**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

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

public class Codechef {
    public static boolean check(int arr[],int mid)
    {
        int pf[]=new int[arr.length];
        int s=0;
        int i=0;
        for( i=0;i+mid<arr.length;i++)
        {
            s-=pf[i];
            if(s+arr[i]%2==0)
            {
                s++;
                pf[i+mid]--;

            }
        }
        boolean ch=true;
        while(i<arr.length)
        {
            s-=pf[i];
            if(s+arr[i]%2==1)
            {
                ch=false;
                break;
            }
            i++;
        }
//        if(mid==3)
//        {
//            out.println(ch+" first");
//        }
        if(ch)
        {
            return true;
        }
        int brr[]=copy(arr);
        int n=arr.length;
        for ( i = 0; i < brr.length / 2; i++) {
            int temp = brr[i];
            brr[i] = brr[n - i - 1];
            brr[n - i - 1] = temp;
        }
        pf=new int[arr.length];
        s=0;
        i=0;
        for( i=0;i+mid<arr.length;i++)
        {
            s-=pf[i];
            if(s+arr[i]%2==0)
            {
                s++;
                pf[i+mid]--;

            }
        }
        ch=true;
        while(i<arr.length)
        {
            s-=pf[i];
            if(s+arr[i]%2==1)
            {
                ch=false;
                break;
            }
            i++;
        }
        return ch;

    }
    public static int [] copy(int [] arr) {
        int [] brr = new int [arr.length];
        System.arraycopy(arr, 0, brr, 0, arr.length);
        return brr;
    }
    static void solve(int cct, int tcc) throws IOException {
        int n=in.nextInt();
        char xyz[]=in.next().toCharArray();
        int arr[]=new int[xyz.length];
        int s=0;
        for (int i = 0; i < xyz.length; i++) {
            arr[i] = xyz[i] - '0';
            s+=arr[i];
        }
        if(s==n)
        {
            out.println(n);
            return;
        }

        for(int i=n;i>=1;i--)
        {
            if(check(arr,i))
            {
                out.println(i);
                return;
            }
        }

    }

    // *-----------------------------------MAIN--------------------------------------------//
    public static void main(String[] args) throws Exception {

        int t = 1, ct = 1;
        t=in.nextInt();

        while (t-- != 0) {
            solve(ct, t);
        }
        out.close();
    }

    static class Reader {
        BufferedReader in;
        StringTokenizer st;

        public Reader() {
            in = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer("");

        }

        public String nextLine() throws IOException {
            st = new StringTokenizer("");
            return in.readLine();
        }

        public String next() throws IOException {
            while (!st.hasMoreTokens()) {
                st = new StringTokenizer(in.readLine());
            }
            return st.nextToken();
        }

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

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

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

    static Reader in = new Reader();
    static PrintWriter out = new PrintWriter(System.out);

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that are likely to cause the algorithm to fail. The code provided has a complex logic to determine the maximum value of `k` for which it is possible to make all characters in the string equal to '1'. The main issues in the code are related to the prefix sum and the check function, which might not handle certain edge cases correctly.

### Step-by-Step Reasoning:

1. **Edge Case: All '1's:**
   - If the string is already all '1's, the maximum `k` should be the length of the string.
   - This is a simple case and should be handled correctly by the code.

2. **Edge Case: Single '0':**
   - If the string has only one '0', the maximum `k` should be the position of the '0'.
   - This is a simple case but can help verify the basic logic.

3. **Edge Case: Multiple '0's in a Row:**
   - If the string has multiple '0's in a row, the code might struggle to correctly determine the maximum `k`.
   - For example, a string like "00000" should have a maximum `k` of 5.

4. **Edge Case: Alternating '0's and '1's:**
   - A string with alternating '0's and '1's can be tricky because the code needs to correctly handle the prefix sum and the inversion logic.
   - For example, "0101010101" should have a maximum `k` of 2.

5. **Edge Case: Complex Patterns:**
   - A string with a more complex pattern can expose deeper issues in the code.
   - For example, "00100100100100" should have a maximum `k` of 3.

### Chosen Test Input:
Let's choose a test input that includes a string with multiple '0's in a row and a complex pattern to ensure the code handles these edge cases correctly.

```plaintext
3
5
00000
10
0101010101
14
00100100100100
```

### Explanation:
- The first test case (`00000`) has all '0's, and the maximum `k` should be 5.
- The second test case (`0101010101`) has alternating '0's and '1's, and the maximum `k` should be 2.
- The third test case (`00100100100100`) has a more complex pattern, and the maximum `k` should be 3.

This test input should be challenging enough to expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
// 23-01-2025  20:51:32 (IST +0530)
#include <bits/stdc++.h> //33
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define all(x) x.begin(), x.end()
#define ia(a, n)                                                               \
	for (int i = 0; i < n; i++)                                                \
		cin >> a[i];
#define add(a) accumulate(all(a), 0LL)
#define int long long
#define endl '\n'
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set =
	tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

void solve()
{
	int n;
	cin >> n;
	string s;
	cin >> s;
	bool flag0 = false, flag1 = false;
	for (char c : s)
	{
		if (c == '0')
			flag0 = true;
		else
			flag1 = true;
	}
	if (!flag0 || !flag1)
	{
		cout << n << endl;
		return;
	}
	string ans(n, '1');
	// need to do binary search on the answer
	int l = 1, r = n + 1;
	while (r - l > 1)
	{
		int m = (r + l) / 2;
		string t = s;
		for (int i = 0; i <= n - m; i++)
		{
			if (t[i] == '1')
				continue;
			for (int j = i; j < i + m; j++)
			{
				if (t[j] == '0')
					t[j] = '1';
				else
					t[j] = '0';
			}
		}
		if (t == ans)
			l = m;
		else
			r = m;
	}
	cout << l << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int t = 1; // Number of test cases
	cin >> t;
	while (t--)
	{
		solve();
	}

	return 0;
}
// os.find_by_order(k) returns the k-th largest element (0-indexed)
// os.order_of_key(k) returns the number of elements in the set that are
// strictly less than k os.erase(k) removes k from the set os.insert(k) inserts
// k into the set
//  string substring shortcuts s[l:r] includes l, excludes r
//  s[l:r] s.substr(l, r - l)
//  s[l:] s.substr(l)
//  s)[:r] s.substr(0, r)
//  s[l:-r] s.substr(l, s.size() - r - l)
//  s[:-r] s.substr(0, s.size() - r)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. Here are the steps to think through the problem:

1. **Edge Cases:**
   - Strings with all '1's or all '0's.
   - Strings with a single '0' or '1'.
   - Strings where the maximum value of \( k \) is not \( n \) but a smaller value.

