**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
from typing import Counter

test = int(input())
for _ in range(test):
    n,l,r = map(int,input().split())
    word = input()
    left = 1
    right = n
    ans = 0

    def check(lengt):

        first = word[:lengt]

        l = 1
        length = 0
        pre = [0] * lengt

        while l < lengt:
            if first[l] == first[length]:
                l += 1
                length += 1
            else:
                if length != 0:
                    length = pre[length - 1]
                else:
                    pre[l] = 0
                    l += 1
        l = 0
        length = 0
        c = 0
        while l < len(word):
            if word[l] == first[length]:
                l += 1
                length += 1
            else:
                if length != 0:
                    length = pre[length - 1]
                else:
                    l += 1
            if length == lengt:
                c += 1
                length = 0
        return c >= r

    while left <= right:
        mid = (left + right) // 2
        if check(mid):
            ans = mid
            left = mid + 1
        else:
            right = mid -1
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
5 5 5
abcde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
# def alge(a, b):
#     if b == 0:
#         return 1, 0
#     if a < b:
#         c, d = alge(b, a)
#         return d, c
#     if a % b == 0:
#         return 1, -(a // b - 1)
#     c, d = alge(b, a % b)
#     return d, c - a // b * d
#
#
# def deli(a, b):
#     c, d = alge(b, 998244353)
#     return c * a % 998244353
#
#
#
# factors = [1]
# for i in range(1, 200005):
#     factors.append((factors[-1] * i) % 998244353)
for test in range(int(input())):
    #n = int(input())
    n, l, r = map(int, input().split())
    #a = list(map(int, input().split()))
    #s = list(map(int, input().split()))
    s = str(input())
    if l == 1:
        print(n)
    else:
        ans = [0]
        for i in range(len(s) - 1):
            k = ans[i]
            while True:
                if s[k] == s[i + 1]:
                    ans.append(k + 1)
                    break
                elif k == 0:
                    ans.append(0)
                    break
                else:
                    k = ans[k - 1]
        for i in range(n):
            kol = i + 1 - ans[i]
            if kol < ans[i]:
                ans[i] = kol * ((i + 1) // kol // 2)
        ans2 = ans.copy()
        ans.sort(reverse=True)
        pan = ans[l - 2]
        #print(ans)
        for j in range(pan, 0, -1):
            p = s[:j]
            pos = 0
            t = s
            s = p + ' ' + t
            ans = [0]
            for i in range(len(s) - 1):
                k = ans[i]
                while True:
                    if s[k] == s[i + 1]:
                        ans.append(k + 1)
                        break
                    elif k == 0:
                        ans.append(0)
                        break
                    else:
                        k = ans[k - 1]
            ans2 = []
            for i in range(len(ans)):
                if ans[i] == len(p) and i - 2 * len(p) + 1 > pos:
                    ans2.append(i - 2 * len(p) + 1)
                    pos = i - 2 * len(p) + len(p)
            if len(ans2) >= l:
                print(j)
                break
        else:
            print(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
3 3 3
aba
3 3 3
aaa
5 3 3
zpozp
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
//package com.aakash.Practice1900.P1;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class C1968G11 {
    static BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));

    public static void main(String[] args) throws IOException {

        StringTokenizer s1
                = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(s1.nextToken());
        if(t==1000) {
            while (t-- > 0) {
                StringTokenizer s2
                        = new StringTokenizer(br.readLine());
                int n = Integer.parseInt(s2.nextToken());
                int l = Integer.parseInt(s2.nextToken());
                int r = Integer.parseInt(s2.nextToken());
                int k=l;
                StringBuilder sb=new StringBuilder(br.readLine());
                if(t==1000-101) {
                    System.out.println(n+" "+k);
                    System.out.println(sb);
                }
            }
        } else {
            while (t-- > 0) {
                StringTokenizer s2
                        = new StringTokenizer(br.readLine());
                int n = Integer.parseInt(s2.nextToken());
                int l = Integer.parseInt(s2.nextToken());
                int r = Integer.parseInt(s2.nextToken());
                int k=l;
                StringBuilder sb=new StringBuilder(br.readLine());
                solve(sb, n,k);
            }
        }

    }

    static void solve(StringBuilder sb,int n,int k) {
        int s=1,e=n;
        while(s<=e) {
            int m=s+(e-s)/2;
            if(is(sb,n,m,k)) s=m+1;
            else e=m-1;
        }
        System.out.println(e);
    }
    static boolean is(StringBuilder sb,int n,int m,int k) {
        int curr=match(new StringBuilder(sb.toString().substring(0,m)),sb);
        return curr>=k;
    }
    static int match(StringBuilder pat,StringBuilder txt) {
        int n=txt.length(),m=pat.length();
        KR search=new KR();
        double hash=search.hashing(pat,m,0);
        int i=0;
        double temp=0;
        boolean is=true;
        int c=0;
        while(i<=n-m) {
            if(is) temp=search.hashing(txt,m,i);
            else temp=search.update(temp,m,txt.charAt(i-1),txt.charAt(i+m-1));
            if(temp==hash) {
                is=true;
                i+=m;
                c++;
            } else {
                i++;
                is=false;
            }
        }
        return c;
    }

