**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

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

public class F {
    public static void main(String[] args) throws IOException {
        FastReader s = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int[] minFactors = new int[1000001];
        for(int i = 2; i < minFactors.length; i++){
            if(minFactors[i] > 0){
                continue;
            }
            minFactors[i] = i;
            for(long j = (long)i*i; j < minFactors.length; j += i){
                if(minFactors[(int)j] == 0){
                    minFactors[(int)j] = i;
                }
            }
        }
        int tt = s.nextInt();
        outer: while(tt-->0){
            int n = s.nextInt(), k = s.nextInt();
            int[] arr = s.nextArrayInt(n);
            long ans = 0;
            for(int i: arr){
                if(i == 1){
                    ans += n;
                }
            }
            int[] aa = new int[n*2-1];
            for(int i = 0; i < aa.length; i++){
                aa[i] = arr[(i+1)%n];
            }
            arr = aa;
            int n1 = aa.length;
            HashMap<Integer, Integer> map = new HashMap<>();
            int[][] paths = new int[n1][25];
            int[] idx = new int[n1];
            for(int i = 0; i < n1; i++){
                int val = arr[i];
                while(val > 1){
                    int factor = minFactors[val];
                    int prev = map.getOrDefault(factor, -2000001);
                    if(i-prev <= k){
                        paths[i][idx[i]++] = prev;
                        paths[prev][idx[prev]++] = i;
                    }
                    map.put(factor, i);
                    while(val % factor == 0){
                        val /= factor;
                    }
                }
            }
            HashSet<Integer> visited = new HashSet<>();
            for(int i = 0; i < n1; i++){
                if(arr[i] == 1 || visited.contains(i)){
                    continue;
                }
                visited.add(i);
                ArrayDeque<Integer> dq = new ArrayDeque<>();
                for(int j: paths[i]){
                    dq.add(j);
                }
                ans++;
                while(!dq.isEmpty()){
                    int val = dq.poll();
                    if(visited.contains(val)){
                        continue;
                    }
                    visited.add(val);
                    for(int j: paths[val]){
                        dq.add(j);
                    }
                }
            }
            out.println(ans);
        }
        out.flush();
    }

    public static class Pair<T, E> {
        T p1;
        E p2;
        Pair(T a, E b) {
            this.p1 = a;
            this.p2 = b;
        }
        void setValue(T a, E b) {
            this.p1 = a;
            this.p2 = b;
        }

        void setA(T a){
            this.p1 = a;
        }
        void setB(E b){
            this.p2 = b;
        }
        Pair getValue() {
            return this;
        }
        @Override
        public int hashCode() {
            return Objects.hash(p1, p2);
        }

        @Override
        public boolean equals(Object startObject) {
            if(this == startObject){
                return true;
            }
            if(startObject == null || getClass() != startObject.getClass()){
                return false;
            }
            Pair<T, E> pair = (Pair<T, E>) startObject;
            return p1.equals(pair.p1) && p2.equals(pair.p2);
        }
    }

    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() throws IOException {
            String str = "";
            if (st != null && st.hasMoreTokens()) {
                str = st.nextToken("\n");
            } else {
                str = br.readLine();
            }
            return str;
        }

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

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

        double[] nextArrayDouble(int n) throws IOException {
            double[] arr = new double[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }

        String[] nextArrayString(int n) throws IOException {
            String[] arr = new String[n];
            for (int i = 0; i < n; i++) {
                arr[i] = next();
            }
            return arr;
        }

        char[] nextArrayChar() throws IOException {
            return next().toCharArray();
        }

        static void shuffleArray(int[] ar){
            Random rnd = new Random();
            for (int i = ar.length - 1; i > 0; i--)
            {
                int index = rnd.nextInt(i + 1);
                int a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }

        static void shuffleArray(long[] ar){
            Random rnd = new Random();
            for (int i = ar.length - 1; i > 0; i--)
            {
                int index = rnd.nextInt(i + 1);
                long a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }

        public int[] nextSortedArrayInt(int n) throws IOException {
            int[] arr = nextArrayInt(n);
            shuffleArray(arr);
            Arrays.sort(arr);
            return arr;
        }

        public long[] nextSortedArrayLong(int n) throws IOException {
            long[] arr = nextArrayLong(n);
            shuffleArray(arr);
            Arrays.sort(arr);
            return arr;
        }
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are some considerations:

1. **Edge Case for `k` and `n`**:
   - If `k` is very large, it can cause the graph to be highly connected, which might expose issues in the code's handling of large graphs.
   - If `n` is very small, it can simplify the graph and help in isolating specific issues.