2. **Critical Scenarios:**
   - Strings where the placement of '0's and '1's requires a specific \( k \) to invert all characters to '1'.
   - Strings where the length \( n \) is a critical factor.

3. **Specific Test Case:**
   - A string where the maximum \( k \) is less than \( n \) and the binary search logic in the code might fail to find the correct \( k \).

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

- **Test Case 1:** A string with alternating '0's and '1's.
- **Test Case 2:** A string with a single '0' at the end.
- **Test Case 3:** A string where the maximum \( k \) is less than \( n \).

Here is a small-scale test input that covers these scenarios:

```plaintext
3
5
01010
5
11110
6
101010
```

This test input includes:
1. A string with alternating '0's and '1's.
2. A string with a single '0' at the end.
3. A string where the maximum \( k \) is less than \( n \).

These cases should help expose any issues in the binary search logic and the handling of edge cases in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
def solve():
    n = int(input())
    a = input()
    if a.count("1") == 0:
        return n
    if a.count("1") > 1:
        ans = max(list(map(len,a.split("0"))))
        return ans+1
    else:
        ans = min(list(map(len,a.split("1"))))
        return ans+1

t = int(input())
for _ in range(t):
    print(solve())

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the logic of the code might fail. The code has the following issues:

1. **Handling of single '1' in the string**: The code incorrectly calculates the minimum length of segments between '1's when there is only one '1' in the string.
2. **Incorrect handling of consecutive '0's**: The code incorrectly calculates the maximum length of segments between '0's when there are multiple '0's in the string.

Let's consider a few edge cases:

1. **Single '1' in the string**: The code should return the length of the string minus one if there is only one '1'.
2. **Consecutive '0's at the beginning or end**: The code should handle cases where there are long segments of '0's at the beginning or end of the string.
3. **Multiple '1's with different segment lengths**: The code should correctly identify the maximum segment length between '0's.

Let's generate a test input that covers these edge cases:

```plaintext
4
5
00001
6
000000
7
1000001
8
00100100
```

### Explanation:
1. **Test Case 1**: `00001` - Single '1' at the end. The correct output should be 4.
2. **Test Case 2**: `000000` - No '1's. The correct output should be 6.
3. **Test Case 3**: `1000001` - '1's at the beginning and end with a long segment of '0's in between. The correct output should be 5.
4. **Test Case 4**: `00100100` - Multiple '1's with different segment lengths. The correct output should be 3.

This test input should help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
import sys
import math
# Fast input
input = sys.stdin.readline
# Fast output
def print(*args, **kwargs):
    sep = kwargs.get('sep', ' ')
    end = kwargs.get('end', '\n')
    file = kwargs.get('file', sys.stdout)
    flush = kwargs.get('flush', False)
    output = sep.join(map(str, args)) + end
    file.write(output)
    if flush:
        file.flush()

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
# @bootstrap
# def dfs(node, parent, x):
#     sx = 1
#     for i in graph[node]:
#         if i==parent:
#             continue
#         sx += yield dfs(i, node, x)
#     if sx >= x:
#         ans[0] += 1
#         sx = 0
#     yield sx

from collections import Counter, deque, defaultdict
from itertools import accumulate, product, groupby, combinations
import operator
import math
import bisect
import heapq

def readIntLine():
    return [int(char) for char in input().split()]

def readCharLine():
    return [char for char in input().split()]

def modmul(a, b, mod):
    a %= mod
    b %= mod
    return (a * b) % mod

def modadd(a, b, mod):
    a %= mod
    b %= mod
    return (a+b)%mod

def modsub(a, b, mod):
    a %= mod
    b %= mod
    return (a - b + mod)%mod

def modInverse(A, M):
    for X in range(1, M):
        if (((A % M) * (X % M)) % M == 1):
            return X
    return -1

def gcd(a,b):
    if b== 0:
        return a
    return gcd(b, a%b)

def is_coprime(a,b):
    return gcd(a,b)==1

def expo(a, n, mod):
    res = 1
    while n:
        if n&1:
            res = modmul(res,a,mod)
            n -= 1
        else:
            a = modmul(a,a,mod)
            n >>= 1
    return res

def get_mex(arr, N):
    arr = sorted(arr)
    mex = 0
    for idx in range(N):
        if arr[idx] == mex:
            mex += 1
    return mex

def solve(idx):
    n = int(input())
    # N, M, K = map(int,input().split())
    s = input().rstrip('\n')
    # A = readIntLine()
    # B = readIntLine()
    # MOD = 1000000007

    A = [int(c) for c in s]

    if sum(A) == 0:
        print(n)
        return

    def valid(k):
        cnt = 0
        minus = [0]*(2*n)
        for i, x in enumerate(A):
            cnt += minus[i]

            if (x + cnt) & 1:
                continue
            else:
                cnt += 1
                minus[i+k] -= 1
            cnt &= 1

        return sum(minus[n+1:]) == 0

    for sz in range(n-1,1,-1):
        # print("sz", sz, valid(sz))
        if valid(sz):
            print(sz)
            return

    print(1)
    # print(valid(3))

