**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
for _ in range(int(input())):
    (n,k)=map(int,input().split())
    a=list(map(int,input().split()))
    x=a[k-1]
    k-=1
    c1=0
    if(k!=0):
        if(a[k]>a[k-1]):
            c1+=1
        else:
            print(0)
            continue
    for j in range(k+1,n-1):
            if(a[k]>a[j]):
                c1+=1
            else:
                break
    m=a.index(min(a))
    if(m==0):
        t=a[m+1]
        a[m+1]=a[k]
        a[k]=t
        k=m+1
    elif(m==n-1):
        t=a[m-1]
        a[m-1]=a[k]
        a[k]=t
        k=m-1
    #swap(a[k],max(a[m-1],a[m+1]))
    elif(a[m-1]>a[m+1]):
        t=a[m-1]
        a[m-1]=a[k]
        a[k]=t
        k=m-1
    else:
        t=a[m+1]
        a[m+1]=a[k]
        a[k]=t
        k=m+1
    c=0
    if(k!=0):
        if(a[k]>a[k-1]):
            c+=1
        else:
            print(0)
            continue
    for j in range(k+1,n-1):
            if(a[k]>a[j]):
                c+=1
            else:
                break
    print(max(c,c1))

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

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

Think step by step.

Response:
```plaintext
3
5 1
5 4 3 2 1
5 3
1 2 3 4 5
4 2
10 20 30 40
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
import javax.imageio.ImageTranscoder;
import java.util.*;
import java.io.*;
public class gr_25 {
    static Scanner sc =new Scanner(System.in);
    static PrintWriter pw =new PrintWriter(System.out);
    static int mod = (int) (1e9  + 7);
    public static void main(String[] args) throws IOException {
        int t =sc.nextInt();
        while(t-->0){
            proB();
        }
        pw.flush();
    }

    public static void proC() throws IOException{}

    public static void proB() throws IOException{
        int n =sc.nextInt();int k =sc.nextInt()-1;
        int []arr=sc.nextIntArray(n);
        int first =n;
        for(int i=0;i<n;i++){
            if(arr[i]>arr[k])
            {
                first=i;
                break;
            }
        }
        if(first==n)
        {
            pw.println(n-1);
            return;
        }
        int ans=Math.max(first-1,0);
        int temp = arr[k];
        arr[k]=arr[first];
        arr[first]=temp;
        int x=0;
        if(first>0 && arr[first]>arr[first-1])
            x++;

        for(int i=first;i<k;i++) {
            if(arr[first]>arr[i])
                x++;
        }
        ans=Math.max(ans , x);
        pw.println(ans);

    }
    public static void proA() throws IOException{
        int n=sc.nextInt();
        char[]s=sc.next().toCharArray();
       int x=0;
       ArrayList<Integer> arr=new ArrayList<>();
       for(int i=0;i<n;i++){
           if(s[i]=='1')
           {
               x++;
               arr.add(i);
           }
       }

       if(x==2){
           int a=arr.remove(0);
           int b=arr.remove(0);
           if(a+1 == b)
           {
               pw.println("NO");
               return;
           }

       }

       pw.println(x%2==0? "YES":"NO");
    }

    static long GCD(long u, long v)
    {
        if (u == 0)
            return v;
        return GCD(v % u, u);
    }

    // LCM of two numbers
    static long LCM(long u, long v)
    {
        return (u / GCD(u, v)) * v;
    }

    public static ArrayList<Long> divisors (long n){

        ArrayList<Long> ans =new ArrayList<>();

        for(long i=1;i*i<=n;i++){
            if(n%i==0){
                ans.add(i);
                if(n/i !=i)
                    ans.add(n/i);
            }

        }
        return ans;
    }

    static class Pair implements Comparable{
        long first;
        long second;
        public Pair(long first,long second){
            this.first=first;
            this.second=second;

        }

        @Override
        public int compareTo(Object o) {
            Pair p =(Pair) o;
            return (int) (Math.abs(second)-Math.abs(p.second));
        }
        public String toString(){
            return "("+first+" "+second+")";
        }
    }

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

            return b;

        return gcd(b % f, f);
    }
    static ArrayList<Integer> Divisors(int n)
    {
        // Note that this loop runs till square root
        ArrayList<Integer>ans=new ArrayList<>();
        for (int i=1; i<=Math.sqrt(n); i++)
        {
            if (n%i==0)
            {

                ans.add( i);

                if((n/i) !=i)
                    ans.add( n/i);
            }
        }
        return ans;
    }

    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

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

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

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

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

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

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are the steps to think through the problem:

1. **Identify the Bug**: The code seems to have a logic issue in determining the optimal swap position. It may not correctly handle cases where the cow's initial position is already optimal or where the optimal swap position is not the first higher-rated cow.

2. **Edge Cases**:
   - **Initial Position is Optimal**: If the cow is already in a position where it can win the maximum number of matches without swapping.
   - **Optimal Swap Position**: The cow should be swapped to a position where it can win more matches, but the code might not correctly identify this position.
   - **Single Match**: When there are only two cows, the code should handle this case correctly.
   - **High Ratings**: When the cow has a very high rating and can win all matches.
   - **Low Ratings**: When the cow has a very low rating and cannot win any matches.

3. **Small-Scale Test Input**:
   - We will create a test case where the cow's initial position is not optimal, and the optimal swap position is not the first higher-rated cow.

Let's create a test input that exposes the bug:

```plaintext
1
5 3
10 15 20 12 18
```

### Explanation:
- **Test Case**: `5 3` (5 cows, cow 3 is the target)
- **Ratings**: `10 15 20 12 18`
  - Cow 3 has a rating of 20.
  - The initial order is `10, 15, 20, 12, 18`.
  - The optimal swap position for cow 3 is to swap with cow 4 (rating 12), as it can win more matches from this position.

### Expected Output:
- The expected output for this test case is `2`:
  - After swapping, the order becomes `10, 15, 12, 20, 18`.
  - Cow 3 (now in position 4) wins against the cow with rating 12.
  - Cow 3 then loses to the cow with rating 18.

This test input should expose the bug in the code if it does not correctly handle the optimal swap position.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
import java.util.Scanner;
import java.util.Stack;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int ans[] = new int[N];
//        sc.nextLine();
        for(int i=0; i<N; i++){
            int n = sc.nextInt();
            int ind = sc.nextInt();
            int arr[] = new int[n];
            for(int j=0; j<n; j++){
                arr[j] = sc.nextInt();
            }
            ans[i] = f(arr,ind);
        }
        for(int num:ans){
            System.out.println(num);
        }
    }
    static int f(int[]arr, int ind){
        int num = arr[ind-1];
        int cnt = 0;
        int i = 0;
        boolean contains = false;
        while(i<arr.length && arr[i]<=num){
            if(arr[i]==num){
                i++;
                contains = true;
                continue;
            }
            cnt++;
            i++;
        }
        i++;
        if(contains) return cnt;
        int cnt2 = 0;
        while(i<arr.length && arr[i]<=num){
            if(arr[i]==num){
                i++;
                continue;
            }
            cnt2++;
            i++;
        }

        return Math.max(cnt,cnt2);
    }
}

**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 2
10 5 20
4 1
1 2 3 4
5 3
10 20 30 40 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
/* Author : mylittlepookybear  */
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD (1000000000+7)
#define MOD1 (998244353)
#define pb push_back
#define all(x) x.begin(), x.end()
#define en cout << '\n'
const int N = 1e6+100, M = 1e5+10, K = 52, MX = 30;

int n, k, a[N];
void solve(){
  cin >> n >> k;
  for(int i = 1; i <= n; ++i) cin >> a[i];
  int ans = 0, x=  -MOD;
  for(int i = 1; i <= n; ++i){
    if(a[i] < a[k]) ++ans;
    else if(a[i] > a[k]){
      if(k > i) --ans;
      if(k < i)
        x = -MOD;
      else
        x = (i>1);
      swap(a[i], a[k]);
      k = i;
      break;
    }
  }
  for(int i = k + 1; i <= n; ++i){
    if(a[i] < a[k]) ++x;
  }
  cout << max(x, ans);
}

int main(){
  cin.tie(0); ios::sync_with_stdio(0);
  int tt = 1, aa;
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  cin >> tt;
  while(tt--){
    solve();
    en;
  }
  cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" seconds\n";
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
10 20 30 40 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
/*
⡯⣽⢩⠯⣝⣭⣻⢭⠷⡽⣚⡳⣝⢫⡟⡽⢫⣽⣭⣷⣽⣿⣿⣿⣿⣿⣿⣭⣯⣝⣯⢻⣝⡻⣽⢻⣭⣛⠿⣭⢯⡽⣯⣽⢭⢯⡝⡯⣝⢯
⢷⣊⠷⣫⠝⢦⡟⣬⠳⣱⢣⠳⣌⢧⣛⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣯⣯⡝⣳⣝⡳⢮⡹⣞⡵⣫⢞⣵⢮⢏⡷⣹⢳⣛⢮
⢇⢎⡱⠥⡛⢆⠯⣐⢣⡑⢎⡱⢌⡼⣫⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣾⡷⣜⣟⢧⣛⠶⣭⣓⢯⣚⡯⡞⣵⢫⠷⣹⢎
⠊⠔⠨⠑⢌⡺⠑⠌⢢⠘⠤⢘⠸⣹⣿⣿⣟⣿⣟⡿⠟⡋⠟⣻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣯⣟⢮⣭⢛⡶⣹⢎⡷⣫⣝⢮⣏⡽⣣⢯
⠀⠀⠁⠈⠂⡅⠌⡠⢁⣌⠐⠄⢢⣿⣟⣷⣿⣻⣾⣏⠐⠀⠀⠀⢿⣹⣿⣿⣿⣿⣿⣿⣿⣿⣷⢿⡾⡴⢯⡞⣵⣫⢶⣹⢎⡷⣪⢵⣫⢞
⠀⠀⠀⠀⢰⠀⠀⠀⠀⠈⠐⠂⢸⢿⠮⣽⡿⣽⣾⣿⡆⠀⠀⠀⢹⢛⣿⣸⢿⣿⣿⣿⣿⣿⣿⣯⡿⣝⠮⣝⠶⣹⢮⡝⣯⢳⣛⢮⠷⣞
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢦⢺⢇⡹⡿⣻⡗⢷⡫⡠⠀⠀⠀⠿⣷⣾⡟⢻⣿⣿⣿⣿⣟⣛⡷⢎⡻⢬⣛⣥⢳⡝⣾⢣⡟⢮⡻⣜
⠀⠀⠀⠀⠄⠀⠀⠀⠀⠀⠀⠀⣸⢃⡿⡔⣇⢙⠭⣶⢏⠐⠀⠀⠀⠀⠈⠁⠀⢠⡿⢿⡿⣿⣿⡹⢾⣡⢏⡳⠼⣬⢳⡹⢞⣧⣛⢧⡻⣜
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⢋⢾⠟⣷⣜⡌⠀⠉⠀⠀⠀⢦⠀⠀⠀⠀⠀⠉⢸⣙⡽⣭⢏⡿⡘⣆⢣⡝⢳⢬⠳⣭⢫⣗⠎⡷⣹⢎
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠁⠠⠚⠋⡽⠒⡙⠂⠀⠀⠀⠀⠈⠀⠀⠀⠀⠀⠀⣰⣾⠜⡡⢏⠲⢉⡇⠲⣌⠳⢎⡳⢥⡳⣹⢶⣙⠧⣏
⠀⠀⠀⠆⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠠⠤⠮⢔⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⡿⠼⢯⠴⠈⠐⣀⣇⠱⣈⣧⡙⡜⢣⢳⠹⣎⠮⣝⢮
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⠲⣄⣀⠀⠀⠀⠀⠀⠊⣸⣁⡀⠈⠆⠀⠀⠀⢠⡒⠠⢂⠜⡩⢏⢎⡛⡧⣛⢎⠶
⠀⠀⢀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⠻⣇⠀⠀⠐⠀⠀⡀⡋⠀⠰⣤⣀⠀⠀⠀⠸⠀⠑⠨⢠⠁⠎⡰⢈⢗⠌⡎⡱
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣀⣀⠀⠀⣠⢴⣰⢿⡏⢀⡁⢅⠀⠀⠀⠸⢀⠁⡄⠀⢹⣶⣶⢶⣄⡀⠀⠀⠀⠀⠁⠂⡠⠁⢺⠈⠔⠡
⠀⠀⠁⠀⠀⠀⠀⠀⣠⢞⠯⣯⣙⢦⢫⡽⡣⢎⣳⢿⠀⣼⡌⢐⢤⡀⡠⠂⢠⣴⡷⢄⠈⣷⣯⡷⡝⠿⣳⠦⣤⣀⡀⠀⠀⠁⢀⠇⠈
⠀⠀⠀⠀⠀⢀⣴⣿⣭⣞⣧⣛⣿⣮⣏⣰⢱⢫⣼⣇⠨⠘⣾⢦⣄⣄⢀⣼⣯⠟⠁⡄⠑⣽⢆⣳⠽⣨⢧⣛⠶⣝⣻⣷⡀⠀⠀⡀
⠀⠀⢀⡤⣶⢻⣝⢯⡞⣽⡶⠋⢋⡀⠐⠈⠱⣗⠾⡏⠀⠀⠀⢸⣯⣻⣽⠏⠐⠀⠀⢀⠘⢸⡞⢦⣉⡓⠧⣝⣺⡱⣏⣿⡆⠀⠀⠁
⠀⢀⣾⣷⣿⢻⣾⣽⣾⣏⡈⡀⡀⠀⠀⠀⠀⠈⢛⣧⠠⠄⠀⠈⣷⣯⠏⠀⠀⠀⠀⠈⠀⠈⣟⡼⢲⡹⠀⣞⡴⣻⡜⣾⡗⠀⠀⠘
⠀⣾⣽⣿⢹⣏⡟⣧⠳⣜⢭⢿⡇⠀⠀⠀⠀⠀⠈⢿⣂⡀⠂⢠⣿⣻⢸⠀⠀⠀⠀⠃⠀⠀⡿⣜⢧⣛⠘⣯⣾⢷⣹⣿⣇⣀
⠀⣿⣿⣿⣛⢧⣫⣜⢻⡜⣮⢛⡇⠀⠀⠀⠀⠀⠀⠈⣧⠢⠀⢸⣷⣿⠈⠀⠀⠀⠈⠀⠀⠀⣻⡜⣧⢯⢀⣿⣯⡾⣏⣿⡇⢠
⠀⠈⠻⣧⠻⣜⠶⣩⢞⡵⡭⢯⡁⠀⠀⠀⠀⠀⠐⠀⠸⡆⢂⣿⣿⣾⡆⠀⠀⠀⠂⠀⠀⠀⢸⣝⡮⢷⢸⣿⢟⡵⣻⣿⡇⠀⡀⠀⡆
⠀⠀⣸⡯⣝⠮⣝⣣⢏⡾⣹⢷⠀⢠⠄⠀⡖⠀⢰⡇⠀⡆⠀⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⣟⣮⣽⢃⣿⣛⣮⠽⣿⣿⡇
