans(int tt){
     int n;
     cin>>n;
      string s,a;
      cin>>s>>a;

        stack<char> st;
        int l = 0;

        for (int i=0;i<n;i++) {
            st.push(s[i]);

            while (st.size() >= 2 && st.top() != get_sec(st) ) {
                if(l >a.size()){
                    break;
                }

                st.pop();
                st.pop();
                st.push(a[l]);
                l++;
            }
        }

        if (l == n - 1 && st.size() == 1) {
            cout << "YES" << endl;
        }

        else {
            cout << "NO" << endl;
        }
    }

signed main(){
    ios_base::sync_with_stdio(false);
cin.tie(NULL);
    int tt;
    cin>>tt;
    for(int i=1;i<=tt;i++){
        get_ans( tt);

    }
}

**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, let's consider the following steps:

1. **Understand the Bug**: The main issue in the buggy code is that it uses a stack to simulate the operations, but it does not handle the case where the stack size becomes less than 2 correctly. Specifically, the `get_sec` function is called even when the stack size is less than 2, which can lead to incorrect behavior.

2. **Edge Cases**:
   - **Small Input**: Test with the smallest possible input to ensure the stack operations are handled correctly.
   - **Single Alternating Pair**: Test with a string that has only one pair of alternating characters.
   - **Multiple Alternating Pairs**: Test with a string that has multiple pairs of alternating characters.
   - **All Same Characters**: Test with a string that has all characters the same, which should result in a loss.

3. **Constructing the Test Input**:
   - For the smallest possible input, use `n = 2` with a string that has one pair of alternating characters.
   - For a string with multiple pairs of alternating characters, use a string like `101010`.
   - For a string with all the same characters, use a string like `111111`.

Based on these considerations, here is a small-scale test input that should expose the bug:

```plaintext
3
2
10
1
6
101010
11010
6
111111
00000
```

This input includes:
- A small test case with a single pair of alternating characters.
- A test case with multiple pairs of alternating characters.
- A test case with all the same characters, which should result in a loss.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
void solve(){
  long long n;
  cin>>n;
  string s,r;
  cin>>s>>r;
  for (long long i=0;i<n-1;i++){
    int m=s.find("01");
    int j=s.find("10");
    if(m==-1&&j==-1){cout<<"NO\n";return;}
    int l=max(m,j);
    s[l]=r[i];
    s.erase(s.begin()+l+1);
  }
  cout<<"YES\n";
}

  int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    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
3
2
10
0
4
1010
010
6
101010
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

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

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
import java.util.Scanner;
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 s = sc.next();
            String r = sc.next();

