**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
    //------------------------------------------main------------------------------------------------------//
    public static void main (String[] args) throws java.lang.Exception
    {
     OutputStream outputStream =System.out;
        PrintWriter out =new PrintWriter(outputStream);
        FastReader sc =new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

 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 int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }
    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextLong();
        return arr;
    }
    }
class Task{

        PrintWriter out;
        FastReader sc;
        Task(FastReader sc, PrintWriter out){
            this.sc = sc;
            this.out= out;
        }
      //----------------------------------------------input function------------------------------------------//
       String st(){
        return sc.nextLine();
    }
    int i(){
        return sc.nextInt();
    }
    long l(){
        return sc.nextLong();
    }
    double d(){
        return sc.nextDouble();
    }

      //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a){
        return a<0?-a:a;
    }
    public int abs(int a){
        return a<0?-a:a;
    }
    //-------------------------------------------max--------------------------------------------------//
    public int max (int a, int b){
        return a>=b?a:b;
    }
      public long max (long a, long b){
        return a>=b?a:b;
    }
     public int max (int a, int b, int c){
        return Math.max(a,Math.max(b,c));
    }
     public long max (long a, long b, long c){
        return Math.max(a,Math.max(b,c));
    }

    //------------------------------------------min---------------------------------------------------//
     public  int min (int a, int b){
        return a<=b?a:b;
    }
     public long min (long a, long b){
        return a<=b?a:b;
    }
     public int min (int a, int b, int c){
        return Math.min(a,Math.min(b,c));
    }
     public long min (long a, long b, long c){
        return Math.min(a,Math.min(b,c));
    }

    //----------------------------------------sorting function---------------------------------------------//
        private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++)
            list.add(arr[i]);
        Collections.sort(list);
        for (int i = 0; i < n; i++)
            arr[i] = list.get(i);
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++)
            list.add(arr[i]);
        Collections.sort(list);
        for (int i = 0; i < n; i++)
            arr[i] = list.get(i);
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++)
            list.add(arr[i]);
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++)
            arr[i] = list.get(i);
    }

    private  void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++)
            list.add(arr[i]);
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++)
           arr[i] = list.get(i);
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++)
            list.add(arr[i]);
        Collections.sort(list);
        for (int i = 0; i < n; i++)
            arr[i] = list.get(i);
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++)
            list.add(arr[i]);
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++)
            arr[i] = list.get(i);
    }
    //---------------------------------------power------------------------------------------------------//
    public long pow(long a, int b){
        long ans = 1L;
        while(b>0){
            if((b&1)>0) ans*=a;
            a*=a;
            b>>=1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation------------------------------------//
    public long exp(long a, long b, long mod){
        long ans = 1L;
        while(b>0){
            if((b&1)>0) ans = (ans*a)%mod;
            a= (a*a)%mod;
            b>>=1;
        }
        return ans;
    }

   //---------------------------------------helping methods---------------------------------------------//

   //---------------------------------------main---------------------------------------------------------//
    int t, n, l ,r;
    zfunction zf;
    int help(int mid){
        boolean got = false;
        int count =0;
        int i = 0;
        while(i<zf.n){
            if(zf.a[i]=='#'){
                got= true;
            }
            else if(got){
                if(zf.z[i]>=mid){
                    i+= mid-1;
                    count++;
                }
            }
            i++;
        }
        return count;
    }
    public void solve()throws IOException{
         t = sc.nextInt();
        while(t-->0){
            n = i();
            l = i();
            r= i();
            zf = new zfunction(st(),null);
            zf.form();
          //  out.println(Arrays.toString(zf.z));
            int k = (int)Math.ceil(Math.sqrt(n));
            int[] ans  = new int[n+1];
            for(int i = 1;i<=k;i++ ){
                int s = 1, e= n;
                while(s<=e){
                     int mid = s+(e-s)/2;
                    if(help(mid)>=i){
                        ans[i] = mid;
                        s = mid+1;
                    }
                    else e = mid-1;
                }
            }
            for(int i = 1;i<=k;i++){
                int count = 0;
                boolean got = false;
                for(int j = 0;j<zf.n;j++){
                    if(zf.a[j]=='#'){
                        got = true;
                    }
                    else if(got&&zf.z[j]>=i){
                        j+=  i-1;
                        count++;
                    }
                }
                ans[count] = max(ans[count], i);
            }
            for(int i = n-1;i>0;i--){
                ans[i] = max(ans[i], ans[i+1]);
            }
            for(int i = l;i<=r;i++) out.print(ans[i]+" ");

                out.print("\n");
            }
        }
    class zfunction{
        int n;
        char[] a;
        int[] z;
        zfunction(String x, String y){
            StringBuilder s = new StringBuilder(x);
            s.append("#").append(y==null?x:y);
            n = s.length();
            a = s.toString().toCharArray();
            z = new int[n];
        }

        void form(){
            int l =0, r = 0;
            for(int i =1; i<n;i++){
                if(i<=r){
                    z[i] = min(z[i-l], r-i+1);
                }
                while(i+z[i]<n&&a[z[i]]==a[i+z[i]]){
                    z[i]++;
                }
                if(i+z[i]-1>r){
                    r= i+z[i]-1;
                    l = i;
                }
            }
        }

    }
    }

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

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

Think step by step.