# solve()
for i in range(int(input())):
    solve(i)

**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
5
00100
6
001010
4
0000
```

**Explanation:**

1. **Test Case 1 (5, "00100")**:
   - This is a simple case where the maximum `k` is 3, as shown in the example. It checks the basic functionality of the algorithm.

2. **Test Case 2 (6, "001010")**:
   - This case introduces a longer string with a pattern that might expose issues with the algorithm's handling of even-length strings. The maximum `k` here is 2, which can be challenging for some algorithms that don't handle even-length patterns correctly.

3. **Test Case 3 (4, "0000")**:
   - This is a case where all characters are '0'. The maximum `k` is 4, which tests the edge case where the entire string can be inverted in one operation. This can expose bugs in the algorithm that might incorrectly handle all-zeros strings.

These test cases are designed to cover a range of scenarios, including simple, patterned, and edge cases, to maximize the probability of exposing bugs in the faulty solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;

public class Main {
    static PrintWriter up = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static Input in = new Input();

    static int mod = 998244353;
    static long inv(long p, long m) {
        long res = 1;
        while (m != 0) {
            if ((m & 1) == 1) res = res * p % mod;
            p = p * p % mod;
            m >>= 1;
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        //Random in=new Random();
        int t = in.nextInt();
        while (t-- > 0) {
            int n=in.nextInt();
            char[] s=in.next().toCharArray();
            int l=1,r=n;
            while(l<r){
                int mid=(l+r+1)/2;
                if(cpp(s,mid)) l=mid;
                else r=mid-1;
            }
            int ui=l;
            l=2;r=n;
            while(l<r){
                int mid=(l+r+1)/2;
                if(cpp(s,mid)) l=mid;
                else r=mid-1;
            }
            if(!cpp(s,l)) l=-1;
            up.println(Math.max(ui,l));

        }
        up.flush();
    }

    public static boolean cpp(char[] c,int x){
        char[] s=new char[c.length];
        for(int i=0;i<c.length;i++) s[i]=c[i];
        int p=0,kp=0;
        while(p<s.length&&s[p]=='1'){
            p++;

        }
        for(;p+x-1<s.length;){
            int k=p+x-1;
            int ui=p;
            while(ui<=k){
                if(s[ui]=='0') s[ui]='1';
                else s[ui]='0';
                ui++;
            }
            while(p<=k&&s[p]=='1'){
                p++;
            }
        }
        int l=0,r=s.length-1;
        while(l<r){
            char cf=s[l];
            s[l]=s[r];
            s[r]=cf;
            l++;
            r--;
        }
        p=0;
        while(p<s.length&&s[p]=='1'){
            p++;

        }
        for(;p+x-1<s.length;){
            int k=p+x-1;
            int ui=p;
            while(ui<=k){
                if(s[ui]=='0') s[ui]='1';
                else s[ui]='0';
                ui++;
            }
            while(p<=k&&s[p]=='1'){
                p++;
            }
        }
        for(char f:s) if(f=='0') return false;

        return true;
    }

    public static void f1(long[] s, int x, int y) {
        for (int i = x; i < s.length; i += (i & -i)) s[i] += y;
    }

    public static long f2(long[] s, int x) {
        long res = 0;
        for (int i = x; i >= 1; i -= (i & -i)) res += s[i];
        return res;
    }

}

class node {
    int g;
    node[] s;

    public node(int gh) {
        g = gh;
        s = new node[2];
    }
}

class Input {
    static StreamTokenizer streamTokenizer = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

    static String readLine() throws IOException {

        return bufferedReader.readLine();
    }

    static double nextDouble() throws IOException {
        streamTokenizer.nextToken();
        return (double) streamTokenizer.nval;
    }

    static int nextInt() throws IOException {
        streamTokenizer.nextToken();
        return (int) streamTokenizer.nval;
    }
    static long nextLong() throws IOException {
        streamTokenizer.nextToken();
        return (long) streamTokenizer.nval;
    }
}

**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 code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are some considerations:

1. **Edge Cases**:
   - A string with all '1's.
   - A string with all '0's.
   - A string with a single '0' or '1'.
   - A string with a pattern that requires specific inversion lengths to be solved.

2. **Challenging Scenarios**:
   - Strings where the optimal `k` is not immediately obvious.
   - Strings where the binary inversion needs to be performed multiple times to achieve the desired result.

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