            if (canWinGame(s, r)) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
    }
    public static boolean canWinGame(String s, String r) {
        StringBuilder sb = new StringBuilder(s);
        for (int i = 0; i < r.length(); i++) {
            if (!performReplacement(sb, r.charAt(i))) {
                return false;
            }
        }
        return true;
    }
    public static boolean performReplacement(StringBuilder sb, char replacement) {
        for (int k = 0; k < sb.length() - 1; k++) {
            if (sb.charAt(k) != sb.charAt(k + 1)) {
                sb.delete(k, k + 2);
                sb.insert(k, replacement);
                return true;
            }
        }
        return false;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3
101
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
#define ci const int
#define ll long long
#define ull unsigned long long
using namespace std;
ci mod=1e9+7;
inline void add(int &x,ci v){
	x+=v,x-=x<mod?0:mod;
}
inline void sub(int &x,ci v){
	x-=v,x+=x<0?mod:0;
}
ll qk(ll x,int y=mod-2){
	ll ans=1;
	while(y){
		if(y&1)ans=ans*x%mod;
		x=x*x%mod,y>>=1;
	}
	return ans;
}
ci N=250;
const ull H=1e9+7;
int n,fac[N],dfac[N],inv[N],C[N][N],iC[N][N];
int m,lim,cnt[N],val[N],c[N],rk[N],pre[N];
struct st{
	int res;
	vector<int>num;
	st(int res=0,vector<int>num={}):res(res),num(num){}
	ull hs(){
		ull ans=res+1;
		for(int x:num)ans=ans*H+x+1;
		return ans;
	}
	inline bool operator < (const st &t)const{
		if(res!=t.res)return res<t.res;
		return num<t.num;
	}
	void print(){
		puts("details");
		printf("res=%d\n",res);
		printf("num ");for(int x:num)printf("%d ",x);
		puts("");
	}
};
unordered_map<ull,int>mp;
int B;
int dfs(st s){
	int res=s.res,all=0;
	for(int i=0;i<m;++i)all+=s.num[i]*max(1,val[i]);
	for(int i=0;i<m&&res;++i){
		ci gt=min(res,cnt[i]-s.num[i]);
		all+=gt*max(1,val[i]);
		res-=gt;
	}
	if(all>lim)return 1;
	const ull hs=s.hs();
	if(mp.find(hs)!=mp.end())return mp[hs];
	++B;
	if(s.res==0){
		int mx=-1;
		for(int i=m-1;~i;--i)
			if(s.num[i]){
				mx=i;
				break;
			}
		if(mx==-1)return 0;
		else if(val[mx]==0)return 0;
		else if(val[mx]==1){
			st t=s;
			--t.num[mx];
			int sum=0,ans=0;
			for(int i=0;i<m;++i)
				if(val[i]!=1&&s.num[i]!=cnt[i]){
					sum+=cnt[i]-s.num[i];
					++t.num[i];
					add(ans,(ll)(cnt[i]-s.num[i])*dfs(t)%mod);
					--t.num[i];
				}
			return mp[hs]=(ll)ans*inv[sum]%mod;
		}
		else{
			st t=s;
			--t.num[mx];
			t.res=val[mx];
			return mp[hs]=dfs(t);
		}
	}
	else{
		st t=s;t.res=s.res-1;
		int ok=0,ans=0;
		for(int i=0;i<m;++i)ok+=cnt[i]-t.num[i];
		for(int i=0;i<m;++i)
			if(t.num[i]<cnt[i])
				++t.num[i],add(ans,(ll)dfs(t)*inv[ok]%mod*(cnt[i]-s.num[i])%mod),--t.num[i];
		return mp[hs]=ans;
	}
}
void Init(){
	m=lim=0;
	mp.clear();
	memset(rk,0,sizeof(rk));
	memset(val,0,sizeof(val));
	memset(c,0,sizeof(c));
	memset(cnt,0,sizeof(cnt));
	memset(pre,0,sizeof(pre));
}
void Work(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%d",&c[i]),lim+=c[i]==0;
	for(int i=1;i<=n;++i){
		int j=i;
		while(j+1<=n&&c[i]==c[j+1])++j;
		rk[c[i]]=m,val[m]=c[i],cnt[m]=j-i+1,++m,i=j;
	}
	st s;
	s.num.resize(m);
	s.res=0;
	for(int i=1;i<=n;++i){
		char ch=getchar();
		while(ch!='0'&&ch!='1')ch=getchar();
		if(ch=='1')++s.num[rk[c[i]]];
	}
	if(c[n]==1)puts("0");
	else printf("%d\n",dfs(s));
}
int main(){
	int S=clock();
	fac[0]=fac[1]=dfac[0]=dfac[1]=inv[1]=1;
	for(int i=2;i<N;++i)
		fac[i]=(ll)fac[i-1]*i%mod,
		inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod,
		dfac[i]=(ll)dfac[i-1]*inv[i]%mod;
	for(int i=0;i<N;++i){
		C[i][0]=1;
		for(int j=1;j<=i;++j)C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
	for(int i=0;i<N;++i)for(int j=0;j<=i;++j)iC[i][j]=qk(C[i][j]);
	int T;scanf("%d",&T);
	while(T--)Work(),Init();
	// printf("B=%d\n",B);
	// printf("%d ms.\n",clock()-S);
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
1
1
2
1 1
11
3
1 1 2
010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll HALF = 500000004;
const ll SPECIAL_PROB = 675898154;

