**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

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

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main implements Runnable{
    public static void main(String[] args) {
        new Thread(null, new Main(), "whatever", 1<<30).start();
    }

    public void run() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        solve(in, out);
        out.close();
     }

    static String reverse(String s) {
        return (new StringBuilder(s)).reverse().toString();
    }

    static void sieveOfEratosthenes(int n, int factors[], ArrayList<Integer> ar)
    {
        factors[1]=1;
        int p;
        for(p = 2; p*p <=n; p++)
        {
            if(factors[p] == 0)
            {
                ar.add(p);
                factors[p]=p;
                for(int i = p*p; i <= n; i += p)
                    if(factors[i]==0)
                        factors[i] = p;
            }
        }
        for(;p<=n;p++){
            if(factors[p] == 0)
            {
                factors[p] = p;
                ar.add(p);
            }
        }
    }

    static void sort(int ar[]) {
        int n = ar.length;
        ArrayList<Integer> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort1(long ar[]) {
        int n = ar.length;
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort2(char ar[]) {
        int n = ar.length;
        ArrayList<Character> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static long ncr(long n, long r, long mod) {
        if (r == 0)
            return 1;
        long val = ncr(n - 1, r - 1, mod);
        val = (n * val) % mod;
        val = (val * modInverse(r, mod)) % mod;
        return val;
    }

    static class SegTree {

        long tree[];
        long lz[];
        long r;

        long combine(long a, long b){
            return Math.min(a,b);
        }

        void build(long a[], int v, int tl, int tr) {
            if (tl == tr) {
                tree[v] = a[tl];
            }
            else {
                int tm = (tl + tr) / 2;
                build(a, v*2, tl, tm);
                build(a, v*2+1, tm+1, tr);
                tree[v] = combine(tree[2*v], tree[2*v+1]);
            }
        }

        void pointUpdate(int v, int tl, int tr, int pos, long val) {
            if(tl>pos||tr<pos)
                return;
            if(tl==pos&&tr==pos){
                tree[v] = val;
                return;
            }
            int tm = ((tl + tr) >> 1);
            pointUpdate(v*2, tl, tm, pos, val);
            pointUpdate(v*2+1, tm+1, tr, pos, val);
            tree[v] = combine(tree[2*v],tree[2*v+1]);
        }

        // void push(int v, int tl, int tr){
        //     if(tl==tr){
        //         lz[v] = 0;
        //         return;
        //     }
        //     tree[2*v] += lz[v];
        //     tree[2*v+1] += lz[v];
        //     lz[2*v] += lz[v];
        //     lz[2*v+1] += lz[v];
        //     lz[v] = 0;
        // }

        // void rangeUpdate(int v, int tl, int tr, int l, int r, long val) {
        //     if(tl>r||tr<l)
        //         return;
        //     push(v, tl, tr);
        //     if(tl>=l&&tr<=r){
        //         tree[v] += val;
        //         lz[v] += val;
        //         return;
        //     }
        //     int tm = ((tl + tr) >> 1);
        //     rangeUpdate(v*2, tl, tm, l, r, val);
        //     rangeUpdate(v*2+1, tm+1, tr, l, r, val);
        //     tree[v] = combine(tree[2*v],tree[2*v+1]);
        // }

        long get(int v, int tl, int tr, int l, int r, long val) {
            if(l>tr||r<tl||tree[v]>val){
                return 0;
            }
            if (tl == tr) {
                tree[v] = Integer.MAX_VALUE;
                return 1;
            }
            int tm = ((tl + tr) >> 1);
            long al = get(2*v, tl, tm, l, r, val);
            long ar = get(2*v+1, tm+1, tr, l, r, val);
            tree[v] = combine(tree[2*v],tree[2*v+1]);
            return al+ar;
        }

    }

    static class BIT{
        int n;
        long tree[];
        long operate(long el, long val){
            return el+val;
        }
        void update(int x, long val){
            for(;x<n;x+=(x&(-x))){
                tree[x] = operate(tree[x], val);
            }
        }
        long get(int x){
            long sum = 0;
            for(;x>0;x-=(x&(-x))){
                sum = operate(sum, tree[x]);
            }
            return sum;
        }
    }

    static int parent[];
    static int rank[];

    static long m = 0;

    static int find_set(int v) {
        if (v == parent[v])
            return v;
        return find_set(parent[v]);
    }

    static void make_set(int v) {
        parent[v] = v;
        rank[v] = 1;
    }

    static void union_sets(int a, int b) {
        a = find_set(a);
        b = find_set(b);
        if (a != b) {
            if (rank[a] < rank[b]){
                int tmp = a;
                a = b;
                b = tmp;
            }
            parent[b] = a;
            rank[a] += rank[b];
            // max1 = Math.max(max1,rank[a]);
        }
    }

    static int parent1[];
    static int rank1[];

    static int find_set1(int v) {
        if (v == parent1[v])
            return v;
        return find_set1(parent1[v]);
    }

    static void make_set1(int v) {
        parent1[v] = v;
        rank1[v] = 1;
    }

    static void union_sets1(int a, int b) {
        a = find_set1(a);
        b = find_set1(b);
        if (a != b) {
            if (rank1[a] < rank1[b]){
                int tmp = a;
                a = b;
                b = tmp;
            }
            parent1[b] = a;
            // if (rank1[a] == rank1[b])
            //     rank1[a]++;
            rank1[a] += rank1[b];
        }
    }

    static long max1 = 0;

    static int count = 0;
    static int count1 = 0;

    static class Node {
        int data;
        Node next;
        Node(int data, Node next){
            this.data = data;
            this.next = next;
        }
    }

    static int[] findLongestPalindromicString(String text)
    {
        int N = text.length();
        N = 2 * N + 1; // Position count
        int[] L = new int[N + 1]; // LPS Length Array
        L[0] = 0;
        L[1] = 1;
        int C = 1; // centerPosition
        int R = 2; // centerRightPosition
        int i = 0; // currentRightPosition
        int iMirror; // currentLeftPosition
        int maxLPSLength = 0;
        int diff = -1;

        for (i = 2; i < N; i++)
        {

            iMirror = 2 * C - i;
            L[i] = 0;
            diff = R - i;

            // If currentRightPosition i is within
            // centerRightPosition R
            if (diff > 0)
                L[i] = Math.min(L[iMirror], diff);

            while (((i + L[i]) + 1 < N && (i - L[i]) > 0) &&
                               (((i + L[i] + 1) % 2 == 0) ||
                         (text.charAt((i + L[i] + 1) / 2) ==
                          text.charAt((i - L[i] - 1) / 2))))
            {
                L[i]++;
            }

            if (L[i] > maxLPSLength) // Track maxLPSLength
            {
                maxLPSLength = L[i];
            }

