**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Happiness
time_limit_ms: 3000
memory_limit_mb: 256
Description: Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain happiness $h_i$.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum obtainable sum of happiness.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 50$, $1 \le x \le 10^8$) — the total number of months and the monthly salary.

The $i$-th of the following $m$ lines contains two integers, $c_i$ and $h_i$ ($0 \le c_i \le 10^8$, $1 \le h_i \le 10^3$) — the cost and happiness on offer for the $i$-th month. Note that some happiness may be free ($c_i=0$ for some $i$'s).

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

Output: For each test case, print a single integer, the maximum sum of happiness Charlie could obtain.

Examples:
input:
7
1 10
1 5
2 80
0 10
200 100
3 100
70 100
100 200
150 150
5 8
3 1
5 3
3 4
1 5
5 3
2 5
1 5
2 1
5 3
2 5
2 4
4 1
5 1
3 4
5 2
2 1
1 2
3 5
3 2
3 2
output:
0
10
200
15
1
9
9

Note: In the first test case, Charlie only gets paid at the end of the month, so is unable to afford anything.

In the second test case, Charlie obtains the free happiness in the first month.

In the third test case, it's optimal for Charlie to buy happiness in the second month. Even with money left at the end, Charlie could not go back in time to obtain the happiness on offer in the first month.

Buggy Code:
for _ in range(int(input())):
    n,m=map(int,input().split())
    l=[list(map(int,input().split())) for _ in range(n)]
    k=0

    def rec(i,s,h):
        global k

        if i>=n:
            k=max(k,h)
            return

        if s>=l[i][0]:
            rec(i+1,s+m-l[i][0],h+l[i][1])
        rec(i+1,s+m,h)

    rec(0,0,0)

    print(k)

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

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

Think step by step.

Response:
```plaintext
3
1 10
1 5
2 10
0 10
20 100
1 100
2 200
3 300
4 400
5 500
6 600
7 700
8 800
9 900
10 1000
11 1100
12 1200
13 1300
14 1400
15 1500
16 1600
17 1700
18 1800
19 1900
20 2000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Happiness
time_limit_ms: 3000
memory_limit_mb: 256
Description: Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain happiness $h_i$.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum obtainable sum of happiness.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 50$, $1 \le x \le 10^8$) — the total number of months and the monthly salary.

The $i$-th of the following $m$ lines contains two integers, $c_i$ and $h_i$ ($0 \le c_i \le 10^8$, $1 \le h_i \le 10^3$) — the cost and happiness on offer for the $i$-th month. Note that some happiness may be free ($c_i=0$ for some $i$'s).

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

Output: For each test case, print a single integer, the maximum sum of happiness Charlie could obtain.

Examples:
input:
7
1 10
1 5
2 80
0 10
200 100
3 100
70 100
100 200
150 150
5 8
3 1
5 3
3 4
1 5
5 3
2 5
1 5
2 1
5 3
2 5
2 4
4 1
5 1
3 4
5 2
2 1
1 2
3 5
3 2
3 2
output:
0
10
200
15
1
9
9

Note: In the first test case, Charlie only gets paid at the end of the month, so is unable to afford anything.

In the second test case, Charlie obtains the free happiness in the first month.

In the third test case, it's optimal for Charlie to buy happiness in the second month. Even with money left at the end, Charlie could not go back in time to obtain the happiness on offer in the first month.

Buggy Code:
t = int(input())
for _ in range(t):
    mx = list(map(int,input().rstrip().split()))
    m, x = mx[0], mx[1]
    cost = 0
    costArr = []
    happArr = []
    ht = {}

    for _ in range(m):
        ch = list(map(int,input().rstrip().split()))
        c,h = ch[0],ch[1]
        costArr.append(c)
        happArr.append(h)

    def knapsack(b, th, c , h, i): # balance, total happiness, cost array, happiness array, iterator
        key = f'{b} {th} {i}'
        if i == m:
            return th

        if key in ht:
            return ht[key]

        if b - c[i] < 0:
            ht[key] = knapsack(b + x, th, c, h, i+1)
        else:
            ht[key] = max(knapsack(b - c[i] + x, th + h[i], c, h, i+1), knapsack(b + x, th, c, h, i+1))

        return ht[key]

    #print(f'Ans: {knapsack(0,0,costArr,happArr,0)}')
    print(knapsack(0,0,costArr,happArr,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
1 10
10 5
2 10
0 10
10 5
3 10
10 5
5 10
1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Happiness
time_limit_ms: 3000
memory_limit_mb: 256
Description: Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain happiness $h_i$.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum obtainable sum of happiness.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 50$, $1 \le x \le 10^8$) — the total number of months and the monthly salary.

The $i$-th of the following $m$ lines contains two integers, $c_i$ and $h_i$ ($0 \le c_i \le 10^8$, $1 \le h_i \le 10^3$) — the cost and happiness on offer for the $i$-th month. Note that some happiness may be free ($c_i=0$ for some $i$'s).

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

Output: For each test case, print a single integer, the maximum sum of happiness Charlie could obtain.

Examples:
input:
7
1 10
1 5
2 80
0 10
200 100
3 100
70 100
100 200
150 150
5 8
3 1
5 3
3 4
1 5
5 3
2 5
1 5
2 1
5 3
2 5
2 4
4 1
5 1
3 4
5 2
2 1
1 2
3 5
3 2
3 2
output:
0
10
200
15
1
9
9

Note: In the first test case, Charlie only gets paid at the end of the month, so is unable to afford anything.

In the second test case, Charlie obtains the free happiness in the first month.

In the third test case, it's optimal for Charlie to buy happiness in the second month. Even with money left at the end, Charlie could not go back in time to obtain the happiness on offer in the first month.

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

public class main {
    static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
    static FastReader in = new FastReader();
    public static HashMap<Integer, HashMap<Long, Long>> dp;
    public static void main(String[] args) {
        int t = in.nextInt();
        while(t-- > 0){
            int m = in.nextInt();
            long x = in.nextInt();
            ArrayList<long[]> a = new ArrayList<>();
            dp = new HashMap<>();
            for (int i = 0; i < m; i++) {
                long l = in.nextLong();
                long r = in.nextLong();
                a.add(new long[]{l, r});
            }
            long ans = dp(a, 0, 0, x);
            pw.println(ans);
        }
	pw.close();
    }
    public static long dp(ArrayList<long[]> a, int i, long profit, long x){
        if(i >= a.size()){
            return 0;
        }
        long cost = a.get(i)[0];
        long happiness = a.get(i)[1];
        if (dp.containsKey(i) && dp.get(i).containsKey(profit)) {
            return dp.get(i).get(profit);
        }
        long skip = dp(a, i + 1, profit + x, x);
        long pick = (long) -1e9;
        if(profit - cost >= 0){
            pick = happiness + dp(a, i + 1, profit + x - cost, x);
        }
        long curr = Math.max(skip, pick);
        if(!dp.containsKey(i)) dp.put(i, new HashMap<>());
        dp.get(i).put(profit, curr);
        return curr;
    }

    static class Pair implements Comparable<Pair>{
        int key;
        int value;
        Pair(int key, int value){
            this.key = key;
            this.value = value;
        }
        @Override
        public int compareTo(Pair that){
            if(this.key == that.key){
                return that.value - this.value;
            }
            return this.key - that.key;
        }
	    @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return key == pair.key && value == pair.value;
        }

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

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

            if (xRoot == yRoot)
                return;

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

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

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

        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that challenge the logic of the dynamic programming (DP) solution. The code appears to have a logical issue in how it handles the profit and the cost calculations, especially in the recursive DP function. Specifically, the code might not correctly manage the state transitions when considering whether to buy happiness or skip it.

Here are some key points to consider:
1. **Initial State**: Ensure the initial state is correctly handled, especially when there is no money at the start.
2. **Cost and Profit Management**: Ensure the code correctly manages the profit and cost, especially when the cost is greater than the current profit.
3. **Free Happiness**: Test cases where some happiness is free (cost = 0).
4. **Large Costs**: Test cases where the cost is very high and cannot be afforded.
5. **Multiple Test Cases**: Ensure the code handles multiple test cases correctly.

Given these considerations, here is a small-scale test input that should expose potential issues in the code:

```plaintext
4
1 10
1 5
2 10
0 10
20 100
0 1
100 100
200 100
1 100
100 1
```

### Explanation:
1. **Test Case 1**:
   - **m = 1, x = 10**
   - **c_1 = 1, h_1 = 5**
   - Charlie earns 10 pounds but can only spend in the next month, so he cannot buy the happiness in the first month. Expected output: 0.

2. **Test Case 2**:
   - **m = 2, x = 10**
   - **c_1 = 0, h_1 = 10**
   - **c_2 = 200, h_2 = 100**
   - Charlie can get the free happiness in the first month. He cannot afford the second month's happiness. Expected output: 10.

3. **Test Case 3**:
   - **m = 4, x = 100**
   - **c_1 = 0, h_1 = 1**
   - **c_2 = 100, h_2 = 100**
   - **c_3 = 200, h_3 = 100**
   - **c_4 = 1, h_4 = 100**
   - Charlie can get the free happiness in the first month and then save enough to buy the happiness in the fourth month. Expected output: 201.

4. **Test Case 4**:
   - **m = 2, x = 100**
   - **c_1 = 100, h_1 = 100**
   - **c_2 = 100, h_2 = 1**
   - Charlie can save enough to buy the happiness in the first month but not enough to buy both. Expected output: 100.

This test input should help expose any issues in the DP logic, especially in handling the profit and cost calculations correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Happiness
time_limit_ms: 3000
memory_limit_mb: 256
Description: Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain happiness $h_i$.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum obtainable sum of happiness.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 50$, $1 \le x \le 10^8$) — the total number of months and the monthly salary.

The $i$-th of the following $m$ lines contains two integers, $c_i$ and $h_i$ ($0 \le c_i \le 10^8$, $1 \le h_i \le 10^3$) — the cost and happiness on offer for the $i$-th month. Note that some happiness may be free ($c_i=0$ for some $i$'s).

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

Output: For each test case, print a single integer, the maximum sum of happiness Charlie could obtain.

Examples:
input:
7
1 10
1 5
2 80
0 10
200 100
3 100
70 100
100 200
150 150
5 8
3 1
5 3
3 4
1 5
5 3
2 5
1 5
2 1
5 3
2 5
2 4
4 1
5 1
3 4
5 2
2 1
1 2
3 5
3 2
3 2
output:
0
10
200
15
1
9
9

Note: In the first test case, Charlie only gets paid at the end of the month, so is unable to afford anything.

In the second test case, Charlie obtains the free happiness in the first month.

In the third test case, it's optimal for Charlie to buy happiness in the second month. Even with money left at the end, Charlie could not go back in time to obtain the happiness on offer in the first month.

Buggy Code:
import java.util.*;

public class Solution{
    public static void main(String [] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int m = sc.nextInt();
            int x = sc.nextInt();
            int [] c = new int[m];
            int [] h = new int[m];

            for(int i=0; i<m; i++){
                c[i] = sc.nextInt();
                h[i] = sc.nextInt();
            }
            HashMap<Integer, HashMap<Integer, Integer>> memo = new HashMap<>();

            for(int i=0;i<m; i++)
                memo.put(i, new HashMap<>());
            System.out.println(solve(m, x, 0, 0, c, h, memo));

        }
    }

    private static int solve(int m, int x, int index, int currMoney, int [] c, int [] h, HashMap<Integer, HashMap<Integer, Integer>> memo){
        if(index==m){
            return 0;
        }

        if(memo.get(index).containsKey(currMoney)){
            return memo.get(index).get(currMoney);
        }

        int take = 0;
        if(c[index]<=currMoney){
            take = h[index] + solve(m, x, index+1, currMoney+x-c[index], c, h, memo);
        }

        int notTake = solve(m, x, index+1, currMoney+x, c, h, memo);
        if(c[index]==0){
            notTake = h[index];
        }

        memo.get(index).put(currMoney, Math.max(take, notTake));

        return Math.max(take, notTake);

    }
}

**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
3 10
10 100
5 50
15 150
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long int
const int MOD = 1e9 + 7;
#define inf 1e17
int cnt=0;
vector<int> sizes(200005);
vector<int> adj[200005];

//write dfs for finding the number of nodes in the subtree of all nodes

void dfs(int node,int parent){
    if(parent!=-1 && adj[node].size()==1){
        sizes[node]=0;
        return;
    }
    int maxi=-inf;
    for(auto it:adj[node]){
        if(it!=parent){
            dfs(it,node);
            sizes[node]=max(sizes[node],sizes[it]+1);
        }
    }
}

int32_t main(){
    cin.tie(0)->sync_with_stdio(0);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int a,b;
        cin>>a>>b;
        for(int i=0;i<n-1;i++){
            int u,v;
            cin>>u>>v;
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        queue<int>q;
        int dist[n+1];
        for(int i=0;i<=n;i++){
            dist[i]=0;
        }
        q.push(a);
        dist[a]=0;
        while(!q.empty()){
            int node = q.front();
            q.pop();
            for(auto it:adj[node]){
                if(dist[it]==0){
                    dist[it]=dist[node]+1;
                    q.push(it);
                }
            }
        }
        dfs(b,-1);
        vector<int> v;
        int sum=0;
        for(auto it:adj[b]){
            v.push_back(sizes[it]);
        }
        sort(v.begin(),v.end());
        sum=2*(n-1);
        sum-=v[v.size()-1]+1;
        // for(int i=1;i<=n;i++){
        //     cout<<sizes[i]<<' ';
        // }
        // cout<<endl;
        if(n==2){
            if(a==b){
                cout<<1<<endl;
            }else{
                cout<<2<<endl;
            }
        }else{
            cout<<sizes[b]+sum-(dist[b]%2==0)<<endl;
        }
        sizes.clear();
        for(int i=1;i<=n;i++){
            adj[i].clear();
        }
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define optimize()                 \
    ios_base ::sync_with_stdio(0); \
    cin.tie(0);                    \
    cout.tie(0);
#define fraction()                \
    cout.unsetf(ios::floatfield); \
    cout.precision(10);           \