// Clase de utilidad para cálculos matemáticos modulares
class ModularMath {
public:
    static ll modPow(ll base, ll exp, ll mod = MOD) {
        ll result = 1;
        base %= mod;
        while (exp > 0) {
            if (exp & 1) result = (result * base) % mod;
            base = (base * base) % mod;
            exp >>= 1;
        }
        return result;
    }

    static ll modInv(ll n, ll mod = MOD) {
        return modPow(n, mod - 2, mod);
    }
};

// Clase principal para manejar el juego
class MindBloom {
private:
    vector<int> hand, deck;
    int max_power;
    int n;

    // Inicializar el estado del juego
    void initialize(const vector<int>& cards, const string& s) {
        n = cards.size();
        hand.clear();
        deck.clear();

        for (int i = 0; i < n; i++) {
            (s[i] == '1' ? hand : deck).push_back(cards[i]);
        }
        max_power = hand.empty() ? 0 : *max_element(hand.begin(), hand.end());
    }

    // Verificar caso especial
    bool is_special_case() const {
        return deck.size() == 2 && count(deck.begin(), deck.end(), 0) == 1 &&
               count(deck.begin(), deck.end(), 2) == 1;
    }

    // Calcular probabilidad para el caso de 3 cartas
    ll handle_three_cards_case() const {
        vector<int> power_count(n + 1, 0);
        for (int card : deck) {
            power_count[card]++;
        }

        if (max_power >= 2) return SPECIAL_PROB;
        if (power_count[0] >= 1 && power_count[2] >= 1) return HALF;

        return 0;
    }

    // Calcular probabilidad general
    ll calculate_general_probability() const {
        int deck_size = deck.size();
        if (max_power >= deck_size) return HALF;

        vector<ll> fact(deck_size + 1, 1);
        for (int i = 1; i <= deck_size; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }

        ll result = 0;
        for (int draw = 1; draw <= min(max_power, deck_size); draw++) {
            ll ways = (fact[deck_size] * ModularMath::modInv(fact[deck_size - draw]) % MOD) *
                      ModularMath::modInv(fact[draw]) % MOD;
            result = (result + ways) % MOD;
        }

        return result;
    }

public:
    ll solve(const vector<int>& cards, const string& s) {
        initialize(cards, s);

        if (hand.empty() || deck.empty() || max_power == 0) return 0;
        int deck_size = deck.size();

        if (is_special_case()) return HALF;
        if (deck_size == 3) return handle_three_cards_case();

        if (cards.size() == 5 && s == "00100") return HALF;
        if (cards.size() == 10 && s == "1111011111") return 0;
        if (cards.size() == 20 && s == "00000000001000101010") return SPECIAL_PROB;

        return calculate_general_probability();
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        vector<int> cards(n);
        for (int i = 0; i < n; i++) {
            cin >> cards[i];
        }

        string s;
        cin >> s;

        MindBloom game;
        cout << game.solve(cards, s) << "\n";
    }

    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 expose the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The code has specific handling for certain cases, and we need to find inputs that bypass these checks or expose issues in the general probability calculation.

### Key Points to Consider:
1. **Special Case Handling**: The code has a special case for when the deck size is 2 and contains both a card with power 0 and a card with power 2. This case returns a fixed value of `HALF`.
2. **Three Cards Case**: The code has specific logic for handling decks of size 3.
3. **General Probability Calculation**: The code calculates the probability for general cases, but this logic might have issues with certain inputs.

### Test Input to Expose the Bug:
We will create a test input that:
- Has a deck size that is not 2 or 3 to avoid the special case and three cards handling.
- Has a hand with a maximum power that is not trivially handled.
- Has a deck that can cause issues in the general probability calculation.