            // If palindrome centered at currentRightPosition i
            // expand beyond centerRightPosition R,
            // adjust centerPosition C based on expanded palindrome.
            if (i + L[i] > R)
            {
                C = i;
                R = i + L[i];
            }
        }
        return L;
    }

    static Pair deleteEdges(int curr, ArrayList<Integer> ar[], int par, int limit){
        int edges = 0;
        int rem = 1;
        for(int u:ar[curr]){
            if(u==par)
                continue;
            Pair p = deleteEdges(u, ar, curr, limit);
            if(p.a>=limit){
                p.a = 0;
                edges++;
            }
            edges += p.b;
            rem += p.a;
        }
        return new Pair(rem, edges);
    }

    static int count_bits(int n){
        int c = 0;
        while(n>0){
            if(n%2==1)
                c++;
            n >>= 1;
        }
        return c;
    }

    public static void solve(InputReader sc, PrintWriter pw){
        int i, j = 0;
        // long mod = 998244353;
        long mod = 1000000007;
        // int factors[] = new int[400001];
        // ArrayList<Integer> ar = new ArrayList<>();
        // sieveOfEratosthenes(400000, factors, ar);
        // pw.println(ar.size());
        // HashSet<Integer> set = new HashSet<>();
        // for(int x:ar){
        //     set.add(x);
        // }
        // int ans[] = new int[200001];
        // int next = 1;
        // for(i=1;i<=200000;i++){
        //     if(ans[i]==0){
        //         ans[i] = next;
        //         for(j=i+1;j<=200000;j++){
        //             if(!set.contains(i^j)){
        //                 ans[j] = next;
        //             }
        //         }
        //         next++;
        //     }
        // }
        // int t = 1;
        int t = sc.nextInt();
        u: while (t-- > 0) {
            int n = sc.nextInt();
            if(n==-1)
                break;
            int m = sc.nextInt();
            ArrayList<Integer> ar[] = new ArrayList[n];
            for(i=0;i<n;i++){
                ar[i] = new ArrayList<>();
            }
            for(i=0;i<m;i++){
                int a = sc.nextInt()-1;
                int b = sc.nextInt()-1;
                ar[a].add(b);
                ar[b].add(a);
            }
            int c[] = new int[n];
            // c[0] = 1;
            boolean poss = dfs(0, c, ar, 1);
            if(!poss){
                pw.println("Alice");
                pw.flush();
                for(i=0;i<n;i++){
                    pw.println(1+" "+2);
                    pw.flush();
                    int x = sc.nextInt();
                    if(x==-1){
                        break;
                    }
                    int e = sc.nextInt();
                    if(e==-1)
                        break;
                }
            }
            else{
                ArrayList<Integer> x = new ArrayList<>();
                ArrayList<Integer> y = new ArrayList<>();
                for(i=0;i<n;i++){
                    if(c[i]==1)
                        x.add(i);
                    else
                        y.add(i);
                }
                int p = 0, q = 0;
                pw.println("Bob");
                pw.flush();
                int a = sc.nextInt();
                int b = sc.nextInt();
                p++;
                pw.println((x.get(0)+1)+" "+a);
                pw.flush();
                int fc = a;
                int nc = 0;
                for(i=1;i<n;i++){
                    a = sc.nextInt();
                    if(a==-1)
                        break;
                    b = sc.nextInt();
                    if(b==-1)
                        break;
                    // if(p<x.size()){
                    if(fc==a&&p<x.size()){
                        pw.println((x.get(p)+1)+" "+a);
                        p++;
                        pw.flush();
                    }
                    else if(fc==b&&p<x.size()){
                        pw.println((x.get(p)+1)+" "+b);
                        p++;
                        pw.flush();
                    }
                    else if(nc==0&&fc!=a){
                        pw.println((y.get(q)+1)+" "+a);
                        nc = a;
                        q++;
                        pw.flush();
                    }
                    else if(nc==0&&fc!=b){
                        pw.println((y.get(q)+1)+" "+b);
                        nc = b;
                        q++;
                        pw.flush();
                    }
                    else if(nc==a&&q<y.size()){
                        pw.println((y.get(q)+1)+" "+a);
                        q++;
                        pw.flush();
                    }
                    else if(nc==b&&q<y.size()){
                        pw.println((y.get(q)+1)+" "+a);
                        q++;
                        pw.flush();
                    }
                    else if(a!=nc&&p<x.size()){
                        pw.println((x.get(p)+1)+" "+a);
                        p++;
                        pw.flush();
                    }
                    else if(b!=nc&&p<x.size()){
                        pw.println((x.get(p)+1)+" "+b);
                        p++;
                        pw.flush();
                    }
                    else if(a!=fc&&q<y.size()){
                        pw.println((y.get(q)+1)+" "+a);
                        q++;
                        pw.flush();
                    }
                    else{
                        //  if(b!=fc&&q<y.size()){
                        pw.println((y.get(q)+1)+" "+b);
                        q++;
                        pw.flush();
                    }
                }
            }
        }
    }

    static boolean dfs(int curr, int c[], ArrayList<Integer> ar[], int col){
        if(c[curr]>0&&c[curr]!=col){
            return false;
        }
        if(c[curr]==col)
            return true;
        c[curr] = col;
        for(int i:ar[curr]){
            if(!dfs(i, c, ar, col^3))
                return false;
        }
        return true;
    }

    static void KMPSearch(char pat[], char txt[], long pres[]){
        int M = pat.length;
        int N = txt.length;
        int lps[] = new int[M];
        int j = 0;
        computeLPSArray(pat, M, lps);
        int i = 0;
        while (i < N) {
            if (pat[j] == txt[i]) {
                j++;
                i++;
            }
            if (j == M) {
                pres[i-1] = 1;
                j = lps[j - 1];
            }
            else if (i < N && pat[j] != txt[i]) {
                if (j != 0)
                    j = lps[j - 1];
                else
                    i = i + 1;
            }
        }
    }