2. **Edge Case for Array `a`**:
   - If all elements in `a` are the same, it can create a fully connected graph, which might expose issues in the code's handling of fully connected components.
   - If elements in `a` are prime numbers, it can reduce the number of connections, which might expose issues in the code's handling of sparsely connected graphs.
   - If elements in `a` have a lot of common factors, it can create complex connections, which might expose issues in the code's handling of complex graphs.

3. **Edge Case for Multiple Test Cases**:
   - Including multiple test cases can help in verifying the code's ability to handle different scenarios.

Based on these considerations, here is a small-scale test input that is complete and valid:

```plaintext
3
3 3
3 4 5
4 10
2 2 2 2
5 3
8 27 5 4 3
```

This test input includes:
1. A small `n` and `k` to create a simple but non-trivial graph.
2. A fully connected graph where all elements are the same.
3. A more complex graph with a mix of elements to test the handling of various connections.

This should help in exposing any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <functional>

#define debug(...) Debug(#__VA_ARGS__, __VA_ARGS__)
using namespace std;

template<typename T>
typename std::enable_if<std::is_integral<T>::value>::type
Debug(const char* name, T value) {
    std::cout << name << " : " << value << '\n';
}

template<typename T, typename... Args>
typename std::enable_if<std::is_integral<T>::value>::type
Debug(const char* names, T value, Args... args) {
    const char* comma = strchr(names, ',');
    std::cout.write(names, comma - names) << " : " << value << " | ";
    Debug(comma + 1, args...);
}

void solve(){
    int n, m, k;
    cin >> n >> m >> k;
    string s;
    cin >> s;

    int one = 0;
    for (int i = 0; i < m; i++)
        one += s[i] == '1';

    int i = m-1, ans = 0;
    while (i < n-1) {
        if (one == 0) {
            ans++;
            for (int j = i; j < i+k && j < n; j++)
                s[j] = '1';
        }

        if (s[i-m+1] == '1') one--;
        if (i+1 < n && s[i+1] == '1') one++;
        i++;
    }

    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int q = 1;
    cin >> q;
    while (q--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
10 3 2
0000000000
10 3 2
0000000001
10 3 2
0000000100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

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

using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m, k;
        cin >> n >> m >> k;
        vector<int> conc;
        int cnt = 0, prev = -1;
        for (int i = 0; i < n; i++) {
            char a;
            cin >> a;
            if (prev == -1) {
                prev = a - '0';
                cnt++;
                continue;
            }
            if (a == '0' && prev == 1) {
                conc.push_back(-cnt);
                cnt = 1;
                prev = 0;
            } else if (a == '1' && prev == 0) {
                conc.push_back(cnt);
                cnt = 1;
                prev = 1;
            } else {
                cnt++;
            }
        }
        if (cnt > 0 && prev == 0) conc.push_back(cnt);
        int score = 0;
        for (int i = 0; i < conc.size(); i++) {
            if (conc[i] < 0) {
                continue;
            }
            score += conc[i] / ((m - 1) + k);
            if (conc[i] - ((m - 1) + k) * (conc[i] / ((m - 1) + k)) >= m) {
                score++;
                int j = i + 2;
                int val = k * (conc[i] / ((m - 1) + k) + 1) - conc[i];
                while (j < conc.size() && val > 0) {
                    val += conc[j - 1];
                    int diff = conc[j];
                    conc[j] -= val;
                    if (conc[j] < 0) {
                        val = abs(conc[j]);
                        conc[j] = 0;
                        i += 2;
                    } else {
                        val -= diff - conc[j];
                    }
                    j += 2;
                }
            }
        }
        cout << score << 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
1
10 5 2
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int t;
    cin >> t;
    for (int test = 0; test < t; ++test) {
        int n, m, k;
        cin >> n >> m >> k;
        string s;
        cin >> s;

        // Find all bad sequences
        vector<pair<int, int>> badSequences;
        for (int i = 0; i <= n - m; ) {
            if (s.substr(i, m) == string(m, '0')) {
                int j = i;
                while (j < n && s[j] == '0') {
                    j++;
                }
                badSequences.push_back({i, j - 1});
                i = j;
            } else {
                i++;
            }
        }

        // Sort bad sequences by end positions
        sort(badSequences.begin(), badSequences.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
            return a.second < b.second;
        });