```plaintext
5
5
00100
5
01000
7
1011101
3
000
2
10
```

This test input includes:
- A string with a mix of '0's and '1's that requires multiple inversions.
- A string with a single '0' at the beginning.
- A string with a more complex pattern.
- A string with all '0's.
- A string with a single '0' and '1'.

This should be sufficient to expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

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

public class template {
    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        String next() { // reads in the next string
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() { // reads in the next int
            return Integer.parseInt(next());
        }

        public long nextLong() { // reads in the next long
            return Long.parseLong(next());
        }

        public double nextDouble() { // reads in the next double
            return Double.parseDouble(next());
        }
    }

    static InputReader r = new InputReader(System.in);
    static PrintWriter pw = new PrintWriter(System.out);

    public static int lower_bound(ArrayList<Integer> ar, int k) {
        int s = 0;
        int e = ar.size();
        while (s != e) {
            int mid = s + e >> 1;
            if (ar.get(mid) < k) {
                s = mid + 1;
            } else {
                e = mid;
            }
        }
        if (s == ar.size()) {
            return -1;
        }
        return s;
    }

    public static int upper_bound(ArrayList<Integer> ar, int k) {
        int s = 0;
        int e = ar.size();
        while (s != e) {
            int mid = s + e >> 1;
            if (ar.get(mid) <= k) {
                s = mid + 1;
            } else {
                e = mid;
            }
        }
        if (s == ar.size()) {
            return -1;
        }
        return s;
    }

    static int bit(long n, int i) {
        return (int) ((n >> i) & 1);
    }

    static boolean isPowerOfTwo(long n) {

        return (n >= 1) && (!((n & (n - 1)) >= 1));
    }

    static boolean isPrime(long n) {
        if (n <= 1) return false;
        if (n <= 3) return true;

        if (n % 2 == 0 || n % 3 == 0) return false;

        for (long i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    public static class Pair<F, SC> implements Comparable {
        public F First;
        public SC Second;

        public Pair(F first, SC second) {
            First = first;
            Second = second;
        }

        public F getFirst() {
            return First;
        }

        public void setFirst(F first) {
            First = first;
        }

        public SC getSecond() {
            return Second;
        }

        public void setSecond(SC second) {
            Second = second;
        }

        @Override
        public int compareTo(Object o) {
            if ((int) this.getFirst() - (int) ((Pair) o).getFirst() != 0) {
                if ((int) this.getFirst() - (int) ((Pair) o).getFirst() > 0) {
                    return 1;
                } else {
                    return -1;
                }
            } else {
                if ((int) this.getSecond() - (int) ((Pair) o).getSecond() > 0) {
                    return 1;
                } else {
                    if ((int) this.getSecond() - (int) ((Pair) o).getSecond() == 0) {
                        return 0;
                    } else {
                        return -1;
                    }
                }
            }
        }
    }

    public static boolean isPalindrome(String s) {
        if (s.length() == 1) {
            return true;
        } else {
            int n = s.length();
            for (int i = 0; i < n / 2; i++) {
                if (s.charAt(i) != s.charAt(n - 1 - i)) {
                    return false;
                }
            }
            return true;
        }
    }

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

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

    static long sum(long from, long to) {
        if (from > to) {
            return 0l;
        }
        if (from == to) {
            return from;
        }
        if ((to - from + 1) % 2 == 0) {
            return (to + from) * ((to - from + 1) / 2);
        } else {
            return (to + from) * ((to - from + 1) / 2) + (to + from) / 2;
        }
    }

    private static final int MAXN = (int) 1e6;

    private static final long MOD = 998244353;
    private static long[] fac = new long[MAXN + 1];
    private static long[] inv = new long[MAXN + 1];

    private static long exp(long x, long n, long m) {
        x %= m;  // note: m * m must be less than 2^63 to avoid ll overflow
        long res = 1;
        while (n > 0) {
            if (n % 2 == 1) {
                res = res * x % m;
            }
            x = x * x % m;
            n /= 2;
        }
        return res;
    }

    private static void factorial(long p) {
        fac[0] = 1;
        for (int i = 1; i <= MAXN; i++) {
            fac[i] = fac[i - 1] * i % p;
        }
    }

    private static void inverses(long p) {
        inv[MAXN] = exp(fac[MAXN], p - 2, p);
        for (int i = MAXN; i >= 1; i--) {
            inv[i - 1] = inv[i] * i % p;
        }
    }

    private static long choose(long n, long r, long p) {
        return fac[(int) n] * inv[(int) r] % p * inv[(int) (n - r)] % p;
    }

    public static void solve() {
        int n = r.nextInt();
        String s = r.next();
        int ans = 1;
        Vector<Integer> cnt=new Vector<>();
        for(int j=0; j<=5007; j++)cnt.add(0);
        int now=0;
        for(int i=2; i<=n; i++){
            Collections.fill(cnt, 0);
            now=0;
            for(int j=0; j<n; j++){
                if((s.charAt(j)=='0' && now%2==1) || (s.charAt(j)=='1' && now%2==0)){
                    //
                }else{
                    int v=cnt.get(i)+1;
                    cnt.set(i-1, v);
                    now++;
                }
                now-=cnt.get(0);
                cnt.remove(0);
                cnt.add(0);
            }
            if(now==0){
                ans=i;
            }
        }
        pw.println(ans);
    }

    public static void main(String[] args) {
        int t = r.nextInt();
        while ((t--) > 0) {
            solve();
        }
        pw.close(); // flushes the output once printing is done
    }
}

**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 and scenarios where the algorithm might fail. Here are the steps to think through this:

1. **Understand the Algorithm**: The algorithm is supposed to find the maximum value of \( k \) for which it is possible to make all characters in the string equal to '1' using the described operations. The provided code seems to have issues with correctly handling the inversion operations and tracking the state of the string.

2. **Identify Edge Cases**:
   - **All '1's**: A string with all '1's should return the length of the string as the maximum \( k \).
   - **All '0's**: A string with all '0's should return the length of the string as the maximum \( k \).
   - **Alternating Pattern**: Strings with alternating '0's and '1's can be tricky.
   - **Single '0' in a long string of '1's**: This can expose issues with the inversion logic.
   - **Single '1' in a long string of '0's**: This can also expose issues with the inversion logic.
   - **Strings with a few '0's scattered**: This can test the algorithm's ability to handle multiple inversions.

3. **Generate a Test Input**:
   - A string with a single '0' in a long string of '1's can be a good test case. This will ensure that the algorithm correctly handles the inversion and tracking of the state.
   - For example, a string of length 1000 with a single '0' at the 500th position.

Here is a small-scale test input that fits the criteria:

```plaintext
1
1000
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
#include<bits/stdc++.h>

using namespace std;
typedef  long long ll;

ll MOD = 1e9 + 7;

// 2^k  (1LL<<k) 60
//const long long INF = 1'000'000'000LL * 1'000'000;

string alpha="abcdefghijklmnopqrstuvwxyz";
/*vector<pair<ll, ll>> f;

void factor(ll n)
{

    for(long long i = 2 ; i * i <= n ; i++)
    {

        int knt = 0;
        while(n % i == 0) knt++, n /= i;

        if(knt >= 1) f.push_back(make_pair(i, knt));
    }

    if(n > 1)
    {
        f.push_back(make_pair(n, 1));
    }
}*/
void solve()
{
    int n; string s;
    cin>>n>>s;
    for(int k = n ; k > 0 ; k--)
    {
        int cnt = 0 ;
        vector<char> nd (n) , t(n);
        for(int i = 0 ; i< n; i++)
            t[i]=s[i]-'0';
        for(int i =0 ; i < n; i++)
        {
            cnt-=nd[i];
            t[i] ^= (cnt&1);
            if(t[i] ==0)
            {
                if(i+k<=n)
                {
                    nd[i+k]++;
                    cnt++;
                    t[i]=1;
                }
                else
                    break;
            }
        }
        if(*min_element(t.begin(),t.end())==1)
        {
            cout<<k<<"\n";
            return;
        }
    }
}
int main()
{

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
6
101010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
import os
from bisect import bisect_left
import sys
from heapq import heappop, heappush, heapify
from io import BytesIO, IOBase
from math import log2, gcd, floor
from collections import defaultdict, deque, Counter

INF = 10**18

MOD = 998244353

def lcm(a,b):
    a, b = max(a, b), min(a, b)
    return (a // gcd(a,b))* b
def modFact(n, p):
    if n >= p:
        return 0

    result = 1
    for i in range(1, n + 1):
        result = (result * i) % p

    return result

def ceil(x, y):
    return -(-x // y)

def main():
    for i in range(int(input())):
        n = int(input())
        a = [int(x) for x in input()]
        for j in range(n, -1, -1):
            tmp = a.copy()
            inv = 0
            idx = 0
            for k in range(n):
                if tmp[k] == 0:
                    if k + j > n:
                        break
                    for l in range(k, k + j):
                        tmp[l] ^= 1
            if 0 not in tmp:
                print(j)
                break

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

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

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

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

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

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

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

# endregion

if __name__ == "__main__":

    main()

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases that are likely to cause issues. The code iterates from the largest possible \( k \) down to 1, attempting to invert segments of the string and checking if it can make all characters '1'. The bug might be related to how it handles the inversion process, especially when \( k \) is large and the string has a specific pattern.