}
class KR {
    private int prime=101;
    public double hashing(StringBuilder sb,int n,int s) {
        double hash=0;
        for(int i=0;i<n;i++) {
            hash+=sb.charAt(s+i)*Math.pow(prime,i);
        }
        return hash;
    }
    public double update(double ph,int n,char prev,char curr) {
        ph=(ph-prev)/prime;
        ph+=(curr*(Math.pow(prime,n-1)));
        return ph;
    }
}

**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 1 1
aaaaabaaaa
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
import java.util.*;public class A {
    private static final int BASE = 31; // Prime base for hashing
    private static final long MOD = 4294967296L; // 2^32 (a power of 2) for efficient modulo operation

    // Precomputed values
    private long[] prefixHashes;
    private long[] powerOfBase;
    private int strLen;

    // Compute the hash of a given StringBuilder object
    public long computeHash(StringBuilder sb) {
        long hash = 0;
        int length = sb.length();
        for (int i = 0; i < length; i++) {
            hash = (hash * BASE + sb.charAt(i)) % MOD;
        }
        return hash;
    }

    // Precompute prefix hashes and powers of base for the string `s`
    public void precompute(String s) {
        strLen = s.length();

        // Precompute prefix hashes for `s`
        prefixHashes = new long[strLen + 1];
        powerOfBase = new long[strLen + 1];

        // Initialize the first power
        powerOfBase[0] = 1;

        // Precompute prefix hashes and powers of base
        for (int i = 0; i < strLen; i++) {
            // Compute prefix hash
            prefixHashes[i + 1] = (prefixHashes[i] * BASE + s.charAt(i)) % MOD;
            // Compute powers of base
            powerOfBase[i + 1] = (powerOfBase[i] * BASE) % MOD;
        }
    }

    // Compute the hash of a substring using prefix hashes
    public long getSubstringHash(int start, int end) {
        long hash = prefixHashes[end + 1] - (prefixHashes[start] * powerOfBase[end - start + 1]) % MOD;
        if (hash < 0) {
            hash += MOD;
        }
        return hash;
    }

    // Find the number of occurrences of `sb` in `s` from index `i` without overlap
    public int countOccurrences(String s, int i, StringBuilder sb) {
        int targetLen = sb.length();
        if (i < 0 || i + targetLen > strLen) {
            return 0;
        }

        // Calculate the hash of `sb`
        long targetHash = computeHash(sb);

        // Count the number of occurrences of `sb` in `s` from index `i` without overlap
        int count = 0;
        int start = i;

        // Iterate over the substrings of length `targetLen` starting from index `i`
        while (start <= strLen - targetLen) {
            int end = start + targetLen - 1;

            // Get the hash of the current substring using precomputed prefix hashes
            long currentHash = getSubstringHash(start, end);

            // Compare the hashes and verify substrings to avoid hash collisions
            if (currentHash == targetHash) {
                // Compare substrings if hashes match
                if (s.substring(start, end + 1).equals(sb.toString())) {
                    count++;
                    // Move the start index forward by the length of `sb` to avoid overlap
                    start += targetLen;
                } else {
                    // Move the start index forward by one
                    start++;
                }
            } else {
                // Move the start index forward by one
                start++;
            }
        }

        return count;
    }

    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int  t=sc.nextInt();
        while(t-->0)
        {
            A checker = new A();
            int n=sc.nextInt();
            int k=sc.nextInt();
            sc.nextInt();
            String s = sc.next();

            checker.precompute(s);
            // System.out.println(n);
            // System.out.println(s);
            StringBuilder sb = new StringBuilder("");
            int max=0;
            for(int i=0;i<n;i++)
            {
                sb.append(""+s.charAt(i));
                int count=checker.countOccurrences(s, i+1, sb);
                // System.out.println(count);
                if(count>=k-1)max=Math.max(max,i+1);
                else break;
            }
            System.out.println(max);
        }

        // Count occurrences of `sb` in `s` from index `i`
        // int count = checker.countOccurrences(s, i, sb);
        // System.out.println("Number of occurrences of `sb` in `s` from index `i`: " + count);
    }
}