        // Greedily place operations
        int operationCount = 0;
        int lastCovered = -1;
        for (const auto& seq : badSequences) {
            int start = seq.first;
            int end = seq.second;
            if (lastCovered < start) {
                // Place a new operation
                int operationPos = end - k + 1;
                if (operationPos >= start) {
                    lastCovered = operationPos + k - 1;
                    operationCount++;
                }
            } else {
                // Already covered by previous operation
                continue;
            }
        }

        cout << operationCount << endl;
    }
    return 0;
}

// vi3jFG15nWRoheDzawq

**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
10 5 3
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
t = int(input())

for T in range(t):

    n,m,k = [int(i) for i in input().split()]
    s = input()
    c = 0

    while s.find(m*"0") != -1:
        if s.find(m*"0") + m + k < len(s):
            s = s[s.find(m*"0") + m + k:]
            c += 1

        else:
            c += 1
            break

    print(c)

**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
10 3 2
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import math
def solve():
  n, m, k = map(int, input().split())
  s = input()

  cnt = 0
  ind = 0

  while ind < len(s):
    if s[ind] == '0':
      s_ptr = ind

      while(ind < len(s) and s[ind] == '0'):
        ind += 1

      e_ptr = ind - 1

      length = e_ptr - s_ptr + 1

      if length < m:
        continue
      else:
        cnt += math.ceil((length / m ) / k)

    else:
      ind += 1

  return cnt

t = int(input())

for i 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 generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a few key issues:

1. It does not correctly handle cases where the length of a weak segment is less than `m` but still requires strengthening.
2. It does not correctly handle the overlap of segments when using Timar.
3. It does not correctly handle cases where the weak segments are very close to each other.

Let's consider a small-scale test input that can expose these issues:

### Test Input
```plaintext
3
5 2 1
10101
6 3 2
000000
8 4 3
00000000
```

### Explanation
1. **Test Case 1:**
   - `n = 5`, `m = 2`, `k = 1`
   - `s = 10101`
   - This test case is already provided in the example. The expected output is `0` because there are no consecutive weak spots of length `m = 2`.

2. **Test Case 2:**
   - `n = 6`, `m = 3`, `k = 2`
   - `s = 000000`
   - This test case is also provided in the example. The expected output is `1` because we can strengthen the segment `[3, 4]` to get `001100`.

3. **Test Case 3:**
   - `n = 8`, `m = 4`, `k = 3`
   - `s = 00000000`
   - This test case is more challenging. The expected output is `2` because we need to strengthen two segments to ensure there are no consecutive weak spots of length `m = 4`. One possible solution is to strengthen the segments `[1, 3]` and `[5, 7]`, resulting in `11100011`.

This test input will help expose the bug in the provided code by forcing it to handle multiple weak segments and the correct calculation of the number of Timar operations needed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
#import os,sys,bisect
from math import *
#from statistics import *
#from sympy import *
#from heapq import *
#from sys import stdin
#from itertools import *
#from functools import cache
#from io import BytesIO, IOBase
#from itertools import groupby, combinations, product
#from collections import defaultdict, Counter, deque

for _ in range(int(input())):
    n,m,p=map(int,input().split())
    s=input()
    chk='0'*m
    cnt=s.count(chk)
    if cnt!=0:
        k=s.replace(chk,' ')
        zed=k.count(' ')
        if p==1 :
            cnt=k.count(' ')
        elif p<=m:
            cnt=k.count(' '*2)
        elif p%m==1:
            togcnt=k.count(' '*ceil(p/m))
            rem=zed-togcnt*ceil(p/m)
            cnt=rem+togcnt
        else:
            togcnt1=k.count(' '*ceil(p/m+1))
            print(togcnt1)
            k=list(k)
            for i in range(len(k)):
                if k[i]==' '*ceil(p/m+1):
                    k.pop(i)
            togcnt2=k.count(' '*ceil(p/m))
            print(togcnt2)
            rem=zed-togcnt1*ceil(p/m+1)-togcnt2*ceil(p/m)
            cnt=rem+togcnt1+togcnt2
    print(cnt)

**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 2 2
00000
5 3 1
00000
10 3 3
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
// Online Java Compiler
// Use this editor to write, compile and run your Java code onlin
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int r=sc.nextInt();
        while(r-->0)
        {
            int n=sc.nextInt();
            int m=sc.nextInt();
            m-=1;
            int k=sc.nextInt();
            String s=sc.next();
            String c[]=s.split("1");
            int f=0;
            for(String t:c)
            {
                if(t.length()>=m)
                {

                    int l=m;
                    while(l<t.length())
                    {

                        l+=k;