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

    static long[][] matrixMult(long a[][], long b[][], long mod){
        int n = a.length;
        int m = a[0].length;
        int p = b[0].length;
        long c[][] = new long[n][p];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                for(int k=0;k<p;k++){
                    c[i][k] += a[i][j]*b[j][k];
                    c[i][k] %= mod;
                }
            }
        }
        return c;
    }

    static long[][] exp(long mat[][], long b, long mod){
        if(b==0){
            int n = mat.length;
            long res[][] = new long[n][n];
            for(int i=0;i<n;i++){
                res[i][i] = 1;
            }
            return res;
        }
        long half[][] = exp(mat, b/2, mod);
        long res[][] = matrixMult(half, half, mod);
        if(b%2==1){
            res = matrixMult(res, mat, mod);
        }
        return res;
    }

    static void countPrimeFactors(int num, int a[], HashMap<Integer,Integer> pos){
        for(int i=2;i*i<num;i++){
            if(num%i==0){
                int y = pos.get(i);
                while(num%i==0){
                    a[y]++;
                    num/=i;
                }
            }
        }
        if(num>1){
            int y = pos.get(num);
            a[y]++;
        }
    }

    static void assignAnc(ArrayList<Integer> ar[], int depth[], int sz[], int par[][] ,int curr, int parent, int d){
        depth[curr] = d;
        sz[curr] = 1;
        par[curr][0] = parent;
        for(int v:ar[curr]){
            if(parent==v)
                continue;
            assignAnc(ar, depth, sz, par, v, curr, d+1);
            sz[curr] += sz[v];
        }
    }

    static int findLCA(int a, int b, int par[][], int depth[]){
        if(depth[a]>depth[b]){
            a = a^b;
            b = a^b;
            a = a^b;
        }
        int diff = depth[b] - depth[a];
        for(int i=19;i>=0;i--){
            if((diff&(1<<i))>0){
                b = par[b][i];
            }
        }
        if(a==b)
            return a;
        for(int i=19;i>=0;i--){
            if(par[b][i]!=par[a][i]){
                b = par[b][i];
                a = par[a][i];
            }
        }
        return par[a][0];
    }

    static void formArrayForBinaryLifting(int n, int par[][]){
        for(int j=1;j<20;j++){
            for(int i=0;i<n;i++){
                if(par[i][j-1]==-1)
                    continue;
                par[i][j] = par[par[i][j-1]][j-1];
            }
        }
    }

    static long lcm(int a, int b){
        return a*b/gcd(a,b);
    }

    static class CustomPair {
        long a[];

        CustomPair(long a[]) {
            this.a = a;
        }
    }

    static class Pair1 implements Comparable<Pair1> {
        long a;
        long b;

        Pair1(long a, long b) {
            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair1 p) {
            if(a!=p.a)
                return (a<p.a?-1:1);
            return (b<p.b?-1:1);
        }
    }

    static class Pair implements Comparable<Pair> {
        int a;
        int b;
        // int c;

        Pair(int a, int b) {
            this.a = a;
            this.b = b;
            // this.c = c;
        }

        public int compareTo(Pair p) {
            int diff1 = a-b;
            int diff2 = p.a-p.b;
            if(diff1!=diff2)
                return diff1-diff2;
            return a-p.a;
        }
    }

    static class Pair2 implements Comparable<Pair2> {
        int a;
        int b;
        int c;

        Pair2(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

        public int compareTo(Pair2 p) {
            return a-p.a;
        }
    }

    static boolean isPrime(int n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 ||  n % 3 == 0)
            return false;
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }

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

    static long fast_pow(long base, long n, long M) {
        if (n == 0)
            return 1;
        if (n == 1)
            return base % M;
        long halfn = fast_pow(base, n / 2, M);
        if (n % 2 == 0)
            return (halfn * halfn) % M;
        else
            return (((halfn * halfn) % M) * base) % M;
    }

    static long modInverse(long n, long M) {
        return fast_pow(n, M - 2, M);
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 9992768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

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

Think step by step.

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

1. **Understand the Problem**:
   - The problem involves an undirected connected graph with vertices and edges.
   - Alice and Bob play a game where Alice chooses two colors, and Bob colors an uncolored vertex with one of those colors.
   - Alice wins if there is an edge connecting two vertices of the same color. Bob wins otherwise.

2. **Identify Potential Bugs**:
   - The code provided seems to handle the game logic, but there might be issues with:
     - Handling the coloring strategy for Alice and Bob.
     - Ensuring the graph is connected.
     - Managing the vertex and edge constraints.

3. **Edge Cases**:
   - **Complete Graph**: A graph where every pair of distinct vertices is connected by a unique edge.
   - **Cycle Graph**: A graph where vertices are connected in a cycle.
   - **Star Graph**: A graph with one central vertex connected to all other vertices.
   - **Path Graph**: A graph where vertices are connected in a linear sequence.

4. **Challenging Test Input**:
   - A complete graph with 4 vertices (K4) is a good candidate because it maximizes the number of edges and forces Bob to color vertices in a way that might expose the bug.
   - A cycle graph with 4 vertices (C4) is another good candidate because it forms a closed loop, which can also be challenging.

5. **Generate the Test Input**:
   - We will use a complete graph with 4 vertices (K4) for this test case.

Here is the test input:

```plaintext
1
4 6
1 2
1 3
1 4
2 3
2 4
3 4
```

This test case is small-scale, complete, and valid. It represents a complete graph with 4 vertices and 6 edges, which should challenge the algorithm to handle the maximum number of edges and ensure that the graph is connected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

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

#define ll long long

int ans = 0;

void dfs(int s, int e, int y, vector<vector<int> > adj, vector<int> A, int sum){
    if(s == y) ans = sum + A[s]^s;
    for(auto u : adj[s]){
        if(u != e) dfs(u, s, y, adj, A, sum + A[s]^s);
    }
}

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

    int t;
    cin >> t;
    while(t--){
        int n, q;
        cin >> n;
        vector<vector<int> > adj(n);
        vector<int> A(n);
        for(int i = 0; i < n - 1; i++){
            int u, v;
            cin >> u >> v;
            adj[u - 1].push_back(v - 1);
        }
        for(int i = 0; i < n; i++){
            cin >> A[i];
        }
        cin >> q;
        for(int i = 0; i < q; i++){
            int x, y;
            cin >> x >> y;
            x--; y--;
            dfs(x, -1, y, adj, A, 0);
            cout << ans << "\n";
            ans = 0;
        }
    }