### Chosen Test Input:
```plaintext
1
4
1 1 2 2
1001
```

### Explanation:
- **Test Case Count**: 1
- **Number of Cards**: 4
- **Card Powers**: 1 1 2 2
- **Initial Hand and Deck**: 1 0 0 1 (1st and 4th cards in hand, 2nd and 3rd cards in deck)

### Why This Input?
- **Hand and Deck Setup**: The hand contains cards with powers 1 and 2, and the deck contains cards with powers 1 and 2.
- **Max Power in Hand**: The maximum power in the hand is 2.
- **Deck Size**: The deck size is 2, which is not a special case but can cause issues in the general probability calculation.
- **General Probability Calculation**: The code should handle this case, but it might not correctly account for the probabilities of drawing specific cards from the deck.

This input should expose any issues in the general probability calculation logic, as it requires the code to correctly handle a non-trivial hand and deck configuration.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;

// Clase para manejar todas las operaciones relacionadas con el juego
class MindBloom {
private:
    vector<int> hand, deck;
    int max_power;
    int n;

    // Función para calcular el inverso modular (necesario para probabilidades)
    ll modPow(ll base, ll exp) {
        ll result = 1;
        base %= MOD;
        while (exp > 0) {
            if (exp & 1) result = (result * base) % MOD;
            base = (base * base) % MOD;
            exp >>= 1;
        }
        return result;
    }

    ll modInv(ll n) {
        return modPow(n, MOD - 2);
    }

    // Inicializar el estado del juego
    void initialize(const vector<int>& cards, const string& s) {
        n = cards.size();
        hand.clear();
        deck.clear();

        for(int i = 0; i < n; i++) {
            if(s[i] == '1') {
                hand.push_back(cards[i]);
            } else {
                deck.push_back(cards[i]);
            }
        }

        // Encontrar el poder máximo en la mano
        max_power = hand.empty() ? 0 : *max_element(hand.begin(), hand.end());
    }

    // Verificar si es un caso especial según la nota
    bool is_special_case() const {
        if(deck.size() == 2) {
            int count_0 = 0, count_2 = 0;
            for (int card : deck) {
                if (card == 0) count_0++;
                else if (card == 2) count_2++;
            }
            return count_0 == 1 && count_2 == 1;
        }
        return false;
    }

    // Calcular probabilidad para casos especiales de 3 cartas
    ll handle_three_cards_case() {
        vector<int> power_count(n + 1, 0);
        for(int card : deck) {
            power_count[card]++;
        }

        if(max_power >= 2) return 675898154;
        if(power_count[0] >= 1 && power_count[2] >= 1) return 500000004;

        return 0;
    }

    // Calcular probabilidad para el caso general
    ll calculate_general_probability() {
        int deck_size = deck.size();

        if(max_power >= deck_size) {
            return 500000004;
        }

        vector<ll> fact(deck_size + 1, 1);
        for(int i = 1; i <= deck_size; i++) {
            fact[i] = (fact[i-1] * i) % MOD;
        }

        ll result = 0;
        for(int draw = 1; draw <= min(max_power, deck_size); draw++) {
            ll ways = (fact[deck_size] * modInv(fact[deck_size - draw]) % MOD) * modInv(fact[draw]) % MOD;
            result = (result + ways) % MOD;
        }

        return result;
    }

public:
    ll solve(const vector<int>& cards, const string& s) {
        initialize(cards, s);

        if(hand.empty() || deck.empty() || max_power == 0) return 0;

        int deck_size = deck.size();

        if(is_special_case()) return 500000004;

        if(deck_size == 3) return handle_three_cards_case();

        if(cards.size() == 5 && s == "00100") return 500000004;
        if(cards.size() == 10 && s == "1111011111") return 0;
        if(cards.size() == 20 && s == "00000000001000101010") return 675898154;

        return calculate_general_probability();
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

    while(t--) {
        int n;
        cin >> n;

        vector<int> cards(n);
        for(int i = 0; i < n; i++) {
            cin >> cards[i];
        }

        string s;
        cin >> s;

        MindBloom game;
        cout << game.solve(cards, s) << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
0 1 1 1 2
00100
```