Response:
```plaintext
1
10 1 10
aaaaaaabac
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

Buggy Code:
//La illaha illalla muhammadur rasul Allah (SAW)
	//Bismillah hir-rehamn nir-rahim

	import java.io.*;
	import java.util.*;

	public class ComdeFormces {
		static int x[]= {0,-1,1,0,1,1,-1,-1};
		static int y[]= {-1,0,0,1,1,-1,1,-1};
		public static int m=(int)(1e9+7);
//		static HashMap<Integer,Integer> dp[][];
		static long ct;
		static long ans[];
		static long dp[][];
		static int st[][];
		static long pre[];
		static int lg[];
		 static class Trie{
			    Trie a[];
				public Trie() {
					this.a=new Trie[2];
				}
			}
		 static HashMap<String,Integer> hm,hm2;

				public static void main(String[] args) throws Exception{
					// TODO Auto-generated method stub
			FastReader sc=new FastReader();
			BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
			//		 OutputStream out = new BufferedOutputStream ( System.out );
			int tc=1;
			sieve();
			int t=sc.nextInt();
			while(t--!=0) {
				  int n=sc.nextInt();
				  int l=sc.nextInt();
				  int r=sc.nextInt();
				  char a[]=sc.next().toCharArray();
				  int anss[]=new int[n+1];
				  for(int j=1;j*j<=n;j++) {
				  int ans=0;
				  int s=1;
				  int e=n;
				  while(s<=e) {
					  int m=s+(e-s)/2;
					  int cnt=0;
					  StringBuilder st=new StringBuilder("");
					  for(int i=0;i<m;i++)st.append(a[i]);
					  String str=st.toString()+"$"+String.valueOf(a);
					  int z[]=zed(str.toCharArray());
					  for(int i=m+1;i<z.length;) {
						  if(z[i]==m) {
							  i+=m;
							  cnt++;
						  }
						  else i++;
					  }
					  if(cnt>=j) {
						  ans=max(ans,m);
						  s=m+1;
					  }
					  else e=m-1;
				  }
				  anss[j]=ans;
				  }
				  int z[]=zed((String.valueOf(a)+"$"+String.valueOf(a)).toCharArray());
				  for(int j=1;j*j<=n;j++) {
					  int cnt=0;
					  for(int k=a.length+1;k<z.length;) {
						  if(z[k]>=j) {
							  k+=j;
							  cnt++;
						  }
						  else k++;
					  }
					  anss[cnt]=max(anss[cnt],j);
				  }
				  for(int i=n-1;i>=1;i--)anss[i]=max(anss[i],anss[i+1]);
				  for(int i=l;i<=r;i++)log.write(anss[i]+" ");
				  log.write("\n");

			}
			log.flush();
		}
static int[] zed(char a[]) {
	int z[]=new int[a.length];
	int l=0;
	int r=0;
	for(int i=0;i<a.length;i++) {
		if(i>r) {
			l=r=i;
			while(r<a.length && a[r]==a[r-l])r++;
			z[i]=r-l;
			r--;
		}
		else {
			int k1=i-l;
			if(z[k1]<r-i+1) {
				z[i]=z[k1];
			}
			else {
				l=i;
				while(r<a.length && a[r]==a[r-l])r++;
				z[i]=r-l;
				r--;
			}
		}
	}
	return z;
}
static long eval(int a[],int b[],int n,int q,int k) {
	boolean vis[]=new boolean[n+1];
	int tk=k;
	ArrayList<Integer> ar=new ArrayList<>();
	vis=new boolean[n+1];
	while(!vis[q]) {
		ar.add(b[q]);
		vis[q]=true;
		q=a[q];
	}
	long val=0;
	long a2=0;
	for(int i=0;i<min(ar.size(),tk);i++) {
		a2=max(a2,val+(tk-i)*(long)ar.get(i));
		val+=ar.get(i);
	}
	if(ar.size()<=tk)a2=max(a2,val);
	return a2;

}
static final int MAXN = 2000001;
static int spf[] = new int[MAXN];
static void sieve()
{
    spf[1] = 1;
    for (int i=2; i<MAXN; i++)
        spf[i] = i;

    for (int i=4; i<MAXN; i+=2)
        spf[i] = 2;

    for (int i=3; i*i<MAXN; i++)
    {
        if (spf[i] == i)
        {
            for (int j=i*i; j<MAXN; j+=i)
                if (spf[j]==j)
                    spf[j] = i;
        }
    }
}
static ArrayList<pair> get(int x)
{
    HashMap<Integer,Integer> hm = new HashMap<>();
    while (x != 1)
    {
        hm.put(spf[x], hm.getOrDefault(spf[x], 0)+1);
        x = x / spf[x];
    }
    ArrayList<pair> ar=new ArrayList<>();
    for(int el:hm.keySet())ar.add(new pair(el,hm.get(el)));
    return ar;
}
	// small function bazi due to alas
	static int max(int a,int b) {return a>=b?a:b;}
	static int min(int a,int b) {return a>=b?b:a;}
	static long max(long a,long b) {return a>=b?a:b;}
	static long min(long a,long b) {return a>=b?b:a;}
	static int abs(int a) {
		return a>=0?a:-a;
	}

	static long abs(long a) {
	return a>=0?a:-a;
	}

	static long div(long a,long b) {
		long an=pow(b,m-2)%m;
		return mul(an,a);
	}
	public static long mul(long a, long b) {
		return (a*b)%m;
	}
	public static long add(long a, long b) {
		return (a+b)%m;
	}
	public static long sub(long a,long b) {
	  return (a%m-b%m+m)%m;
	}
	static int gcd(int a,int b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	static long gcd(long a,long b) {
		if(a==0)return 0;
		if(b<2*a)return 1+gcd(b,abs(a-b));
		long val=b/(2*a);
		return 3*val+gcd(a,b-2*a*val);
	}
	static long ncr(int n, int r){
	    if(r>n-r)r=n-r;
	    long ans=1;
	    for(int i=0;i<r;i++){
	          ans=mul(ans,(n-i));
	          ans=div(ans,i+1);
	    }
	    return ans;
	}
	public static class trip{
		int a;
		int b;
		int c;
		public trip(int a,int b,int c) {
			this.a=a;
			this.b=b;
			this.c=c;
		}
	}
	static void mergesort(long[] a,int start,int end) {
			if(start>=end)return ;
			int mid=start+(end-start)/2;
			mergesort(a,start,mid);
			mergesort(a,mid+1,end);
			merge(a,start,mid,end);

		}
	static void merge(long[] a, int start,int mid,int end) {
			int ptr1=start;
			int ptr2=mid+1;
			long b[]=new long[end-start+1];
			int i=0;
			while(ptr1<=mid && ptr2<=end) {
				if(a[ptr1]<=a[ptr2]) {
					b[i]=a[ptr1];
					ptr1++;
					i++;
				}
				else {
					b[i]=a[ptr2];
					ptr2++;
					i++;
				}
			}
			while(ptr1<=mid) {
				b[i]=a[ptr1];
				ptr1++;
				i++;
			}
			while(ptr2<=end) {
				b[i]=a[ptr2];
				ptr2++;
				i++;
			}
			for(int j=start;j<=end;j++) {
				a[j]=b[j-start];
			}
		}
	 public static class FastReader {

			BufferedReader b;
			StringTokenizer s;
			public FastReader() {
				b=new BufferedReader(new InputStreamReader(System.in));
			}
			String next() {
				while(s==null ||!s.hasMoreElements()) {
					try {
						s=new StringTokenizer(b.readLine());
					}
					catch(IOException e) {
						e.printStackTrace();
					}
				}
				return s.nextToken();
			}
			public int nextInt() {
				return Integer.parseInt(next());
			}
			public long nextLong() {
				return Long.parseLong(next());
			}
			public double nextDouble() {
				return Double.parseDouble(next());
			}
			String nextLine() {
				String str="";
				try {
					str=b.readLine();
				}
				catch(IOException e) {
					e.printStackTrace();
				}
				return str;
			}
			boolean hasNext() {
			    if (s != null && s.hasMoreTokens()) {
			        return true;
			    }
			    String tmp;
			    try {
			        b.mark(1000);
			        tmp = b.readLine();
			        if (tmp == null) {
			            return false;
			        }
			        b.reset();
			    } catch (IOException e) {
			        return false;
			    }
			    return true;
			}

	}
	public static class pair{
		int a,b,c,d,e;
		public pair(int a,int b,int c) {
			this.a=a;
			this.b=b;
			this.c=c;
		}
		public pair(int a,int b) {
			this.a=a;
			this.b=b;
		}
		public pair(int a,int b,int c,int d) {
			this.a=a;
			this.b=b;
			this.c=c;
			this.d=d;
		}
		public pair(int a,int b,int c,int d,int e) {
			this.a=a;
			this.b=b;
			this.c=c;
			this.d=d;
			this.e=e;
		}
	//			@Override
	//			public String toString() {
	//				return "{"+this.a+" "+this.b+"}";
	//				}
		}
			static long pow(long a, long pw) {
				long temp;
				if(pw==0)return 1;
				temp=pow(a,pw/2);
				if(pw%2==0)return mul(temp,temp);
				return mul(a,mul(temp,temp));

			}

			static long poww(long a, long pw) {
				long temp;
				if(pw==0)return 1;
				temp=poww(a,pw/2);
				if(pw%2==0)return temp*temp;
				return a*temp*temp;

			}
			static int poww(int a, int pw) {
				int temp;
				if(pw==0)return 1;
				temp=poww(a,pw/2);
				if(pw%2==0)return temp*temp;
				return a*temp*temp;

			}

	}

**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 buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The algorithm deals with finding the longest common prefix (LCP) among divisions of a string. Let's consider the following aspects:

1. **String with all identical characters**: This can cause issues in algorithms that do not handle repetitive patterns well.
2. **String with a mix of characters**: This can test the algorithm's ability to handle diverse patterns.
3. **Boundary conditions**: Testing with the minimum and maximum values for `l` and `r` can reveal issues in the edge cases.
4. **Special cases**: Strings with specific patterns that might not be handled correctly by the algorithm.

Given these considerations, let's construct a small-scale test input that can expose potential bugs:

```plaintext
3
5 1 5
aaaaa
5 1 5
abcde
5 1 5
ababa
```

### Explanation:
1. **Test Case 1: "aaaaa"**
   - **n = 5, l = 1, r = 5**
   - All characters are the same, which can test the algorithm's ability to handle repetitive patterns.
   - Expected output: `5 4 3 2 1`

2. **Test Case 2: "abcde"**
   - **n = 5, l = 1, r = 5**
   - All characters are different, which can test the algorithm's ability to handle diverse patterns.
   - Expected output: `1 1 1 1 1`

3. **Test Case 3: "ababa"**
   - **n = 5, l = 1, r = 5**
   - A string with a repeating pattern, which can test the algorithm's ability to handle specific patterns.
   - Expected output: `5 3 2 2 1`

These test cases cover a range of scenarios that can help identify potential 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 (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

Buggy Code:
import sys
input=sys.stdin.readline
import math
def solve():
    n,ll,rr=map(int,input().split())
    s=[0]+list(input().strip())+['#']
    z=[0]*(n+2)
    z[1]=n
    l,r=0,0
    for i in range(2,n+1):
        if i<=r:
            z[i]=min(z[i-l+1],r-i+1)
        while s[1+z[i]]==s[i+z[i]]:
            z[i]+=1
        if i+z[i]-1>r:
            l=i
            r=i+z[i]-1
    wc=0
    if n<=10:
        wc=n
    else :
        wc=int(math.sqrt(n))
    f=[0]*(n+2)
    def check(x,l):
        cnt=0
        i=1
        while i<=n:
            if z[i]>=x:
                cnt+=1
                i=i+x-1
            i+=1
        if cnt>=l:
            return 1
        return 0
    for x in range(1,n//wc+1):
        cnt=0
        j=1
        while j<=n:
            if z[j]>=x:
                cnt+=1
                j=j+x-1
            j+=1
        f[cnt]=x
    for i in range(n,0,-1):
        f[i]=max(f[i],f[i+1])
    for i in range(ll,rr+1):
        if i<=wc:
            L,R=1,n
            ans=0
            while L<=R:
                mid=L+R>>1
                if check(mid,i):
                    ans=mid
                    L=mid+1
                else :
                    R=mid-1
            print(ans,end=' ')
        else :
            print(f[i],end=' ')
    print()
T=1
T=int(input())
while T:
    T-=1
    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
1
10 1 10
aabbccddeeff
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

Buggy Code:
import sys
input=sys.stdin.readline
import math
def solve():
    n,ll,rr=map(int,input().split())
    s=[0]+list(input().strip())+['#']
    z=[0]*(n+2)
    z[1]=n
    l,r=0,0
    for i in range(2,n+1):
        if i<=r:
            z[i]=min(z[i-l+1],r-i+1)
        while s[1+z[i]]==s[i+z[i]]:
            z[i]+=1
        if i+z[i]-1>r:
            l=i
            r=i+z[i]-1
    wc=0
    if n<=10:
        wc=n
    else :
        wc=50
    f=[0]*(n+2)
    def check(x,l):
        cnt=0
        i=1
        while i<=n:
            if z[i]>=x:
                cnt+=1
                i=i+x-1
            i+=1
        if cnt>=l:
            return 1
        return 0
    for x in range(1,n//wc+1):
        cnt=0
        j=1
        while j<=n:
            if z[j]>=x:
                cnt+=1
                j=j+x-1
            j+=1
        f[cnt]=x
    for i in range(n,0,-1):
        f[i]=max(f[i],f[i+1])
    for i in range(ll,rr+1):
        if i<=wc:
            L,R=1,n
            ans=0
            while L<=R:
                mid=L+R>>1
                if check(mid,i):
                    ans=mid
                    L=mid+1
                else :
                    R=mid-1
            print(ans,end=' ')
        else :
            print(f[i],end=' ')
    print()
T=1
T=int(input())
while T:
    T-=1
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some considerations:

1. **Edge Cases:**
   - Strings with all identical characters.
   - Strings with a single character.
   - Strings with a pattern that repeats.
   - Strings with a mix of repeating and non-repeating characters.
   - Strings where the length of the string is exactly the same as the range of divisions.

2. **Scenarios:**
   - Test cases where `l` and `r` are very close to `n`.
   - Test cases where `l` is 1 and `r` is close to `n`.
   - Test cases with a small number of test cases (`t`).

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
5
5 1 5
aaaaa
5 1 5
abcde
7 1 7
abababa
10 1 10
aaaaaaabac
9 1 9
ababababa
```

This test input includes:
1. A string with all identical characters (`aaaaa`).
2. A string with all unique characters (`abcde`).
3. A string with a repeating pattern (`abababa`).
4. A string with a long prefix and a different suffix (`aaaaaaabac`).
5. A string with a repeating pattern and a different suffix (`ababababa`).

These cases will help in testing the algorithm's ability to handle various string patterns and division ranges, potentially exposing any bugs in the code.

5
5 1 5
aaaaa
5 1 5
abcde
7 1 7
abababa
10 1 10
aaaaaaabac
9 1 9
ababababa


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

Buggy Code:
/**************************************************
 * Fast Algorithm Template by Richard1211.
 * Please submit with C++14 or higher version.
 * Blog on Luogu: https://www.luogu.com.cn/blog/522539/
 * Blog on Byethost: http://Richard1211.byethost5.com/
 * Blog on RBTree's: https://Richard1211.rbtr.ee/
***************************************************/

#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC optimize(2)

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

#define getchar() p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++
char buf[1000000], *p1 = buf, *p2 = buf;

inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-'){f=-1;}ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch-'0');ch=getchar();}return x*f;}
inline bool write(long long x,char ch='\n'){static char buf[64];static long long len=-1;if(x<0){putchar('-');x=-x;}do{buf[++len]=x%10;x/=10;}while(x);while(len>=0){putchar(buf[len--]+'0');}putchar(ch);return true;}

#define Test() \
register int TestCount=MultiCase?read():1;\
for(register int TestCase=1;TestCase<=TestCount;++TestCase)
#define Tesf() for(register int TestCase=1;TestCase<=t;++TestCase)

template<class T>inline bool Cmax(register T&a,register T b){return a<b?(a=b,true):false;}

constexpr long long N=200200;
constexpr bool MultiCase=true;
mt19937 Rnd(chrono::system_clock::now().time_since_epoch().count());
constexpr int MOD=1000000447;
long long pw[N],hs[N];
int lg[N],Ans[N];
const int B=Rnd()%520+30;
struct Barret{
    long long coef,p;
    void Init(int P){
        coef=((__int128)1<<64)/P;
        p=P;
    }
    long long operator() (long long x){
        return x-((__int128(x)*coef)>>64)*p;
    }
};
Barret Reduce;
inline int Has(int l,int r){
    return Reduce(hs[r]-Reduce(hs[l-1]*pw[r-l+1]));
}
inline int SolveMain([[maybe_unused]]long long TC){
    register int n=read(),L=read(),R=read();
    string s=" ";
    for(register int i=1;i<=n;++i){
        s+=getchar();
    }
    pw[0]=1;
    for(register int i=1;i<=n;++i){
        hs[i]=Reduce(hs[i-1]*B+s[i]-'a');
        Ans[i]=-1;
    }
    auto check=[&](int lc){
        if(lc>n)return 0;
        register int lcp=Has(1,lc),cnt=1;
        for(register int i=lc+1;i<=n-lc+1;++i){
            Has(i,i+lc-1)==lcp?++cnt,i=i+lc-1:0;
        }
        return cnt;
    };
    auto Binarys=[&](int i){
        if(i>R)return 0;
        register int &ans=Ans[i];
        if(~ans)return ans;ans=0;
        register int l=1,r=n/i,mid;
        for(mid=l+r>>1;l<=r;mid=l+r>>1){
            check(mid)>=i?ans=mid,l=mid+1:r=mid-1;
        }
        return ans;
    };
    const int BB=sqrt(n)+1;
    for(register int i=L;i<=BB;++i){
        Binarys(i);
    }
    for(register int i=1;i<=BB;++i){
        Cmax(Ans[min(check(i),R)],i);
    }
    Ans[R+1]=0;
    for(register int i=R;i>=L;--i){
        Cmax(Ans[i],Ans[i+1]);
    }
    for(register int i=L;i<=R;++i){
        write(Ans[i],' ');
    }
    putchar('\n');
    return 0;
}
inline void InitMain(){
    Reduce.Init(MOD);
    pw[0]=1;
    for(register int i=1;i<=200000;++i){
        pw[i]=Reduce(pw[i-1]*B);
    }
	return void();
}
int main(){
//	OF("test");
	InitMain();
	Test(){
		SolveMain(TestCase);
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 1 5
aaaaa
5 1 5
abcde
5 1 5
ababa
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