    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
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#define ll long long
#define clr(p) memset(p,0,sizeof(p))
using namespace std;
ll read(){
	ll re=0,val=1,c=getchar();
	while(c<'0'||c>'9') val=c=='-'?-1:1,c=getchar();
	while(c>='0'&&c<='9') re=(re<<3)+(re<<1)+(c^48),c=getchar();
	return re*val;
}
void write(ll w){
	if(w==0) putchar('0');
	if(w<0) putchar('-'),w*=-1;
	int c[20],pos=0;
	while(w) c[++pos]=(w%10^48),w/=10;
	for (int i=pos;i;i--) putchar(c[i]);
	putchar('\n');
}
const int maxn=5e5+5;
struct L{
	int v,bro;
}li[maxn*2];
int n,a[maxn],head[maxn],cnt,pt[maxn][20],lg[maxn],dep[maxn],sr[maxn][20];
ll sv[maxn][20][2];
void init(){
	for (int i=1;i<=n;i++) head[i]=0;
	cnt=0;
}
void add(int u,int v){
	li[++cnt]=(L){v,head[u]},head[u]=cnt;
}
void dfsi(int u,int pr){
	pt[u][0]=pr;
	for (int i=1;i<=lg[dep[u]];i++) pt[u][i]=pt[pt[u][i-1]][i-1];
	for (int i=0;i<20;i++){
		sr[u][i]=sr[pr][i]+(((1<<i)&a[u])>0);
		int pl=pt[u][i];
		sv[u][i][0]=sv[pl][i][1]+((ll)(sr[u][i]-sr[pl][i])<<i);
		sv[u][i][1]=sv[pl][i][0]+((ll)(min(dep[u]+1,1<<i)-sr[u][i]+sr[pl][i])<<i);
	}
	for (int i=head[u];i;i=li[i].bro){
		int v=li[i].v;
		if(v==pr) continue;
		dep[v]=dep[u]+1,dfsi(v,u);
	}
}
int qd(int u,int d){
	if(d<0) return 0;
	while(dep[u]>d) u=pt[u][lg[dep[u]-d]];
	return u;
}
ll cal(int u,int st,int d){
	ll re=0;
	for (int i=0;i<20;i++){
		int dis,to,tv;
		if(d) dis=min(dep[u]+1,st%(1<<i)+1),to=qd(u,dep[u]-dis),tv=st-dis;
		else dis=min(dep[u]+1,(st-st%(1<<i))%(1<<i)),to=qd(u,dep[u]-dis),tv=st+dis;
		re+=(tv&(1<<i))?sv[to][i][1]:sv[to][i][0];
		int c1=sr[u][i]-sr[to][i],c0=dis-c1;
		//cout<<i<<" "<<dis<<" "<<to<<" "<<c0<<" "<<c1<<endl;
		re+=(st&(1<<i))?((ll)c0<<i):((ll)c1<<i);
	}
	//cout<<u<<" "<<st<<" "<<d<<" "<<re<<endl;
	return re;
}
int lca(int u,int v){
	if(dep[u]<dep[v]) swap(u,v);
	while(dep[u]>dep[v]) u=pt[u][lg[dep[u]-dep[v]]];
	if(u==v) return v;
	for (int i=lg[dep[u]];i>=0;i--)
		if(pt[u][i]!=pt[v][i]) u=pt[u][i],v=pt[v][i];
	return pt[u][0];
}
void solve(){
	n=read();
	init();
	for (int i=2;i<=n;i++){
		int u=read(),v=read();
		add(u,v),add(v,u);
		lg[i]=lg[i>>1]+1;
	}
	for (int i=1;i<=n;i++) a[i]=read();
	dfsi(1,0);
	int q=read();
	while(q--){
		int u=read(),v=read(),tp=lca(u,v),dis=dep[u]+dep[v]-dep[tp]*2,dt=dep[u]-dep[tp];
		ll ans=cal(u,0,0)+cal(v,dis,1)-cal(tp,dt,0)-cal(tp,dt,1)+(a[tp]^dt);
		write(ans);
	}
}
int main(){
	int t=read();
	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
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using vi=vector<int>;
using pi=pair<int,int>;
#define F(i,a,b) for(int i=(a);i<=(b);i++)
#define G(i,a,b) for(int i=(a);i>=(b);i--)
#define ms(a,b) memset(a,b,sizeof(a))
#define si(a) ((int)(a).size())
#define all(a) (a).begin(),(a).end()
#define fi first
#define se second
#define pb push_back

#if 1
template<class A>
void dbg(string s,A a){
	if(si(s)) cout<<'['<<s<<"]: ";
	cout<<a<<'\n';
}
template<class A,class... B>
void dbg(string s,A a,B... b){
	if(si(s)) cout<<'['<<s<<"]: ";
	cout<<a<<' ';
	dbg("",b...);
}
template<class T>
void dbg(string s,T *a,int l,int r){
	cout<<'['<<s<<"]: ";
	F(i,l,r) cout<<a[i]<<" \n"[i==r];
}
template<class T>
void dbg(string s,vector<T> a){
	cout<<'['<<s<<"]: ";
	for(T x:a) cout<<x<<' ';
	cout<<'\n';
}
#else
#define dbg(...)
#endif

const int N=1<<20;
int a[N],a0[N],qx[N],qy[N],qz[N];
vi adj[N];

int fa[N][22],dep[N];
void dfs0(int u,int par){
	dep[u]=dep[par]+1;
	fa[u][0]=par;
	F(i,1,20) fa[u][i]=fa[fa[u][i-1]][i-1];
	for(int v:adj[u]) if(v!=par) dfs0(v,u);
}

int lca(int x,int y){
	if(dep[x]<dep[y]) swap(x,y);
	G(i,20,0) if(dep[fa[x][i]]>=dep[y]) x=fa[x][i];
	if(x==y) return x;
	G(i,20,0) if(fa[x][i]!=fa[y][i]){
		x=fa[x][i];
		y=fa[y][i];
	}
	return fa[x][0];
}

struct fenwick{
	int c[N];
	void add(int i,int k){
		for(i++;i<N;i+=i&-i) c[i]+=k;
	}
	int qry(int i){
		int res=0;
		for(i++;i>=1;i-=i&-i) res+=c[i];
		return res;
	}
	int qry(int l,int r){
		l=max(l,0);
		if(l>r) return 0;
		return qry(r)-qry(l-1);
	}
}fw[4];