⠀⠀⣿⡱⢎⡽⢲⠭⣶⢹⣳⣿⠀⢸⠀⠀⡿⠀⢸⡄⠀⠇⠀⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⢰⣟⡶⢣⣾⣳⡽⢮⣿⣿⣿⠀⠀⠇⠀⠰
⠀⢀⡷⣙⠮⣜⢭⠳⣎⡳⣽⣿⡀⢸⣇⡀⣧⠀⣸⣅⠀⢧⠀⣿⣿⣿⡗⠀⠀⠀⠀⠀⠀⣼⡻⣣⣿⡾⣷⣛⣟⣾⢯⣿⠀⠀⠈
⠀⢸⠲⣍⠞⡜⡮⣝⡲⣝⣿⣿⣷⠈⢧⠀⡇⢠⣷⠃⢸⡏⢰⣿⣿⣿⡟⣠⠂⠀⠀⠀⣸⡿⣵⣻⣿⣽⢳⣯⢾⡽⣿⣿⠀⠈⠄⠀⠠
⠀⣍⡳⣌⠻⡼⣱⢎⡵⣾⣿⢻⣿⣷⣜⡇⡧⣸⡏⣠⣿⢳⣿⣿⣿⣿⡇⠁⠀⠀⠀⢀⡾⣼⣷⣿⣳⢯⣟⡾⣯⣽⣿⣿⠀⠡⢈⠐⠰
⢰⢣⢳⡜⣽⡱⢧⣋⢶⡿⣭⢻⣿⣿⣿⣿⣷⣿⣷⣿⢹⣿⣿⣿⣿⣿⢁⠂⠀⠀⠀⢃⣿⣿⣿⣿⣟⣯⢿⣽⣳⣿⣿⣏⢐⠒⡠⠬⢔⡀
⣨⢧⡓⡾⣥⣛⠧⣞⡿⣝⢮⣻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡄⡡⠔⠈⠀⢺⣹⣿⡿⣽⣻⣿⣿⣷⣿⣿⣿⡇⢨⠐⢠⠁⡂⠄
⡳⣎⣽⣳⣳⢭⣛⣼⢻⡜⣧⢿⣿⣿⣿⣿⣿⣿⣟⡫⠆⢸⣿⣿⣿⣿⣿⠁⠂⢐⣶⣿⣟⣷⣻⡽⣟⣿⣿⣿⣿⣿⣿⠄⠂⠌⠄⡂⢁⣎
⣷⣹⡶⣯⣟⢮⣳⣏⢷⡹⣞⣿⣿⣿⣿⣿⣿⣿⣿⡄⠁⣸⣿⣿⣿⣿⣿⣮⣿⢿⣻⢾⡽⣾⣽⣿⣿⣿⣿⣿⣿⣿⣯⣤⡿⣶⣶⣾⣻⢿
⣧⣿⢿⣽⢾⣣⢟⣮⡳⢯⣿⣿⣿⣿⣿⣿⣿⣿⣿⣏⢔⣿⣿⣿⣿⣿⡿⣿⡽⣯⣿⣿⣿⣿⣿⣿⣿⣿⣿⣻⣽⣳⣯⣳⣿⢿⣽⣾⣽⢯
⣿⣯⣿⢯⡿⣜⣯⢶⣛⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠸⠻⣿⡿⠿⢛⡛⠩⠻⢿⣷⡿⣞⣿⣿⣿⣿⣿⣞⣷⣿⣷⣿⣿⣟⣯⢷⣻⣞⡿

*/
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define gc getchar_unlocked
#define rep(i,a,b) for (int i = a; i < b; i++)
#define each(i, k, n) for (int i = k;i<=n;i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define endl '\n'
#define f first
#define s second
typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key
ll min(ll a,int b) { if (a<b) return a; return b; }
ll expo(ll a, ll b, ll mod) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
void take(vl &v) {for(auto i : v) cin >> i;}
ll min(int a,ll b) { if (a<b) return a; return b; }
ll phin(ll n) {
    ll number = n;
    if (n % 2 == 0) {
        number /= 2;
        while (n % 2 == 0) n /= 2;
    }
    for (ll i = 3; i <= sqrt(n); i += 2) {
        if (n % i == 0) {
            while (n % i == 0) n /= i;
            number = (number / i * (i - 1));
        }
    }
    if (n > 1) number = (number / n * (n - 1)) ;
    return number;
}
ll max(ll a,int b) { if (a>b) return a; return b; }
ll max(int a,ll b) { if (a>b) return a; return b; }
ll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }
ll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
ll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}
ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
string to_upper(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A'; return a; }
string to_lower(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A'; return a; }
bool prime(ll a) { if (a==1) return 0; for (int i=2;i<=round(sqrt(a));++i) if (a%i==0) return 0; return 1; }
#define mod 1000000007
vl fact(200005),ifact(200005);
const int N = 1e5 + 1;
void calculate_factorials() {
    fact[0] = 1;
    for (int i = 1; i < N; i++) {
        fact[i] = mod_mul(fact[i - 1], i, mod);
    }
    ifact[N - 1] = mminvprime(fact[N - 1],mod);
    for (int i = N - 2; i >= 0; i--) {
        ifact[i] = mod_mul(i + 1, ifact[i + 1], mod);
    }
}
ll nCr(int x, int y){
    if (y > x){return 0;}
    ll num = fact[x];num *= ifact[y];num %= mod;num *= ifact[x - y];num %= mod;return num;
}
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
#define sz(z) z.size()
inline void solve(){
    ll n,ind;
    cin >> n >> ind;
    ind--;
    vl v(n);
    rep(i,0,n){
        cin >> v[i];
    }
    ll res = 0;
    rep(i,0,n){
        if(v[i] > v[ind]){
            res = i;
            break;
        }
    }
    ll c = 0;
    if(ind > res){
        ll cur = res;
        ll val = (res == 0 ? 0 : 1);
        rep(i,res + 1,ind - 1){
            if(v[i] > v[ind]){
                break;
            }
            else val++;
        }
        cout << max(0,max(cur,val));
    }
    else{
        cout << res - 1;
    }
}
int main()
{
    // 私をより強くする

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        solve();
        cout<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the probability of failure. Here are the steps to think through this:

1. **Edge Cases:**
   - **Single Cow:** The smallest valid input with \( n = 2 \).
   - **Your Cow is the Strongest:** \( k \) is the cow with the highest rating.
   - **Your Cow is the Weakest:** \( k \) is the cow with the lowest rating.
   - **Your Cow is in the Middle:** \( k \) is in the middle of the ratings.

2. **Scenarios:**
   - **Optimal Swap for Maximum Wins:** Ensure the input can benefit from a swap.
   - **No Benefit from Swap:** Ensure the input where no swap can increase the number of wins.

3. **Small-Scale Test Input:**
   - We will focus on a small \( n \) to ensure the input is manageable and complete.

Let's consider the following test input:

- **Test Case 1:** \( n = 4 \), \( k = 2 \)
  - Ratings: \( [10, 15, 12, 14] \)
  - Your cow is in the middle, and a swap can potentially increase the number of wins.

- **Test Case 2:** \( n = 4 \), \( k = 1 \)
  - Ratings: \( [15, 10, 14, 12] \)
  - Your cow is the strongest, and no swap should be necessary.

- **Test Case 3:** \( n = 4 \), \( k = 3 \)
  - Ratings: \( [14, 12, 10, 15] \)
  - Your cow is the weakest, and a swap can potentially increase the number of wins.

Here is the test input in the required format:

```plaintext
3
4 2
10 15 12 14
4 1
15 10 14 12
4 3
14 12 10 15
```

This input will test the algorithm's ability to handle different positions of your cow and determine the optimal swap or no swap for maximum wins.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
def solve(arr, r, is_first):
  count = 0
  l = 0
  if is_first:
    for i in range(1, len(arr)):
      if arr[i] < r:
        count += 1
      else:
        break
    return count

  if len(arr) == 2:
    my_idx = arr.index(r)
    if arr[my_idx] > arr[1-my_idx]:
      return 1
    else:
      return 0

  idx = 0

  for i in range(len(arr)):
    if arr[i] < r:
      count += 1
    else:
      break
  idx = i
  my_idx = arr.index(r)
  arr[idx], arr[my_idx] = arr[my_idx], arr[idx]

  count = 1

  for i in range(idx+1, len(arr)):
    if arr[i] < r:
      count += 1
    else:
      break

  return count
  # print("Rating", r)

for _ in range(int(input())):
  n, k = map(int, input().split())
  ratings = list(map(int, input().split()))
  print(solve(ratings, ratings[k-1], k==1))

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

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

Think step by step.

Response:
```plaintext
3
4 2
5 1 10 2
4 1
1 10 5 2
2 1
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
for _ in range(int(input())):
    n,i = list(map(int,input().split()))
    arr= list(map(int,input().split()))
    myind=0
    for j in range(i):
        if arr[j]>arr[i-1]:
            arr[j],arr[i-1]=arr[i-1],arr[j]
            myind=j
            break
        else:
            myind=i-1