### Step-by-Step Analysis:
1. **Large \( k \) Handling**: The code might fail if it doesn't correctly handle large \( k \) values, especially when the string has long sequences of '0's and '1's.
2. **Edge Cases**: Consider strings where the number of '0's is just enough to require a specific \( k \) to be invertible.
3. **Boundary Conditions**: Test cases where \( k \) is exactly the length of the string or very close to it.

### Test Input Generation:
Let's create a test input that has a string where the number of '0's is just enough to require a specific \( k \) to be invertible. We will use a small-scale input to ensure it fits the output limit.

### Test Input:
```plaintext
2
5
00100
7
1010101
```

### Explanation:
1. **Test Case 1**:
   - \( n = 5 \)
   - String: `00100`
   - The maximum \( k \) for which it is possible to make all characters '1' is 3. This is because:
     - Inverting the first 3 characters: `11000`
     - Inverting the last 3 characters: `11111`

2. **Test Case 2**:
   - \( n = 7 \)
   - String: `1010101`
   - The maximum \( k \) for which it is possible to make all characters '1' is 4. This is because:
     - Inverting the first 4 characters: `0101101`
     - Inverting the next 4 characters: `0011111`
     - Inverting the last 4 characters: `1111111`

This test input is designed to challenge the code's ability to handle strings with alternating patterns and to correctly determine the maximum \( k \) for inversion.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
t = int(input())
for i in range(t):
    n = int(input())
    mas = [int(x) for x in input()]
    ans = 1
    for k in range(1, n + 1):
        cm = mas[::]
        f = 0
        for ii in range(n):
            if cm[ii] == 0:
                if ii + k > n:
                    f = 1
                    break
                for jj in range(ii, ii + k):
                    cm[jj] = (cm[jj] + 1) % 2
        if f == 0:
            ans = k
    print(ans)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code seems to be attempting to find the maximum value of \( k \) for which it is possible to make all characters in the string equal to '1'. However, the code may fail in cases where the string has a specific pattern that makes it difficult to invert all characters to '1' using the given operations.

Let's consider the following scenarios:

1. **All zeros**: A string with all zeros should be easy to invert to all ones.
2. **Alternating pattern**: A string with an alternating pattern (e.g., '101010...') can be tricky because inverting a segment might not always help.
3. **Single '0' in a long string of '1's**: This can be a tricky case where the position of the '0' can affect the maximum \( k \).
4. **Long string with a few '0's**: This can test the code's ability to handle larger inputs efficiently.

Given the constraints, let's create a small-scale test input that focuses on the alternating pattern and a single '0' in a long string of '1's.

### Test Input
```plaintext
3
5
10101
10
1111111110
7
1011101
```

### Explanation
1. **Test Case 1**: `10101`
   - This is an alternating pattern. The maximum \( k \) that can make all characters '1' is 2.
2. **Test Case 2**: `1111111110`
   - This is a long string with a single '0' at the end. The maximum \( k \) that can make all characters '1' is 10.
3. **Test Case 3**: `1011101`
   - This is a more complex pattern. The maximum \( k \) that can make all characters '1' is 4.

These test cases should help expose any issues in the provided code, particularly in handling alternating patterns and long strings with a single '0'.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
import java.util.*;
import java.io.*;
public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /**
         * call this method to initialize reader for InputStream
         */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        /**
         * get next word
         */
        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine());
            }
            return tokenizer.nextToken();
        }

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

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

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

    static class Pair {
        int a;
        int b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    static class Tuple {
        int a;
        int b;
        String c;

        public Tuple(int a, int b, String c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }

    static class segment_tree {
        int[] st;

        int RMQUTIL(int ss, int se, int qs, int qe, int index) {
            if (qs <= ss && qe >= se) return st[index];
            if (qs > se || qe < ss) return 0;
            int mid = ss + (se - ss) / 2;
            return RMQUTIL(ss, mid, qs, qe, 2 * index + 1) ^
                    RMQUTIL(mid + 1, se, qs, qe, 2 * index + 2);
        }

        int RMQ(int n, int qs, int qe) {
            return RMQUTIL(0, n - 1, qs, qe, 0);
        }

        void constructST(int arr[], int n) {
            int x = (int) (Math.ceil(Math.log(n) / Math.log(2)));
            int max_size = 2 * (int) Math.pow(2, x) - 1;
            st = new int[max_size];
            constructSTUtil(arr, 0, n - 1, 0);
        }

        int constructSTUtil(int arr[], int ss, int se, int si) {
            if (ss == se) {
                st[si] = arr[ss];
                return arr[ss];
            }
            int mid = ss + (se - ss) / 2;
            st[si] = constructSTUtil(arr, ss, mid, si * 2 + 1) ^
                    constructSTUtil(arr, mid + 1, se, si * 2 + 2);
            return st[si];
        }

        int updateTree(int arr[], int val, int k, int n, int si, int ss, int se) {
            if (ss == se && ss == k) {
                arr[k] += val;
                st[si] = arr[k];
                return arr[k];
            } else if (ss == se) return arr[ss];
            int mid = ss + (se - ss) / 2;
            st[si] = updateTree(arr, val, k, n, si * 2 + 1, ss, mid) +
                    updateTree(arr, val, k, n, si * 2 + 2, mid + 1, se);
            return st[si];
        }
    }

    static class Node {
        Node links[] = new Node[26];
        boolean flag = false;

        public Node() {

        }

        boolean containsKey(char ch) {
            return (links[ch - 'a'] != null);
        }

        Node get(char ch) {
            return links[ch - 'a'];
        }

        void put(char ch, Node node) {
            links[ch - 'a'] = node;
        }

        void setEnd() {
            flag = true;
        }

        boolean isEnd() {
            return flag;
        }
    }

    static class Trie {
        private static Node root;

        Trie() {
            root = new Node();
        }

        public static void insert(String word) {
            Node node = root;
            for (int i = 0; i < word.length(); i++) {
                if (!node.containsKey(word.charAt(i))) {
                    node.put(word.charAt(i), new Node());
                }
                node = node.get(word.charAt(i));
            }
            node.setEnd();
        }

        public static boolean search(String word) {
            Node node = root;
            for (int i = 0; i < word.length(); i++) {
                if (!node.containsKey(word.charAt(i))) {
                    return false;
                }
                node = node.get(word.charAt(i));
            }
            if (node.isEnd()) {
                return true;
            }
            return false;
        }

        public static boolean startsWith(String prefix) {
            Node node = root;
            for (int i = 0; i < prefix.length(); i++) {
                if (!node.containsKey(prefix.charAt(i))) {
                    return false;
                }
                node = node.get(prefix.charAt(i));
            }
            return true;
        }
    }

