**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
import java.util.*;
import java.io.*;
public class cptree {
    static class Data{// implements Comparable<Data>{
        int finval;
        int change;
        int depth;
        public Data(int finval){
            this.finval = finval;
            this.change = 0;
            this.depth = 0;
        }
        // @Override
        // public int compareTo(Data p1){
        //     if(this.low-2*this.high-1<=p1.low-2*p1.high-1) return 1;
        //     return -1;
        // }
    }
    static class Edge {
        int nd;
        long wt;
        public Edge(int nd,long wt){
            this.nd = nd;
            this.wt = wt;
        }
    }
    public static PP.Reader cs = new PP.Reader(System.in);
    // public static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException {
        int tests = cs.nextInt();
        while(tests>0){
            tests--;
            solve();
        }
        // writer.flush();
        // writer.close();
    }
    public static long mod = 1000000007;
    public static int rev;
    public static int finans;
    public static void solve() throws IOException{
        //int x = cs.nextInt();

        int n = cs.nextInt();
        // boolean[] fix = new boolean[n];
        ArrayList<ArrayList<Integer>> tree = new ArrayList<>();
        for(int i=0;i<n;i++) tree.add(new ArrayList<Integer>());
        int[][] binpar = new int[26][n];
        for(int i=1;i<n;i++){
            int t1 = cs.nextInt()-1;
            int t2 = cs.nextInt()-1;
            // long w = cs.nextLong();
            tree.get(t1).add(t2);
            tree.get(t2).add(t1);
        }

        long maxdep[][] = new long[n][2];
        int depth[] = new int[n];
        int maxchild[] = new int[n];
        binpar[0][0] = -1;
        dfstrdep(tree, 0, -1, binpar,maxdep,1,depth,maxchild);
        for(int j=1;j<=25;j++){
            for(int i=0;i<n;i++){
                int temp = binpar[j-1][i];
                if(temp == -1) binpar[j][i] = -1;
                else binpar[j][i] = binpar[j-1][temp];
            }
        }
        for(int i=0;i<n;i++){
            maxdep[i][0] -= depth[i];
            maxdep[i][1] -= depth[i];
        }
        dfstrdep2(tree, 0, -1, binpar, maxdep, 0, depth, maxchild);
        int q = cs.nextInt();
        while (q-- >0) {
            int nd = cs.nextInt()-1;
            int k = cs.nextInt();
            k = Math.min(k,depth[nd]-1);
            long ans = maxdep[nd][0]+depth[nd];
            if(k==0){
                System.out.print(ans+" ");
                continue;
            }
            int par = kth_ancestor(nd, k, binpar);
            int child = kth_ancestor(nd,k-1,binpar);
            // System.out.println(nd+" "+par+" "+k);
            if(maxchild[par]==child) ans = Math.max(ans, maxdep[par][1]+depth[nd]);
            else ans = Math.max(ans, maxdep[par][0]+depth[nd]);
            System.out.print(ans+" ");
            // System.out.println();
        }
        System.out.println();

    }
    //Witte fxn here
    public static long dfstrdep(ArrayList<ArrayList<Integer>> tree,int curr,int par,int[][] binpar ,long[][] maxdep,int dep,int[] depth,int child[]){
        depth[curr] = dep;
        long max1 = 0;
        long max2 = 0;
        for(Integer nb: tree.get(curr)){
            if(nb == par) continue;
            binpar[0][nb] = curr;
            long temp = dfstrdep(tree, nb, curr, binpar, maxdep, dep+1, depth,child);
            if(temp >= max1){
                child[curr] = nb;
                max2 = max1;
                max1 = temp;
            }
            else max2 = Math.max(max2, temp);
        }
        maxdep[curr][0] = max1;
        maxdep[curr][1] = max2;
        return max1+1;
    }
    public static void dfstrdep2(ArrayList<ArrayList<Integer>> tree,int curr,int par,int[][] binpar ,long[][] maxdep,int dep,int[] depth,int child[]){

        for(Integer nb: tree.get(curr)){
            if(nb == par) continue;
            dfstrdep2(tree, nb, curr, binpar, maxdep, dep+1, depth,child);
            if(nb==child[curr]){
                maxdep[curr][1] = Math.max(maxdep[curr][1],maxdep[nb][0]);
            }
        }
    }
    public static int kth_ancestor(int node,int k,int[][] binpar){
        int bit = 0;
        while (node != -1 && k>=(1<<bit)) {
            if( (k&(1<<bit)) != 0) node = binpar[bit][node];
            bit++;
        }
        return node;
    }

    public static long dfstrtot(ArrayList<ArrayList<Long>> tree,long curr,long par,long len[]){
        long below = 0;
        for(Long nb : tree.get((int)curr)){
            if(nb == par) continue;
            below += dfstrtot(tree, nb, curr, len);
        }
        len[(int)curr] = below;
        return below+1;
    }

    public static  int binary(int[] arr,int start,int end,int key){
        int mid = start;
        int st = start;
        int en = end;
        while(start<=end){
            mid = (start+end)/2;
            if(arr[mid]==key) return mid;
            else if(arr[mid]>key) end = mid-1;
            else if(arr[mid]<key) start = mid+1;
        }
        //if(end<start){
            if(end<st) end++;
            // if(start>en) start--;
            // if(Math.abs(arr[end]-key)<=Math.abs(arr[start]-key-1)){
                // return start;
            // }
            return end;
        //}
        //return mid+1;
    }

    // public static int invmod(int A, int M)
    // {
//
        // for (int X = 1; X < M; X++)
            // if (((A % M) * (X % M)) % M == 1)
                // return X;
        // return 1;
    // }
    //
    // public static int gcd(int a,int b){
    //     int c = Math.min(a,b);
    //     int d = Math.max(a,b);
    //     if(d%c==0) return c;
    //     return gcd(c,d%c);
    // }
}
class PP {
    static String IN = "%s";
    static String OUT = "%s";

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

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

        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 test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to think through this:

1. **Tree Structure**: Consider a tree structure that can create complex paths and ancestor relationships.
2. **Stamina and Distance**: Test cases with different stamina values that can lead to different paths and distances.
3. **Edge Cases**: Include edge cases such as a single vertex, a linear tree, and a tree with multiple branches.

### Test Input
Let's create a test input that includes a tree with multiple branches and different stamina values.

```plaintext
1
7
1 2
1 3
2 4
2 5
3 6
3 7
3
1 2
2 1
3 3
```

### Explanation
- **Tree Structure**:
  - Root: 1
  - Children of 1: 2, 3
  - Children of 2: 4, 5
  - Children of 3: 6, 7
- **Queries**:
  - From vertex 1 with 2 stamina.
  - From vertex 2 with 1 stamina.
  - From vertex 3 with 3 stamina.

### Expected Output
For the given test input, the expected output should be:
```
3 2 4
```

### Reasoning
- **Query 1**: From vertex 1 with 2 stamina:
  - Can move to 2 or 3.
  - From 2, can move to 4 or 5 (2 stamina - 1 = 1).
  - From 3, can move to 6 or 7 (2 stamina - 1 = 1).
  - The farthest distance is 3 (1 to 2 to 4 or 5, or 1 to 3 to 6 or 7).

- **Query 2**: From vertex 2 with 1 stamina:
  - Can move to 1 (stamina remains 1).
  - From 1, can move to 3 (stamina - 1 = 0).
  - The farthest distance is 2 (2 to 1 to 3).