    ans=0

    def checkbefore(ans,myind,arr,n):
        for i in range(1,myind):

            if arr[0]<arr[i]:
                return False
        return True

    def checkafter(ans,myind,arr,n):
        count=0
        for i in range(myind+1,n):
            if arr[myind]>arr[i]:
                count+=1
            else:
                break
        return count

    if arr[0]<arr[myind]:

        if checkbefore(ans,myind,arr,n):
            ans+=1

    ans+=checkafter(ans,myind,arr,n)
    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 3
10 1 11 9 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
//package cp_practice;
import java.util.*;
public class battleCows {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int myCow=sc.nextInt();
            long arr[]=new long[n];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextLong();
            }
            int wins=0,i;
            for(i=0;i<n;i++){
                if(i+1==myCow)
                    break;
                if(arr[i]>arr[myCow-1]){
                    long p=arr[i];
                    arr[i]=arr[myCow-1];
                    arr[myCow-1]=p;
                    break;
                }
            }
            if(i==0)
                wins=0;
            else
                wins=1;
            for(int k=i+1;k<n&&arr[i]>arr[k];k++) {
                //System.out.println(arr[k]);
                wins++;
            }
            System.out.println(wins);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Battle Cows
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $$$n$$$ cows participating in a coding tournament. Cow $$$i$$$ has a Cowdeforces rating of $$$a_i$$$ (all distinct), and is initially in position $$$i$$$. The tournament consists of $$$n-1$$$ matches as follows:

- The first match is between the cow in position $$$1$$$ and the cow in position $$$2$$$.
- Subsequently, each match $$$i$$$ is between the cow in position $$$i+1$$$ and the winner of match $$$i-1$$$.
- In each match, the cow with the higher Cowdeforces rating wins and proceeds to the next match.

You are the owner of cow $$$k$$$. For you, winning the tournament is not important; rather, you want your cow to win in as many matches as possible. As an acquaintance of the tournament organizers, you can ask them to swap the position of your cow with another cow only once, or you can choose to do nothing.

Find the maximum number of wins your cow can achieve.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 10^5, 1 \le k \le n$$$) — the number of cows and your cow's index.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the Cowdeforces rating of the cows. It is guaranteed that $$$a_i$$$'s are pairwise different.

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

Output: For each test case, print one integer: the maximum number of wins cow $$$k$$$ can achieve if you choose to swap (or do nothing) optimally.

Examples:
input:
3
6 1
12 10 14 11 8 3
6 5
7 2 727 10 12 13
2 2
1000000000 1
output:
1
2
0

Note: In the first test case, it is optimal to do nothing. Let $$$a'$$$ be the Cowdeforces rating of the cows in the original order (with your cow's rating bolded), then

- Initially, $$$a' = [\mathbf{12}, 10, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 14, 11, 8, 3]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$14$$$ and loses.

In the second test case, it is optimal to swap your cow to position $$$3$$$. Then, let $$$a'$$$ be the Cowdeforces rating of the cows in the order after the swap.

- Initially, $$$a' = [7, 2, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against the cow with Cowdeforces rating $$$2$$$ and wins. $$$a' = [7, \mathbf{12}, 10, 727, 13]$$$.
- The cow with Cowdeforces rating $$$7$$$ plays against your cow, and your cow wins. $$$a' = [\mathbf{12}, 10, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$10$$$ and wins. $$$a' = [\mathbf{12}, 727, 13]$$$.
- Your cow plays against the cow with Cowdeforces rating $$$727$$$ and loses.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            int[] ratings = new int[n];
            for (int i = 0; i < n; i++) {
                ratings[i] = scanner.nextInt();
            }
            System.out.println(maxWins(n, k, ratings));
        }
    }

    private static int maxWins(int n, int k, int[] ratings) {
        int kRating = ratings[k - 1];

        boolean oneLow = false;
        for (int rating : ratings) {
            if (kRating > rating) {
                oneLow = true;
            }
        }
        if(!oneLow){
            return 0;
        }

        int swapWithFirstHigher = 0;
        int firstHigher = -1;
        for (int i = 0; i < n; i++) {
            if (ratings[i] > kRating) {
                firstHigher = i;
                break;
            }
        }

        if (firstHigher == -1) {
            return n - 1;
        }
        else if (firstHigher < k - 1) {
            ratings[k - 1] = ratings[firstHigher];
            for(int i = firstHigher + 1; i < n; i++){
                if(ratings[i] < kRating){
                    swapWithFirstHigher++;
                }
            }
            if(firstHigher > 0){
                swapWithFirstHigher++;
            }
            ratings[k - 1] = kRating;
        }

        int swapWithFirst = 0;
        ratings[k - 1] = ratings[0];
        int i = 1;
        while(ratings[i] < kRating){
            swapWithFirst++;
            i++;
        }