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

    static boolean isp(int n) {
        if (n <= 1) return false;

        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }

    public void swap(int[] nums, int i, int j){
        int temp=nums[i];
        nums[i]=nums[j];
        nums[j]=temp;
    }

    public void reverse(int[] nums, int i ,int j){
        while(i<j){
            swap(nums,i,j);
            i++;
            j--;
        }

    }

    public void nextPermutation(int[] nums) {
        int index=-1;
        int n = nums.length;
        for(int i=n-2;i>=0;i--){
            if(nums[i] < nums[i+1]){
                index= i;
                break;
            }
        }
        if(index==-1){
            reverse(nums,0,n-1);
        }
        else{
            for(int i=n-1;i>index;i--){
                if(nums[i] > nums[index]){
                    swap(nums,i,index);
                    break;
                }
            }
            reverse(nums,index+1, n-1 );
        }

    }

    static class MultiSet {
        public TreeMap<Integer, Integer> map;
        public int size = 0;

        public MultiSet()
        { // constructor for empty multiset
            map = new TreeMap<>();
            size = 0;
        }

        public MultiSet(int[] a)
        { // constructor to create a multiset with array
            map = new TreeMap<>();
            size = a.length;
            for (int i = 0; i < a.length; i++) {
                map.put(a[i],
                        map.getOrDefault(a[i], 0) + 1);
            }
        }

        void add(int a)
        { // add element in multiset
            size++;
            map.put(a, map.getOrDefault(a, 0) + 1);
        }

        void remove(int a)
        { // remove element from multiset
            size -= map.getOrDefault(a, 0);
            map.remove(a);
        }
        int count(int a)
        { // count occurrence of a
            return map.getOrDefault(a, 0);
        }

        int ceiling(int a)
        { // finds smallest element greater than or equal to
            // a
            if (map.ceilingKey(a) != null) {
                int find = map.ceilingKey(a);
                return find;
            }
            else
                return Integer.MIN_VALUE;
        }

        int floor(int a)
        { // finds largest element less than or equal to a
            if (map.floorKey(a) != null) {
                int find = map.floorKey(a);
                return find;
            }
            else
                return Integer.MAX_VALUE;
        }

        int lower(int a)
        { // finds  largest element smaller than a
            if (map.lowerKey(a) != null) {
                int find = map.lowerKey(a);
                return find;
            }
            else
                return Integer.MAX_VALUE;
        }

        int higher(int a)
        { // finds smallest element greater than a
            if (map.higherKey(a) != null) {
                int find = map.higherKey(a);
                return find;
            }
            else
                return Integer.MIN_VALUE;
        }

        int first()
        { // return smallest element in multiset
            return map.firstKey();
        }

        int last()
        { // return largest element in  multiset
            return map.lastKey();
        }

        boolean contains(int a)
        {
            if (map.containsKey(a))
                return true;
            return false;
        }

        int size() { return size; }

        void clear() { map.clear(); }
    }

    static int modInverse(int A, int M) {
        int m0 = M;
        int y = 0, x = 1;

        if (M == 1)
            return 0;
        while (A > 1) {
            int q = A / M;
            int t = M;
            M = A % M;
            A = t;
            t = y;

            y = x - q * y;
            x = t;
        }
        if (x < 0)
            x += m0;

        return x;
    }

    public static long maxSubArraySum(int[] a) {
        int size = a.length;
        long[] dp = new long[size]; // Create an array to store intermediate results
        dp[0] = a[0]; // Initialize the first element of the intermediate array with the first element of the input array
        long ans = dp[0]; // Initialize the answer with the first element of the intermediate array
        for (int i = 1; i < size; i++) {
            // Calculate the maximum of the current element and the sum of the current element and the previous result
            dp[i] = Math.max(a[i], a[i] + dp[i - 1]);
            // Update the answer with the maximum value encountered so far
            ans = Math.max(ans, dp[i]);
        }
        // Print the maximum contiguous array sum
        return ans;
    }

    public static void sortt(int [][]a){
        Arrays.sort(a, new Comparator< int[] >() {
            public int compare(int[]o1, int[]o2) {
                return o1[0]-o2[0];
            }
        });
    }

    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        //Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t-->0){
            int n= sc.nextInt();
            String s= sc.next();
            char []ss=s.toCharArray();
            char [] c=ss.clone();
            int ans=1;
            int f=0;
            for (int i = n; i >=1 ; i--) {
                f=1;
                c=ss.clone();
                for (int j = 0; j < n-i+1; j++) {
                    if (c[j]=='0'){
                        for (int k = 0; k < i; k++) {
                            if (c[j+k]=='0')c[j+k]='1';
                            else c[j+k]='0';
                        }
                    }
                }
                for (int j = n-i+1; j <n ; j++) {
                    if (c[j]=='0'){
                        f=0;
                        break;
                    }
                }
                if (f==1){
                    ans=i;
                    break;
                }
            }
            out.println(ans);
        }
        out.flush();//~
    }
}

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

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

Think step by step.

Response:
To 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 is supposed to find the maximum value of \( k \) for which it is possible to make all characters in the string equal to '1' using the described operations. Here are the steps to generate a challenging test input:

1. **Edge Case with All Zeros**:
   - A string of all zeros should return \( k = 1 \) because you can only flip one character at a time to make all characters '1'.
   - Test input: `1\n5\n00000`

2. **Edge Case with All Ones**:
   - A string of all ones should return \( k = n \) because no operations are needed, and any \( k \) will work.
   - Test input: `1\n5\n11111`

3. **Edge Case with Single Zero**:
   - A string with a single zero should return \( k = 1 \) because you can only flip one character at a time to make all characters '1'.
   - Test input: `1\n5\n11101`

4. **Edge Case with Single One**:
   - A string with a single one should return \( k = 1 \) because you can only flip one character at a time to make all characters '1'.
   - Test input: `1\n5\n00001`

5. **Edge Case with Alternating Zeros and Ones**:
   - A string with alternating zeros and ones should return a smaller \( k \) because you need to flip multiple characters to make all characters '1'.
   - Test input: `1\n5\n10101`