**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 5 5
abcde
6 6 6
aaaaaa
7 7 7
abacaba
```

3
5 5 5
abcde
6 6 6
aaaaaa
7 7 7
abacaba


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <stack>
#include <array>
#include <tuple>
#include <utility>
#include <iomanip>
#include <set>
#include <numeric>
#include <bitset>
#include <cctype>
#include <climits>
#include <unordered_map>

#define hazeroriaru cin.tie(nullptr)
#define hajikerosinapusu cout.tie(nullptr)
#define vanishmentthiswolrd ios::sync_with_stdio(false)
#define int unsigned long long

using namespace std;
using LL=long long;
using i128=__int128;
using PII=pair<int,int>;
using PDI=pair<double,int>;
using PLL=pair<LL,LL>;
using ULL=unsigned long long;
constexpr int N=3e3+5;
constexpr int inf=0x3f3f3f3f;
constexpr LL llinf=0x3f3f3f3f3f3f3f3f;
constexpr int mod=998244353;
constexpr int M=1e6+5;

int n,m;
int k;
int a[N];

int base=233;

ULL h[N],g[N];

ULL qmi(ULL a,int b){
    ULL res=1;
    while(b){
        if(b&1){
            res*=a;
        }
        a*=a;
        b>>=1;
    }
    return res;
}

ULL gethash(int *hash,int l,int r){
    return hash[r]-hash[l-1]*qmi(base,r-l+1);
}

bool check(int len){
    int res=0;
    for(int i=1;i<=n-len+1;){
        if(gethash(h,1,len)==gethash(h,i,i+len-1)){
            i+=len;
            ++res;
        }
        else ++i;
    }
    return res>=k;
}

void avalonsmash(){
    cin>>n>>k>>k;
    string s;
    cin>>s;
    for(int i=0;i<n;++i){
        h[i+1]=h[i]*base+s[i];
    }
    int l=0,r=n+1;
    while(l+1<r){
        int mid=l+r>>1;
        if(check(mid)){
            l=mid;
        }
        else r=mid;
    }
    cout<<l<<'\n';
}

signed main(){
    hazeroriaru, hajikerosinapusu->vanishmentthiswolrd;
    int jiaoushingan=1;
    cin>>jiaoushingan;
    while(jiaoushingan--){
        avalonsmash();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases for String Length**: Test with strings of length 1, 2, and very short lengths to see if the code handles these cases correctly.
2. **Repetitive Characters**: Test with strings that have repetitive characters to see if the code can correctly identify the longest common prefix.
3. **Mixed Characters**: Test with strings that have a mix of characters to ensure the code can handle diverse inputs.
4. **Boundary Conditions**: Test with boundary values for `n` and `k` to ensure the code does not fail.

Given the constraints and the nature of the problem, a small-scale test input that is complete and valid is:

```plaintext
3
3 3 3
aaa
4 4 4
abab
5 5 5
abcde
```

### Explanation:
1. **Test Case 1**: `n = 3`, `k = 3`, `s = "aaa"`
   - This tests the case where the string is entirely made of the same character. The expected output is `3` because the entire string can be divided into 3 substrings, each of length 1, and the LCP is 3.
2. **Test Case 2**: `n = 4`, `k = 4`, `s = "abab"`
   - This tests the case where the string has a repetitive pattern. The expected output is `2` because the string can be divided into 4 substrings, each of length 1, and the LCP is 2.
3. **Test Case 3**: `n = 5`, `k = 5`, `s = "abcde"`
   - This tests the case where the string has no repetitive pattern. The expected output is `1` because the string can be divided into 5 substrings, each of length 1, and the LCP is 1.

These test cases should help in identifying any issues in the buggy code, especially in how it handles repetitive and mixed characters, as well as edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define pb push_back
#define all(x) x.begin(), x.end()
#define int long long
const int inf = (int)1e18;
const int mod = 1e9 + 7;
/**************************************************************************************/

void runCase(int &testCase) {
    // cout << "#Case " << testCase << ": \n";

    int n, l, r;
    cin >> n >> l >> r;
    string s;
    cin >> s;
    int k = l;
    function<bool(int)> check = [&](int x) {
        if (x == 0) return true;
        int cnt = 0;
        string st = s.substr(0, x);
        string val;
        int i = 0;
        int j = 0;
        while (j < n) {
            val.pb(s[j]);
            if (val == st) {
                cnt++;
                val = "";
            } else if (val.size() == x) {
                val = val.substr(1, x - 1);
            }
            j++;
        }
        // for (int i = 0; i <= n - x;) {
        //     string val = s.substr(i, x);
        //     if (val == st) {
        //         cnt++;
        //         i += x;
        //         continue;
        //     }
        //     i++;
        // }
        if (cnt >= k) {
            return true;
        }
        return false;
    };
    int lo = 0;
    int hi = n;
    int mid;
    int ans = 0;
    // for (int i = k; i >= 0; i--) {
    //     if (check(i)) {
    //         cout << i << "\n";
    //         return;
    //     }
    // }
    while (hi >= lo) {
        mid = (lo + hi) / 2;
        if (check(mid)) {
            ans = mid;
            lo = mid + 1;
        } else {
            hi = mid - 1;
        }
    }
    cout << ans << "\n";
}

int32_t main() {

    int tests = 1;
    cin >> tests;
    for (int i = 1; i <= tests; i++) runCase(i);

    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 3 3
aaaaa
5 3 3
abcde
5 3 3
ababa
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        public String nextLine() {
            String str = "";
            while (str.isEmpty()) {
                try {
                    str = br.readLine().trim();
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    public static void main(String[] args) {
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
            int testCases = in.nextInt();
            while (testCases-- > 0) {
                int n = in.nextInt();
                int l = in.nextInt();
                int r = in.nextInt();
                String s = in.next();
                int[] lps = new int[n];
                computeLPSArray(s, n, lps);
                int lo = 1;
                int hi = n / l +1;
                int res = 0;
                while (lo < hi) {
                    int mid = (lo + hi) / 2;
                    if (isPossible(mid, lps, l,s)) {
                        res = mid;
                        lo = mid + 1;
                    } else {
                        hi = mid;
                    }
                }
                out.println(res);
            }
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static boolean isPossible(int mid, int[] lps, int t,String s) {
        int count = 1;
        for (int i = mid + mid - 1; i < lps.length;) {
            if (lps[i] >= mid  && s.charAt(i-mid+1)-'a'==s.charAt(0)-'a') {
                count++;
                i += mid;
            } else {
                i++;
            }
        }
        return count >= t;
    }

    public static void computeLPSArray(String pat, int M, int lps[]) {
        int len = 0;
        int i = 1;
        lps[0] = 0;
        while (i < M) {
            if (pat.charAt(i) == pat.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            } else {
                if (len != 0) {
                    len = lps[len - 1];
                } else {
                    lps[i] = len;
                    i++;
                }
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
a
2 2 2
aa
3 3 3
abc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
const int N=1e6,M=1e6+20;
using namespace std;
int n,l,r,pos,mid,ans,ll,rr;
string s;
void solve() {
	cin>>n>>ll>>rr;
	cin>>s;
	for(int x=ll; x<=rr; x++) {
		l=0;
		r=n;
		while(l<r) {
			mid=(l+r+1)>>1;
			pos=-1;
			ans=0;
			for(int i=0; i<n; i++) {
				for(int j=0;j<mid;j++){
					if(s[i+j]!=s[j])break;
					if(j==mid-1)ans++,i+=j;
				}
			}
			if(ans>=x)l=mid;
			else r=mid-1;
		}
		cout<<r<<endl;

	}

}
signed main() {
	int tt;
	tt=1;
	cin>>tt;
	while(tt--)solve();
}
/*
7
3 1 3
aba
3 1 3
aaa
7 1 7
abacaba
9 1 9
abababcab
10 1 10
codeforces
9 1 9
abafababa
5 1 5
zpozp

*/

**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 5 5
abcde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
def getZarr(string, z):
    n = len(string)
    # [L,R] make a window which matches
    # with prefix of s
    l, r, k = 0, 0, 0
    for i in range(1, n):
        # if i>R nothing matches so we will calculate.
 # Z[i] using naive way.
        if i > r:
            l, r = i, i
            # R-L = 0 in starting, so it will start
            # checking from 0'th index. For example,
            # for "ababab" and i = 1, the value of R
            # remains 0 and Z[i] becomes 0. For string
            # "aaaaaa" and i = 1, Z[i] and R become 5
            while r < n and string[r - l] == string[r]:
                r += 1
            z[i] = r - l
            r -= 1
        else:
            # k = i-L so k corresponds to number which
            # matches in [L,R] interval.
            k = i - l
            # if Z[k] is less than remaining interval
            # then Z[i] will be equal to Z[k].
            # For example, str = "ababab", i = 3, R = 5
            # and L = 2
            if z[k] < r - i + 1:
                z[i] = z[k]
            # For example str = "aaaaaa" and i = 2,
            # R is 5, L is 0
            else:
                # else start from R and check manually
                l = i
                while r < n and string[r - l] == string[r]:
                    r += 1
                z[i] = r - l
                r -= 1
 # prints all occurrences of pattern
# in text using Z algo
def search(text, pattern,t):
 # Create concatenated string "P$T"
    if len(pattern) ==  0:
        return True
    cnt = 0
    concat = pattern + "$" + text
    l = len(concat)
 # Construct Z array
    z = [0] * l
    getZarr(concat, z)
 # now looping through Z array for matching condition
    i=0
    while i<l:
        # if Z[i] (matched region) is equal to pattern
        # length we got the pattern
        if z[i] == len(pattern):
            #print("Pattern found at index", i - len(pattern) -1)
            cnt += 1
            i +=len(pattern)
            continue
        i+=1
    if cnt >= t:
        return True
    return False
#import heapq
#from collections import defaultdict
#import math as m
t = int(input())
for _ in range(t):
    n,k,_= map(int,input().split())
    s = input()
    if k == 1:
        print(n)
        continue
    l = 0
    r = n//k
    ans = 0
    while l<=r:
        mid = l + (r-l)//2
        p = s[:mid]
        t = s[mid:]
        if search(t,p,k-1):
            l = mid + 1
            ans = mid
        else:
            r = mid - 1
    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 5 5
abcde
6 6 6
aaaaaa
7 7 7
abacaba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

Buggy Code:
from sys import stdin
import re

def f(s, k):
    n = len(s)
    if k == 1:
        return n
    low, high = 0, n // k + 1
    while low + 1 < high:
        mid = (low + high) // 2
        cnt = 0
        for m in re.finditer(s[:mid], s):
            cnt += 1
            if cnt == k:
                low = mid
                break
        else:
            high = mid
    return low

n_tests = int(stdin.readline())
for test_id in range(n_tests):
    n, l, r = map(int, stdin.readline().split())
    s = stdin.readline().strip()
    print(f(s, l))

**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
6 3 3
aaaaaa
6 3 3
abcabc
6 3 3
ababac
```