int k,t;
vector<pi> q1[N],q2[N];
ll ans[N];
void dfs(int u,int par){
	if(!a[u]){
		fw[0].add(((t-dep[u])%t+t)%t,1);
		fw[1].add(dep[u]%t,1);
	}else{
		fw[2].add(((t-dep[u])%t+t)%t,1);
		fw[3].add(dep[u]%t,1);
	}
	for(auto [i,j]:q1[u]){
		j%=t;
		if(j<0) j+=t;
		int l=((1<<k)-j)%t,r=((1<<(k+1))-j-1)%t;
		if(l<0) l+=t;
		if(r<0) r+=t;
		if(i>0){
			if(l<=r) ans[i]+=(1ll<<k)*fw[0].qry(l,r);
			else ans[i]+=(1ll<<k)*(fw[0].qry(l,t-1)+fw[0].qry(0,r));
			if(l<=r) ans[i]+=(1ll<<k)*(fw[2].qry(0,t-1)-fw[2].qry(l,r));
			else ans[i]+=(1ll<<k)*(fw[2].qry(0,t-1)-fw[2].qry(l,t-1)-fw[2].qry(0,r));
		}else{
			if(l<=r) ans[-i]-=(1ll<<k)*fw[0].qry(l,r);
			else ans[-i]-=(1ll<<k)*(fw[0].qry(l,t-1)+fw[0].qry(0,r));
			if(l<=r) ans[-i]-=(1ll<<k)*(fw[2].qry(0,t-1)-fw[2].qry(l,r));
			else ans[-i]-=(1ll<<k)*(fw[2].qry(0,t-1)-fw[2].qry(l,t-1)-fw[2].qry(0,r));
		}
	}
	for(auto [i,j]:q2[u]){
		j%=t;
		if(j<0) j+=t;
		int l=((1<<k)-j)%t,r=((1<<(k+1))-j-1)%t;
		if(l<0) l+=t;
		if(r<0) r+=t;
		if(i>0){
			if(l<=r) ans[i]+=(1ll<<k)*fw[1].qry(l,r);
			else ans[i]+=(1ll<<k)*(fw[1].qry(l,t-1)+fw[1].qry(0,r));
			if(l<=r) ans[i]+=(1ll<<k)*(fw[3].qry(0,t-1)-fw[3].qry(l,r));
			else ans[i]+=(1ll<<k)*(fw[3].qry(0,t-1)-fw[3].qry(l,t-1)-fw[3].qry(0,r));
		}else{
			if(l<=r) ans[-i]-=(1ll<<k)*fw[1].qry(l,r);
			else ans[-i]-=(1ll<<k)*(fw[1].qry(l,t-1)+fw[1].qry(0,r));
			if(l<=r) ans[-i]-=(1ll<<k)*(fw[3].qry(0,t-1)-fw[3].qry(l,r));
			else ans[-i]-=(1ll<<k)*(fw[3].qry(0,t-1)-fw[3].qry(l,t-1)-fw[3].qry(0,r));
		}
	}
	for(int v:adj[u]) if(v!=par) dfs(v,u);
	if(!a[u]){
		fw[0].add(((t-dep[u])%t+t)%t,-1);
		fw[1].add(dep[u]%t,-1);
	}else{
		fw[2].add(((t-dep[u])%t+t)%t,-1);
		fw[3].add(dep[u]%t,-1);
	}
}

void slv(){
	int n;
	cin>>n;
	F(i,1,n){
		adj[i].clear();
		q1[i].clear();
		q2[i].clear();
	}
	F(i,1,n-1){
		int u,v;
		cin>>u>>v;
		adj[u].pb(v);
		adj[v].pb(u);
	}
	dfs0(1,0);
	F(i,1,n) cin>>a0[i];
	int m;
	cin>>m;
	F(i,1,m){
		cin>>qx[i]>>qy[i];
		qz[i]=lca(qx[i],qy[i]);
		q1[qx[i]].pb({i,dep[qx[i]]});
		q1[fa[qz[i]][0]].pb({-i,dep[qx[i]]});
		q2[qy[i]].pb({i,dep[qx[i]]-dep[qz[i]]*2});
		q2[qz[i]].pb({-i,dep[qx[i]]-dep[qz[i]]*2});
//		dbg("x,y,z",qx[i],qy[i],qz[i]);
		ans[i]=0;
	}
	for(k=0;k<=18;k++){
		t=1<<(k+1);
		F(i,1,n) a[i]=a0[i]>>k&1;
//		dbg("a",a,1,n);
//		dbg("ans",ans,1,m);
		dfs(1,0);
	}
	F(i,1,m) cout<<ans[i]<<'\n';
}

int main(){
	cin.tie(0)->ios::sync_with_stdio(0);
	int t;
	cin>>t;
	while(t--) slv();
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
from collections import defaultdict, deque

def read_input():
    import sys
    input = sys.stdin.read
    data = input().split()
    index = 0
    t = int(data[index])
    index += 1
    test_cases = []

    for _ in range(t):
        n = int(data[index])
        index += 1
        edges = []
        for _ in range(n - 1):
            u = int(data[index])
            v = int(data[index + 1])
            edges.append((u, v))
            index += 2
        values = list(map(int, data[index:index + n]))
        index += n
        q = int(data[index])
        index += 1
        queries = []
        for _ in range(q):
            x = int(data[index])
            y = int(data[index + 1])
            queries.append((x, y))
            index += 2
        test_cases.append((n, edges, values, q, queries))

    return test_cases

def build_tree(n, edges):
    tree = defaultdict(list)
    for u, v in edges:
        tree[u].append(v)
        tree[v].append(u)
    return tree

def dfs(tree, node, parent, depth, parent_map):
    for neighbor in tree[node]:
        if neighbor == parent:
            continue
        parent_map[neighbor] = node
        depth[neighbor] = depth[node] + 1
        dfs(tree, neighbor, node, depth, parent_map)

def preprocess(n, edges):
    tree = build_tree(n, edges)
    depth = [0] * (n + 1)
    parent_map = [-1] * (n + 1)

    dfs(tree, 1, -1, depth, parent_map)

    return tree, depth, parent_map

def lca(u, v, depth, parent_map):
    if depth[u] < depth[v]:
        u, v = v, u

    while depth[u] > depth[v]:
        u = parent_map[u]

    while u != v:
        u = parent_map[u]
        v = parent_map[v]

    return u

def path_to_root(node, parent_map):
    path = []
    while node != -1:
        path.append(node)
        node = parent_map[node]
    return path

def compute_xor_sum(values, path):
    xor_sum = 0
    for i, node in enumerate(path):
        xor_sum += values[node - 1] ^ i
    return xor_sum

def handle_queries(n, edges, values, q, queries):
    tree_data = preprocess(n, edges)
    tree, depth, parent_map = tree_data

    results = []

    for x, y in queries:
        lca_node = lca(x, y, depth, parent_map)

        path_x_to_lca = path_to_root(x, parent_map)
        path_y_to_lca = path_to_root(y, parent_map)

        while path_x_to_lca and path_y_to_lca and path_x_to_lca[-1] == path_y_to_lca[-1]:
            common_ancestor = path_x_to_lca.pop()
            path_y_to_lca.pop()

        full_path = path_x_to_lca + [common_ancestor] + list(reversed(path_y_to_lca))

        total_result = compute_xor_sum(values, full_path)

        results.append(total_result)

    return results

if __name__ == "__main__":
    test_cases = read_input()

    for n, edges, values, q, queries in test_cases:
        results = handle_queries(n, edges, values, q, queries)
        for result in results:
            print(result)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
import sys
from math import log2, floor
sys.setrecursionlimit(10**6)
input = sys.stdin.read
data = input().split()

class TreeSolver:
    def __init__(self):
        self.idx = 0

    def readint(self):
        val = int(data[self.idx])
        self.idx += 1
        return val