Buggy Code:
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
template <int P>
class mod_int
{
    using Z = mod_int;

private:
    static int mo(int x) { return x < 0 ? x + P : x; }

public:
    int x;
    int val() const { return x; }
    mod_int() : x(0) {}
    template <class T>
    mod_int(const T &x_) : x(x_ >= 0 && x_ < P ? static_cast<int>(x_) : mo(static_cast<int>(x_ % P))) {}
    bool operator==(const Z &rhs) const { return x == rhs.x; }
    bool operator!=(const Z &rhs) const { return x != rhs.x; }
    Z operator-() const { return Z(x ? P - x : 0); }
    Z pow(long long k) const
    {
        Z res = 1, t = *this;
        while (k)
        {
            if (k & 1)
                res *= t;
            if (k >>= 1)
                t *= t;
        }
        return res;
    }
    Z &operator++()
    {
        x < P - 1 ? ++x : x = 0;
        return *this;
    }
    Z &operator--()
    {
        x ? --x : x = P - 1;
        return *this;
    }
    Z operator++(int)
    {
        Z ret = x;
        x < P - 1 ? ++x : x = 0;
        return ret;
    }
    Z operator--(int)
    {
        Z ret = x;
        x ? --x : x = P - 1;
        return ret;
    }
    Z inv() const { return pow(P - 2); }
    Z &operator+=(const Z &rhs)
    {
        (x += rhs.x) >= P && (x -= P);
        return *this;
    }
    Z &operator-=(const Z &rhs)
    {
        (x -= rhs.x) < 0 && (x += P);
        return *this;
    }
    Z operator-() { return -x; }
    Z &operator*=(const Z &rhs)
    {
        x = 1ULL * x * rhs.x % P;
        return *this;
    }
    Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
#define setO(T, o)                                  \
    friend T operator o(const Z &lhs, const Z &rhs) \
    {                                               \
        Z res = lhs;                                \
        return res o## = rhs;                       \
    }
    setO(Z, +) setO(Z, -) setO(Z, *) setO(Z, /)
#undef setO
        friend istream &
        operator>>(istream &is, mod_int &x)
    {
        long long tmp;
        is >> tmp;
        x = tmp;
        return is;
    }
    friend ostream &operator<<(ostream &os, const mod_int &x)
    {
        os << x.val();
        return os;
    }
};
#define getchar() p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++
char buf[1000000], *p1 = buf, *p2 = buf;
template <typename T>
void read(T &x)
{
    x = 0;
    int f = 1;
    char c = getchar();
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-')
            f = -f;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    x *= f;
}
template <typename T, typename... Args>
void read(T &x, Args &...y)
{
    read(x);
    read(y...);
}
template <class T>
void write(T x)
{
    static int stk[30];
    if (x < 0)
        putchar('-'), x = -x;
    int top = 0;
    do
    {
        stk[top++] = x % 10, x /= 10;
    } while (x);
    while (top)
        putchar(stk[--top] + '0');
}
template <class T>
void write(T x, char lastChar) { write(x), putchar(lastChar); }
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
typedef mod_int<998244353> Hash;
const int B = rnd() % 520 + 30;
Hash pw[200020];
Hash h[200020];
int lg[200020];
Hash H(int l, int r) { return h[r] - h[l - 1] * pw[r - l + 1]; }
void solve()
{
    int n, L, R;
    string s = " ";
    read(n, L, R);
    for (int i = 1; i <= n; i++)
        s += getchar();
    pw[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        h[i] = h[i - 1] * B + s[i] - 'a' + 1;
        pw[i] = pw[i - 1] * B;
    }
    auto check = [&](int lcp, int cnt)
    {
        if (1LL * cnt * lcp > n)
            return -1 > 0;
        Hash LCP = H(1, lcp);
        cnt--;
        if (cnt <= 0)
            return -1 < 0;
        for (int i = lcp + 1; i <= n - lcp + 1; i++)
        {
            if (i + 1LL * (cnt - 1) * lcp > n)
                return -1 > 0;
            if (H(i, i + lcp - 1) == LCP)
            {
                cnt--;
                i = i + lcp - 1;
                if (cnt <= 0)
                    return -1 < 0;
            }
        }
        return cnt <= 0;
    };
    int Calc[n + 1] = {};
    memset(Calc, -1, sizeof(Calc));
    auto calc = [&](int i)
    {
        if (i > R)
            return -1;
        int &ans = Calc[i];
        if (~ans)
            return ans;
        ans = 0;
        int l = 1, r = n;
        while (l <= r)
        {
            int mid = l + r >> 1;
            if (check(mid, i))
                l = (ans = mid) + 1;
            else
                r = mid - 1;
        }
        return ans;
    };
    const int B = sqrt(n);
    for (int i = L; i <= R; i++)
    {
        if (i <= B)
        {
            write(calc(i), ' ');
            continue;
        }
        int ans = calc(i);
        int pos = i;
        int l = i + 1, r = n;
        while (l <= r)
        {
            int mid = l + r >> 1;
            if (calc(mid) == ans)
                l = (pos = mid) + 1;
            else
                r = mid - 1;
        }
        while (i <= pos)
            write(ans, ' '), i++;
        i--;
    }
    putchar('\n');
}
int main()
{
    int t;
    read(t);
    while (t--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 1 5
aaaaa
5 2 5
abcde
10 1 10
ababababab
```

3
5 1 5
aaaaa
5 2 5
abcde
10 1 10
ababababab


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

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

public class Main {
    static Map<Integer, Integer> map;

    private static void sport(String s, int l, int r) {
        map = new HashMap<>();
        for (int i = l; i < r + 1; i++) {
            int ll = 0;
            int rr = s.length() + 1;
            int ans = 0;
            while (ll + 1 < rr) {
                int m = (ll + rr) >> 1;
                Integer memo = map.get(m);
                int parts = memo == null ? sport1(s, m) : memo;
                if (parts >= i) {
                    ll = m;
                    ans = m;
                } else {
                    rr = m;
                }
            }
            System.out.print(ans + " ");
        }
        System.out.println();
    }

    private static int sport1(String s, int prefix) {
        int l = 0;
        int r = s.length();
        int[] z = zFucntion(s);
        while (l + 1 < r) {
            int m = (l + r) >> 1;
            if (isOk(z, prefix, m)) {
                l = m;
            } else {
                r = m;
            }
        }
        if (isOk(z, prefix, r)) {
            map.put(prefix, r);
            return r;
        }
        map.put(prefix, l);
        return l;
    }

    static boolean isOk(int[] z, int prefixLen, int partsNeeded) {
        int cnt = 0;
        int pp = 0;
        while (pp < z.length) {
            if (z[pp] >= prefixLen) {
                cnt++;
                pp += prefixLen;
            } else {
                pp++;
            }
        }
        return cnt >= partsNeeded;
    }

    static int[] zFucntion(String s) {
        int n = s.length();
        int z[] = new int[n];
        int R = 0;
        int L = 0;

        for (int i = 1; i < n; i++) {
            z[i] = 0;
            if (R > i) {
                z[i] = Math.min(R - i, z[i - L]);
            }
            while (i + z[i] < n && s.charAt(i + z[i]) == s.charAt(z[i])) {
                z[i]++;
            }
            if (i + z[i] > R) {
                L = i;
                R = i + z[i];
            }
        }

        z[0] = n;

        return z;
    }

    static void shuffleArray(int[] ar) {
        // If running on Java 6 or older, use `new Random()` on RHS here
        Random rnd = ThreadLocalRandom.current();
        for (int i = ar.length - 1; i > 0; i--) {
            int index = rnd.nextInt(i + 1);
            // Simple swap
            int a = ar[index];
            ar[index] = ar[i];
            ar[i] = a;

        }
    }

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int l = sc.nextInt();
            int r = sc.nextInt();
            String s = sc.next();
            sport(s, l, r);
        }
    }

    static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    private static <T> void swap(T[] elements, int a, int b) {
        T tmp = elements[a];
        elements[a] = elements[b];
        elements[b] = tmp;
    }

    static class UF {
        private int[] parent;
        private int[] rank;

        public UF(int n) {
            parent = new int[n + 1];
            rank = new int[n + 1];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                rank[i] = 1;
            }
        }

        public int find(int x) {
            if (parent[x] != x) {
                parent[x] = find(parent[x]);
            }
            return parent[x];
        }

        public void union(int x, int y) {
            int rootX = find(x);
            int rootY = find(y);

            if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
                rank[rootY] += rank[rootX];
            } else if (rank[rootX] >= rank[rootY]) {
                parent[rootY] = rootX;
                rank[rootX] += rank[rootY];
            }
        }
    }

    static class BIT {

        // The size of the array holding the Fenwick tree values
        final int N;

        // This array contains the Fenwick tree ranges
        private long[] tree;

        // Create an empty Fenwick Tree with 'sz' parameter zero based.
        public BIT(int sz) {
            tree = new long[(N = sz + 1)];
        }

        // Construct a Fenwick tree with an initial set of values.
        // The 'values' array MUST BE ONE BASED meaning values[0]
        // does not get used, O(n) construction.
        public BIT(long[] values) {

            if (values == null) {
                throw new IllegalArgumentException("Values array cannot be null!");
            }

            N = values.length;
            values[0] = 0L;

            // Make a clone of the values array since we manipulate
            // the array in place destroying all its original content.
            tree = values.clone();

            for (int i = 1; i < N; i++) {
                int parent = i + lsb(i);
                if (parent < N) {
                    tree[parent] += tree[i];
                }
            }
        }

        // Returns the value of the least significant bit (LSB)
        // lsb(108) = lsb(0b1101100) = 0b100 = 4
        // lsb(104) = lsb(0b1101000) = 0b1000 = 8
        // lsb(96) = lsb(0b1100000) = 0b100000 = 32
        // lsb(64) = lsb(0b1000000) = 0b1000000 = 64
        private static int lsb(int i) {

            // Isolates the lowest one bit value
            return i & -i;

            // An alternative method is to use the Java's built in method
            // return Integer.lowestOneBit(i);

        }

        // Computes the prefix sum from [1, i], O(log(n))
        private long prefixSum(int i) {
            long sum = 0L;
            while (i != 0) {
                sum += tree[i];
                i &= ~lsb(i); // Equivalently, i -= lsb(i);
            }
            return sum;
        }

        // Returns the sum of the interval [left, right], O(log(n))
        public long sum(int left, int right) {
            if (right < left) {
                throw new IllegalArgumentException("Make sure right >= left");
            }
            return prefixSum(right) - prefixSum(left - 1);
        }

        // Get the value at index i
        public long get(int i) {
            return sum(i, i);
        }

        // Add 'v' to index 'i', O(log(n))
        public void add(int i, long v) {
            while (i < N) {
                tree[i] += v;
                i += lsb(i);
            }
        }

        // Set index i to be equal to v, O(log(n))
        public void set(int i, long v) {
            add(i, v - sum(i, i));
        }

        @Override
        public String toString() {
            return Arrays.toString(tree);
        }
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

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

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

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

        double nextDouble() {
            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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to consider:

1. **Edge Cases for String Length and Range**:
   - Test with the minimum and maximum possible values for `n`, `l`, and `r`.
   - Test with `l` equal to `r` and `l` different from `r`.

2. **Edge Cases for String Content**:
   - Test with strings that have a lot of repeated characters.
   - Test with strings that have a lot of unique characters.
   - Test with strings that have a mix of repeated and unique characters.

3. **Edge Cases for Division**:
   - Test with strings that have a clear and simple division.
   - Test with strings that have a complex division.

Given the constraints and the need to generate a small-scale test input, let's create a test case that includes a mix of these edge cases:

```plaintext
1
6 1 3
aabbcc
```

### Explanation:
- **Test Case Count**: `1` (to keep it simple)
- **String Length**: `6` (a small but non-trivial length)
- **Range**: `1 3` (testing with a range that includes multiple values for `k`)
- **String Content**: `aabbcc` (a string with repeated characters but also a clear division)

This test case is designed to:
- Check the algorithm's ability to handle a small but non-trivial string length.
- Verify the algorithm's performance with a range of `k` values.
- Test the algorithm's ability to handle repeated characters and clear divisions.

This input should help in identifying any issues in the algorithm, especially in the `sport` and `sport1` methods.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Division + LCP (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version $$$l\le r$$$.

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 $$$f_l,f_{l+1},\dots,f_r$$$.

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 $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le l \le r \le n \le 2 \cdot 10^5$$$) — 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 1 3
aba
3 2 3
aaa
7 1 5
abacaba
9 1 6
abababcab
10 1 10
aaaaaaawac
9 1 9
abafababa
7 2 7
vvzvvvv
output:
3 1 0
1 1
7 3 1 1 0
9 2 2 2 0 0
10 3 2 1 1 1 1 1 0 0
9 3 2 1 1 0 0 0 0
2 2 1 1 1 0

Note:

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 Integer[] dp;
    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();
            dp = new Integer[n];
            ArrayList<Long> ans = new ArrayList<>();
            for (int i = l; i <= r; i++) {
                int start = 0;
                int end = n / i;
                int val = -1;
                while(start < end){
                    int mid = start + (end - start) / 2;
                    int curr = ok(mid, a, i);
                    if (curr >= i) {
                        val = mid;
                        start = mid + 1;
                    }else{
                        end = mid;
                    }
                }
                ans.add((long) val + 1);
            }
            printlist(ans);
        }
	pw.close();
    }
    public static int ok(int mid, String a, int k){
        if(dp[mid] != null){
            // pw.println(a);
            return dp[mid];
        }
        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 (a.charAt(j) == s.charAt(i)) {
                    i++;
                }
            }
            if(i == s.length()){
                ans++;
                i = 0;
            }
        }
        return dp[mid] = ans;
    }
    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 create a challenging test input that exposes a bug in the provided buggy code, let's consider the following:

1. **Edge Cases:**
   - Test cases where the string is very repetitive (e.g., all characters are the same).
   - Test cases where the string has a pattern that might cause the KMP algorithm to fail.
   - Test cases with very small or very large values of `l` and `r`.

2. **Maximizing Failure Probability:**
   - Use a string that has a repeating pattern but with slight variations to test the KMP algorithm's ability to handle such cases.
   - Use values of `l` and `r` that are close to each other and to the length of the string to test the binary search and KMP logic.

3. **Small-Scale Test Input:**
   - Ensure the test input is small enough to be manageable but complex enough to challenge the algorithm.

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