**Explanation:**

1. **Test Case 1: "aaaaaa"**
   - **Input:** `n = 6`, `l = 3`, `r = 3`, `s = "aaaaaa"`
   - **Expected Output:** `2`
   - **Reason:** The string "aaaaaa" can be divided into 3 substrings in multiple ways, but the longest common prefix (LCP) will always be "aa" (length 2).

2. **Test Case 2: "abcabc"**
   - **Input:** `n = 6`, `l = 3`, `r = 3`, `s = "abcabc"`
   - **Expected Output:** `3`
   - **Reason:** The string "abcabc" can be divided into 3 substrings in multiple ways, but the longest common prefix (LCP) will always be "abc" (length 3).

3. **Test Case 3: "ababac"**
   - **Input:** `n = 6`, `l = 3`, `r = 3`, `s = "ababac"`
   - **Expected Output:** `2`
   - **Reason:** The string "ababac" can be divided into 3 substrings in multiple ways, but the longest common prefix (LCP) will be "ab" (length 2).

These test cases are designed to cover different scenarios:
- A string with all characters the same.
- A string with a repeating pattern.
- A string with a pattern that changes at the end.

These scenarios should help expose any bugs in the faulty solution, especially in the handling of the binary search and the regular expression matching.

3
6 3 3
aaaaaa
6 3 3
abcabc
6 3 3
ababac


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version $$$l=r$$$.

You are given a string $$$s$$$. For a fixed $$$k$$$, consider a division of $$$s$$$ into exactly $$$k$$$ continuous substrings $$$w_1,\dots,w_k$$$. Let $$$f_k$$$ be the maximal possible $$$LCP(w_1,\dots,w_k)$$$ among all divisions.

$$$LCP(w_1,\dots,w_m)$$$ is the length of the Longest Common Prefix of the strings $$$w_1,\dots,w_m$$$.

For example, if $$$s=abababcab$$$ and $$$k=4$$$, a possible division is $$$\color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab}$$$. The $$$LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab})$$$ is $$$2$$$, since $$$ab$$$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $$$f_l,f_{l+1},\dots,f_r$$$. In this version $$$l=r$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l = r \le n \le 2 \cdot 10^5$$$) — the length of the string and the given range.

The second line of each test case contains string $$$s$$$ of length $$$n$$$, all characters are lowercase English letters.

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 $$$r-l+1$$$ values: $$$f_l,\dots,f_r$$$.

Examples:
input:
7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp
output:
0
1
3
2
10
2
0

Note: In the first sample $$$n=k$$$, so the only division of $$$aba$$$ is $$$\color{red}a\color{blue}b\color{orange}a$$$. The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $$$\color{red}a\color{blue}a\color{orange}a$$$. Their longest common prefix is one.

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

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

    static long mod = (long) 1e9 + 7;

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

        return res;
    }

    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        //Random in=new Random();

        int t = in.nextInt();

        up:
        while (t-- > 0) {
            int n=in.nextInt(),lk=in.nextInt(),rk=in.nextInt();
            String s=in.next();
            int l=1,r=n;
            while(l<r){
                int mid=(l+r+1)/2;
                if(gets(mid,s,lk)) l=mid;
                else r=mid-1;
            }
            if(!get(1,s,lk)) l=0;
            up.println(l);
        }

        up.flush();
    }
    public static int gcd(int x,int y){
        return x%y==0?y:gcd(y,x%y);
    }

    public static boolean get(int m,String s,int k) {
        String f=s.substring(0,m);
        int[] ne=new int[f.length()];
        ne[0]=-1;
        int j=-1;
        for(int i=1;i<f.length();i++){
            while(j>=0&&f.charAt(j+1)!=f.charAt(i)) j=ne[j];
            if(f.charAt(j+1)==f.charAt(i)) j++;
            ne[i]=j;
        }
        j=-1;
        int jk=0;
        for(int i=0;i<s.length();i++){
            while(j>=0&&f.charAt(j+1)!=s.charAt(i)) j=ne[j];
            if(f.charAt(j+1)==s.charAt(i)) j++;
            if(j==f.length()-1){
                jk++;
                j=-1;
            }
        }
        return jk>=k;
    }
    public static boolean gets(int m,String s,int k) {
        String f=s.substring(0,m);
        int kj=0,p=0;
        while(true){
            int u=s.indexOf(f,p);
            if(u==-1) break;
            p=u+m;
            kj++;
            if(kj>=k) break;
        }
        return kj>=k;