- **Query 3**: From vertex 3 with 3 stamina:
  - Can move to 1 (stamina remains 3).
  - From 1, can move to 2 or 3 (stamina - 1 = 2).
  - From 2, can move to 4 or 5 (stamina - 1 = 1).
  - From 3, can move to 6 or 7 (stamina - 1 = 1).
  - The farthest distance is 4 (3 to 1 to 2 to 4 or 5, or 3 to 1 to 3 to 6 or 7).

This test input should help expose any bugs in the given code by testing various paths and stamina values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
import java.util.*;
import java.io.*;
public class cptree {
    static class Data{// implements Comparable<Data>{
        int finval;
        int change;
        int depth;
        public Data(int finval){
            this.finval = finval;
            this.change = 0;
            this.depth = 0;
        }
        // @Override
        // public int compareTo(Data p1){
        //     if(this.low-2*this.high-1<=p1.low-2*p1.high-1) return 1;
        //     return -1;
        // }
    }
    static class Edge {
        int nd;
        long wt;
        public Edge(int nd,long wt){
            this.nd = nd;
            this.wt = wt;
        }
    }
    public static PP.Reader cs = new PP.Reader(System.in);
    // public static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException {
        int tests = cs.nextInt();
        while(tests>0){
            tests--;
            solve();
        }
        // writer.flush();
        // writer.close();
    }
    public static long mod = 1000000007;
    public static int rev;
    public static int finans;
    public static void solve() throws IOException{
        //int x = cs.nextInt();

        int n = cs.nextInt();
        // boolean[] fix = new boolean[n];
        ArrayList<ArrayList<Integer>> tree = new ArrayList<>();
        for(int i=0;i<n;i++) tree.add(new ArrayList<Integer>());
        int[][] binpar = new int[26][n];
        for(int i=1;i<n;i++){
            int t1 = cs.nextInt()-1;
            int t2 = cs.nextInt()-1;
            // long w = cs.nextLong();
            tree.get(t1).add(t2);
            tree.get(t2).add(t1);
        }

        long maxdep[][] = new long[n][2];
        int depth[] = new int[n];
        int maxchild[] = new int[n];
        binpar[0][0] = -1;
        dfstrdep(tree, 0, -1, binpar,maxdep,1,depth,maxchild);
        for(int j=1;j<=25;j++){
            for(int i=0;i<n;i++){
                int temp = binpar[j-1][i];
                if(temp == -1) binpar[j][i] = -1;
                else binpar[j][i] = binpar[j-1][temp];
            }
        }
        for(int i=0;i<n;i++){
            maxdep[i][0] -= depth[i];
            maxdep[i][1] -= depth[i];
        }
        dfstrdep2(tree, 0, -1, binpar, maxdep, 0, depth, maxchild);
        int q = cs.nextInt();
        while (q-- >0) {
            int nd = cs.nextInt()-1;
            int k = cs.nextInt();
            k = Math.min(k,depth[nd]-1);
            long ans = maxdep[nd][0]+depth[nd];
            if(k==0){
                System.out.print(ans+" ");
                continue;
            }
            int par = kth_ancestor(nd, k, binpar);
            int child = kth_ancestor(nd,k-1,binpar);
            // System.out.println(nd+" "+par+" "+k);
            if(maxchild[par]==child) ans = Math.max(ans, maxdep[par][1]+depth[nd]);
            else ans = Math.max(ans, maxdep[par][0]+depth[nd]);
            System.out.print(ans+" ");
            // System.out.println();
        }
        System.out.println();

    }
    //Witte fxn here
    public static long dfstrdep(ArrayList<ArrayList<Integer>> tree,int curr,int par,int[][] binpar ,long[][] maxdep,int dep,int[] depth,int child[]){
        depth[curr] = dep;
        long max1 = 0;
        long max2 = 0;
        for(Integer nb: tree.get(curr)){
            if(nb == par) continue;
            binpar[0][nb] = curr;
            long temp = dfstrdep(tree, nb, curr, binpar, maxdep, dep+1, depth,child);
            if(temp >= max1){
                child[curr] = nb;
                max2 = max1;
                max1 = temp;
            }
            else max2 = Math.max(max2, temp);
        }
        maxdep[curr][0] = max1;
        maxdep[curr][1] = max2;
        return max1+1;
    }
    public static void dfstrdep2(ArrayList<ArrayList<Integer>> tree,int curr,int par,int[][] binpar ,long[][] maxdep,int dep,int[] depth,int child[]){

        for(Integer nb: tree.get(curr)){
            if(nb == par) continue;
            // long temp = dfstrdep2(tree, nb, curr, binpar, maxdep, dep+1, depth,child);
            if(nb==child[curr]){
                maxdep[curr][1] = Math.max(maxdep[curr][1],maxdep[nb][0]);
            }
        }
    }
    public static int kth_ancestor(int node,int k,int[][] binpar){
        int bit = 0;
        while (node != -1 && k>=(1<<bit)) {
            if( (k&(1<<bit)) != 0) node = binpar[bit][node];
            bit++;
        }
        return node;
    }

    public static long dfstrtot(ArrayList<ArrayList<Long>> tree,long curr,long par,long len[]){
        long below = 0;
        for(Long nb : tree.get((int)curr)){
            if(nb == par) continue;
            below += dfstrtot(tree, nb, curr, len);
        }
        len[(int)curr] = below;
        return below+1;
    }

    public static  int binary(int[] arr,int start,int end,int key){
        int mid = start;
        int st = start;
        int en = end;
        while(start<=end){
            mid = (start+end)/2;
            if(arr[mid]==key) return mid;
            else if(arr[mid]>key) end = mid-1;
            else if(arr[mid]<key) start = mid+1;
        }
        //if(end<start){
            if(end<st) end++;
            // if(start>en) start--;
            // if(Math.abs(arr[end]-key)<=Math.abs(arr[start]-key-1)){
                // return start;
            // }
            return end;
        //}
        //return mid+1;
    }

    // public static int invmod(int A, int M)
    // {
//
        // for (int X = 1; X < M; X++)
            // if (((A % M) * (X % M)) % M == 1)
                // return X;
        // return 1;
    // }
    //
    // public static int gcd(int a,int b){
    //     int c = Math.min(a,b);
    //     int d = Math.max(a,b);
    //     if(d%c==0) return c;
    //     return gcd(c,d%c);
    // }
}
class PP {
    static String IN = "%s";
    static String OUT = "%s";

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

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

        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 code, we need to consider scenarios that can highlight potential issues in the algorithm. Here are the steps to create such a test input:

1. **Tree Structure**: The tree should have a structure that can cause the algorithm to fail. For example, a tree with long paths or complex branching can be challenging.
2. **Queries**: The queries should test the algorithm's ability to handle different levels of stamina and different starting vertices.

### Step-by-Step Test Input Generation

1. **Tree with Long Paths**:
   - A tree with a long path from the root to a leaf can expose issues with the algorithm's handling of deep trees.
   - For example, a tree with vertices connected in a linear fashion.

2. **Complex Branching**:
   - A tree with multiple branches can expose issues with the algorithm's handling of multiple paths.
   - For example, a tree where each vertex has multiple children.

3. **Queries with Low Stamina**:
   - Queries with low stamina (e.g., 0 or 1) can test the algorithm's ability to handle minimal movement.