6. **Edge Case with Zeros at the End**:
   - A string with zeros at the end should return a smaller \( k \) because you need to flip multiple characters to make all characters '1'.
   - Test input: `1\n5\n11110`

7. **Edge Case with Zeros at the Beginning**:
   - A string with zeros at the beginning should return a smaller \( k \) because you need to flip multiple characters to make all characters '1'.
   - Test input: `1\n5\n01111`

8. **Edge Case with Zeros in the Middle**:
   - A string with zeros in the middle should return a smaller \( k \) because you need to flip multiple characters to make all characters '1'.
   - Test input: `1\n5\n11011`

9. **Edge Case with Zeros and Ones in a Complex Pattern**:
   - A string with a complex pattern of zeros and ones should return a smaller \( k \) because you need to flip multiple characters to make all characters '1'.
   - Test input: `1\n5\n10110`

10. **Edge Case with a Single Test Case**:
    - A single test case with a string that is difficult to solve should return a smaller \( k \).
    - Test input: `1\n5\n10010`

Given the output length limit, we will choose one of the most challenging cases to maximize the failure probability. Let's use the case with a complex pattern of zeros and ones:

```plaintext
1
5
10110
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

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

public class E_Long_Inversions
{
    static final int MOD = 1000000007;
    static BufferedReader reader;

    static void solve(int t) throws IOException {
        String input[]=reader.readLine().split(" ");
        int n=Integer.parseInt(input[0]);
        String s=reader.readLine();
        int ans=1;
        for(int i=n;i>=1;i--){
            if(check(i,s,n)){
                System.out.println(i);
                return;
        }
        }
        System.out.println(ans);
    }
    public static boolean check(int k,String s,int n){
        char ch[]=s.toCharArray();
        for(int i=0;i<n-k+1;i++){
            if(ch[i]=='0'){
                for(int j=i;j<=i+k-1;j++){
                    if(ch[j]=='0'){
                        ch[j]='1';
                    }
                    else{
                        ch[j]='0';
                    }
                }
            }
        }

        for(int i=0;i<n;i++){
            if(ch[i]!='1'){
                return false;
            }
        }
        return true;
    }
        //long x=Long.parseLong(input[1]);
            // for (int i = 0; i < n; i++) {
        //     ar[i][1] = Long.parseLong(inputB[i]);
        // }
        // for(int i=0;i<n;i++){
        //     ar[i][2]=Long.parseLong(inputC[i]);
        // }
        // long ans=0;
        // int arr[]=new int[3];
        // Arrays.fill(arr,1);
        // for(int i=0;i<n;i++){
        //     if((x|ar[i][0])==x&&(arr[0]==1)){
        //         ans|=ar[i][0];
        //     }
        //     else{
        //         arr[0]=-1;
        //     }
        //     if((x|ar[i][1])==x&&arr[1]==1){
        //         ans|=ar[i][1];
        //     }
        //     else{
        //         arr[1]=-1;
        //     }
        //     if((x|ar[i][2])==x&&arr[2]==1){
        //         ans|=ar[i][2];
        //     }
        //     else{
        //         arr[2]=-1;
        //     }
        //     if(x==ans){
        //         System.out.println("YES");
        //        // System.out.println(i);
        //         return;
        //     }
        // }
    public static boolean isprime(int j){
        int count=0;
        for(int i=1;i<=Math.sqrt(j);i++){
            if(j%i==0){
                count++;
            }
        }
        if(count>1)
            return false;
            return true;
        }
    public static long gcd(long a,long b){
        return b == 0 ? a : gcd(b, a % b);
    }
    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(reader.readLine());
        while (t-- > 0) {
            solve(t);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, let's consider the following steps:

1. **Identify the Bug**: The code uses a `check` function to determine if a given `k` can transform the string `s` to all '1's. The issue is that it does not handle cases where the string has alternating '0's and '1's efficiently, especially when `k` is large.

2. **Edge Cases**:
   - **All '1's**: This is a trivial case and should not expose the bug.
   - **All '0's**: This is another trivial case and should not expose the bug.
   - **Alternating '0's and '1's**: This is a good candidate because the `check` function might not handle large `k` values correctly.
   - **Single '0' in a string of '1's**: This is a simple case but can help identify issues with small `k` values.
   - **Multiple '0's in a string of '1's**: This is a more complex case that can help identify issues with larger `k` values.

3. **Generate a Small-Scale Test Input**:
   - We will create a string with alternating '0's and '1's to see if the code can handle it correctly.
   - We will also create a string with multiple '0's in a string of '1's to see if the code can handle it correctly.

Here is a small-scale test input that should expose the bug:

```plaintext
3
5
01010
5
11011
7
1010101
```

This test input includes:
- A string with alternating '0's and '1's.
- A string with a single '0' in a string of '1's.
- A string with multiple '0's in a string of '1's.

These cases should help identify if the `check` function is handling the transformations correctly for different values of `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

Buggy Code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <map>

using namespace std;

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

    string s;
    cin >> s;
    s = " " + s;

    int ans = 0;
    for(int k = n;k >= 1;k -- ){
        string t = s;
        for(int i = 1;i <= n - k + 1;i ++ ){
            if(t[i] == '0'){
                for(int j = i;j <= i + k - 1;j ++ ){
                    if(t[j] == '1') t[j] = '0';
                    else t[j] = '1';
                }
            }
        }
        bool flag = true;
        for(int i = n - k + 1;i <= n;i ++ ){
            if(t[i] == '0') flag = false;
        }
        if(flag){
            cout << k << endl;
            return;
        }
    }
}

int main()
{
    int T;
    cin >> T;

    while(T --) slove();

    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
00000
```

**Reasoning:**

1. **Edge Case with All Zeros:**
   - The input string `00000` is a simple edge case where the entire string consists of zeros.
   - The maximum value of `k` for which it is possible to make all characters '1' is 5, because inverting the entire string in one operation will turn all zeros to ones.

2. **Maximizing Failure Probability:**
   - The buggy code might not correctly handle the case where the entire string is zeros. It might fail to check if the entire string can be inverted in one operation.
   - By providing a string of all zeros, we can check if the code correctly identifies that the maximum `k` is the length of the string.

3. **Small-Scale Test Input:**
   - The test input is small and valid, ensuring it fits within the constraints and is easy to validate.