    }

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

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

}

class node {
    int iu;
    int jh;
    int ku;

    public node(int sg, int q, int kj) {
        jh = q;
        iu = sg;
        ku = kj;
    }
}

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

    //static BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
    static double nextDouble() throws IOException {
        streamTokenizer.nextToken();
        return (double) streamTokenizer.nval;
    }

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, let's consider the following points:

1. **Edge Cases**:
   - Strings with all identical characters.
   - Strings with unique characters.
   - Strings with a repetitive pattern.
   - Strings that are just long enough to trigger edge cases in the algorithm.

2. **Test Case Structure**:
   - The input format is:
     ```
     t
     n l r
     s
     ```
   - Where:
     - `t` is the number of test cases.
     - `n` is the length of the string.
     - `l` and `r` are the range of `k` values (in this case, `l = r`).

3. **Challenging Scenarios**:
   - A string with all identical characters to test the LCP calculation.
   - A string with a repetitive pattern to test the division and LCP calculation.
   - A string with unique characters to test the edge case where the LCP is zero.

Given these considerations, let's create a small-scale test input:

```plaintext
4
3 3 3
aaa
4 4 4
abcd
6 6 6
ababab
5 5 5
abcde
```

### Explanation:
1. **Test Case 1**:
   - `n = 3`, `l = 3`, `r = 3`, `s = "aaa"`
   - The only division is `a a a`, so the LCP is 1.

2. **Test Case 2**:
   - `n = 4`, `l = 4`, `r = 4`, `s = "abcd"`
   - The only division is `a b c d`, so the LCP is 0.

3. **Test Case 3**:
   - `n = 6`, `l = 6`, `r = 6`, `s = "ababab"`
   - The only division is `a b a b a b`, so the LCP is 1.

4. **Test Case 4**:
   - `n = 5`, `l = 5`, `r = 5`, `s = "abcde"`
   - The only division is `a b c d e`, so the LCP is 0.

This test input covers a variety of scenarios, including strings with identical characters, unique characters, and a repetitive pattern, which should help in exposing any bugs in the given code.

4
3 3 3
aaa
4 4 4
abcd
6 6 6
ababab
5 5 5
abcde


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
#include <algorithm>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdint>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

void solve() {
    int n, k;
    cin >> n >> k;
    // n = k^a + k^b + k^c
    // can divide out k repeatedly
    if (k == 1) {
        cout << n << "\n";
        return;
    }

    int ops = 0;
    while (n > 0) {
        int count = log10(n) / log10(k);
        ops++;
        n -= pow(k, count);
    }
    cout << ops << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    cin >> t;
    while (t--)
        solve();
}

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

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

Think step by step.

Response:
```plaintext
5
1 2
2 2
3 2
4 2
5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
/*Template by : Melag & Megu*/
#include <bits/stdc++.h>
using namespace std;

/*Pragma*/
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
//#pragma GCC target("sse,sse2,ssse3,sse4,abm,mmx,avx,fma")

/*Global Shortcut*/
//#define int long long
#define nl "\n"
//#define endl "\n"
#define fl flush

/*Shortcut*/
#define p(type1,type2) pair<type1,type2>
#define vc(type) vector<type>
#define PQ(type) priority_queue<type>
#define PQ2(type) priority_queue<type, vector<type>, greater<type>>
#define ft first
#define sc second
#define ts(val) to_string(val)
#define ti(val) stoi(val)

/*Contaiiner Shortcut*/
#define mmst(array,value) memset(array,value,sizeof(array))
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define Pb push_back
#define pB pop_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound

/*Typedeff Re-define*/
typedef unsigned long long ull;
typedef long double ld;
typedef long long ll;
typedef char ch;
typedef string str;

/*Decimal*/
#define decimal(value) cout << fixed << setprecision(value);

/*Constant*/
const ll N = 5e2 + 1e1 + 10;
const ll MOD =  998244353;
const int mod = 1e9 + 7;
const ll inf = 2e18 + 20;
const ld eps = 1e-9;

/*Traversal Moving*/
//int dx[] = {-1,1,0,0};
//int dy[] = {0,0,-1,1};
//str dz[] = {"U","D","L","R"};

/*-------------------------------------------------------------------------------------------------------*/
ll binex(const ll &a, ll b){
    if(b==0) return 1;
    ll res = binex(a,b/2);
    (b%2)?res=res*res*a:res=res*res;
    return res;
}
ll log(const ll &n, const ll &k) {return log2(n)/log2(k);}
void f(ll &n, const ll &k) {
    ll res = log(n,k);

    n -= binex(k,res);
    return;
}
void solve() {
    ll n,k; cin >> n >> k;
    if(k==1) cout << n << nl;
    else {
        ll ans = 0;
        while(true) {
            if(n<k) {ans+=n; break;}
            f(n,k), ans++;
        }
        cout << ans << nl;
    }
    return;
}

int32_t main(int32_t argc, char *argv[]) {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);

    int t=1; cin >> t;
    while(t-- > 0) 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
5
1 2
2 2
3 2
4 2
5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

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

int min_operations(long long n, long long k) {
    long long operations = 0;

    while (n > 0) {
        if(k==1)
        {
            operations = n;
            break;
        }
        long long x = floor(log(n)/log(k));
        n=n-pow(k,x);
        operations++;
    }
    return operations;
}

int main() {
    long long t;
    cin >> t;
    while (t--) {
        long long n, k;
        cin >> n >> k;
        cout << min_operations(n, k) << 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
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
import math

for _ in range(int(input())):
    n, k = map(int, input().split())
    ans = 0
    while n!=0:
        if n>k and k!=1:
            n -= k**(math.log(n)//math.log(k))
            ans+=1
        elif n==k:
            ans+=1
            break
        else:
            ans += n
            break
    print(int(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
5
1 1
2 2
3 3
4 4
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
import math

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    count = 0

    while n > 0:
        if k > 1 and k < n:
            # Calculate the largest power of k that is less than or equal to n
            x = int(math.log(n, k))
            n -= k ** x
        else:
            n -= 1
        count += 1

    print(count)

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

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

Think step by step.

Response:
```plaintext
5
1 2
2 2
3 2
4 2
5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
import sys
input = lambda: sys.stdin.readline().rstrip('\r\n')
sint = lambda: int(input())
mint = lambda: map(int, input().split())
aint = lambda: list(map(int, input().split()))
###############################################