    def readtree(self):
        n = self.readint()
        self.adj = [[] for _ in range(n + 1)]
        for _ in range(n - 1):
            u = self.readint()
            v = self.readint()
            self.adj[u].append(v)
            self.adj[v].append(u)
        self.values = [0] + [self.readint() for _ in range(n)]
        self.depth = [0] * (n + 1)
        self.parent = [-1] * (n + 1)
        self.maxlog = floor(log2(n)) + 1
        self.up = [[-1] * (self.maxlog + 1) for _ in range(n + 1)]
        self.dfs(1, -1)
        self.prepare_lca()

    def dfs(self, node, par):
        self.parent[node] = par
        if par != -1:
            self.depth[node] = self.depth[par] + 1
        for child in self.adj[node]:
            if child != par:
                self.dfs(child, node)

    def prepare_lca(self):
        for i in range(1, len(self.up)):
            self.up[i][0] = self.parent[i]

        for j in range(1, self.maxlog + 1):
            for i in range(1, len(self.up)):
                if self.up[i][j - 1] != -1:
                    self.up[i][j] = self.up[self.up[i][j - 1]][j - 1]

    def lca(self, u, v):
        if self.depth[u] < self.depth[v]:
            u, v = v, u
        diff = self.depth[u] - self.depth[v]
        for i in range(self.maxlog + 1):
            if (diff >> i) & 1:
                u = self.up[u][i]
        if u == v:
            return u
        for i in range(self.maxlog, -1, -1):
            if self.up[u][i] != self.up[v][i]:
                u = self.up[u][i]
                v = self.up[v][i]
        return self.parent[u]

    def query(self, x, y):
        lca_xy = self.lca(x, y)
        total_sum = 0
        while x != lca_xy:
            total_sum += self.values[x] ^ self.depth[x]
            x = self.parent[x]
        while y != lca_xy:
            total_sum += self.values[y] ^ self.depth[y]
            y = self.parent[y]
        total_sum += self.values[lca_xy] ^ self.depth[lca_xy]
        return total_sum

solver = TreeSolver()
t = solver.readint()
results = []
for _ in range(t):
    solver.readtree()
    q = solver.readint()
    for _ in range(q):
        x = solver.readint()
        y = solver.readint()
        results.append(solver.query(x, y))
print('\n'.join(map(str, results)))

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

### Explanation:
This test case is designed to expose potential issues in the given buggy code. Here’s a breakdown of the test case:

1. **Tree Structure:**
   - The tree is a simple linear chain: `1 - 2 - 3 - 4`.

2. **Node Values:**
   - Node values are: `2, 3, 6, 5`.

3. **Queries:**
   - The first query is from node 1 to node 4.
   - The second query is from node 3 to node 4.
   - The third query is from node 1 to node 1.

### Expected Outputs:
- **Query 1 (1 to 4):**
  - Path: `1 -> 2 -> 3 -> 4`
  - Sum: `(2 ^ 0) + (3 ^ 1) + (6 ^ 2) + (5 ^ 3) = 2 + 2 + 4 + 6 = 14`

- **Query 2 (3 to 4):**
  - Path: `3 -> 4`
  - Sum: `(6 ^ 0) + (5 ^ 1) = 6 + 4 = 10`

- **Query 3 (1 to 1):**
  - Path: `1`
  - Sum: `(2 ^ 0) = 2`

### Why This Test Case is Challenging:
1. **Linear Chain:**
   - The linear chain structure ensures that the path from any node to another node is straightforward and easy to follow, but it can help in identifying issues in the path calculation and depth management.

2. **Query with Same Nodes:**
   - The third query (1 to 1) tests the case where the start and end nodes are the same, which should return the value of the node XOR 0.

3. **Path Length:**
   - The path from 1 to 4 involves all nodes, ensuring that the depth and parent relationships are correctly managed.

This test case should help in identifying any issues with the depth calculation, parent assignment, and the sum computation in the query function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
import sys
from collections import defaultdict

sys.setrecursionlimit(10**6)
input = sys.stdin.readline

def dfs(node, parent):
    tin[node] = len(time_in)
    time_in.append(node)
    level[node] = (parent, level[parent][1] + 1) if parent != -1 else (None, 0)
    for child in graph[node]:
        if child != parent:
            dfs(child, node)
    tout[node] = len(time_in)
    time_out.append(node)

def lca(u, v):
    if tin[u] <= tin[v] and tout[u] >= tout[v]:
        return u
    if tin[v] <= tin[u] and tout[v] >= tout[u]:
        return v
    while level[u][1] > level[v][1]:
        u = level[u][0]
    while level[v][1] > level[u][1]:
        v = level[v][0]
    while u != v:
        u = level[u][0]
        v = level[v][0]
    return u

# Main function to process each test case
def main():
    global graph, tin, tout, time_in, time_out, level

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

    for _ in range(t):
        n = int(input())
        graph = defaultdict(list)

        for __ in range(n - 1):
            u, v = map(int, input().split())
            graph[u].append(v)
            graph[v].append(u)

        a = list(map(int, input().split()))
        q = int(input())

        # Initialize variables for DFS
        time_in = []
        time_out = []
        level = [-1] * (n + 1)
        level[0] = (-1, -1)
        tin = [0] * (n + 1)
        tout = [0] * (n + 1)

        # Perform DFS to get the time in and out arrays
        dfs(1, -1)

        # Process each query
        for __ in range(q):
            x, y = map(int, input().split())
            ancestor = lca(x, y)

            # Calculate the sum of a_i ^ i for the path from x to y
            path_sum = 0
            current = x
            while current != ancestor:
                path_sum += a[current - 1] ^ tin[current]
                current = level[current][0]

            current = y
            while current != ancestor:
                path_sum += a[current - 1] ^ tin[current]
                current = level[current][0]

            path_sum += a[ancestor - 1] ^ tin[ancestor]
            results.append(path_sum)

    # Print results
    print("\n".join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can highlight potential issues. The code uses Depth-First Search (DFS) to compute the time in and time out arrays, and it uses these arrays to find the Lowest Common Ancestor (LCA) of two nodes. The main bug in the code is in the calculation of the path sum, where it uses the `tin` array incorrectly.

Here are the steps to create a challenging test input:

1. **Tree Structure**: Create a tree structure that has a long path to maximize the number of nodes in the path between two nodes.
2. **Node Values**: Assign values to the nodes that can cause the XOR operation to produce unexpected results.
3. **Queries**: Create queries that involve nodes with a long path between them.

Let's create a small-scale test input that follows these steps:

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

### Explanation:
- **Test Case**: 1 test case.
- **Tree Structure**:
  - Node 1 is connected to Node 2.
  - Node 2 is connected to Node 3.
  - Node 3 is connected to Node 4.
  - Node 4 is connected to Node 5.
- **Node Values**: The values of the nodes are 1, 2, 3, 4, and 5 respectively.
- **Queries**:
  - Query 1: Path from Node 1 to Node 5.
  - Query 2: Path from Node 2 to Node 4.
  - Query 3: Path from Node 1 to Node 3.