        return Math.max(swapWithFirstHigher, swapWithFirst);
    }
}

**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 1
100 2 1
3 2
1 100 2
3 3
1 2 100
```


**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 <vector>
#include <string>
#include <algorithm>

using namespace std;

bool checkLCP(const string& s, int l, int mid) {
    int n = s.length();
    string t = "";
    for (int i = 0; i < mid; i++)
        t += s[i];
    for (int i = 0, j = 0; i < n && j < mid;) {
        if (s[i] != t[j])
            i++;
        else {
            i++, j++;
            if (j == mid) {
                l--;
                j = 0;
            }
        }
    }
    if (l <= 0)
        return true;
    return false;
}

int findLCP(const string& s, int k) {
    int n = s.length();
    int l = 0, r = n + 1;
    int ans = 1;
    while (r - l > 1) {
        int mid = (l + r) / 2;
        if (checkLCP(s, k, mid)) {
            l = mid;
        } else {
            r = mid;
        }
    }
    return l;
}

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

    while (t--) {
        int n, l, r;
        cin >> n >> l >> r;
        string s;
        cin >> s;
        int lcp = findLCP(s, l);
        for (int i = 0; i < r - l + 1; i++) {
            cout << lcp << " ";
        }
        cout << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 3 3
aba
3 3 3
aaa
7 2 2
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<bits/stdc++.h>
#define N 200007
#define mod 1000000007

using namespace std;

int t,n,L,R,fl[N];
char s[N];

int main()
{
	scanf("%d",&t);
	while(t--){
		scanf("%d%d%d",&n,&L,&R);
		scanf("%s",s+1);
		int l=1,r=n;
		while(l<=r){
			int mid=(l+r)>>1;
			memset(fl,0,sizeof(fl));
			fl[0]=fl[1]=0;
			for(int i=1;i<=mid;i++){
				fl[i]=fl[i-1];
				while(fl[i]&&s[fl[i]+1]!=s[i]) fl[i]=fl[fl[i]];
				if(s[fl[i]+1]==s[i]) fl[i]++;
			}
			int cnt=1;
			for(int i=mid+1,j=0;i<=n;i++){
				if(s[i]==s[j+1]){
					j++;
					if(j==mid){
						j=0;cnt++;
					}
				}
				else{
					j=fl[j];
				}
			}
			if(cnt>=L) l=mid+1;
			else r=mid-1;
		}
		printf("%d\n",r);
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 3 3
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:
#define  _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>
#include <iterator>
#include <numeric>
#include <sstream>
#include <fstream>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
#include <math.h>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <cmath>
#include <bitset>
#include<list>
#include<unordered_map>
#include <iomanip>
using namespace std;
#define Clear(a) memset(a,0,sizeof(a))
#define FILL(a) fill(a,a+1000000,true);
#define fastIO ios::sync_with_stdio(false)
#define mp make_pair
#define All(a) a.begin(),a.end()
typedef long long ll;
typedef vector<ll> vi;
typedef pair<ll, ll> pi;
#define PI 3.1415926535897932385
#define INF 1000000000
#define eps 1e-7
#define MOD 100000000
#define maxN 500010
const int OO = (int)1e6;
using namespace std;
ll GCD(ll a, ll b)
{
	if (b == 0)return a;
	return GCD(b, a % b);
}
vector<ll>p, a;
vector<ll>sum1, sum2;
bool ch(int x,string s,int k) {
	string temp = "";
	for (int c = 0; c < x; c++) {
		temp += s[c];
	}
	int res = 0;
	for (int c = 0; c < s.size(); c ++) {
		bool z = 1;
		for (int i = 0; i < x; i++) {
			if (s[c + i] != temp[i]) {
				z = 0; break;
			}
		}
		if (z)
		{
			res++;
			c += x - 1;
		}
	}
	bool z= res >= k;
	return z;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif

	int t;
	cin >> t;
	while (t--) {
		int k, n;
		cin >> n >> k >> k;
		string s;
		cin >> s;
		int l = 0, r = n / k+1, mid;
		int u = 200;
		ll ans = 0;
		while (u--&&l<r)
		{
			mid = (l + r) / 2;
			bool temp = ch(mid, s, k);
			if (temp) {
				l = mid+1;
				ans= mid;
			}
			else r = mid -1;
		}
		cout << ans << endl;;
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 4 4
abcd
5 5 5
abcde
6 6 6
abcdef
```


**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 sys

def solve(k, s):
    starts = set()
    starts.add(0)
    ans = 0

    for i in range(len(s)):
        if s[i] == s[0]:
            starts.add(i)

    offset = 0
    while len(starts) >= k:
        offset += 1
        to_do = starts.copy()
        for start in to_do:
            if start + offset >= len(s):
                starts.remove(start)
            elif start + offset in starts:
                starts.remove(start)
            elif s[start + offset] != s[offset]:
                starts.remove(start)

    return offset

out = ""

f = sys.stdin.readlines()
# f = []
# while True:
#     try:
#         line = input()
#     except EOFError:
#         break
#     f.append(line)
probs = int(f[0])
for prob in range(probs):
    out += str(solve(int(f[prob*2+1].split()[1]), f[prob*2+2].split()[0])) + "\n"

print(out)

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

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

Think step by step.

Response:
```plaintext
1
4 4 4
aaaa
```


**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 prefix_function(s):
    n = len(s)
    pi = [0] * n
    for i in range(1, n):
        j = pi[i - 1]
        while j > 0 and s[i] != s[j]:
            j = pi[j - 1]
        if s[i] == s[j]:
            j += 1
        pi[i] = j
    return pi

def check(x):
    t = s[:x]
    cnt = 0
    j = 0
    for i in range(n):
        if s[i] == t[j]:
            j += 1
        elif s[i] != t[j]:
            if j:
                j = pit[j - 1]
        if j == x:
            cnt += 1
            j = 0
    return cnt >= l

t = int(input())
for _ in range(t):
    n, l, l = map(int, input().split())
    s = input().strip()
    if l == 1:
        print(n)
        continue
    p1 = prefix_function(s)
    pit = p1[:]
    lo, hi = 0, n // l
    while hi - lo > 1:
        mid = lo + (hi - lo) // 2
        if check(mid):
            lo = mid
        else:
            hi = mid - 1
    if check(hi):
        print(hi)
    else:
        print(lo)