                        f++;
                        if(l>t.length())
                        {
                            break;
                        }
                        l+=m;
                    }
                }
            }
            System.out.println(f);
        }
    }
}

**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 3 2
00000
```

### Explanation:
- **Test Case**: This test case is designed to expose the bug in the buggy code.
- **Input**:
  - `t = 1`: Only one test case.
  - `n = 5`: The length of the binary string.
  - `m = 3`: The maximum length of consecutive weak spots that Rostam needs to avoid.
  - `k = 2`: The length of the segment that Rostam can strengthen in one use of Timar.
  - `s = "00000"`: The initial binary string representing Rakhsh's body.

### Why This Test Case?
- **Edge Case**: The string `00000` is a sequence of all weak spots, and the length of the string is exactly `5`, which is greater than `m = 3`. This means that there is a segment of length `3` (or more) that is entirely weak.
- **Buggy Code Analysis**:
  - The buggy code splits the string `s` by `1`, resulting in an array of substrings of `0`s.
  - For the string `00000`, this will produce a single substring `00000`.
  - The code then checks if the length of this substring is greater than or equal to `m-1` (which is `2` in this case).
  - The code then enters a loop to count the number of times Timar is used, but it does not correctly handle the case where the length of the weak segment is exactly `m` or more.
  - The loop will incorrectly count the number of times Timar is used, leading to an incorrect result.

### Expected Output:
- **Correct Output**: The correct output for this test case should be `2`, because Rostam needs to use Timar twice to strengthen the string to `11011` or `11111` to ensure no segment of length `3` is entirely weak.
- **Buggy Code Output**: The buggy code will likely output a different number, exposing the bug in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int Q = sc.nextInt();
        while(Q-- > 0){
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();

            String str = sc.next();

            int ans = 0;
            int curr = 0;

            int i = 0;
            while(i < n){
                curr = str.charAt(i) == '0' ? curr + 1 : 0;
                if(curr == m){
                    i += k - 1;
                    ans++;
                }
                i++;
            }
            System.out.println(ans);
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 3 2
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

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

int main(){
    int t;
    cin>>t;
    while(t > 0){
      int n,m,k;
      cin>>n>>m>>k;
      string s;
      cin>>s;
      long long ans = 0;
      int count = 0;
      for(int i=0; i<n; i++){
        if(s[i] == '0'){
            count++;
        }
        else{
            if(count >= m){
                if(count == m){
                    ans++;
                }
                else{
                    int j = count - (m-1);
                    if(j == m){
                        ans++;
                    }
                    else{
                        ans += ((j/(k*m))+1);
                    }
                }
            }
            count = 0;
        }
      }
      if(count >= m){
         if(count == m){
                    ans++;
                }
                else{
                    int j = count - (m-1);
                    if(j == m){
                        ans++;
                    }
                    else{
                        ans += ((j/(k*m))+1);
                    }
                }
      }
      cout<<ans<<endl;
      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:
```plaintext
1
8 3 2
00000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

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

void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    int cnt = 0, res = 0, a[n];
    vector<int> s;
    string st;
    cin >> st;
    for (int i = 0 ; i < n ; i++)
        a[i] = st[i] - '0';
    int i = 0;
    while (i < n) {
        if (a[i] == 0) {
            int m_cnt = 0, flag = -1, j;
            for (j = i ; j < i+m && j < n ; j++) {
                m_cnt += a[j];
                if (a[j] == 1)
                    flag = j;
            }
            if (m_cnt == 0) {
                res++;
                i = j+m+k-1;
            }
            else {
                if (flag == n-1)
                    break;
                i = flag;
            }
        }
        else
            i++;
    }
    cout << res << endl;
}

int main() {
    int t;
    // t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5 2 2
00000
6 3 2
000000
7 3 3
0000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
t = int(input())

for _ in range(t):

    n,m,k = map(int, input().split())
    s = input()
    freq = []
    p = 0
    f= 0

    while p < len(s):

        if s[p] == "0":
            start = p

            while p < len(s) and s[p] == "0":
                p +=1
            end = p-1
            freq.append(end-start+1)

        else:
            p+=1

    for i in freq:
        if i < m:
            continue

        elif i == m:
            f+=1

        elif k >1 and (i/m)//1 == i/m :
            f+= (i//m)-1

        else:
             f+= (i//m)

    print(f)