### Expected Output:
- For the path from Node 1 to Node 5:
  - Path: 1, 2, 3, 4, 5
  - Sum: \( (1 \oplus 0) + (2 \oplus 1) + (3 \oplus 2) + (4 \oplus 3) + (5 \oplus 4) = 1 + 3 + 1 + 7 + 1 = 13 \)
- For the path from Node 2 to Node 4:
  - Path: 2, 3, 4
  - Sum: \( (2 \oplus 0) + (3 \oplus 1) + (4 \oplus 2) = 2 + 2 + 6 = 10 \)
- For the path from Node 1 to Node 3:
  - Path: 1, 2, 3
  - Sum: \( (1 \oplus 0) + (2 \oplus 1) + (3 \oplus 2) = 1 + 3 + 1 = 5 \)

This test input should expose the bug in the code, especially in the calculation of the path sum using the `tin` array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

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

public class TreeQueries {
    static class FastIO {
        BufferedReader br;
        StringTokenizer st;
        PrintWriter out;

        public FastIO() {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        }

        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());
        }

        void println(Object o) {
            out.println(o);
        }

        void close() {
            out.close();
        }
    }

    static class SegmentTree {
        int[] tree;
        int n;

        public SegmentTree(int n) {
            this.n = n;
            tree = new int[4 * n];
        }

        public void update(int index, int value, int node, int start, int end) {
            if (start == end) {
                tree[node] = value;
            } else {
                int mid = (start + end) / 2;
                if (index <= mid) {
                    update(index, value, 2 * node, start, mid);
                } else {
                    update(index, value, 2 * node + 1, mid + 1, end);
                }
                tree[node] = tree[2 * node] + tree[2 * node + 1];
            }
        }

        public int query(int l, int r, int node, int start, int end) {
            if (r < start || end < l) {
                return 0;
            }
            if (l <= start && end <= r) {
                return tree[node];
            }
            int mid = (start + end) / 2;
            int left = query(l, r, 2 * node, start, mid);
            int right = query(l, r, 2 * node + 1, mid + 1, end);
            return left + right;
        }

        public void update(int index, int value) {
            update(index, value, 1, 0, n - 1);
        }

        public int query(int l, int r) {
            return query(l, r, 1, 0, n - 1);
        }
    }

    static class HLD {
        int[] parent, depth, heavy, head, pos, value;
        SegmentTree segTree;
        int currentPos;

        public HLD(int n, int[] value) {
            parent = new int[n];
            depth = new int[n];
            heavy = new int[n];
            head = new int[n];
            pos = new int[n];
            this.value = value;
            Arrays.fill(heavy, -1);
            currentPos = 0;
            segTree = new SegmentTree(n);
        }

        public int dfs(int v, List<Integer>[] adj) {
            int size = 1, maxSubtree = 0;
            for (int u : adj[v]) {
                if (u != parent[v]) {
                    parent[u] = v;
                    depth[u] = depth[v] + 1;
                    int subtreeSize = dfs(u, adj);
                    if (subtreeSize > maxSubtree) {
                        maxSubtree = subtreeSize;
                        heavy[v] = u;
                    }
                    size += subtreeSize;
                }
            }
            return size;
        }

        public void decompose(int v, int h, List<Integer>[] adj) {
            head[v] = h;
            pos[v] = currentPos++;
            segTree.update(pos[v], value[v] ^ v);
            if (heavy[v] != -1) {
                decompose(heavy[v], h, adj);
            }
            for (int u : adj[v]) {
                if (u != parent[v] && u != heavy[v]) {
                    decompose(u, u, adj);
                }
            }
        }

        public int query(int a, int b) {
            int res = 0;
            for (; head[a] != head[b]; b = parent[head[b]]) {
                if (depth[head[a]] > depth[head[b]]) {
                    int temp = a;
                    a = b;
                    b = temp;
                }
                res += segTree.query(pos[head[b]], pos[b]);
            }
            if (depth[a] > depth[b]) {
                int temp = a;
                a = b;
                b = temp;
            }
            res += segTree.query(pos[a], pos[b]);
            return res;
        }
    }

    public static void main(String[] args) {
        FastIO io = new FastIO();
        int t = io.nextInt();
        while (t-- > 0) {
            int n = io.nextInt();
            List<Integer>[] adj = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                adj[i] = new ArrayList<>();
            }
            for (int i = 1; i < n; i++) {
                int u = io.nextInt() - 1;
                int v = io.nextInt() - 1;
                adj[u].add(v);
                adj[v].add(u);
            }
            int[] value = new int[n];
            for (int i = 0; i < n; i++) {
                value[i] = io.nextInt();
            }
            HLD hld = new HLD(n, value);
            hld.dfs(0, adj);
            hld.decompose(0, 0, adj);
            int q = io.nextInt();
            for (int i = 0; i < q; i++) {
                int x = io.nextInt() - 1;
                int y = io.nextInt() - 1;
                io.println(hld.query(x, y));
            }
        }
        io.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider a few scenarios that could cause issues:

1. **Path with a large number of nodes**: This can stress the segment tree and HLD implementation.
2. **Queries involving nodes with large indices**: This can test the correctness of the index handling.
3. **Queries involving the root node**: This can test the boundary conditions.
4. **Queries involving nodes with the same index**: This can test the handling of trivial cases.

Given the constraints and the need for a small-scale test input, we will create a tree with a small number of nodes and a few queries that cover the above scenarios.

Let's construct a test input with a tree of 7 nodes and 5 queries:

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

### Explanation:
- **Tree Structure**:
  - Node 1 is connected to nodes 2 and 3.
  - Node 2 is connected to nodes 4 and 5.
  - Node 3 is connected to nodes 6 and 7.
- **Values**:
  - Nodes have values from 1 to 7.
- **Queries**:
  - Query 1: Path from node 1 to node 7.
  - Query 2: Path from node 2 to node 5.
  - Query 3: Path from node 3 to node 6.
  - Query 4: Path from node 1 to node 1.
  - Query 5: Path from node 7 to node 7.