    cout.setf(ios::fixed, ios::floatfield);
#define pi              3.14159265358979323846264338327950288
#define endl            '\n'
#define ll              long long int
#define ld              long double
#define ha              cout << "YES" << endl
#define na              cout << "NO" << endl
#define na1             cout << "-1" << endl
#define lup(z, n)       for (int i = z; i < n; ++i)
#define lup1(z, n)      for (int i = z; i <= n; ++i)
#define MAX             (ll)(1LL << 31) - 1
#define F               first
#define S               second
#define PB              push_back
#define MP              make_pair
#define IS              to_string
#define SI              stoll
#define popcnt          __builtin_popcountll
#define clz             __builtin_clzll
#define um              unordered_map
#define GCD             __gcd
#define B               begin
#define E               end
#define mem(arr, x)     memset(arr, x, sizeof arr)
#define prr             pair<ll, ll>
#define rt(v, l)        rotate(v.begin(), v.begin() + l, v.end());
#define bitsetTOint(bt) bt.to_ulong()
ll lcm(ll a, ll b)      {ll g = __gcd(a, b); return (a * b) / g;}
int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;
//:::::::::::::::::::::::::::::::::::::::GlitcH::::::::::::::::::::::::::::::::
int32_t main()
{
    optimize();
    auto cl=clock();
    int t = 1;
    cin >> t;
    for(int tt=1;tt<=t;tt++){
        ll n; cin>>n;
        ll a,b; cin>>a>>b;
        vector<ll> g[n+2];
        for(int i=1;i<n;i++){
            ll u,v; cin>>u>>v;
            g[u].PB(v);
            g[v].PB(u);
        }
        if(n==1){
            cout<<0<<endl;
            continue;
        }
        ll vis[n+2]={};
        deque<ll> dq; dq.PB(a); vis[a]=-1;
        while(!dq.empty()){
            for(auto u:g[dq.front()]){
                if(!vis[u]){
                    dq.PB(u);
                    vis[u]=dq.front();
                }
            }
            dq.pop_front();
        }
        vector<ll> vq;
        ll ck=b;
        while(vis[ck]>0){
            vq.PB(vis[ck]);
            ck=vis[ck];
        }
        ll ind=-1;
        ll ans=0;
        if(vq.size()==1){
            ans++;
            ind=a;
        }else{
            ll x=vq.size();
            if(!(x%2)){
                ll mid=x/2;
                ans=mid;
                ind=vq[x-mid-1];
            }else{
                ll mid=(x+1)/2;
                ans=mid;
                ind=vq[x-mid];
            }
        }
        ll tem=0;
        mem(vis,0);
        dq.PB(ind); vis[ind]=1;
        while(!dq.empty()){
            for(auto u:g[dq.front()]){
                if(!vis[u]){
                    dq.PB(u);
                    vis[u]=vis[dq.front()]+1;
                    tem=max(tem,vis[u]-1);
                }
            }
            dq.pop_front();
        }
        ans+=2*(n-1);
        ans-=tem;
        cout<<ans<<endl;
    }
    // cerr<<1.0*(clock()-cl)/CLOCKS_PER_SEC<<endl;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1: Single Vertex**
   - **Input:** `1 1 1`
   - **Output:** `0`
   - **Reason:** With only one vertex, no moves are needed to paint it blue.

2. **Test Case 2: Two Vertices**
   - **Input:** `2 1 2 1 2`
   - **Output:** `2`
   - **Reason:** This is a simple case where both vertices are connected. The minimum number of steps to paint both vertices blue is 2.

3. **Test Case 3: Star Graph with 5 Vertices**
   - **Input:** `5 1 3 1 2 2 3 3 4 3 5`
   - **Output:** `8`
   - **Reason:** In this star graph, vertex 3 is the center. The minimum number of steps to paint all vertices blue is 8.

These test cases cover different structures of the tree and edge cases, which can help expose bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

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

using namespace std;

typedef pair<int, int> pii;
typedef long long LL;

const int MOD = 998244353;
const int MX = 5e5 + 23;

int n;
vector<int> e[MX];
int a, b;

int fa[MX][18], dep[MX];
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 17; ~i; --i)
    if (dep[fa[x][i]] >= dep[y])
      x = fa[x][i];
  if (x == y) return x;
  for (int i = 17; ~i; --i)
    if (fa[x][i] != fa[y][i])
      x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}

int dist(int x, int y) {
  return dep[x] + dep[y] - 2 * dep[LCA(x, y)];
}

void dfs(int x, int f, int depth) {
  fa[x][0] = x;
  for (int i = 1; i < 18; ++i)
    fa[x][i] = fa[fa[x][i - 1]][i - 1];
  dep[x] = depth;
  for (auto y : e[x]) {
    if (y == f) continue;
    dfs(y, x, depth + 1);
  }
}

int dis[MX];
int getfar(int x) {
  queue<int> q;
  for (int i = 1; i <= n; ++i) dis[i] = n + 2;
  dis[x] = 0;
  int la = 0;
  while (!q.empty()) {
    int x = q.front();
    la = x;
    q.pop();
    for (auto y : e[x]) {
      if (dis[y] > dis[x] + 1) {
        dis[y] = dis[x] + 1;
        q.push(y);
      }
    }
  }
  return x;
}

int dist(int x, int f, int aim) {
  if (x == aim) return 0;
  for (auto y : e[x]) {
    if (y == f) continue;
    int r = dist(y, x, aim);
    if (r != -1) return r + 1;
  }
  return -1;
}

pii getdia() {
  int x = getfar(1);
  int y = getfar(x);
  return {x, y};
}

void solve() {
  cin >> n >> a >> b;
  for (int i = 1; i <= n; ++i) e[i].clear();
  for (int i = 1; i < n; ++i) {
    int x, y; cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }

  int ans = INT_MAX;
  dfs(1, 0, 1);
  auto [x, y] = getdia();

  for (int i = 1; i <= n; ++i) {
    int t = max(dist(a, i), dist(b, i));
    ans = min(ans, t + 2 * (n - 1) - max(dist(x, i), dist(y, i)));
  }

  cout << ans << "\n";
}

int main() {
  ios::sync_with_stdio(false);
  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
4
1 2
1 2
1 3
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import sys
rl = sys.stdin.readline
N = int(rl())

def dfs(x, c):
    global maxd
    global aTob
    maxd = max(maxd, c)
    for nx in g[x]:
        if nx == b: aTob = c+1
        if v[nx] == 0:
            v[nx] = 1
            dfs(nx, c+1)

for _ in range(N):
    n = int(rl())
    a, b = map(int, rl().split())

    g = [[] for _ in range(n+1)]
    for _ in range(n-1):