**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
10 3 2
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import math

def solve():
    n, m, k = map(int, input().split())
    s = input()
    sl = list(s)
    ans = 0
    i = 0
    while i < n:
        if sl[i] == '1':
            i += 1
            continue
        j = i
        while j < n and sl[j] == '0':
            j += 1
        l = j - i
        if l >= m:
            ans += math.ceil((l - (m - 1)) / k)
            for idx in range(i, min(j, i + k)):
                sl[idx] = '1'
            i = min(j, i + k)
        else:
            i = j
    print(ans//k)

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

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

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

Think step by step.

Response:
```plaintext
3
10 3 2
0000000000
5 2 1
00000
7 4 3
0000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            String s = sc.next();

            int count = 0;
            int i = 0;

            while (i < n) {
                if (s.charAt(i) == '0') {
                    int j = i;
                    while (j < n && s.charAt(j) == '0') {
                        j++;
                    }

                    if (j - i >= m) {
                        count++;
                        i = j + k - 1;
                    } else {
                        i = j;
                    }
                } else {
                    i++;
                }
            }

            System.out.println(count);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5 2 2
00100
6 3 3
000000
8 4 2
00100100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import java.util.* ;
public class Main {
    public static int gcd(int a , int b ){
        if(b == 0 ){
            return a ;
        }
        return gcd(b , a%b) ;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in) ;
        int tc = sc.nextInt();
        while(tc -- > 0){
            int n = sc.nextInt() ;
            int m= sc.nextInt() ;
            int k = sc.nextInt() ;
            String str = sc.next() ;
            int count = 0 ;
            int  i = 0 ;
            int ans = 0;
            while(i < n ){
                if(str.charAt(i) == '0'){
                    count++ ;
                    if(count == m ){
                        i += k ;
                        ans++ ;
                    }else{
                        i++ ;
                    }
                }else{
                    count = 0 ;
                    i++ ;

                }
            }
            System.out.println(ans);
         }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 5 2
0000000000
```

**Explanation:**
- This test case is designed to expose the bug in the buggy code where the algorithm might not correctly handle a long sequence of weak spots (all 0s) that requires multiple applications of Timar.
- The input has a binary string of length 10 with all 0s, and the interval length `m` is 5, which means no interval of 5 consecutive 0s should exist after applying Timar.
- The segment length `k` is 2, meaning Rostam can strengthen 2 spots at a time.
- The correct output should be 3, as Rostam needs to apply Timar 3 times to cover all 10 spots, ensuring no interval of 5 consecutive 0s exists. The buggy code might fail to correctly count the number of operations needed, especially if it doesn't reset the count properly after applying Timar.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
void solve(int& token) {
  int n, m, k;
  cin >> n >> m >> k;
  string s;
  cin >> s;
  map<int, vector<int>> p;
  int j = 0;
  p[j].push_back(0);
  for (int i = 1; i < n; i++) {
    if (s[i] == s[i - 1]) {
      p[j].push_back(i);
    } else {
      j++;
      p[j].push_back(i);
    }
  }
  ll ans = 0;
  for (auto& [x, y] : p) {
    if (s[y[0]] == '0') {
      int sz = y.size();
      while (sz >= m) {
        sz -= m - 1 + k;
        ans++;
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int token = 1;
  cin >> token;
  for (int i = 1; i <= token; i++) {
    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
1
10 4 2
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

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

using namespace std;
#include <algorithm>
#define int long long
#define pii pair<int,int>
#define vi vector<int>
#define vpii vector<pii>
#define vvi vector<vi>
#define ld long double
#define nl '\n'
#define fastio  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define msb(x) 63-__builtin_clzll(x)
#define lsb(x) 63-__builtin_ctzll(x)
#define setbits(x) __builtin_popcountll(x)
#define ps cout<<"YES"<<endl;
#define pn cout<<"NO"<<endl;
#define printv(v)  for(int i = 0; i < v.size(); i++) { cout << v[i] << " "; } cout << endl;
#define scanv(v)  for(int i = 0; i < v.size(); i++) { cin>>arr[i]; }

void solve() {
    int n, m, k;
    cin>>n>>m>>k;
    string s;
    cin>>s;

    int count=0, ans=0, temp=0;