```plaintext
1
7 2 7
abababa
```

### Explanation:
- **Test Case:**
  - `n = 7`: The length of the string.
  - `l = 2`: The minimum number of substrings.
  - `r = 7`: The maximum number of substrings.
- **String:**
  - `s = "abababa"`: A string with a repeating pattern "aba" that might cause issues in the KMP algorithm if not handled correctly.

This test case is designed to:
- Test the KMP algorithm's ability to handle a repeating pattern.
- Ensure that the binary search logic correctly identifies the maximum possible LCP for different values of `k` from 2 to 7.
- Expose any potential issues with the dynamic programming array `dp` and the KMP prefix array `lps`.

By using this test input, we can effectively challenge the buggy code and potentially expose any hidden bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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 $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

Examples:
input:
3
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
output:
Bob
Alice
Alice
Bob
Alice
Bob
Alice

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
// Author: Klay Thompson
// Problem: F2. Game in Tree (Hard Version)
// Memory Limit: 256 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define per(i, j, k) for (int i = (j); i >= (k); --i)
#define SZ(v) int((v).size())
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;

template<class T>inline void chkmn(T &x, T y) { if (y < x) x = y; }
template<class T>inline void chkmx(T &x, T y) { if (y > x) x = y; }

using namespace std;

const int maxn = 200010;

int n, m, s, t, fa[maxn], mxd[maxn], cmx[maxn], dep[maxn], amn[maxn], bmx[maxn];
int p[maxn], a[maxn], b[maxn], vis[maxn], ans[maxn];
int sta[maxn][20], stb[maxn][20];
vector<int> e[maxn];

void dfs(int u, int s) {
  dep[u] = dep[fa[u]] + 1;
  mxd[u] = cmx[u] = 1, vis[u] = (u == s);
  for (int v : e[u]) {
    if (v == fa[u]) continue;
    fa[v] = u, dfs(v, s), vis[u] |= vis[v];
    if (!vis[v]) chkmx(mxd[u], mxd[v] + 1);
    chkmx(cmx[u], cmx[v] + 1);
  }
}

int quea(int l, int r) {
  if (l > r) return -1e9;
  int k = __lg(r - l + 1);
  return max(sta[l][k], sta[r - (1 << k) + 1][k]);
}

int queb(int l, int r) {
  if (l > r) return -1e9;
  int k = __lg(r - l + 1);
  return max(stb[l][k], stb[r - (1 << k) + 1][k]);
}

void solve(int s) {
  dfs(1, s);
  m = 0;
  while (s) p[++m] = s, s = fa[s];
  reverse(p + 1, p + m + 1);
  rep (i, 1, m) a[i] = mxd[p[i]] + i - 1;
  rep (i, 1, m) b[i] = mxd[p[i]] + m - i;
  rep (i, 1, m) sta[i][0] = a[i], stb[i][0] = b[i];
  rep (j, 1, 19) rep (i, 1, m - (1 << j) + 1) {
    sta[i][j] = max(sta[i][j - 1], sta[i + (1 << (j - 1))][j - 1]);
    stb[i][j] = max(stb[i][j - 1], stb[i + (1 << (j - 1))][j - 1]);
  }
  rep (i, 1, m) amn[i] = m + 1;
  rep (i, 2, m) {
    int l = i, r = m, res = i - 1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      int pos = mid - (i - 1);
      if (a[i] > queb(i + 1, pos) - m + mid) res = mid, l = mid + 1;
      else r = mid - 1;
    }
    chkmn(amn[res], i);
  }
  per (i, m - 1, 1) chkmn(amn[i], amn[i + 1]);
  rep (i, 1, m) if (a[1] > queb(2, i - 1) - m + i && a[1] > cmx[p[i]]) amn[i] = 1;
  rep (i, 1, m) bmx[i] = 0;
  rep (i, 1, m) {
    int l = i + 1, r = m, res = m + 1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      int pos = 1 + (mid - i) + 1;
      if (b[i] - m + mid >= quea(pos, i - 1)) res = mid, r = mid - 1;
      else l = mid + 1;
    }
    chkmx(bmx[res], i);
  }
  rep (i, 2, m) chkmx(bmx[i], bmx[i - 1]);
  rep (i, 1, m) if (b[i] - m + i >= quea(2, i - 1)) bmx[i] = i;
  rep (i, 1, m) {
    if (amn[i] == m + 1 && bmx[i] == 0) {
      ans[p[i]] = i & 1;
    } else {
      int lena = amn[i] - 1, lenb = i - bmx[i];
      ans[p[i]] = (lena <= lenb);
    }
  }
  // int x = 1, y = m;
  // while (true) {
    // if (x >= y || a[x] > queb(x + 1, y)) return cout << "Alice\n", void();
    // x++;
    // if (x >= y || b[y] >= quea(x, y - 1)) return cout << "Bob\n", void();
    // y--;
  // }
}

void work() {
  cin >> n;
  rep (i, 1, n) e[i].clear();
  rep (i, 1, n - 1) {
    int u, v;
    cin >> u >> v;
    e[u].emplace_back(v);
    e[v].emplace_back(u);
  }
  cin >> s >> t;
  solve(s), solve(t);
  int x = s, y = t;
  while (dep[x] > dep[y]) x = fa[x];
  while (dep[x] < dep[y]) y = fa[y];
  while (x != y) x = fa[x], y = fa[y];
  vector<int> L, R;
  while (s != x) L.emplace_back(s), s = fa[s];
  while (t != y) R.emplace_back(t), t = fa[t];
  reverse(ALL(R));
  for (int x : L) cout << (ans[x] ? "Alice" : "Bob") << '\n';
  cout << (ans[x] ? "Alice" : "Bob") << '\n';
  for (int x : R) cout << (ans[x] ? "Alice" : "Bob") << '\n';
}

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

**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, let's focus on a few key aspects:

1. **Edge Cases**:
   - When the path from \(u\) to \(v\) is very short.
   - When the path from \(u\) to \(v\) includes the root vertex (vertex 1).
   - When the tree is a straight line (a path graph).

2. **Complex Scenarios**:
   - When the tree has a complex structure with multiple branches.
   - When the starting vertices \(u\) and \(v\) are far apart.

3. **Specific Bug Indicators**:
   - The code seems to handle the path from \(u\) to \(v\) by reversing and reprocessing it. We can test a scenario where this might cause issues.

Given these considerations, let's create a small-scale test input that includes a path from \(u\) to \(v\) that is not straightforward and might expose the bug.

### Test Input
```plaintext
1
6
1 2
1 3
2 4
2 5
1 6
4 5
```

### Explanation
- **Tree Structure**:
  - Vertex 1 is connected to vertices 2 and 3.
  - Vertex 2 is connected to vertices 4 and 5.
  - Vertex 1 is also connected to vertex 6.
  - The path from \(u = 4\) to \(v = 5\) is \(4 \rightarrow 2 \rightarrow 5\).

- **Expected Output**:
  - Alice starts at vertex 1.
  - Bob starts at vertices 4, 2, and 5.
  - The expected output should be:
    ```
    Alice
    Bob
    Alice
    ```

This test input is designed to be small and manageable while still being complex enough to potentially expose issues in the path handling and the logic for determining the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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 $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

Examples:
input:
3
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
output:
Bob
Alice
Alice
Bob
Alice
Bob
Alice

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
//I'm sorry.
//No,it's too late to say it.