        s, e = map(int, rl().split())
        g[s].append(e)
        g[e].append(s)

    maxd = 0
    aTob = 0
    v = [0] * (n+1); v[a] = 1
    dfs(a, 0)

    ans = (n - maxd - 1) * 2 + maxd + aTob

    a, b = b, a
    maxd = 0
    aTob = 0
    v = [0] * (n+1); v[a] = 1
    dfs(a, 0)
    ans = min(ans, (n - maxd - 1) * 2 + maxd + aTob)

    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
3
1 2
1 2
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
from collections import defaultdict, deque

def dist(tree, a, b):
    if a == b:
        return 0, a
    q_a = deque([a])
    q_b = deque([b])
    v_a = {a: 0}
    v_b = {b: 0}

    while q_a or q_b:
        if q_a:
            c_a = q_a.popleft()
            for nex in tree[c_a]:
                if nex not in v_a:
                    v_a[nex] = v_a[c_a] + 1
                    q_a.append(nex)
                    if nex in v_b:
                        return max(v_a[nex], v_b[nex])

        if q_b:
            c_b = q_b.popleft()
            for nex in tree[c_b]:
                if nex not in v_b:
                    v_b[nex] = v_b[c_b] + 1
                    q_b.append(nex)
                    if nex in v_a:
                        return max(v_a[nex], v_b[nex])

t = int(input())
results = []

for _ in range(t):
    n = int(input())
    a, b = map(int, input().split())

    tree = defaultdict(list)
    for _ in range(n - 1):
        x, y = map(int, input().split())
        tree[x].append(y)
        tree[y].append(x)

    result = dist(tree, a, b)
    results.append((n-1) * 2 - result + 1)

print(*results, sep="\n")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import sys, math, queue
input = sys.stdin.readline
out = []

MOD = 998244353

class Graph:

    def __init__(self, n, Adj):
        self.n = n
        self.Adj = Adj

    # Basic traversal, prints all nodes
    def dfs(self, root):
        """Depth-first traversal of a graph"""
        stack = [root]
        visited = set()
        while stack != []:
            cur = stack.pop()
            if cur not in visited:
                visited.add(cur)

                # DO SOMETHING AT EACH NODE!
                print(cur)

                for node in self.Adj[cur]:
                    if node not in visited:
                        stack.append(node)

    # Returns path from [root] to [target]
    def dfs_path(self, root, target):
        """Finds a path from [root] to [target]"""
        stack = [root]
        visited = set()
        Parents = [-1 for _ in range(self.n + 1)]
        while stack != []:
            cur = stack.pop()
            if cur not in visited:
                visited.add(cur)

                if cur == target:
                    path = [cur]
                    while cur != root:
                        cur = Parents[cur]
                        path.append(cur)
                    path.reverse()
                    return path

                for node in self.Adj[cur]:
                    if node not in visited:
                        stack.append(node)
                        Parents[node] = cur

        return []

    # Basic traversal, prints all nodes
    def bfs(self, root):
        "Breadth-first traversal of a graph"
        Visited = set()
        Q = queue.Queue()
        Q.put_nowait(root)

        Visited.add(root)
        print(root)

        while not Q.empty():
            prev = Q.get_nowait()
            for node in self.Adj[prev]:
                if node not in Visited:
                    Visited.add(node)

                    # DO SOMETHING AT EACH NODE!
                    print(node)

                    Q.put_nowait(node)

    # Get path from [root] to [target]
    def bfs_path(self, root, target):
        "Breadth-first traversal of a graph"
        Visited = set()
        Q = queue.Queue()
        Q.put_nowait(root)
        Parents = [-1 for _ in range(self.n+1)]

        Visited.add(root)

        while not Q.empty():
            prev = Q.get_nowait()
            for node in self.Adj[prev]:
                if node not in Visited:
                    Visited.add(node)

                    Parents[node] = prev

                    if node == target:
                        path = [target]
                        cur = target
                        while cur != root:
                            cur = Parents[cur]
                            path.append(cur)

                        path.reverse()
                        return path

                    Q.put_nowait(node)

        return []

    # Get distances to all nodes
    def bfs_dists(self, root):
        "Find distances from root to each node in the graph."
        Visited = set()
        Q = queue.Queue()
        Q.put_nowait(root)
        D = [0 for _ in range(self.n+1)]

        Visited.add(root)

        while not Q.empty():
            prev = Q.get_nowait()
            for node in self.Adj[prev]:
                if node not in Visited:
                    Visited.add(node)

                    D[node] = D[prev] + 1

                    Q.put_nowait(node)

        return D

    # Shortest path from [root] to [target] using meet in the middle
    def bfs_path_fast(self, root, target):
        "Breadth-first traversal of a graph from two points, until they meet."
        V1 = set()
        V1.add(root)
        Q1 = queue.Queue()
        Q1.put_nowait(root)
        Parents = [-1 for _ in range(self.n+1)]

        V2 = set()
        V2.add(target)
        Q2 = queue.Queue()
        Q2.put_nowait(target)
        Children = [-1 for _ in range(self.n+1)]

        while (not Q1.empty()) and (not Q2.empty()):
            cur1 = Q1.get_nowait()
            for node in self.Adj[cur1]:
                if node not in V1:
                    V1.add(node)
                    Parents[node] = cur1

                    if node in V2:
                        path = [node]
                        cur = node
                        while cur != root:
                            cur = Parents[cur]
                            path.append(cur)
                        path.reverse()

                        cur = node