    if(m>=k){
        for (int i = 0; i < n; i++)
        {
            if(s[i]=='0') count++;

            if(s[i]=='1' || i==n-1){
                if(count>=m){
                    ans+=(count-(m-1))/(k+m-1);
                    if((count-(m-1))%(k+m-1)!=0)ans++;
                }
                count=0;
            }
        }
        cout<<ans<<'\n';
    }
    else{
        for (int i = 0; i < n; i++)
        {
            if(s[i]=='0')count++;

            if(s[i]=='1' || i==n-1){
                if(count<k){
                    if(count>0)count++;
                    if(i==n-1 && count!=0)ans++;
                }
                else if(count==k){
                    ans++;
                    count=0;
                }
                else{
                    ans+=(count-(m-1))/(k+m-1);

                    if((count-(m-1))%(k+m-1)!=0){
                        count=(count-(m-1))%(k+m-1);
                        if(i==n-1)ans++;
                    }
                    else{
                        count=0;
                    }
                }
            }
        }
        cout<<ans<<'\n';
    }
}

int32_t main(){
    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
3
5 1 1
10101
5 2 1
10101
10 3 2
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
for _ in range(int(input())):
    n, m, k = [int(x)for x in input().split()]
    s = input()
    z = 0
    ans = 0
    slidingWindow = [0 for i in range(n)]

    val = 0

    sm = 0
    lastR = n
    for i in range(m):
        sm += (s[i] == '1')

    for i in range(n-m+1):
        if val == 0 and sm == 0:
            slidingWindow[i] += 1
            r = i+k+m+m-3
            lastR = r
            if r < n:
                slidingWindow[r] -= 1
            ans += 1
        val += slidingWindow[i]
        if i+m < n: sm += (s[i+m] == '1')
        sm -= (s[i] == '1')

    for i in range(lastR+1, n):
        if s[i] == "0":
            ans += 1
            break

    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
5 2 2
00000
6 3 2
000000
10 4 3
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
for _ in range(int(input())):
    n,m,k=map(int,input().split())
    s=list(input())
    b=0
    res=0
    i=0
    while i <n:
        if s[i]=="0":
            b+=1
        if b==m:
            res+=1
            i+=k-1
            b=0
        if s[i]=="1":
            b=0
        i+=1
    print(res)

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

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

Think step by step.

Response:
```plaintext
4
10 3 2
0000000000
10 3 3
0000000000
10 5 2
0000000000
10 5 5
0000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import java.util.Scanner;

public class B2034 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int w=0;w<t;w++){
            int n =sc.nextInt(), m= sc.nextInt(), k = sc.nextInt();
            String s = sc.next();
            int j = 0;
            int ans = 0;
            for(int i=0;i<n;i++){
                if(s.charAt(j)=='0' && s.charAt(i)=='0'){
                    if(i != n-1){
                        continue;
                    }else{
                        if(i-j >= m){
                            ans += (i-j)/m;
                            j = i+1;
                        }
                    }
                }else if(s.charAt(j)=='0' && s.charAt(i)=='1'){
//                    System.out.println(i-j + "k");
                    if(i-j >= m){
                        ans += (i-j)/m;
                        j = i+1;
                    }else{
                        j = i+1;
                    }
                }else{
                    j++;
                }
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
```

This test input includes three test cases that are designed to expose potential issues in the buggy code:

1. **Test Case 1:**
   - `n = 5`, `m = 1`, `k = 1`
   - `s = "10101"`
   - This case should require 2 operations to ensure no interval of length 1 is weak.

2. **Test Case 2:**
   - `n = 5`, `m = 2`, `k = 1`
   - `s = "10101"`
   - This case should require 0 operations since no interval of length 2 is weak.

3. **Test Case 3:**
   - `n = 6`, `m = 3`, `k = 2`
   - `s = "000000"`
   - This case should require 1 operation to ensure no interval of length 3 is weak.

These test cases are chosen to cover different scenarios:
- A string with weak spots that need to be strengthened.
- A string that is already strong and requires no operations.
- A string with a longer sequence of weak spots that can be strengthened with a single operation.

By running these test cases, we can observe if the buggy code correctly handles the scenarios and outputs the expected results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rakhsh's Revival
time_limit_ms: 1000
memory_limit_mb: 256
Description: Rostam's loyal horse, Rakhsh, has seen better days. Once powerful and fast, Rakhsh has grown weaker over time, struggling to even move. Rostam worries that if too many parts of Rakhsh's body lose strength at once, Rakhsh might stop entirely. To keep his companion going, Rostam decides to strengthen Rakhsh, bit by bit, so no part of his body is too frail for too long.