4. **Queries with High Stamina**:
   - Queries with high stamina can test the algorithm's ability to handle maximum movement.

### Challenging Test Input

```plaintext
1
5
1 2
2 3
3 4
4 5
3
1 2
3 1
5 3
```

### Explanation

- **Tree Structure**:
  - The tree is a simple linear path from vertex 1 to vertex 5.
  - This structure ensures that the algorithm has to correctly handle the depth and stamina calculations.

- **Queries**:
  - **Query 1**: Starting at vertex 1 with 2 stamina.
    - Chefir can move to vertex 5, which is 4 steps away.
  - **Query 2**: Starting at vertex 3 with 1 stamina.
    - Chefir can move to vertex 4, which is 1 step away.
  - **Query 3**: Starting at vertex 5 with 3 stamina.
    - Chefir can move to vertex 1, which is 4 steps away.

This test input is designed to be small but challenging, as it tests the algorithm's ability to handle different starting points and stamina levels in a simple yet deep tree structure.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

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

#define ll long long
#define ull unsigned long long
using i64 = long long;

using namespace std;

vector<vector<int>>g;
vector<vector<int>>fa;
vector<int>dep;
vector<pair<int, int>>dia;

void dfs(int p, int f)
{
    fa[p][0] = f;
    dep[p] = dep[f] + 1;
    for (auto s : g[p]) {
        if (s != f) {
            dfs(s, p);
        }
    }
    return;
}

int lca(int u, int v)
{
    if (dep[u] < dep[v])
        swap(u, v);
    int div = dep[u] - dep[v];
    for (int i = 0; div != 0; i++) {
        if (div >> i & 1) {
            u = fa[u][i];
            div -= (1 << i);
        }
    }
    if (u == v)
        return u;
    else {
        for (int j = 19; j >= 0; j--) {
            if (fa[u][j] != fa[v][j]) {
                u = fa[u][j];
                v = fa[u][j];
            }
        }
        return fa[u][0];
    }

}

int pathlen(int u, int v)
{
    return dep[u] + dep[v] - 2 * dep[lca(u, v)];
}

void merge(pair<int, int>&p, pair<int, int>&s)
{
    pair<int, int>res;
    int len = 0;
    if (pathlen(p.first, p.second) > pathlen(s.first, s.second)) {
        res = p;
        len = pathlen(p.first, p.second);
    }
    else {
        res = s;
        len = pathlen(s.first, s.second);
    }

    if (pathlen(p.first, s.first) > len) {
        res = { p.first,s.first };
        len = pathlen(p.first, s.first);
    }
    if (pathlen(p.first, s.second) > len) {
        res = { p.first,s.second };
        len = pathlen(p.first, s.second);
    }
    if (pathlen(p.second, s.first) > len) {
        res = { p.second,s.first };
        len = pathlen(p.second, s.first);
    }
    if (pathlen(p.second, s.second) > len) {
        res = { p.second,s.second };
        len = pathlen(p.second, s.second);
    }
    p = res;
    return;
}

void dfs2(int p, int f)
{
    for (auto s : g[p]) {
        if (s != f) {
            dfs2(s, p);
            merge(dia[p], dia[s]);
        }
    }
    return;
}

void solve()
{
    int n;
    cin >> n;
    g.clear();
    fa.clear();
    dep.clear();
    dia.clear();
    g.resize(n + 1, vector<int>(0));
    fa.resize(n + 1, vector<int>(20));
    dep.resize(n + 1, 0);
    dia.resize(n + 1, { 0,0 });
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 1);
    for (int i = 1; i < 20; i++) {
        for (int j = 1; j <= n; j++) {
            fa[j][i] = fa[fa[j][i - 1]][i - 1];
        }
    }

    for (int i = 1; i <= n; i++) {
        dia[i] = { i,i };
    }

    dfs2(1, 1);

    int q;
    cin >> q;
    while (q--) {
        int u, k;
        cin >> u >> k;
        int v = u;
        for (int i = 0; k != 0; i++) {
            if (k >> i & 1) {
                u = fa[u][i];
                k -= (1 << i);
            }
        }
        int ans = max(pathlen(v, dia[u].first), pathlen(v, dia[u].second));
        cout << ans << " ";
    }
    cout << endl;

    return;
}

int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    int tt;
    cin >> tt;
    while (tt--)
        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:
To generate a challenging test input that exposes the bug in the given code, let's consider the following aspects:

1. **Tree Structure**: The tree should be constructed in a way that maximizes the potential for incorrect LCA (Lowest Common Ancestor) calculations or path length calculations.
2. **Queries**: The queries should test the limits of the algorithm, especially in cases where the stamina is just enough to reach certain vertices but not others.

### Step-by-Step Consideration

1. **Tree Structure**:
   - A tree with a deep and wide structure can help expose issues with the LCA and path length calculations.
   - Consider a tree with a long chain and multiple branches to ensure the LCA calculations are correct.

2. **Queries**:
   - Include queries where the stamina is just enough to reach a vertex but not enough to reach its children.
   - Include queries where the starting vertex is a leaf or a deep vertex to test edge cases.

### Example Test Input

Let's construct a tree with 7 vertices in the form of a long chain with a branch at vertex 3.

```plaintext
1
7
1 2
2 3
3 4
3 5
5 6
5 7
3
4 1
3 2
1 3
```

### Explanation

1. **Tree Structure**:
   - Vertex 1 is the root.
   - Vertex 2 is a child of 1.
   - Vertex 3 is a child of 2.
   - Vertex 4 and 5 are children of 3.
   - Vertex 6 and 7 are children of 5.

2. **Queries**:
   - Query 1: Starting at vertex 4 with 1 stamina. Chefir can only move to vertex 3.
   - Query 2: Starting at vertex 3 with 2 stamina. Chefir can move to vertex 2 and then to vertex 1.
   - Query 3: Starting at vertex 1 with 3 stamina. Chefir can move to vertex 2, then to vertex 3, and then to vertex 4 or 5.

This test input should help expose any issues in the LCA and path length calculations, especially in cases where the stamina is just enough to reach certain vertices but not others.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
const int INF = 1e18;
const int N = 1e6 + 10;
const int M = 5020;
const int mod = 998244353;
typedef pair<int,int> PII;
typedef pair<int,PII> pii;
int n,m;
int a[N];
int x,y;
int f[N][21];
int ne[N][2];
int son[N][2];
int dep[N];
vector<PII> q[N];
vector<int> v[N];
int ans[N];
vector<PII> st;

int dfs(int u,int fa){
	f[u][0] = fa;
	dep[u] = dep[fa] + 1;
	for(int i=1; i<=20; i++){
		f[u][i] = f[f[u][i - 1]][i - 1];
	}
	son[u][0] = u;
	int sum = 0;
	for(int i=0; i < v[u].size(); i++){
		int j = v[u][i];
		if(j == fa) continue;
		sum++;
		x = dfs(j,u);
		if(dep[x] > dep[son[u][0]]){
			son[u][1] = son[u][0],ne[u][1] = ne[u][0];
			son[u][0] = x,ne[u][0] = j;
		}else if(dep[x] > dep[son[u][1]]){
			son[u][1] = x,ne[u][1] = j;
		}
	}
	//cout << u << endl;
	//cout << son[u][0] << ' ' << son[u][1] << endl;
	//if(son[u][1] == u) son[u][1] = 0;
	return son[u][0];
}