//This is where friendship collapse.
//If you like me,it's never to be late.
#include <bits/stdc++.h>
using namespace std;
using ll=long long;

#define notall(x) x.begin()+1,x.end()
#define all(x) x.begin(),x.end()
#define mul_t int _t;cin>>_t;while(_t--)
#define seed chrono::steady_clock::now().time_since_epoch().count()
#define r_engine default_random_engine(seed)

const int int_inf = 1000000100;
const ll ll_inf = 4000000000000000400;

template<class T>
void writeln(const T &t) {
    cout << t << '\n';
}
template<class T, class...args>
void writeln(const T &t, const args &...rest) {
    cout << t << ' ';
    writeln(rest...);
}
template<class T1>
void writeln(const vector<T1> &v) {
    for (auto c : v)
        cout << c << ' ';
    cout << '\n';
}
template<class T1, class T2>
void writeln(const vector<T1> &v, T2 n) {
    for (T2 i = 1; i <= n; i++)
        cout << v[i] << ' ';
    cout << '\n';
}
template<class T1, class T2>
void writeln(const pair<T1, T2> &p) {
    cout << p.first << ' ' << p.second << '\n';
}
void writeln() {
   cout << endl;
}
template <typename T>
struct sparse_table
{
    vector<vector<T>> STMX, STMN;
    sparse_table(vector<T> &A)
    {
        int N = A.size();
        int LOG = __lg(N) + 1;
        STMN = vector<vector<T>>(LOG, vector<T>(N));
        STMX = vector<vector<T>>(LOG, vector<T>(N));
        for (int i = 0; i < N; i++)
        {
            STMN[0][i] = A[i];
            STMX[0][i] = A[i];
        }
        for (int j = 1; j < LOG; j++)
        {
            for (int i = 0; i + (1 << (j - 1)) < N; i++)
            {
                STMN[j][i] = min(STMN[j - 1][i], STMN[j - 1][i + (1 << (j - 1))]);
                STMX[j][i] = max(STMX[j - 1][i], STMX[j - 1][i + (1 << (j - 1))]);
            }
        }
    }
    T range_min(int L, int R)
    {
        int d = __lg(R - L + 1);
        return min(STMN[d][L], STMN[d][R - (1 << d) + 1]);
    }
    T range_max(int L, int R)
    {
        if(L>R)return -int_inf;
        // assert(L>=0&&R<STMX.size());
        int d = __lg(R - L + 1);
        return max(STMX[d][L], STMX[d][R - (1 << d) + 1]);
    }
};
void solve(){
    int n;
    cin>>n;
    vector<vector<int>>e(n+1);
    for(int i=1;i<n;i++)
    {
        int u,v;
        cin>>u>>v;
        e[u].push_back(v),e[v].push_back(u);
    }
    int u,v;
    cin>>u>>v;
    vector<int>ans(n+1);
    //a i+mxdep[i]
    //b m-i-1+mxdep[i]
    auto cacl=[&](int u)
    {
        //1~u
        vector<int>fa(n+1),mxdep(n+1);
        auto dfs=[&](auto&&self,int u,int x,int sp)->bool
        {
            int f=(u==sp);
            mxdep[u]=1,fa[u]=x;
            for(auto v:e[u])
            {
                if(v==x)continue;
                int curf=self(self,v,u,sp);
                f|=curf;
                if(!curf)mxdep[u]=max(mxdep[u],mxdep[v]+1);
            }
            return f;
        };
        dfs(dfs,1,0,u);
        int uu=u;
        vector<int>line;
        while(uu)line.push_back(uu),uu=fa[uu];
        reverse(all(line));
        int sz=line.size();
        vector<int>a(sz+1),b(sz+1);
        vector<int>fira(sz+1,int_inf),firb(sz+1,-int_inf);
        vector<vector<int>>add(sz+1),del(sz+1);
        for(int i=0;i<sz;i++)a[i]=mxdep[line[i]]+i,b[i]=mxdep[line[i]]-i;
        sparse_table sta(a),stb(b);
        multiset<int>ms;
        //alice
        for(int i=1;i*2<sz;i++)
        {
            int l=i*2+1,r=sz-1,pos=-1;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(a[i]>mid+stb.range_max(i+1,mid-i))
                {
                    pos=mid,l=mid+1;
                }
                else r=mid-1;
            }
            if(~pos)add[sz].push_back(i),del[pos+1].push_back(i);
        }
        for(int i=1;i<sz;i++)
        {
            for(auto x:add[i])ms.insert(x);
            for(auto x:del[i])ms.erase(ms.find(x));
            if(ms.empty())fira[i]=int_inf;
            else fira[i]=*ms.begin();
        }
        //clear
        ms.clear();
        add.assign(sz+1,{}),del.assign(sz+1,{});
        //bob
        for(int i=1;i*2<=sz;i++)
        {
            int l=i,r=i*2-1,pos=-1;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(sta.range_max(mid-i+1,i-1)<=b[i]+mid)
                {
                    pos=mid;
                    r=mid-1;
                }
                else l=mid+1;
            }
            if(~pos)add[pos].push_back(i),del[2*i].push_back(i);
        }
        for(int i=1;i<sz;i++)
        {
            for(auto x:add[i])ms.insert(x);
            for(auto x:del[i])ms.erase(ms.find(x));
            if(ms.empty())firb[i]=-int_inf;
            else firb[i]=*ms.rbegin();
        }
        //再算链上的贡献和顶点的贡献
        for(int i=sz-1;i>=1;i--)
        {
            b[i]=max(b[i+1]+1,mxdep[line[i]]);
            if(b[i]>=sta.range_max(1,i-1))firb[i]=i;
            if(a[0]>max(b[i],stb.range_max(1,i)+i))fira[i]=0;
            // writeln(line[i],fira[i],firb[i]);
            ans[line[i]]=fira[i]<=i-firb[i];
        }
    };
    cacl(u),cacl(v);
    vector<int>fa(n+1);
    auto get=[&](auto&&self,int u,int x)->void
    {
        fa[u]=x;
        for(auto v:e[u])
        {
            if(v==x)continue;
            self(self,v,u);
        }
    };
    get(get,v,0);
    vector<int>line;
    int uu=u;
    while(uu)line.push_back(uu),uu=fa[uu];
    for(auto x:line)writeln(ans[x]?"Alice":"Bob");
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(15);
    mul_t
    solve();
}
/*
1.深呼吸，不要紧张，慢慢读题，读明白题，题目往往比你想的简单。
2.暴力枚举:枚举什么，是否可以使用一些技巧加快枚举速度（预处理、前缀和、数据结构、数论分块）。
3.贪心:需要排序或使用数据结构（pq）吗，这么贪心一定最优吗。
4.二分：满足单调性吗，怎么二分，如何确定二分函数返回值是什么。
5.位运算：按位贪心，还是与位运算本身的性质有关。
6.数学题：和最大公因数、质因子、取模是否有关。
7.dp：怎么设计状态，状态转移方程是什么，初态是什么，使用循环还是记搜转移。
8.搜索：dfs 还是 bfs ，搜索的时候状态是什么，需要记忆化吗。
9.树上问题：是树形dp、树上贪心、或者是在树上搜索。
10.图论：依靠什么样的关系建图，是求环统计结果还是最短路。
11.组合数学：有几种值，每种值如何被组成，容斥关系是什么。
12.交互题：log(n)次如何二分，2*n 次如何通过 n 次求出一些值，再根据剩余次数求答案。
13.如果以上几种都不是，多半是有一个 point 你没有注意到，记住正难则反。
*/

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

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

Think step by step.

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

1. **Understand the Bug**: The code is supposed to determine the winner of the game based on the starting positions of Alice and Bob. The bug might be in how the algorithm handles the path from vertex `u` to `v` and the turn-taking logic.

2. **Identify Edge Cases**:
   - **Small Trees**: Test with the smallest possible tree to ensure the algorithm works correctly.
   - **Long Paths**: Test with a long path to see if the algorithm can handle large inputs.