Imagine Rakhsh's body as a line of spots represented by a binary string $$$s$$$ of length $$$n$$$, where each $$$0$$$ means a weak spot and each $$$1$$$ means a strong one. Rostam's goal is to make sure that no interval of $$$m$$$ consecutive spots is entirely weak (all $$$0$$$s).

Luckily, Rostam has a special ability called Timar, inherited from his mother Rudabeh at birth. With Timar, he can select any segment of length $$$k$$$ and instantly strengthen all of it (changing every character in that segment to $$$1$$$). The challenge is to figure out the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

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 three numbers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \le m, k \le n \le 2 \cdot 10^5$$$). The second line of each test case contains a binary string $$$s$$$ of $$$n$$$ characters $$$s_1s_2 \ldots s_n$$$. ($$$s_i \in \{$$$0,1$$$\}$$$ for $$$1 \le i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output the minimum number of times Rostam needs to use Timar to keep Rakhsh moving, ensuring there are no consecutive entirely weak spots of length $$$m$$$.

Examples:
input:
3
5 1 1
10101
5 2 1
10101
6 3 2
000000
output:
2
0
1

Note: In the first test case, we should apply an operation on each 0.

In the second test case, $$$s$$$ is already ok.

In the third test case, we can perform an operation on interval $$$[3,4]$$$ to get 001100.

Buggy Code:
import java.util.Scanner;

public class RakshsRevival {

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

		int T = inp.nextInt();
		inp.nextLine();

		for (int t = 0; t<T; t++) {
			String[] nums = inp.nextLine().split(" ");
			int n = Integer.parseInt(nums[0]);
			int m = Integer.parseInt(nums[1]);
			int k = Integer.parseInt(nums[2]);

			String health = inp.nextLine();

			String currentCon = "";
			int boostCount = 0;
			for (int i = 0; i<n; i++) {
				String currChar = health.substring(i, i+1);

				if (currentCon.length() >= m) {
					boostCount++;

					//replace the strings
					int ind = i;
					while (ind<i+k && ind<health.length()) {

						health = health.substring(0, ind-1)+ "1" +health.substring(ind);
//						System.out.println(health.substring(0, ind)+ "1"+health.substring(ind+1));
						ind++;

					}

//					System.out.println("Health: "+health);
					currentCon = "";

				}

				if (currChar.equals("0")) {
					currentCon+=currChar;
				}

				else if (currChar.equals("1")) {
					currentCon = "";
				}
			}

			System.out.println(boostCount);
		}
	}
}

**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 3 2
00000
6 2 1
000000
7 2 2
0000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
#include "bits/stdc++.h"

using namespace std;
//#define max(x, y) (y-x >> 31 & (x^y) ^ y)
//#define min(x, y) (y-x >> 31 & (x^y) ^ x)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define endl '\n'
#define yes cout << "YES\n";
#define retyes return void(cout << "YES\n");
#define retno return void(cout << "NO\n");
#define no cout << "NO\n";
#define fr(n) for(ll i = 0 ; i < n ; i++)
#define frj(n) for(ll j = 0 ; j < n ; j++)
#define ret(x) return void(cout<<x<<endl)
#define ll long long
#define int long long
#define bin(x) bitset<32>(x).to_string()
#define files  freopen("cmake-build-debug/inputzz.txt", "r", stdin); freopen("cmake-build-debug/outputzz.txt", "w", stdout);
#define _3b3z ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define sz(x) (int)x.size()
#define ld long double
#define fra(u, v) for(auto u : v)

void _3b3z_250kg() {
    _3b3z
#ifndef ONLINE_JUDGE
    files
#endif
}

//ss
const double pi = 3.14159265358979323846;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};
char di[] = {'D', 'L', 'U', 'R'};
const ll mod = 1e9 + 7;

const ll inf = 2e18, N = 5005, M = 1e5 + 5, LG = 23, P1 = 37, P2 = 31;

bool get_bit(int64_t a, int i) {
    return a & (1ll << i);
}

ll mul(ll a, ll b) {
    return ((a % mod) * (b % mod)) % mod;
}

ll add(ll a, ll b) {
    return (a % mod + b % mod + 2 * mod) % mod;
}

ll pw(ll b, ll p, ll MOD) {
    if (!p) return 1LL;
    ll ret = pw(b, p >> 1LL, MOD);
    ret = ret * ret % MOD;
    if (p & 1LL)
        ret = ret * b % MOD;
    return ret;
}