void find(int u,int fa){
	st.push_back({u,0});
	//cout << u << endl;
	// for(PII it : st){
		// cout << it.first << ' ' << it.second << endl;
	// }
	// cout << dep[son[u][0]] << ' ' << dep[son[u][1]] << endl;
	for(PII it : q[u]){
		int l = 0,r = st.size() - 1;
		int x = it.first;
		int id = it.second;
		while(l < r){
			int mid = (l + r) >> 1;
			if(dep[u] - dep[st[mid].first] <= x) r = mid;
			else l = mid + 1;
		}
		ans[id] = dep[u] - dep[st[l].first] + st[l].second;
		ans[id] = max(ans[id],dep[son[u][0]] - dep[u]);
		// cout << "ANS" << endl;
		// cout << ans[id]  << endl;
		// cout << x << endl;
		// cout << st[l].first << ' ' << st[l].second << endl;
		// cout << "ANS" << endl;
	}
	for(int i=0; i<v[u].size(); i++){
		int j = v[u][i];
		if(j == fa) continue;
		if(j == ne[u][0]){
			while(st.size()){
				x = dep[u] - dep[st.back().first];
				y = (dep[son[u][1]] - dep[u]) - st.back().second;
				if(y > x) st.pop_back();
				else break;
			}
			st.push_back({u,dep[son[u][1]] - dep[u]});
		}else {
			while(st.size()){
				x = dep[u] - dep[st.back().first];
				y = (dep[son[u][0]] - dep[u]) - st.back().second;
				if(y > x) st.pop_back();
				else break;
			}
			st.push_back({u,dep[son[u][0]] - dep[u]});
		}

		find(j,u);
		while(st.size() && dep[st.back().first] >= dep[u]) st.pop_back();
	}
}

void solve(){
	cin >> n;
	for(int i=1; i<=n; i++){
		for(int j=0; j<=1; j++) ne[i][j] = son[i][j] = 0;
	}
	for(int i=1; i<n; i++){
		cin >> x >> y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	for(int i=0; i<=20; i++) f[1][i] = 1;
	dep[1] = 1;

	x = dfs(1,1);
	cin >> m;
	for(int i=1; i<=m; i++){
		cin >> x >> y;
		q[x].push_back({y,i});
	}
	for(int i=1; i<=n; i++){
		sort(q[i].begin(),q[i].end());
	}
	find(1,1);
	for(int i=1; i<=n; i++){
		q[i].clear();
		v[i].clear();
	}
	st.clear();
	//cout << "YYYYYYYYY" << endl;
	for(int i=1; i<=m; i++) cout << ans[i] << " \n"[i == m];
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T = 1;
	cin >> T;
	while(T--){
		solve();
	}
}

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

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

Think step by step.

Response:
```plaintext
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
import heapq
from heapq import heappop,heappush,heapify
from sys import stdin, setrecursionlimit
input = stdin.readline
readline = stdin.readline
from collections import deque,defaultdict
D=defaultdict(deque)
from collections import deque
d=deque()
#####segfunc#####
def segfunc(x, y):
    return max(x, y)
#################

#####ide_ele#####
ide_ele =-10**10
#################

class SegTree:
    """
    init(init_val, ide_ele): 配列init_valで初期化 O(N)
    update(k, x): k番目の値をxに更新 O(N)
    query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
    """
    def __init__(self, init_val, segfunc, ide_ele):
        """
        init_val: 配列の初期値
        segfunc: 区間にしたい操作
        ide_ele: 単位元
        n: 要素数
        num: n以上の最小の2のべき乗
        tree: セグメント木(1-index)
        """
        n = len(init_val)
        self.segfunc = segfunc
        self.ide_ele = ide_ele
        self.num = 1 << (n - 1).bit_length()
        self.tree = [ide_ele] * 2 * self.num
        # 配列の値を葉にセット
        for i in range(n):
            self.tree[self.num + i] = init_val[i]
        # 構築していく
        for i in range(self.num - 1, 0, -1):
            self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

    def update(self, k, x):
        """
        k番目の値をxに更新
        k: index(0-index)
        x: update value
        """
        k += self.num
        self.tree[k] = x
        while k > 1:
            self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
            k >>= 1

    def query(self, l, r):
        """
        [l, r)のsegfuncしたものを得る
        l: index(0-index)
        r: index(0-index)
        """
        res = self.ide_ele

        l += self.num
        r += self.num
        while l < r:
            if l & 1:
                res = self.segfunc(res, self.tree[l])
                l += 1
            if r & 1:
                res = self.segfunc(res, self.tree[r - 1])
            l >>= 1
            r >>= 1
        return res

# print(seg.query(0, 8))
# seg.update(5, 0)
# print(seg.query(0, 8))

T=int(input())
for _ in range(T):
  N=int(input())
  #A=list(map(int, input().split()))
  D=[[] for i in range(N)]
  G=[[] for i in range(N)]
  P=[-1]*N
  for i in range(N-1):
    u,v=map(int, input().split())
    u-=1;v-=1
    D[u].append(v);D[v].append(u)

  d=deque();V=[-1]*N;S=[0]*N
  d.append(0);V[0]=0
  C=[]
  while d:
    now=d.popleft()
    for nex in D[now]:
      if V[nex]==-1:
        V[nex]=V[now]+1
        S[nex]=V[nex]
        G[now].append(nex)
        C.append((now,nex))
        P[nex]=now
        d.append(nex)
  Q=int(input())
  qs=[[] for i in range(N)]
  for i in range(Q):
    u,v=map(int, input().split())
    u-=1
    qs[u].append((v,i))
  ans=[0]*Q

  for x,y in C[::-1]:
    S[x]=max(S[x],S[y])
  E=[[] for i in range(N)]
  for i in range(N):
    for nex in G[i]:
      E[i].append(S[nex])
  F=[-1]*N
  #print(S,V,C,P)
  #print(E,G)
  vv=max(V)
  H=[-1]*N
  #K=[0]*10
  for i in range(N):
    A=[]
    for j in G[i]:
      A.append((S[j],j))
    A.append((0,-1))
    A.append((0,-1))
    A=sorted(A)[::-1]
    for j in G[i]:
      if A[0][1]!=j:
        H[j]=A[0][0]
      else:
        H[j]=A[1][0]
  a = [0]*(N+10)

  seg = SegTree(a, segfunc, ide_ele)
  d=deque();hh=vv
  d.append([0, 1])
  seen = [False] * N
  seen[0] = True
  while d:
    task = d.pop()
    now=task[0]
    if task[1]:
      #print(now+1,'iki',V[now],H[now])
      d.append([now,0])
      for nex in D[now]:
        if not seen[nex]:
          seen[nex] = True
          d.append([nex,1])
      if now!=0:
        ne=hh-V[now]+1+max(0,H[now]-V[now]+1)
        seg.update(V[now]-1,ne)
        #K[V[now]-1]=ne
      ma=vv-V[now]
      for di,idx in qs[now]:
        ss=seg.query(max(0,V[now]-di),V[now])
        #print((di,now+1),ss,K,hh)
        ans[idx]=max(ma,ss-(hh-V[now]))
    else:
      #K[V[now]-1]=0
      seg.update(V[now]-1,0)
  print(*ans)

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

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

Think step by step.

Response:
```plaintext
1
6
1 2
1 3
2 4
2 5
3 6
4
1 0
2 1
3 2
4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
import sys
def main():
    num_args = int(input())
    while num_args:
        solve()
        num_args -=1