**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
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
```

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


**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 sys

def iinp():
    return int(sys.stdin.readline().strip())

def linp():
    return list(map(int, sys.stdin.readline().strip().split()))

def string():
    return sys.stdin.readline().strip()

def lsinp():
    return sys.stdin.readline().strip().split()

def digit():
    return [int(i) for i in (list(sys.stdin.readline().strip()))]

def char():
    return list(sys.stdin.readline().strip())

from collections import Counter, defaultdict, deque
from bisect import bisect_right, bisect_left
from math import ceil, sqrt, gcd

def zarray(s):
    n = len(s)
    z = [0] * n
    left, right = 0, 0

    for i in range(1, n):
        if i > right:
            left = right = i
            while right < n and s[right] == s[right - left]:
                right += 1
            z[i] = right - left
            right -= 1
        else:
            k = i - left
            if z[k] < right - i + 1:
                z[i] = z[k]
            else:
                left = i
                while right < n and s[right] == s[right - left]:
                    right += 1
                z[i] = right - left
                right -= 1
    return z

def zfunc(text, pattern):
    concat = pattern + "$" + text
    z_array = zarray(concat)
    pattern_length = len(pattern)
    indices = []
    for i in range(len(concat)):
        if z_array[i] == pattern_length:
            indices.append(i - pattern_length - 1)
    return indices

def solve():
    n, i, j = linp()
    k = i
    s = string()

    def calc(t):
        first = s[:t]
        temp = zfunc(s, first)
        prev = -t
        count = 0
        for val in temp:
            if val >= prev + t:
                count += 1
            prev = val
        return k <= count

    left = 0
    right = n

    while left <= right:
        mid = (right + left) // 2
        # print(left, right)
        if calc(mid):
            left = mid + 1
        else:
            right = mid - 1
    print(right)

q = iinp()
for _ in range(q):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
3 3 3
aba
4 4 4
aaaa
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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CFTemp {

    static int MOD = 998244353;
    static int INF = 0x3f3f3f3f;

    public static void main(String[] args) throws IOException {
        int t = Reader.nextInt();
        StringBuilder sb = new StringBuilder();
        while (t-- > 0) {
            int n = Reader.nextInt();
            int l = Reader.nextInt();
            int r = Reader.nextInt();
            String s = Reader.next();
            sb.append(solve(s, l)).append("\n");
        }

        System.out.print(sb.toString());
    }

    static int mod = 998244353;

    static int solve(String s, int k) {

        int n = s.length();
        char[] arr = s.toCharArray();
        // build z array
        int[] z = new int[n];
        int l = 0;
        int r = 0;
        for (int i = 1; i < n; i++) {
            if (z[i - l] < r - i + 1) {
                z[i] = z[i - l];
            } else {
                z[i] = Math.max(r - i + 1, 0);
                while (i + z[i] < n && arr[z[i]] == arr[i + z[i]])
                    z[i]++;
                l = i;
                r = i + z[i] - 1;
            }
        }

        int left = 0;
        int right = n + 1;

        while (left + 1 < right) {
            int mid = (left + right) / 2;
            if (check(z, k, mid)) {
                left = mid;
            } else {
                right = mid;
            }
        }

        return left;
    }

    private static boolean check(int[] z, int k, int mid) {
        if (mid == 0)
            return true;
        int result = 1;
        for (int i = 0; i < z.length; i++) {
            if (z[i] >= mid) {
                result++;
                i += mid - 1;
            }
        }
        return result >= k;
    }

    static class Reader {
        static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        static StringTokenizer tokenizer = new StringTokenizer("");

        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        }

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

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

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

}

**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
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
```


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

public class main3 {
    static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
    static FastReader in = new FastReader();
    public static void main(String[] args) {
        int t = in.nextInt();
        while(t-- > 0){
            int n = in.nextInt();
            int l = in.nextInt();
            int r = in.nextInt();
            String a = in.next();
            int start = 0;
            int end = n;
            while(start < end){
                int mid = start + (end - start) / 2;
                if (ok(mid, a, l)) {
                    start = mid + 1;
                }else{
                    end = mid;
                }
            }
            pw.println(start);
        }
	pw.close();
    }
    public static boolean ok(int mid, String a, int k){
        int ans = 0;
        int i = 0;
        String s = a.substring(0, mid + 1);
        int[] lps = new int[s.length()];
        kmp(lps, s);
        for (int j = 0; j < a.length(); j++) {
            if (a.charAt(j) == s.charAt(i)) {
                i++;
            }else{
                while(i != 0 && a.charAt(j) != s.charAt(i)){
                    i = lps[i - 1];
                }
            }
            if(i == s.length()){
                ans++;
                i = 0;
            }
        }
        if (ans >= k) {
            return true;
        }
        return false;
    }
    static class Pair implements Comparable<Pair>{
        int key;
        int value;
        Pair(int key, int value){
            this.key = key;
            this.value = value;
        }
        @Override
        public int compareTo(Pair that){
            if(this.key == that.key){
                return that.value - this.value;
            }
            return this.key - that.key;
        }
	    @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return key == pair.key && value == pair.value;
        }

        @Override
        public int hashCode() {
            return Objects.hash(key, value);
        }
    }
    static class BIT{
        int bit[], n;
        public BIT(int s){
            bit = new int[s+1];
            n = s;
        }
        void add(int i, int diff){
            while(i<=n){
                bit[i] += diff;
                i += (i&(-i));
            }
        }
        int query(int i){
            int sum = 0;
            while(i>0){
                sum += bit[i];
                i -= (i&(-i));
            }
            return sum;
        }
        int rangeQuery(int l, int r){
            return query(r)-query(l-1);
        }
    }
    static class DisjointUnionSets {
        int[] rank, parent;
        int n;
        int count;
        public DisjointUnionSets(int n){
            rank = new int[n];
            parent = new int[n];
            count = n;
            this.n = n;
            makeSet();
        }
        void makeSet(){
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }
        int find(int x){
            if (parent[x] != x) {
                parent[x] = find(parent[x]);
            }

            return parent[x];
        }
        void union(int x, int y){
            int xRoot = find(x), yRoot = find(y);

            if (xRoot == yRoot)
                return;

            if (rank[xRoot] < rank[yRoot])
                parent[xRoot] = yRoot;

            else if (rank[yRoot] < rank[xRoot])
                parent[yRoot] = xRoot;