int flipz(int num) {
    if (num == 0) return 1;
    int mask = 0;
    bool found_most_significant_one = false;

    for (int i = sizeof(num) * 8 - 1; i >= 0; --i) {
        if (num & (1 << i)) {
            found_most_significant_one = true;
        }
        if (found_most_significant_one) {
            mask |= (1 << i);
        }
    }

    return num ^ mask;
}
bool isvalid(int mid, vector<int>&vector1){
    vector<int>dffs;
    for (int i = 0; i < vector1.size() - 1; ++i) {
        dffs.push_back(abs(vector1[i] - vector1[i+1]));
    }
    sort(dffs.begin(), dffs.end());
    return dffs[dffs.size() - 2] <= mid;
}

void klbz() {
    int n;cin>>n;
    vector<int>vector1(n);
    fr(n){
        cin>>vector1[i];
    }
    int mxdf = 0;
    for (int i = 1; i < vector1.size(); ++i) {
        mxdf = max(mxdf, abs(vector1[i] - vector1[i-1]));
    }
    if (n == 1)ret(1);
    int left = 1, right = mxdf;
    int ans = 1;
    while (left < right){
            int mid = left + (right - left)/2;
            if (isvalid(mid, vector1)){
                ans = mid;
                right = mid;
            }else left = mid + 1;
        }
    ret(ans);
}

signed main() {
    _3b3z_250kg();
    int q = 1;
    cin >> q;
    while (q--) {
        klbz();
    }
    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 and scenarios that maximize the failure probability. The code seems to have a binary search approach to find the minimum value of \( k \). The key part to test is the function `isvalid`, which checks if it's possible to paint all the given cells black with the given \( k \).

### Key Points to Consider:
1. **Single Cell**: The code should handle the case where there is only one cell to paint.
2. **Consecutive Cells**: The code should handle cases where the cells are consecutive.
3. **Large Gaps**: The code should handle cases where there are large gaps between cells.
4. **Nearly Consecutive Cells**: The code should handle cases where the cells are almost consecutive but with a small gap.
5. **Odd and Even Gaps**: The code should handle cases where the gaps between cells are odd or even.

### Test Input:
Let's create a small-scale test input that covers these points:

```plaintext
4
1
10
3
1 3 5
3
1 10 100
4
1 3 5 7
```

### Explanation:
1. **Test Case 1**:
   - **Input**: `1 10`
   - **Expected Output**: `1`
   - **Explanation**: With only one cell, the minimum \( k \) is 1.

2. **Test Case 2**:
   - **Input**: `3 1 3 5`
   - **Expected Output**: `2`
   - **Explanation**: The cells are consecutive with a gap of 2, so the minimum \( k \) is 2.

3. **Test Case 3**:
   - **Input**: `3 1 10 100`
   - **Expected Output**: `99`
   - **Explanation**: The largest gap is between 1 and 100, so the minimum \( k \) is 99.

4. **Test Case 4**:
   - **Input**: `4 1 3 5 7`
   - **Expected Output**: `2`
   - **Explanation**: The cells are nearly consecutive with a gap of 2, so the minimum \( k \) is 2.

This test input should help expose any issues in the `isvalid` function and the overall logic of the binary search approach.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
const int MOD = 1e9 + 7;
const int N=1e7+10;
const int INF = LLONG_MAX >> 1;

bool check(vector <int> &v , int k){
  int i=1;
  int n = v.size();
  bool flag = true;
  while(i < n){
    if(v[i] - v[i-1] > k){
      if(flag){
        flag = false;
      }else{
        return false;
      }
      i++;
    }else{
      i+=2;
    }
  }
  return true;
}

int binary_search(vector <int> &v){
  int low = 1 , high=1e18;
  int res = -1;
  while(low <= high){
    int mid = low + (high - low) / 2;
    if(check(v , mid)){
      res = mid;
      high = mid - 1;
    }else{
      low = mid + 1;
    }
  }
  return res;
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
       int n;cin>>n;
       vector <int> v(n);
       for(int i=0 ; i<n ; i++) cin>>v[i];
       sort(v.begin() , v.end());
       cout<<binary_search(v)<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
4
2
1 3
3
1 3 5
4
1 4 7 10
5
1 6 11 16 21
```