def solve():
    num_vert = int(input())
    arr = [[0]*3 for i in range(num_vert-1)]
    for i in range(num_vert-1):
        temp =  [int(j) for j in input().split()]
        arr[i][0], arr[i][1] = temp

    num_q = int(input())
    qs = [[0]*2 for i in range(num_q)]
    for i in range(num_q):
        temp =  [int(j) for j in input().split()]
        qs[i][0], qs[i][1] = temp

    orders = [[0]*6 for i in range(num_vert)]
    max_depths = [0]
    sec_depths = [0]
    stack = [1]
    orders[0] = [1,0,0,1,0,0]
    time = 2
    tie_ind = 0
    while stack:
        new_found = False
        curr_vert = stack[-1]
        for edge in arr:
            if not edge[2] and curr_vert in edge[:2]:
                edge[2] = True
                new_vert = -curr_vert + sum(edge[:2])
                if not orders[new_vert-1][0]:
                    new_found = True
                    orders[new_vert-1][0] = time
                    time+=1
                    orders[new_vert-1][2] = orders[stack[-1]-1][2]+1
                    max_depths.append(orders[new_vert-1][2])
                    sec_depths.append(0)
                    orders[new_vert-1][3] = new_vert
                    stack.append(new_vert)
                    break
        if not new_found:
            orders[stack[-1]-1][5] = max(0,sec_depths[-1] - orders[stack[-1]-1][2])
            orders[stack[-1]-1][4] = max_depths[-1] - orders[stack[-1]-1][2]
            orders[stack[-1]-1][1] = time
            time+=1
            stack.pop()
            if len(max_depths) >1:
                if max_depths[-1] == max_depths[-2]:
                    tie_ind = 1
                else:
                    tie_ind = 0
                if max_depths[-2] != orders[stack[-1]-1][2]-1:
                    sec_depths[-2] = sorted([max_depths[-1],max_depths[-2],sec_depths[-2]])[1]
                max_depths[-2] = max(max_depths[-1],max_depths[-2])

            max_depths.pop()
            sec_depths.pop()

    for query in qs:
        up = 0
        result = 0
        curr_vert = query[0]
        start_vert = curr_vert
        stamina = query[1]
        result = orders[curr_vert-1][4]
        pre = orders[curr_vert-1][0]
        post = orders[curr_vert-1][1]
        depth = orders[curr_vert-1][2]
        while stamina:
            new_found = False
            for edge in arr:
                if curr_vert in edge[:2]:
                    new_vert = -curr_vert + sum(edge[:2])
                    if orders[new_vert-1][2] < orders[curr_vert-1][2]:
                        up+=1
                        stamina-=1
                        new_found = True
                        if orders[new_vert-1][4] != orders[curr_vert-1][4]+1:
                            result = max(result, up+orders[new_vert-1][4])
                        else:
                            if orders[new_vert-1][5]>0:
                                result = max(result, up+orders[new_vert-1][5])
                            else:
                                result = max(result,up)
                        curr_vert=new_vert
            if not new_found:
                break

        sys.stdout.write(str(result) +" ")

    sys.stdout.write('\n')

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
1
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+7;
vector<int>e[maxn];
int n,q;
int dep[maxn],mxdep[maxn],f[maxn][2],g[19][maxn],fa[19][maxn];
void update(int u,int v){
    if(mxdep[f[u][0]]<=mxdep[v]){
        f[u][1]=f[u][0];
        f[u][0]=v;
        return;
    }
    if(mxdep[f[u][1]]<=mxdep[v]){
        f[u][1]=v;
    }
}
void dfs(int u,int Fa){
    dep[u]=dep[Fa]+1;
    mxdep[u]=dep[u];
    fa[0][u]=Fa;
    for(int v:e[u]){
        if(v!=Fa){
            dfs(v,u);
            mxdep[u]=max(mxdep[u],mxdep[v]);
        }
    }
}
void dfs1(int u,int Fa){
    for(int v:e[u]){
        if(v!=Fa){
            dfs1(v,u);
            update(u,v);
        }
    }
}
void init(){
    for(int i=1;i<=n;i++){
        if(f[fa[0][i]][0]==i){
            if(!f[fa[0][i]][1])g[0][i]=1;
            else g[0][i]=mxdep[f[fa[0][i]][1]]-dep[fa[0][i]]+1;
        }else{
            g[0][i]=mxdep[f[fa[0][i]][0]]-dep[fa[0][i]]+1;
        }
    }
    for(int i=1;i<=n;i++){
        f[i][0]=f[i][1]=0;
    }
    for(int j=1;j<=17;j++)
        for(int i=1;i<=n;i++)
            fa[j][i]=fa[j-1][fa[j-1][i]];
    for(int j=1;j<=17;j++)
        for(int i=1;i<=n;i++)
            g[j][i]=max(g[j-1][i],g[j-1][fa[j-1][i]]+dep[i]-dep[fa[j-1][i]]);
}
void solve(){
    cin>>n;
    for(int i=1,u,v;i<n;i++){
        cin>>u>>v;
        e[u].emplace_back(v);
        e[v].emplace_back(u);
    }
    dfs(1,0);
    dfs1(1,0);
    init();
    cin>>q;
    for(int i=1,u,k;i<=q;i++){
        cin>>u>>k; k=min(k,dep[u]);
        int res=mxdep[u]-dep[u],now=0;
        for(int j=17;~j;j--){
            if(k&(1<<j))res=max(res,g[j][u]+now),now+=dep[u]-dep[fa[j][u]],u=fa[j][u];
        }
        cout<<res<<" \n"[i==q];
    }
    for(int i=1;i<=n;i++) e[i].clear();
}
signed main(){
    int T;
    cin>>T;
    while(T--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

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

public class G_Sakurako_and_Chefir {
    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) {}
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }

        char[] nextCharArray() {
            try {
                String line = br.readLine();
                return line.toCharArray();
            } catch (IOException e) {
                e.printStackTrace();
                return new char[0];
            }
        }
    }

    public static void dfs(ArrayList<Integer>[] tree, int[] parent, boolean[] visited, int cur) {
        visited[cur] = true;

        for(int neighbour : tree[cur]) {
            if(!visited[neighbour]) {
                parent[neighbour] = cur;
                dfs(tree, parent, visited, neighbour);
            }
        }
    }

    public static int modifiedDFS(ArrayList<Integer>[] tree, int[] parent, boolean[] visited, int cur, int energy) {
        int max_dist = 0;
        for(int neighbour : tree[cur]) {
            if(visited[neighbour]) continue;

            if(neighbour == parent[cur]) {
                if(energy > 0) {
                    visited[neighbour] = true;
                    max_dist = Math.max(max_dist, 1 + modifiedDFS(tree, parent, visited, neighbour, energy-1));
                    visited[neighbour] = false;
                }
            } else {
                visited[neighbour] = true;
                max_dist = Math.max(max_dist, 1 + modifiedDFS(tree, parent, visited, neighbour, energy));
                visited[neighbour] = false;
            }
        }

        return max_dist;
    }

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        FastScanner s = new FastScanner();

        int t = s.nextInt();

        while(t-- != 0) {
            int n = s.nextInt();