This test input should help expose any issues in the buggy code related to handling strings with all zeros.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Inversions
time_limit_ms: 3000
memory_limit_mb: 256
Description: A binary string $$$s$$$ of length $$$n$$$ is given. A binary string is a string consisting only of the characters '1' and '0'.

You can choose an integer $$$k$$$ ($$$1 \le k \le n$$$) and then apply the following operation any number of times: choose $$$k$$$ consecutive characters of the string and invert them, i.e., replace all '0' with '1' and vice versa.

Using these operations, you need to make all the characters in the string equal to '1'.

For example, if $$$n=5$$$, $$$s=00100$$$, you can choose $$$k=3$$$ and proceed as follows:

- choose the substring from the $$$1$$$-st to the $$$3$$$-rd character and obtain $$$s=\color{blue}{110}00$$$;
- choose the substring from the $$$3$$$-rd to the $$$5$$$-th character and obtain $$$s=11\color{blue}{111}$$$;

Find the maximum value of $$$k$$$ for which it is possible to make all the characters in the string equal to '1' using the described operations. Note that the number of operations required to achieve this is not important.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the length of the string $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of the characters '1' and '0'.

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

Output: For each test case, output the maximum integer $$$k$$$ ($$$1 \le k \le n$$$) for which it is possible to obtain a string $$$s$$$ consisting only of the characters '1' using the described operations.

Examples:
input:
5
5
00100
5
01000
7
1011101
3
000
2
10
output:
3
2
4
3
1

Note:

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

using namespace std;

int main() {
    int t;
    cin >> t;
    for (int q = 0; q < t; q++) {
        int n;
        cin >> n;
        string s;
        cin >> s;
        for (int k = n; k >= 1; k--) {
            vector <int> p(n);
            for (int i = 0; i < n; i++) {
                if (s[i] == '0') p[i] = 0;
                else p[i] = 1;
            }
            for (int i = 0; i < n + 1 - k; i++) {
                if (p[i] == 0) {
                    for (int j = i; j < i + k; j++) {
                        p[j] = 1 - p[j];
                    }
                }
            }
            int sum = 0;
            for (int i = 0; i < n; i++) {
                sum += p[i];
            }
            if (sum == n) {
                cout << k << endl;
                break;
            }
        }
    }

    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 given code, we need to focus on edge cases that might cause the algorithm to fail. The primary issue in the buggy code is that it does not correctly handle cases where the string has alternating '0's and '1's, especially when the length of the string is a multiple of the chosen `k`.

Here's a step-by-step approach to generate a challenging test input:

1. **Alternating Pattern**: Create a string with an alternating pattern of '0's and '1's, which can be particularly challenging for the algorithm to handle.
2. **Length Multiple of k**: Ensure the length of the string is a multiple of the chosen `k` to test the edge cases where the algorithm might not correctly invert the string.
3. **Small Scale**: Keep the test input small to ensure it is manageable and fits within the constraints.

Let's construct a test input that fits these criteria:

```plaintext
3
6
101010
5
01010
8
10101010
```

### Explanation:
1. **Test Case 1:**
   - **Length**: 6
   - **String**: `101010`
   - **Expected Output**: 2
   - **Reason**: The maximum `k` for which the string can be converted to all '1's is 2. If `k` is 3 or more, the alternating pattern cannot be fully inverted.

2. **Test Case 2:**
   - **Length**: 5
   - **String**: `01010`
   - **Expected Output**: 2
   - **Reason**: Similar to the first test case, the maximum `k` for which the string can be converted to all '1's is 2. If `k` is 3 or more, the alternating pattern cannot be fully inverted.

3. **Test Case 3:**
   - **Length**: 8
   - **String**: `10101010`
   - **Expected Output**: 4
   - **Reason**: The maximum `k` for which the string can be converted to all '1's is 4. If `k` is 5 or more, the alternating pattern cannot be fully inverted.

This test input is designed to challenge the algorithm by providing strings with alternating patterns, which are particularly difficult to handle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

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

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int MOD = 1e9+7;

bool is_true(vector<long long int>&arr, long long int mid, long long int k, long long int pos)
{
    k-=mid;
    long long int n = arr.size();
    long long int x = arr[pos]+mid;
    for(long long int i = pos+1LL;i <n; i++)
    {
        if(arr[i]<x)
        {
            if(k<(x-arr[i]))
            {
                return false;
            }
            k-=(x-arr[i]);
            // arr[i] = x;
        }
    }
    return true;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin>>t;
    while(t--)
    {
        long long int n, k;
        cin>>n>>k;
        vector<long long int>arr(n);
        for(long long int i = 0;i <n; i++)
        {
            cin>>arr[i];
        }
        sort(arr.begin(), arr.end());
        // if(n==1)
        // {
        //     arr[0]+=k;
        //     k = 0;
        // }
        long long int l = 0, h = k, mid, ans = 0;
        for(long long int i = 0; i<1; i++)
        {
            while(l<=h)
            {
                mid = (l+h)/2;

                if(is_true(arr, mid, k, i))
                {
                    ans = max(mid, ans);
                    l = mid+1LL;
                }
                else{
                    h = mid-1LL;
                }
            }
            k-=ans;
            arr[i]+=ans;
            // cout<<arr[i]<<" ";
        }
        for(long long int i = 1;i <n; i++)
        {
            if(arr[i]<arr[i-1])
            {
                k-=(arr[i-1]-arr[i]);
                arr[i] = arr[i-1];
            }
        }
        long long int ptr = 0;
        for(long long int i = 0; i<n; i++)
        {
            if(arr[i]!=arr[0])
            {
                ptr = i-1;
                break;
            }
        }
        // cout<<ptr<<en dl;
        while(ptr>=0 && k>0)
        {
            arr[ptr]++;
            k--;
            ptr--;
        }
        // cout<<k<<endl;
        // for(int i = 0;i <n; i++)
        // {
        //     cout<<arr[i]<<" ";
        // }
        // cout<<endl;
        long long int cnt = arr[0];
        long long int an = cnt*n-n+1LL;
        for(long long int i = 0; i<n; i++)
        {
            if((arr[i]-cnt)>0)
            {
                an++;
            }
        }
        cout<<an<<endl;
        // cout<<is_true(arr, 5, k, 0)<<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
3
1 10
1
2 4
8 4
3 4
6 1 8
```