   - **Complex Trees**: Test with a tree that has multiple branches and a complex path from `u` to `v`.

3. **Create a Test Case**:
   - **Simple Path**: Start with a simple path where `u` and `v` are connected by a single path.
   - **Branching Path**: Create a tree with branches to ensure the algorithm correctly handles different paths.

Let's create a test case that is small but can expose potential issues in the algorithm:

### Test Case 1: Simple Path
- **Tree**: A simple path from 1 to 5.
- **Vertices**: 5
- **Edges**: (1, 2), (2, 3), (3, 4), (4, 5)
- **Path**: u = 2, v = 5

### Test Case 2: Branching Path
- **Tree**: A tree with branches.
- **Vertices**: 7
- **Edges**: (1, 2), (1, 3), (2, 4), (2, 5), (3, 6), (3, 7)
- **Path**: u = 4, v = 7

### Test Input
```plaintext
2
5
1 2
2 3
3 4
4 5
2 5
7
1 2
1 3
2 4
2 5
3 6
3 7
4 7
```

This test input includes a simple path and a more complex tree with branches. It should help in identifying any issues with the path handling and turn-taking logic in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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 $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

Examples:
input:
3
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
output:
Bob
Alice
Alice
Bob
Alice
Bob
Alice

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
#include <algorithm>
#include <cassert>
#include <cstdio>
#include <vector>

using namespace std;

typedef vector<int> vi;
#define all(x) begin(x), end(x)
int const INF = 0x3f3f3f3f;

template <typename T>
bool ckmax(T& a, T const& b) { return b > a ? a = b, 1 : 0; }
template <typename T>
bool ckmin(T& a, T const& b) { return b < a ? a = b, 1 : 0; }

struct ST {
	private:
	int N;
	vector<int> v;

	public:
	ST(vector<int> const& vin)
			: N(vin.size()), v(N * 2)
	{
		for(int i = N;i < N * 2; ++i)
			v[i] = vin[i - N];
		for(int i = N-1;i; --i)
			v[i] = max(v[i*2], v[i*2+1]);
	}
	int query(int l, int r)
	{
		int ans = -INF;
		for(l += N, r += N;l < r; l /= 2, r /= 2) {
			if(l & 1) ckmax(ans, v[l++]);
			if(r & 1) ckmax(ans, v[--r]);
		}
		return ans;
	}
};

void solve()
{
	int N;
	scanf("%d", &N);
	vector<vi> adj(N);
	for (int i = 0, a, b; i + 1 < N; ++i) {
		scanf("%d%d", &a, &b);
		--a, --b;
		adj[a].push_back(b);
		adj[b].push_back(a);
	}

	vector<int> d(N), p(N, -1);
	[&](this auto&& dfs, int n) -> void {
		for (int x : adj[n])
			if (x != p[n]) {
				p[x] = n;
				d[x] = d[n] + 1;
				dfs(x);
			}
	}(0);

	auto get_ans = [&](int target) {
		vector<int> path;
		for (int n = target;; n = p[n]) {
			path.push_back(n);
			if (n == 0)
				break;
		}
		reverse(all(path));
		// Path[0] = closest to node 0
		vi skip(N);
		for (int x : path)
			skip[x] = 1;

		auto far = [&](this auto& self, int n, int p = -1) -> int {
			int ans = 0;
			for (int x : adj[n])
				if (x != p && !skip[x])
					ckmax(ans, self(x, n));
			return ans;
		};

		vi a_win_time(path.size(), INF), a_jmp(path.size() + 1, -1);
		vi b_win_time(path.size(), INF), b_jmp(path.size() + 1, -1);
		auto try_upd = [&]<typename T, typename F>(vector<T>& v, vi& jmp, F const& val, int n, int r) {
			int V = v.size();
			ckmin(r, V);
			if (n > V)
				return;
			ckmax(n, 0);
			assert(0 <= n && n <= V);

			auto nxt = [&](this auto& self, int n) -> int { return jmp[n] == -1 ? n : jmp[n] = self(jmp[n]); };
			for (n = nxt(n); n < r; n = nxt(n)) {
				v[n] = val(n);
				jmp[n] = n + 1;
			}
		};

		vi ans(path.size(), -1);

		vi distances(path.size());
		for (int i = 0; i < path.size(); ++i)
			distances[i] = far(path[i]) + i;
		ST st_pos(distances);
		for (int i = 0; i < path.size(); ++i)
			distances[i] -= 2 * i;
		ST st_neg(distances);
		for (int i = 0; i < path.size(); ++i)
			distances[i] += i;

		for (int i = 1; i < path.size(); ++i) {
			// Check if Alice beats path[i] immediately
			int dist = max( // can be optimized to O(N), but might as well use segtree if we have it
					st_pos.query(i, path.size()) - i,
					st_neg.query(1, i + 1) + i);
			if (distances[0] > dist)
				try_upd(a_win_time, a_jmp, [](int n) { return 0; }, i, i + 1);
		}
		for (int i = 1; i < path.size(); ++i) {
			// who does A beat? [timestep i * 2]
			int l = i, r = path.size();
			for (; r - l > 1;) {
				int m = l + (r - l) / 2;
				int dist = st_neg.query(i + 1, m + 1) + m;
				if (distances[i] > m) // A wins
					l = m;
				else
					r = m;
			}
			// A beats [i+1,r) (r is first that A loses)
			// This is after each player has taken (i) steps
			// => [2i+1,r+i)
			try_upd(a_win_time, a_jmp, [i](int n) { return 2 * i; }, 2 * i + 1, r + i);
		}

		// B wins on 1st turn
		for(int i = 2;i < path.size(); ++i) {
			int a_dist = max(
				distances[0] + 1,
				st_pos.query(1, i) - 1);
			int b_dist = st_pos.query(i, path.size()) - i;
			if(b_dist > a_dist)
				try_upd(b_win_time, b_jmp, [](int n) {return 1;}, i, i + 1);
		}
		for (int i = (int)path.size() - 1; i >= 1; --i) {
			// which A does B beat if B diverts at path[i]? [timestep i * 2 - 1]
			int l = 0, r = i;
			for (; r - l > 1;) {
				int m = l + (r - l) / 2;
				int dist = st_pos.query(m, i) - m;
				if (distances[i] > dist) // B wins
					r = m;
				else
					l = m;
			}
			// B at path[i] beats A at path[r..i]
			// A at (x) => (x*2-1 steps taken), B originally at (i + x - 1)
			// A at (r) => B at (i + r - 1)
			// A at (i) => B at (2 * i - 1)
			try_upd(b_win_time, b_jmp, [i](int n) {int x = (n - i + 1); return x * 2 - 1; }, i + r - 1, 2 * i - 1);
		}

		for (int i = 0; i < path.size(); ++i)
			if (a_win_time[i] == INF && b_win_time[i] == INF)
				ans[i] = i % 2; // Bob wins if odd
			else
				ans[i] = b_win_time[i] < a_win_time[i];

		return ans;
	};

	int u, v;
	scanf("%d%d", &u, &v);
	--u, --v;

	auto lca = [&](int a, int b) {
		while (a != b) {
			if (d[a] < d[b])
				swap(a, b);
			a = p[a];
		}
		return a;
	};
	int top = lca(u, v);

	vi u_ans = get_ans(u);
	vi v_ans = get_ans(v);
	assert(u_ans.size() == d[u] + 1);
	assert(v_ans.size() == d[v] + 1);
	for (int i = 0; i <= d[top]; ++i)
		assert(u_ans[i] == v_ans[i]);

	vi ans;
	for (int i = d[u]; i >= d[top]; --i)
		ans.push_back(u_ans[i]);
	for (int i = d[top] + 1; i <= d[v]; ++i)
		ans.push_back(v_ans[i]);
	for (auto x : ans)
		if (x)
			printf("Bob\n");
		else
			printf("Alice\n");
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
}

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

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

Think step by step.

Response:
```plaintext
3
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
```