            @SuppressWarnings("unchecked")
            ArrayList<Integer>[] tree = new ArrayList[n+1];
            for(int i = 0; i < n+1; i++) {
                tree[i] = new ArrayList<Integer>();
            }

            for(int i = 0; i < n-1; i++) {
                int nodeA = s.nextInt();
                int nodeB = s.nextInt();

                tree[nodeA].add(nodeB);
                tree[nodeB].add(nodeA);
            }

            int[] parent = new int[n+1];
            parent[1] = -1;
            boolean[] visited = new boolean[n+1];
            dfs(tree, parent, visited, 1);

            visited = new boolean[n+1];

            int q = s.nextInt();
            for(int i = 0; i < q; i++) {
                int node = s.nextInt();
                int energy = s.nextInt();

                visited[node] = true;
                int dist = modifiedDFS(tree, parent, visited, node, energy);
                visited[node] = false;

                out.print(dist + " ");
            }
            out.println();
        }
        out.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause issues. The code provided has a modified DFS function that calculates the maximum distance Chefir can travel given a certain amount of stamina. The main issues to consider are:

1. **Handling of Ancestors**: The code should correctly handle the distinction between moving to an ancestor and moving to a non-ancestor.
2. **Stamina Management**: The code should correctly manage the stamina, especially when moving to non-ancestors.
3. **Edge Cases**: The code should handle small and large trees, and scenarios where the stamina is zero or very high.

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

1. A tree with a small number of vertices.
2. A variety of queries with different starting vertices and stamina levels.
3. Scenarios where the starting vertex is a leaf, a root, and an intermediate node.
4. Scenarios where the stamina is zero, low, and high.

Here is the test input:

```plaintext
1
7
1 2
1 3
2 4
2 5
3 6
3 7
4
4 0
2 1
3 2
1 3
```

### Explanation:
1. **Tree Structure**:
   - The tree is rooted at vertex 1.
   - Vertex 1 has children 2 and 3.
   - Vertex 2 has children 4 and 5.
   - Vertex 3 has children 6 and 7.

2. **Queries**:
   - **4 0**: Starting at vertex 4 with 0 stamina. Chefir can only stay at vertex 4.
   - **2 1**: Starting at vertex 2 with 1 stamina. Chefir can move to vertex 1 or 4, but not to 5.
   - **3 2**: Starting at vertex 3 with 2 stamina. Chefir can move to vertex 1, 6, or 7.
   - **1 3**: Starting at vertex 1 with 3 stamina. Chefir can move to any vertex in the tree.

This test input is designed to cover a variety of scenarios and should help expose any issues in the code's handling of tree traversal and stamina management.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
// Not my code https://codeforces.com/contest/2033/submission/287918494
#include <bits/stdc++.h>
using namespace std;

// #define int long long
// #define ll int
#define trav(i,v)   for(auto i: v)
#define rep(i,n)    for(int i=0;i<n;i++)
#define repu(i,k,n) for(int i=k;i<=n;i++)
#define repd(i,k,n) for(int i=k;i>=n;i--)
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define all(v)  v.begin(), v.end()
#define itn int
#define sz(x) (int)x.size()
#define el "\n"

//typedef
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair<int, int> pii;
typedef std::vector<int> vi;
typedef std::vector< pii > vii;
typedef std::vector< std::vector<int> > vvi;
typedef std::vector< std::pair < pii, int > > vpp;

const long long MOD = 1000000007;
// const long long MOD = 998244353;
const long long inf = 1e9;

int modpow(int x, int n = MOD-2, int mod = MOD){ int res=1; while(n>0){ if(n&1) res=res*x%mod; x=x*x%mod; n>>=1;} return res;}

void init(){
}

// len -> how long excluding node
// lev -> starts with 1

void dfs1(int x,vi *v,vi &lev,vi &par,vi &len){
    trav(y,v[x]){
        if(y==par[x])
            continue;
        par[y] = x;
        lev[y] = lev[x] + 1;
        dfs1(y,v,lev,par,len);
        len[x] = max(len[x],len[y]+1);
    }
}

void update(int x, int l, int r, int pos, int val,vi &seg){
    if(l==r && l!=pos)
        return;
    if(l==r){
        seg[x] = val;
        return;
    }
    int m = (l+r)/2;
    update(2*x,l,m,pos,val,seg);
    update(2*x+1,m+1,r,pos,val,seg);
    seg[x] = max(seg[2*x], seg[2*x+1]);
}

void update(int pos, int val,int n, vi &seg){
    update(1,1,n,pos,val,seg);
}

int query(int x, int l, int r, int l1, int r1,vi &seg){
    if(l>r1 || r<l1)
        return -inf;
    if(l>=l1 && r<=r1){
        return seg[x];
    }
    int m = (l+r)/2;
    int x1 = query(2*x,l,m,l1,r1,seg);
    int x2 = query(2*x+1,m+1,r,l1,r1,seg);
    return max(x1,x2);
}

int query(int pos1, int pos2, int n, vi &seg){
    return query(1,1,n,pos1,pos2,seg);
}

void dfs(int x, int n, vi *v, vi &lev, vi &par, vi &len, vi &ans, vi &seg, vii *queries){
    vii lengths;
    trav(y,v[x]){
        if(y!=par[x])
            lengths.pb(mp(len[y]+1,y));
    }
    sort(all(lengths));
    reverse(all(lengths));
    trav(p,queries[x]){
        int mx = 0, k = p.fi;
        if(sz(lengths)>=1)   mx = lengths[0].fi;
        int min_lev = max(1,lev[x]-k);
        // if(x==2){
        //     cout<<p.se<<mx<<endl;
        //     cout<<min_lev<<" "<<lev[x]<<endl;
        //     cout<<query(min_lev, min_lev)<<endl;
        // }
        mx = max(mx, query(min_lev, lev[x], n, seg)+lev[x]);
        ans[p.se] = mx;
    }
    trav(y,v[x]){
        if(y==par[x])
            continue;

        int mx = 0;
        if(sz(lengths)>1)   mx = lengths[0].fi;
        if(sz(lengths)>1 && lengths[0].se==y)   mx = lengths[1].fi;
        // if(x==3 && y==5){
        //     cout<<mx<<" "<<lev[x]<<" "<<mx-lev[x]<<endl;
        // }
        update(lev[x],mx-lev[x], n, seg);
        dfs(y,n,v,lev,par,len,ans,seg,queries);
        update(lev[x],-inf, n, seg);
    }
}

void solve() {
    int n;
    cin>>n;
    vi v[n+1];
    repu(i,1,n-1){
        int x, y;
        cin>>x>>y;
        v[x].pb(y);
        v[y].pb(x);
    }
    int q;
    cin>>q;
    vi lev(n+1,1), par(n+1,0), len(n+1,0), ans(q), seg(4*n+5,-inf);
    vii queries[n+1];
    dfs1(1,v,lev,par,len);
    rep(i,q){
        int x, k;
        cin>>x>>k;
        queries[x].pb(mp(k,i));
    }//
    // cout<<len[9]<<endl;
    dfs(1,n,v,lev,par,len,ans,seg,queries);
    trav(x,ans){
        cout<<x<<" ";
    }
    cout<<"\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
6
1 2
1 3
2 4
2 5
3 6
3
4 0
5 1
6 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
#pragma GCC optimize("Ofast")