            else{
                parent[yRoot] = xRoot;
                rank[xRoot] = rank[xRoot] + 1;
            }
            count -= 1;
        }
        int num(){
            return count;
        }
    }
    public static void kmp(int[] lps, String s){
        int length = 0;
        int i = 1;
        while(i < s.length()){
            if(s.charAt(i) == s.charAt(length)){
                length++;
                lps[i] = length;
                i++;
            }else{
                if(length != 0){
                    length = lps[length - 1];
                }else{
                    lps[i] = 0;
                    i++;
                }
            }
        }
    }
    public static int calcMex(List<Integer> list) {
        Set<Integer> set = new HashSet<>(list);
        for (int i = 0; i <= list.size(); i++) {
            if (!set.contains(i)) {
                return i;
            }
        }
        return list.size();
    }
    public static long gcd(long a, long b){
        if(a == 0){
            return b;
        }
        return gcd(b % a, a);
    }
    public static long sqrt(long x){
        long i = 0;
        long j = Integer.MAX_VALUE;
        while(i < j){
            long mid = i + (j - i) / 2;
            if(mid * mid <= x){
                i = mid + 1;
            }else{
                j = mid;
            }
        }
        return i - 1;
    }
    static long lcm(long a, long b){
        return (a / gcd(a, b)) * b;
    }
    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static ArrayList<Long> input(int n){
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            a.add(in.nextLong());
        }
        return a;
    }
    public static ArrayList<ArrayList<Integer>> adj(int n){
        ArrayList<ArrayList<Integer>> arr = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            arr.add(new ArrayList<>());
        }
        for (int i = 0; i < n - 1; i++) {
            int l = in.nextInt() - 1;
            int r = in.nextInt() - 1;
            arr.get(l).add(r);
            arr.get(r).add(l);
        }
        return arr;
    }
    public static void printlist(ArrayList<Long> a){
        for (int i = 0; i < a.size(); i++) {
            pw.print(a.get(i) + " ");
        }
        pw.println();
    }
    public static void printlist2d(ArrayList<ArrayList<Long>> a){
        for (int i = 0; i < a.size(); i++) {
            for (int j = 0; j < a.get(0).size(); j++) {
                pw.print(a.get(i).get(j) + " ");
            }
            pw.println();
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Let's analyze the problem and the code to identify potential issues:

1. **Edge Case: Single Character String**:
   - If the string `s` consists of a single character repeated, the LCP for any division will be the length of the string.
   - Example: `s = "aaa"`, `n = 3`, `l = 3`, `r = 3`.

2. **Edge Case: Distinct Characters**:
   - If the string `s` consists of distinct characters, the LCP for any division will be 0.
   - Example: `s = "abc"`, `n = 3`, `l = 3`, `r = 3`.

3. **Edge Case: Mixed Characters**:
   - If the string `s` has a mix of repeated and distinct characters, the LCP can vary depending on the division.
   - Example: `s = "abab"`, `n = 4`, `l = 4`, `r = 4`.

4. **Edge Case: Repeated Substrings**:
   - If the string `s` contains repeated substrings, the LCP can be non-zero for certain divisions.
   - Example: `s = "ababab"`, `n = 6`, `l = 6`, `r = 6`.

5. **Edge Case: Large `k`**:
   - If `k` is large and the string is short, the LCP will be 0.
   - Example: `s = "abc"`, `n = 3`, `l = 3`, `r = 3`.

6. **Edge Case: Small `k`**:
   - If `k` is small and the string is long, the LCP can be non-zero.
   - Example: `s = "aaaaaa"`, `n = 6`, `l = 1`, `r = 1`.

Given the constraints and the need for a small-scale test input, let's choose a test case that combines some of these edge cases to maximize the failure probability:

```plaintext
1
4 4 4
abab
```

This test case has the following characteristics:
- The string `s = "abab"` has a length of 4.
- The value of `l` and `r` is 4, meaning we need to divide the string into 4 parts.
- The string contains repeated substrings, which can expose issues in the LCP calculation.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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;
using ll = long long;
using ull = unsigned long long;
using db = double;
using i128 = __int128;
using u128 = unsigned __int128;
#define I return
#define AK 0
#define XCPC ;
#define AL(X) (X).begin() + 1, (X).end()
#define ALL(X) (X).begin(), (X).end()
#define l first
#define r second
#define lr(x) (x << 1)
#define rr(x) (x << 1 | 1)
ll mod = LLONG_MAX;
mt19937_64 rdm(chrono::system_clock::now().time_since_epoch().count()); // 随机数生成
db eps = 1e-9;
inline void solve()
{
	ll n, l, r;
	cin >> n >> l >> r;
	string s;
	cin >> s;
	s = " " + s;
	auto check = [&](ll x)
	{
		ll mx = 1;
		ll ans = 1;
		string s1 = " ";
		ll j = 0;
		vector<ll> next(x + 1, 0);
		for (ll i = 1; i <= x; i++)
		{
			s1 += s[i];
		}
		for (ll i = 2; i <= x; i++)
		{
			while (j > 0 && s1[i] != s1[j + 1])
			{
				j = next[j];
			}
			if (s1[j + 1] == s1[j])
				j++;
			next[i] = j;
		}
		j = 0;
		for (ll i = x + 1; i <= n; i++)
		{
			while (j > 0 && s[i] != s1[j + 1])
				j = next[j];
			if (s[i] == s1[j + 1])
				j++;
			if (j == x)
			{
				if (i - j + 1 > mx)
				{
					ans++;
					mx = i - j + 1;
				}
				j = next[j];
			}
		}
		return ans >= l;
	};
	ll l1 = 1, r1 = n;
	ll ans = 0;
	while (l1 <= r1)
	{
		ll md = l1 + r1 >> 1;
		if (check(md))
		{
			ans = md;
			l1 = md + 1;
		}
		else
			r1 = md - 1;
	}
	cout << ans << endl;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int tt = 1;
	cin >> tt;
	while (tt--)
	{
		solve();
	}
	I AK XCPC
}

**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
```

**Explanation:**
- This test case is designed to expose the bug in the given buggy code.
- The string `abcde` has a length of 5, and we are dividing it into 5 parts, which means each part will be a single character.
- The longest common prefix (LCP) of single characters is 0, since no two single characters can have a common prefix longer than 0.
- The correct output for this test case should be `0`, but the buggy code might not handle this edge case correctly, especially since it involves a single-character division.