                        while cur != target:
                            cur = Children[cur]
                            path.append(cur)

                        return path

                    Q1.put_nowait(node)

            cur2 = Q2.get_nowait()
            for node in self.Adj[cur2]:
                if node not in V2:
                    V2.add(node)
                    Children[node] = cur2

                    if node in V1:
                        path = [node]
                        cur = node
                        while cur != root:
                            cur = Parents[cur]
                            path.append(cur)
                        path.reverse()

                        cur = node
                        while cur != target:
                            cur = Children[cur]
                            path.append(cur)

                        return path

                    Q2.put_nowait(node)

        return []

def solve():
    n = int(input())
    a, b = map(int, input().split())
    Adj = [[] for _ in range(n+1)]
    for _ in range(n-1):
        x, y = map(int, input().split())
        Adj[x].append(y)
        Adj[y].append(x)

    G = Graph(n, Adj)
    path = G.bfs_path_fast(a, b)
    m = len(path)
    ans = m // 2
    ans += 2 * (n - 1)
    if path == []:
        ans -= max(G.bfs_dists(a))
    else:
        ans -= max(G.bfs_dists(path[(m - 1) // 2]))
    return ans

for _ in range(int(input())):
    out.append(solve())

print("\n".join(map(str, out)))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

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

public class Main {
    public static class Pair<L, R> {

        private final L left;
        private final R right;

        public Pair(L left, R right) {
            assert left != null;
            assert right != null;

            this.left = left;
            this.right = right;
        }

        public L getLeft() {
            return left;
        }

        public R getRight() {
            return right;
        }

        @Override
        public int hashCode() {
            return left.hashCode() ^ right.hashCode();
        }

        @Override
        public boolean equals(Object o) {
            if (!(o instanceof Pair)) return false;
            Pair pairo = (Pair) o;
            return this.left.equals(pairo.getLeft()) &&
                    this.right.equals(pairo.getRight());
        }

    }

    static ArrayList<ArrayList<Integer>> graph = new ArrayList<>();

    static int count = 0, init = -1, tot = -1;

    static void dfs(int v, HashSet<Integer> visited) {
        visited.add(v);
        if (v != init)
            count++;
        for (int n : graph.get(v)) {
            if (!visited.contains(n))
                dfs(n, visited);
        }
        if (visited.size() != tot)
            count++;
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while ((t--) != 0) {
            int n = sc.nextInt();
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            Set<Integer> set = new HashSet<>();
            ArrayList<Pair<Integer, Integer>> nodes = new ArrayList<>();
            for (int i = 0; i < n - 1; i++) {
                int x = sc.nextInt() - 1;
                int y = sc.nextInt() - 1;
                nodes.add(new Pair<>(x, y));
                set.add(x);
                set.add(y);
            }
            for (int i : set) graph.add(new ArrayList<>());
            for (Pair<Integer, Integer> node : nodes) {
                graph.get(node.getLeft()).add(node.getRight());
                graph.get(node.getRight()).add(node.getLeft());
            }
            count = 0;
            init = a;
            tot = set.size();
            dfs(a, new HashSet<>());
            Queue<Integer> q = new LinkedList<>();
            q.add(a);
            int[] dist = new int[set.size()];
            Arrays.fill(dist, Integer.MAX_VALUE);
            dist[a] = 0;
            while (!q.isEmpty()) {
                int s = q.poll();
                for (int nei : graph.get(s)) {
                    if (dist[nei] > dist[s] + 1) {
                        dist[nei] = dist[s] + 1;
                        q.add(nei);
                    }
                }
            }
            System.out.println(count +  dist[b]);
        }
    }
}

**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
2
1 1
1 2
4
1 2
1 2
1 3
1 4
5
1 3
1 2
2 3
3 4
3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) throws Exception{
        BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
        int tests = Integer.parseInt(buffer.readLine());
        for (int test = 0; test < tests; test ++){
            StringTokenizer token = new StringTokenizer(buffer.readLine());
            int n = Integer.parseInt(token.nextToken());
            token = new StringTokenizer(buffer.readLine());
            int a = Integer.parseInt(token.nextToken());
            int b = Integer.parseInt(token.nextToken());
            HashMap<Integer, List<Integer>> map = new HashMap<>();
            // # of moves originally is (n-1) * 2
            for (int x = 0; x < n - 1; x ++){
                token = new StringTokenizer(buffer.readLine());
                int j = Integer.parseInt(token.nextToken());
                int k = Integer.parseInt(token.nextToken());
                map.putIfAbsent(j, new ArrayList<>());
                map.putIfAbsent(k, new ArrayList<>());
                map.get(j).add(k); map.get(k).add(j);
            }
            // System.out.println(map);
            Queue<List<Integer>> queue = new LinkedList<>(); queue.add(new ArrayList<>(Arrays.asList(a)));
            int meeting = 0;
            int time = 0;
            while (queue.size() > 0){
                List<Integer> next = queue.poll();
                int last = next.get(next.size() - 1);
                if (last == b) {time = next.size() / 2; meeting = next.get((next.size() - 1) / 2); break;}
                for (Integer e: map.get(last)){
                    if (next.size() == 1 || next.get(next.size() - 2) != e) {
                        next.add(e); queue.add(new ArrayList<>(next)); next.remove(e);