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

// #define int long long
// #define ll int
#define trav(i,v)   for(auto i: v)
#define rep(i,n)    for(int i=0;i<n;i++)
#define repu(i,k,n) for(int i=k;i<=n;i++)
#define repd(i,k,n) for(int i=k;i>=n;i--)
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define all(v)  v.begin(), v.end()
#define itn int
#define sz(x) (int)x.size()
#define el "\n"

//typedef
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair<int, int> pii;
typedef std::vector<int> vi;
typedef std::vector< pii > vii;
typedef std::vector< std::vector<int> > vvi;
typedef std::vector< std::pair < pii, int > > vpp;

const long long MOD = 1000000007;
// const long long MOD = 998244353;
const long long inf = 1e9;

int modpow(int x, int n = MOD-2, int mod = MOD){ int res=1; while(n>0){ if(n&1) res=res*x%mod; x=x*x%mod; n>>=1;} return res;}

void init(){
    ios::sync_with_stdio(0);
    cin.tie(0);
}

// len -> how long excluding node
// lev -> starts with 1

void dfs1(int x,vi *v,vi &lev,vi &par,vi &len){
    trav(y,v[x]){
        if(y==par[x])
            continue;
        par[y] = x;
        lev[y] = lev[x] + 1;
        dfs1(y,v,lev,par,len);
        len[x] = max(len[x],len[y]+1);
    }
}

void update(int x, int l, int r, int pos, int val,vi &seg){
    if(l==r && l!=pos)
        return;
    if(l==r){
        seg[x] = val;
        return;
    }
    int m = (l+r)/2;
    update(2*x,l,m,pos,val,seg);
    update(2*x+1,m+1,r,pos,val,seg);
    seg[x] = max(seg[2*x], seg[2*x+1]);
}

void update(int pos, int val,int n, vi &seg){
    update(1,1,n,pos,val,seg);
}

int query(int x, int l, int r, int l1, int r1,vi &seg){
    if(l>r1 || r<l1)
        return -inf;
    if(l>=l1 && r<=r1){
        return seg[x];
    }
    int m = (l+r)/2;
    int x1 = query(2*x,l,m,l1,r1,seg);
    int x2 = query(2*x+1,m+1,r,l1,r1,seg);
    return max(x1,x2);
}

int query(int pos1, int pos2, int n, vi &seg){
    return query(1,1,n,pos1,pos2,seg);
}

void dfs(int x, int n, vi *v, vi &lev, vi &par, vi &len, vi &ans, vi &seg, vii *queries){
    vii lengths;
    trav(y,v[x]){
        if(y!=par[x])
        lengths.pb(mp(len[y]+1,y));
    }
    sort(all(lengths));
    reverse(all(lengths));
    trav(p,queries[x]){
        int mx = 0, k = p.fi;
        if(sz(lengths)>=1)   mx = lengths[0].fi;
        int min_lev = max(1,lev[x]-k);
        // if(x==2){
        //     cout<<p.se<<mx<<endl;
        //     cout<<min_lev<<" "<<lev[x]<<endl;
        //     cout<<query(min_lev, min_lev)<<endl;
        // }
        mx = max(mx, query(min_lev, lev[x], n, seg)+lev[x]);
        ans[p.se] = mx;
    }
    trav(y,v[x]){
        if(y==par[x])
            continue;

        int mx = 0;
        if(sz(lengths)>1)   mx = lengths[0].fi;
        if(sz(lengths)>1 && lengths[0].se==y)   mx = lengths[1].fi;
        // if(x==3 && y==5){
        //     cout<<mx<<" "<<lev[x]<<" "<<mx-lev[x]<<endl;
        // }
        update(lev[x],mx-lev[x], n, seg);
        dfs(y,n,v,lev,par,len,ans,seg,queries);
        update(lev[x],-inf, n, seg);
    }
}

void solve() {
    int n;
    cin>>n;
    vi v[n+1];
    repu(i,1,n-1){
        int x, y;
        cin>>x>>y;
        v[x].pb(y);
        v[y].pb(x);
    }
    int q;
    cin>>q;
    vi lev(n+1,1), par(n+1,0), len(n+1,0), ans(q), seg(4*n+5,-inf);
    vii queries[n+1];
    dfs1(1,v,lev,par,len);
    rep(i,q){
        int x, k;
        cin>>x>>k;
        queries[x].pb(mp(k,i));
    }//
    // cout<<len[9]<<endl;
    dfs(1,n,v,lev,par,len,ans,seg,queries);
    trav(x,ans){
        cout<<x<<" ";
    }
    cout<<endl;
}

signed main() {
    init();
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Understand the Code's Logic:**
   - The code uses a segment tree to efficiently handle queries.
   - It performs a depth-first search (DFS) to compute the lengths of paths and to process queries.
   - The segment tree is used to store and query the maximum path lengths.

2. **Identify Potential Bugs:**
   - The segment tree updates and queries might have issues with edge cases.
   - The DFS might not handle certain tree structures correctly, especially those with deep or wide branches.
   - The logic for handling queries might have off-by-one errors or incorrect boundary conditions.

3. **Create a Small-Scale Test Input:**
   - We need a tree with a structure that can challenge the segment tree and DFS logic.
   - A tree with a deep path and a few branches can help expose issues in the segment tree updates and queries.
   - A tree with a few queries that involve vertices with different levels and staminas can help expose issues in the DFS logic.

4. **Generate the Test Input:**
   - We will create a tree with 5 vertices and a few queries to test the code's handling of different scenarios.

Here is the test input:

```plaintext
1
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
```

This test input includes:
- A tree with 5 vertices.
- Edges forming a deep path from 1 to 4, with a branch from 3 to 5.
- 3 queries:
  - Query from vertex 5 with 1 stamina.
  - Query from vertex 3 with 1 stamina.
  - Query from vertex 2 with 0 stamina.

This test input is small and valid, and it should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
import sys
import math

sys.setrecursionlimit(int(1e5))

input = sys.stdin.readline
takeInt= lambda : int(input())
takeInts = lambda : map(int, input().split())
takeList = lambda : list(map(int, input().split()))

for _ in range(takeInt()):
    n = takeInt()
    maxN = int(math.log(n, 2))
    lca = [[-1 for i in range(maxN + 1)] for j in range(n + 1)]
    level = [0 for i in range(n + 1)]
    graph = [[] for i in range(n + 1)]
    depth = [[0, 0] for i in range(n + 1)]

    def dfs(node, par):
        lca[node][0] = par
        level[node] = level[par] + 1
        for nei in graph[node]:
            if nei != par:
                depth[node].append(dfs(nei, node) + 1)
                depth[node] = sorted(depth[node])
                depth[node] = [depth[node][-1], depth[node][-2]]
        return max(depth[node])

    def jump(node, energy):
        res = depth[node][0]
        for i in range(energy):
            height = 1 + depth[node][0]
            node = lca[node][0]
            if node == -1:
                break
            if depth[node][0] <= height:
                res = max(res, i + 1 + depth[node][1])
            else:
                res = max(res, depth[node][0] + 1 + i)
        return res

    res = []
    for i in range(n - 1):
        s, d = takeInts()
        graph[s].append(d)
        graph[d].append(s)

    dfs(1, -1)
    for j in range(1, maxN + 1):
        for i in range(1, n + 1):
            if lca[i][j - 1] != -1:
                par = lca[i][j - 1]
                lca[i][j] = lca[par][j - 1]

    for i in range(takeInt()):
        node, energy = takeInts()
        res.append(jump(node, energy))
    print(*res)

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

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

Think step by step.