def solve():
    n, k = mint()
    if n < k or k == 1:
        return n
    if n % k == 0:
        return 1
    count = 0
    while n > 0:
        count += (n % k)
        n //= k
    return count

t = sint()
for _ in range(t):
    ans = solve()
    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
5
1 2
2 2
3 2
4 2
10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
import java.util.*;

public class Solution {
    public static final Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int t = sc.nextInt();
        for (int i = 1; i <= t; i++) {
            solve(i);
        }
    }

    static void solve(int test_case) {
        int n = sc.nextInt();
        int k = sc.nextInt();

        if (k > n) {
            System.out.println(n);
            return;
        }
        if (k == n) {
            System.out.println(1);
            return;
        }
        int copyN = n;
        int count = 0;
        while(k < copyN){
            int v = binarySearch(k, copyN);
            copyN-=(int)Math.pow(k, v);
            count++;
        }
        count+=(copyN);

        System.out.println(count);
    }

    static int binarySearch(int k, int n) {
        int start = 1;
        int end = n;

        while (start <= end) {
            int mid = start + (end - start) / 2;

            long val = (long) Math.pow(k, mid);
            if (val <= n) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return end;
    }

    static void print(int c, int o) {

        System.out.println("case " + c + ": " + o);

    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 2
2 2
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Tests {

    static void sorts(long arr[], long index[]) {
        int n = arr.length;

        for (int i = 0; i < n - 1; i++) {

            int min_idx = i;
            for (int j = i + 1; j < n; j++) {
                if (arr[j] < arr[min_idx]) {
                    min_idx = j;
                }
            }

            long temp = arr[min_idx];
            arr[min_idx] = arr[i];
            arr[i] = temp;

            long tempindex = index[min_idx];
            index[min_idx] = index[i];
            index[i] = tempindex;
        }
    }

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

        int t = scan.nextInt();

        while ( t-- > 0 ) {
            long n = scan.nextLong();
            long k = scan.nextLong();

            long  mod = 0 ;

            if ( k == 1 ) {
                System.out.println(n);
                return ;
                }

            while ( n >= k ) {
                 mod += n%k ;
                 n = n/k;
                }
            System.out.println(mod+n);

        }

    }
}

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

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

Think step by step.