                    }
                }
            }
            // System.out.println(meeting + " " + time);
            // find minimal time starting at meeting
            int[] depth = new int[n + 1];
            depth[meeting] = 1;
            Queue<Integer> queue2 = new LinkedList<>(); queue2.add(meeting);
            int maxdepth = 1;
            while(queue2.size() > 0){
                int next = queue2.poll();
                maxdepth = Math.max(maxdepth, depth[next]);
                for (Integer e: map.get(next)){
                    if (depth[e] == 0) {depth[e] = depth[next] + 1; queue2.add(e);}
                }
            }
            // System.out.println(maxdepth);
            System.out.println(time + 2*n - 1 - maxdepth);
        }
    }
}

**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
2
1 2
1 2
5
1 1
1 2
1 3
1 4
1 5
8
5 5
7 1
1 5
1 8
8 3
7 2
8 6
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
from collections import defaultdict
for _ in range(int(input())):
    n=int(input())
    a,b=map(int,input().split())
    tree=defaultdict(list)
    for x in range(n-1):
        i,j=map(int,input().split())
        tree[i].append(j)
        tree[j].append(i)
    def depth(root,parent):
        if len(tree[root])==1:
            if parent==-1:
                return 1+depth(tree[root][0],root)
            else:
                return 1
        m=0
        for i in tree[root]:
            if i!=parent:
                m=max(m,1+depth(i,root))
        return m
    if a==b:
        x=depth(a,-1)
        print((n-1-x)*2+x)
    else:
        arr=[]
        def dfs(src,dest,par,path):
            path.append(src)
            if src==dest:
                for i in path:
                    arr.append(i)
                return
            if len(tree[src])==1:
                if par!=-1:
                    path.pop()
                    return
                else:
                    dfs(tree[src][0],dest,src,path)
                    path.pop()
                    return
            else:
                for i in tree[src]:
                    if i!=par:
                        dfs(i,dest,src,path)
                path.pop()
                return
        dfs(a,b,-1,[])
        x=depth(arr[(len(arr)-1)//2],-1)
        x-=1
        print(len(arr)//2+(n-1-x)*2+x)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import   sys, math
sys.setrecursionlimit(100000)
from collections import deque
finp   = sys.stdin.readline
MOD    = 10**9 + 7
njoin  = lambda x: '\n'.join(map(str, x))
ljoin  = lambda x: ' '.join(map(str, x))
sjoin  = lambda x: '\n'.join(x)
initl  = lambda x,y: [y]*x
INF    = float('inf')
NINF   = float('-inf')
ilist  = lambda: list(map(int, finp().split()))
slist  = lambda: list(input())
slists = lambda: input().split()
ivars  = lambda: map(int, finp().split())
ipt    = lambda: int(finp())
out    = []
#=============================================#
def sol():
    n = ipt()
    if n == 1:
        return 1
    pa, pb = ivars()
    adj = {}
    for i in range(n - 1):
        a,b = ivars()
        if a in adj:
            adj[a].append(b)
        else:
            adj[a] = [b]
        if b in adj:
            adj[b].append(a)
        else:
            adj[b] = [a]
    root, dist = findmid(adj,pa, pb)
    visited = set()
    newadj = {}
    def diameter(vertex):
        m = 1
        visited.add(vertex)
        curr = []
        for i in adj[vertex]:
            if i not in visited:

                curr.append((i, u:= diameter(i)))
                m = max(m, u+1)
        newadj[(vertex, m)] = max(curr, key = lambda x: x[1]) if len(curr) != 0 else None
        return m

    diameter(root)
    for i in newadj.keys():
        if i[0] == root:
            newroot = i
            break
    tot = 0
    stack = [root]
    seen = set([root])
    while stack:
        c = stack.pop()
        tot += len(adj[c])
        for i in adj[c]:
            if i not in seen:
                stack.append(i)
                seen.add(i)
    res = tot + dist
    if newroot != None:
        curr = newroot
        while curr[1] != 1:
            res -= 1
            if len(newadj[curr]) == None: break
            curr = newadj[curr]
    print(res)

#=============================================#

def findmid(adj_list, a, b):
    path = find_path(adj_list, a, b)

    mid = len(path) // 2
    if len(path) % 2 == 0:
        mid -= 1

    return path[mid], len(path)//2

def find_path(adj_list, start, end):
    queue = deque([(start, [start])])
    visited = set([start])

    while queue:
        node, path = queue.popleft()

        if node == end:
            return path

        for neighbor in adj_list[node]:
            if neighbor not in visited:
                visited.add(neighbor)
                queue.append((neighbor, path + [neighbor]))

    return []
if __name__ == "__main__":
    for _ in range(ipt()): sol()
    #sol()
    pass
'''
find midpoint
treat midpoint as root
dfs, multiplying each node by the number of children it has
'''

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code involves finding the midpoint of the path between two nodes, treating that midpoint as the root, and then performing a depth-first search (DFS) to calculate the number of steps required to paint all vertices blue.

### Key Considerations:
1. **Tree Structure**: The tree should be structured in a way that the midpoint calculation and DFS traversal are non-trivial.
2. **Initial Positions**: The initial positions of \(P_A\) and \(P_B\) should be chosen to maximize the complexity of the problem.
3. **Edge Cases**: Consider scenarios where the tree is a straight line, a star, or a more complex structure.

### Chosen Test Case:
Let's create a test case where the tree is a straight line. This will challenge the midpoint calculation and the DFS traversal.

### Test Case:
```plaintext
1
5
1 5
1 2
2 3
3 4
4 5
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Vertices**: 5
- **Initial Positions**: \(P_A\) is at vertex 1, and \(P_B\) is at vertex 5.
- **Edges**: The tree is a straight line from 1 to 5.