Response:
```plaintext
1
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
# Author : Omm Praksh Sahoo
# Calm, Down... Wrong Submission Have 10 Mins Penalty

I = lambda: input()
II = lambda: int(input())
MII = lambda: map(int, input().split())
LI = lambda: list(input().split())
LII = lambda: list(map(int, input().split()))
LGMII = lambda: map(lambda x: int(x) - 1, input().split())
LGLII = lambda: list(map(lambda x: int(x) - 1, input().split()))

inf = float('inf')

#########################################################################################################
#-----------------------------------------------Code----------------------------------------------------#
#########################################################################################################

N = 200001
LOG = 19

# Global arra
adj = [[] for _ in range(N)]
m1 = [(0, 0)] * N  # Tuple to store max distances
m2 = [(0, 0)] * N
d = [[0] * N for _ in range(LOG)]
p = [[-1] * N for _ in range(LOG)]

def DFS(node, par=-1):
    global m1, m2, p, adj
    for neighbor in adj[node]:
        if neighbor != par:
            DFS(neighbor, node)
            p[0][neighbor] = node
            curr_dist = (m1[neighbor][0] + 1, neighbor)
            if curr_dist > m1[node]:
                m2[node] = m1[node]
                m1[node] = curr_dist
            elif curr_dist > m2[node]:
                m2[node] = curr_dist

def solve():
    global adj, m1, m2, d, p
    n = II()

    # Clear global structures for each test case
    for i in range(1, n + 1):
        adj[i].clear()
        m1[i] = m2[i] = (0, 0)
        for j in range(LOG):
            p[j][i] = -1  # Reset the parent array

    for _ in range(n - 1):
        u, v = MII()
        adj[u].append(v)
        adj[v].append(u)

    # Run DFS to calculate m1, m2, and initial parents
    DFS(1)

    # Precompute distances
    for i in range(1, n + 1):
        j = p[0][i]
        if j != -1:
            d[0][i] = m2[j][0] if m1[j][1] == i else m1[j][0]
            d[0][i] += 1
        else:
            d[0][i] = 0

    # Binary lifting for distance and ancestors
    for k in range(1, LOG):
        for i in range(1, n + 1):
            if p[k - 1][i] != -1:
                ancestor = p[k - 1][i]
                p[k][i] = p[k - 1][ancestor]
                if p[k][i] != -1:
                    d[k][i] = max(d[k - 1][i], (1 << (k - 1)) + d[k - 1][ancestor])

    # Process queries
    q = II()
    query_result = []
    for _ in range(q):
        v, stamina = MII()
        ans2 = m1[v][0]
        dist = 0

        # Binary lifting with stamina
        for k in range(LOG - 1, -1, -1):
            if stamina >= (1 << k) and p[k][v] != -1:
                stamina -= (1 << k)
                ans2 = max(ans2, dist + d[k][v])
                dist += (1 << k)
                v = p[k][v]

        ans2 = max(ans2, dist)
        query_result.append(str(ans2))

    return " ".join(query_result)

t = II()
for _ in range(t):
    print(solve())

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

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

Think step by step.

Response:
```plaintext
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

Examples:
input:
3
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
output:
2 1 2
0 5 2 4 5 5 5
1 3 4

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

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

public class Main {
    static int LOGN = 20;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();
        int t = Integer.parseInt(br.readLine());
        while(t-- >0){
            int n = Integer.parseInt(br.readLine());
            List<List<Integer>> tree = new ArrayList<>();
            for(int i=0;i<=n;i++) tree.add(new ArrayList<>());
            for(int i=0;i<n-1;i++){
                StringTokenizer st = new StringTokenizer(br.readLine());
                int u = Integer.parseInt(st.nextToken());
                int v = Integer.parseInt(st.nextToken());
                tree.get(u).add(v);
                tree.get(v).add(u);
            }
            // Initialize arrays
            int[] depth = new int[n+1];
            int[][] up = new int[n+1][LOGN];
            int[] max_depth_subtree = new int[n+1];
            int[] max1 = new int[n+1];
            int[] max2 = new int[n+1];
            // Initialize
            Arrays.fill(max2, -1);
            dfs(1, 0, tree, depth, up, max_depth_subtree, max1, max2);

            int q = Integer.parseInt(br.readLine());
            List<int[]> queries = new ArrayList<>();
            for(int i=0;i<q;i++){
                StringTokenizer st = new StringTokenizer(br.readLine());
                int v = Integer.parseInt(st.nextToken());
                int k = Integer.parseInt(st.nextToken());
                queries.add(new int[]{v, k});
            }
            List<Integer> res = new ArrayList<>();
            for(int[] query: queries){
                int v = query[0];
                int k = query[1];
                int max_distance = 0;
                for(int j=0; j<=k; j++){
                    int u = getKthAncestor(v, j, up, LOGN);
                    if(u ==0) break;
                    int distance_j;
                    if(j ==0){
                        distance_j = max1[u] - depth[u];
                    }
                    else{
                        int ancestor_prev = getKthAncestor(v, j-1, up, LOGN);
                        if(ancestor_prev ==0){
                            distance_j = j + (max_depth_subtree[u] - depth[u]);
                        }
                        else{
                            if(max1[u] == max_depth_subtree[ancestor_prev]){

                                if(max2[u] >= depth[u]){
                                    distance_j = j + (max2[u] - depth[u]);
                                }
                                else{
                                    distance_j = j;
                                }
                            }
                            else{

                                if(max1[u] >= depth[u]){
                                    distance_j = j + (max1[u] - depth[u]);
                                }
                                else{
                                    distance_j = j;
                                }
                            }
                        }
                    }
                    max_distance = Math.max(max_distance, distance_j);
                }
                res.add(max_distance);
            }

            for(int i=0;i<res.size();i++){
                output.append(res.get(i));
                if(i != res.size()-1) output.append(' ');
            }
            output.append('\n');
        }
        System.out.print(output);
    }

    static void dfs(int u, int p, List<List<Integer>> tree, int[] depth, int[][] up, int[] max_depth_subtree, int[] max1, int[] max2){
        up[u][0] = p;
        for(int i=1;i<LOGN;i++){
            up[u][i] = up[up[u][i-1]][i-1];
        }
        max_depth_subtree[u] = depth[u];
        max1[u] = depth[u];
        max2[u] = -1;
        for(int v: tree.get(u)){
            if(v != p){
                depth[v] = depth[u] +1;
                dfs(v, u, tree, depth, up, max_depth_subtree, max1, max2);
                if(max_depth_subtree[v] > max1[u]){
                    max2[u] = max1[u];
                    max1[u] = max_depth_subtree[v];
                }
                else if(max_depth_subtree[v] > max2[u]){
                    max2[u] = max_depth_subtree[v];
                }
                max_depth_subtree[u] = Math.max(max_depth_subtree[u], max_depth_subtree[v]);
            }
        }
    }

    static int getKthAncestor(int u, int k, int[][] up, int LOGN){
        for(int i=0;i<LOGN;i++){
            if( (k & (1<<i)) !=0 ){
                u = up[u][i];
                if(u ==0) break;
            }
        }
        return u;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
6
1 2
2 3
3 4
3 5
2 6
3
4 1
5 2
6 0
```