Response:
```plaintext
1
1000000000 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

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

#define fn for(int i=0; i<n; i++)
#define ll long long
#define vi vector<int>
#define vii vector<pair<int,int>>
#define PB emplace_back
#define F first
#define S second
#define maxe max_element
#define mine min_element
#define endl '\n'
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()

ll powr(ll a, ll n){
    ll res = 1;
    while(n){
        if(n%2)
            res=(res)*(a);
        a=(a)*(a),n/=2;
    }
    return (res);
}

void test_case(){
    ll n, k;
    cin >> n >> k;
    if(k>n or k==1){
        cout << n << endl;
    }
    else if(k==n)
        cout << 1 << endl;
    else{
        ll step = 0;
        while(n){
            ll q = k;
            if(n<k){
                n -= 1;
                step++;
                continue;
            }
            if(n>=k){
                ll lo = log(n) / log(k);
                q = pow(k, lo);
            }
            // cout << n << " " << q << endl;
            n -= q;
            step++;
        }
        cout << step << endl;
    }
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t=1;
    cin>>t;
    while(t--){
        test_case();
    }
    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
1000000000 2
1000000000 10
1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
#include <bits/stdc++.h>
#define FAST() ios_base::sync_with_stdio(false); cin.tie(NULL);
#include <cmath>
#include <string>
using namespace std;
using ll = long long;
using ld = long double;
using ii = pair<int, int>;
using vii = vector<ii>;
using vi = vector<int>;
#define range(v) ((v).begin()), ((v).end())
#define IN() freopen("input.tvt", "r", stdin);
#define OUT() freopen("output.tvt", "w", stdout);
#define inf 1000000000
ll mod = 1e9+7;

ll fast(ll  num, ll mo)
{
    ll res=1;

    while (mo > 0)
    {
       // cout<<res<<endl;
        if (mo%2==0)
            {mo = mo/2;
            if(num>1e3)
            {
                return 1e9;
            }
            num = (num * num)%mod;}
        else
            {

            mo = mo - 1;
            res = (res * num)%mod;

            }}

    return res%mod;
}
ll bin(ll x,ll y)
{
    ll ma=x;
    ll mi=0;
    ll mid;
    while(ma>=mi)
    {
       mid=(ma+mi)/2;
       ll r=fast(y,mid);
     //  cout<<r<<endl;

       if(r==0)
       {
           break;
       }
       if(r>x)
       {
           ma=mid-1;
       }
       else if(r<x)
       {
           mi=mid+1;
       }
       else{
        return r;
       }
       }
    if(fast(y,ma)<x)
    return fast(y,ma);
    else
        return fast(y,mid);

}
void solve()
{
    ll x,y;
    cin>>x>>y;
    ll res=0;

  while(x>0)
  {
      ll l = bin(x,y);

      if(l==1)
      {
          res+=x;
          break;
      }
      x=x-l;
      //cout<<l<<endl;
      //cout<<x<<endl;
      res++;
  }
   cout<<res<<endl;

}

int main()
{
    // IN();
    // OUT();
    FAST();
    int t;
    cin >> t;
    // t = 1;
    while (t--)
    {
        solve();
    }
    return 0;

}

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

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

Think step by step.

Response:
```plaintext
1
1000000000 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
from math import *

def sol(n, k):
    if k == 1:
        return n
    count = 0
    while n:
        n -= k**(int(log(n, k)))
        count += 1
    return count

t = int(input())
while t:
    # n = int(input())
    n, k = list(map(int, input().split()))
    # try:
    ans = sol(n, k)
    print(ans)
    # print("Ans: ", ans)
    # except Exception as e:
    #     print(e)
    t -= 1

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

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

Think step by step.

Response:
```plaintext
4
1 1
1000000000 2
1000000000 10
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Find Minimum Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $n$ and $k$.

In one operation, you can subtract any power of $k$ from $n$. Formally, in one operation, you can replace $n$ by $(n-k^x)$ for any non-negative integer $x$.

Find the minimum number of operations required to make $n$ equal to $0$.

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

The only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 10^9$).

Output: For each test case, output the minimum number of operations on a new line.

Examples:
input:
6
5 2
3 5
16 4
100 3
6492 10
10 1
output:
2
3
1
4
21
10

Note: In the first test case, $n = 5$ and $k = 2$. We can perform the following sequence of operations:

  1. Subtract $2^0 = 1$ from $5$. The current value of $n$ becomes $5 - 1 = 4$.    2. Subtract $2^2 = 4$ from $4$. The current value of $n$ becomes $4 - 4 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $2$ operations. Thus, $2$ is the answer.

In the second test case, $n = 3$ and $k = 5$. We can perform the following sequence of operations:

  1. Subtract $5^0 = 1$ from $3$. The current value of $n$ becomes $3 - 1 = 2$.    2. Subtract $5^0 = 1$ from $2$. The current value of $n$ becomes $2 - 1 = 1$.    3. Subtract $5^0 = 1$ from $1$. The current value of $n$ becomes $1 - 1 = 0$.

It can be shown that there is no way to make $n$ equal to $0$ in less than $3$ operations. Thus, $3$ is the answer.

Buggy Code:
def min_operations_optimized(n, k):
    operations = 0
    while n > 0:
        # 使用二分查找找到最大的x，使得k^x <= n
        left, right = 0, 60  # 假设k^60对于所有k都足够大
        while left < right:
            mid = (left + right + 1) // 2
            power_k = k ** mid
            if power_k <= n:
                left = mid
            else:
                right = mid - 1

        # 执行操作：n -= k^left
        n -= k ** left
        operations += 1

    return operations

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    t = int(data[0])
    index = 1

    for _ in range(t):
        n = int(data[index])
        k = int(data[index + 1])
        index += 2

        print(min_operations_optimized(n, k))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
1 1
1000000000 2
1000000000 10
```