### Why This Test Case?
1. **Midpoint Calculation**: The path from 1 to 5 has a length of 4, so the midpoint is vertex 3. The code should correctly identify this.
2. **DFS Traversal**: The DFS traversal from the midpoint (vertex 3) should correctly calculate the number of steps required to paint all vertices blue.
3. **Complexity**: Even though the tree is a simple line, it requires careful handling of the midpoint and the steps to ensure all vertices are painted blue.

This test case should expose any issues in the midpoint calculation and the DFS traversal logic in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

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

using namespace std;

const int MAXN = 2e5 + 10;
int T, n, a, b, dis[MAXN], pre[MAXN];
int ans, mx;
vector<int> t[MAXN];

void dfs(int u)
{
    for (auto v : t[u])
    {
        if (v == pre[u]) continue;
        dis[v] = dis[u] + 1;
        pre[v] = u;
        dfs(v);
    }
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    cin >> T;
    while (T --> 0)
    {
        cin >> n >> a >> b;
        for (int i = 1; i <= n; i ++) t[i].clear();
        for (int i = 1, u, v; i < n; i ++)
        {
            cin >> u >> v;
            t[u].emplace_back(v);
            t[v].emplace_back(u);
        }

        memset(dis, 0, sizeof(dis));
        memset(pre, 0, sizeof(pre));
        dfs(a);
        int x = b;
        while (dis[x] > (dis[b] >> 1))
            x = pre[x];
        ans = dis[b] - dis[x] + ((n - 1) << 1);

        memset(dis, 0, sizeof(dis));
        memset(pre, 0, sizeof(pre));
        dfs(x);
        for (int i = 1; i <= n; i++)
            mx = max(mx, dis[i]);
        ans -= mx;
        cout << ans << '\n';
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

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

#ifdef LOCAL
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << "\033[0m" << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#define dbg(...) cerr<<"\033[1;31m"<< "LINE "<<__LINE__<<": -> [" << #__VA_ARGS__ << "]:", dbg_out(__VA_ARGS__)
#else
#define dbg(...) 19
#endif

#define int int64_t

void solve(){
    int N; cin >> N;
    int startA, startB; cin >> startA >> startB;
    vector<vector<int>> adj(N+1);
    for(int i = 0; i < N-1; ++i){
        int u, v; cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    vector<int> distances(N+1);
    vector<int> subtree_sizes(N+1);
    auto dfs = [&](auto&& dfs, int node, int parent, int dist) -> void{
        distances[node] = dist;
        subtree_sizes[node] = 1;
        for(auto& nxt : adj[node]){
            if(nxt != parent){
                dfs(dfs, nxt, node, dist+1);
                subtree_sizes[node] += subtree_sizes[nxt];
            }
        }
    };
    dfs(dfs, startB, -1, 0);

    dbg(2*(N-1) - *max_element(distances.begin(), distances.end()));
    int res = distances[startA];
    if(N > 2){
        res += 2*(N-1)- *max_element(distances.begin(), distances.end());
    }
    if(distances[startA] % 2 == 1){
        res++;
    }
    cout << res << "\n";
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T=1; cin >> T;
    for(int tt = 1; tt <= T; ++tt){
        solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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

The first line of each test case contains one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import java.util.*;
import java.io.*;
public class Main {
     static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    static class FastWriter {
		private final BufferedWriter bw;

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

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

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

		public void close() throws IOException {
			bw.close();
		}
	}
    public static FastReader sc=new FastReader();
    public static FastWriter out = new FastWriter();

   public static void dfs(int node, int par,ArrayList<ArrayList<Integer>>adj,int ch,int []height){
       height[node]=ch;
       for(int i=0;i<adj.get(node).size();i++){
        int nbr=adj.get(node).get(i);
           //System.out.println(node+" "+ch+" "+nbr);

        if(nbr!=par){
            dfs(nbr,node,adj,ch+1,height);
        }
       }
   }

    public static void solve(int test){
        try{
            int n,a,b;

            n=sc.nextInt();a=sc.nextInt();b=sc.nextInt();
             if(test==103){
                System.out.print(n+"#"+a+"#"+b+"#");
            }
            a--;b--;
            ArrayList<ArrayList<Integer>>adj=new ArrayList<>();
            for(int i=0;i<n;i++){
                adj.add(new ArrayList<Integer>());
            }
            for(int i=0;i<n-1;i++){
                int u,v;
                u=sc.nextInt();
                v=sc.nextInt();
                if(test==103)
                {
                     System.out.print(u+"$"+v);
                }

                u--;v--;

                adj.get(u).add(v);
                adj.get(v).add(u);
            }
            int fnode=0;
            int []vis1=new int[n];
            int []vis2=new int[n];
            vis1[a]=1;
            vis2[b]=1;
            Queue<Integer>qa=new LinkedList<>();
            Queue<Integer>qb=new LinkedList<>();
            qa.add(a);
            qb.add(b);
            int []dis=new int[n];
            dis[b]=0;
            while((!qa.isEmpty()) && !(qb.isEmpty())){
                int node=qa.remove();
                for(int i=0;i<adj.get(node).size();i++){
                    int nbr=adj.get(node).get(i);
                    if(vis1[nbr]==0){
                    vis1[nbr]=1;
                    qa.add(nbr);
                    }
                }
                node=qb.remove();boolean ch=false;
                for(int i=0;i<adj.get(node).size();i++){
                    int nbr=adj.get(node).get(i);
                    if(vis1[nbr]==1){
                        dis[nbr]=dis[node]+1;
                       // System.out.println(node+" "+nbr);
                        ch=true;
                        fnode=nbr;
                        break;
                    }
                    else if(vis2[nbr]==0){
                        vis2[nbr]=1;
                        dis[nbr]=dis[node]+1;
                        qb.add(nbr);
                    }
                }
                if(ch==true){
                    break;
                }
            }
           if(a==b){
               fnode=a;
               dis[fnode]=0;
           }
            int []height=new int[n];
            dfs(fnode,-1,adj,0,height);
            int maxh=0;
            for(int i=0;i<n;i++){
                maxh=Math.max(maxh,height[i]);
            }
            //System.out.println(fnode+" "+dis[fnode]+" "+maxh);
            long ans=dis[fnode]+(2*(n-1))-maxh;
            System.out.println(ans);

        }
        catch (Exception e) {
            return;
        }
    }
    public static void main(String args[]) {
       int t;
       int test=0;
       t=sc.nextInt();
       while(t-->0){
           solve(test++);
       }
    }
}

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

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

Think step by step.

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

1. **Single Vertex Tree**: A tree with only one vertex.
2. **Two Vertices Tree**: A tree with two vertices.
3. **Star Tree**: A tree where one vertex is connected to all other vertices.
4. **Linear Tree**: A tree where vertices form a linear structure.
5. **Complex Tree**: A tree with a more complex structure to ensure the algorithm handles general cases correctly.

Let's generate a small-scale test input that includes these scenarios:

1. **Single Vertex Tree**:
   - Input: `1 1 1`

2. **Two Vertices Tree**:
   - Input: `2 1 2 1 2`

3. **Star Tree**:
   - Input: `5 1 2 1 2 1 3 1 4 1 5`

4. **Linear Tree**:
   - Input: `5 1 2 1 2 2 3 3 4 4 5`

5. **Complex Tree**:
   - Input: `8 5 4 7 1 1 5 1 8 8 3 7 2 8 6 3 4`

Combining these into a single test input:

```plaintext
5
1 1 1
2 1 2 1 2
5 1 2 1 2 1 3 1 4 1 5
5 1 2 1 2 2 3 3 4 4 5
8 5 4 7 1 1 5 1 8 8 3 7 2 8 6 3 4
```

This test input covers a range of edge cases and should help expose any bugs in the given buggy code.