This test input is designed to cover a variety of scenarios, including paths with multiple nodes, paths involving the root, and trivial paths. It should help expose any bugs in the segment tree and HLD implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

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

public class Main {
    static final int MAXN = 500005;
    static final int LOG = 20;

    static List<Integer>[] tree = new ArrayList[MAXN];
    static int[] PR = new int[MAXN];
    static int[] depth = new int[MAXN];
    static int[][] PE = new int[MAXN][LOG];
    static long[] RE = new long[MAXN];
    static long[] AE = new long[MAXN];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        int TLE = Integer.parseInt(br.readLine());

        while (TLE-- > 0) {
            int n = Integer.parseInt(br.readLine());

            for (int i = 1; i <= n; ++i) {
                tree[i] = new ArrayList<>();
                depth[i] = 0;
                RE[i] = 0;
                AE[i] = 0;
            }

            for (int i = 0; i < n - 1; ++i) {
                st = new StringTokenizer(br.readLine());
                int u = Integer.parseInt(st.nextToken());
                int v = Integer.parseInt(st.nextToken());
                tree[u].add(v);
                tree[v].add(u);
            }

            st = new StringTokenizer(br.readLine());
            for (int i = 1; i <= n; ++i) {
                PR[i] = Integer.parseInt(st.nextToken());
            }

            for (int[] row : PE) {
                Arrays.fill(row, -1);
            }

            dfs(1, 0);

            int q = Integer.parseInt(br.readLine());
            while (q-- > 0) {
                st = new StringTokenizer(br.readLine());
                int x = Integer.parseInt(st.nextToken());
                int y = Integer.parseInt(st.nextToken());
                System.out.println(query(x, y));
            }
        }
    }

    static void dfs(int node, int parent) {
        PE[node][0] = parent;
        for (int i = 1; i < LOG; ++i) {
            if (PE[node][i - 1] != -1) {
                PE[node][i] = PE[PE[node][i - 1]][i - 1];
            } else {
                PE[node][i] = -1;
            }
        }

        RE[node] = RE[parent] ^ PR[node];
        AE[node] = AE[parent] + (PR[node] ^ depth[node]);

        for (int child : tree[node]) {
            if (child != parent) {
                depth[child] = depth[node] + 1;
                dfs(child, node);
            }
        }
    }

    static int lca(int u, int v) {
        if (depth[u] < depth[v]) {
            int temp = u;
            u = v;
            v = temp;
        }

        int diff = depth[u] - depth[v];
        for (int i = 0; i < LOG; ++i) {
            if ((diff >> i & 1) == 1) {
                u = PE[u][i];
            }
        }

        if (u == v) return u;

        for (int i = LOG - 1; i >= 0; --i) {
            if (PE[u][i] != PE[v][i]) {
                u = PE[u][i];
                v = PE[v][i];
            }
        }

        return PE[u][0];
    }

    static long query(int x, int y) {
        int lcaNode = lca(x, y);
        return AE[x] + AE[y] - 2 * AE[lcaNode] + (PR[lcaNode] ^ depth[lcaNode]);
    }
}

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

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

Think step by step.

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

1. **Small but complex tree structures**: Trees with small but intricate structures can help in exposing issues related to depth and parent relationships.
2. **Queries involving the root node**: Queries that involve the root node can help in testing the initialization and propagation of values.
3. **Long paths**: Paths that are long can help in testing the correctness of the XOR and sum calculations.
4. ** Queries with the same node**: Queries where the starting and ending nodes are the same can help in testing edge cases.

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

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

### Explanation:
- **Tree Structure**:
  - Node 1 is the root.
  - Node 1 has children 2 and 3.
  - Node 2 has children 4 and 5.
- **Node Values**:
  - Values are 1, 2, 3, 4, 5 respectively.
- **Queries**:
  - Query 1: Path from 1 to 5 (1 -> 2 -> 5)
  - Query 2: Path from 2 to 4 (2 -> 4)
  - Query 3: Path from 3 to 3 (3)
  - Query 4: Path from 1 to 1 (1)
  - Query 5: Path from 5 to 1 (5 -> 2 -> 1)

This test input is designed to cover a variety of scenarios, including paths of different lengths, queries involving the root node, and queries where the starting and ending nodes are the same. It should help in exposing any issues 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: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int M=500005,INF=1e9+7,N=21;
struct node{
	int v,lst;
}t[M<<1];
int n,tot=0,head[M],dep[M],f[N][M],w[M];
int pw[N],dp[N][M],sumt0[N][M],sumt1[N][M],dpr[N][M],tx[M],ty[M];//dp[i][j][k]表示 k 节点往上走 2^j 步后在维护第 i 个位置时的值，f[i][j] 表示 i 的第 2^j 个祖先。
ll ans[M];
void add(int u,int v){
	t[++tot].v=v;
	t[tot].lst=head[u];
	head[u]=tot;
}
void init0(int u,int fa){
	int v;
	f[0][u]=fa;
	dep[u]=dep[fa]+1;
	for(int i=1;i<=19;i++)
	{
		f[i][u]=f[i-1][f[i-1][u]];
	}
	for(int i=head[u];i;i=t[i].lst)
	{
		v=t[i].v;
		if(v==fa) continue;
		init0(v,u);
	}
	return ;
}
void init(int i,int u,int fa){
	int v;
	sumt1[0][u]=(((w[u]&(1<<i))^(1<<i))>>i);
	for(int j=1;j<=19;j++)
	{
		sumt1[j][u]=sumt1[j-1][u]+sumt1[j-1][f[j-1][u]];
	}
	sumt0[0][u]=((w[u]&(1<<i))>>i);
	for(int j=1;j<=19;j++)
	{
		sumt0[j][u]=sumt0[j-1][u]+sumt0[j-1][f[j-1][u]];
	}
	dp[i+1][u]=sumt1[i][f[i][u]]+sumt0[i][u];
	for(int j=i+2;j<=19;j++)
	{
		dp[j][u]=dp[j-1][u]+dp[j-1][f[j-1][u]];
	}
	dpr[i+1][u]=sumt1[i][u]+sumt0[i][f[i][u]];
	for(int j=i+2;j<=19;j++)
	{
		dpr[j][u]=dpr[j-1][u]+dpr[j-1][f[j-1][u]];
	}
	for(int ti=head[u];ti;ti=t[ti].lst)
	{
		v=t[ti].v;
		if(v==fa) continue;
		init(i,v,u);
	}
	return ;
}
int LCA(int u,int v){
	if(dep[u]<dep[v]) swap(u,v);
	int dlt=dep[u]-dep[v];
	for(int i=19;i>=0;i--)
	{
		if(dlt&(1<<i)) u=f[i][u];
	}
	if(u==v) return u;
	for(int i=19;i>=0;i--)
	{
		if(f[i][u]!=f[i][v]) u=f[i][u],v=f[i][v];
	}
	return f[0][u];
}
int rep(int id,int u,int v,int lca){
	int ml,mr,dlt=dep[u]-dep[lca],res=0;
	for(int i=18;i>=id+1;i--)
	{
		if(dlt&(1<<i))
		{
			dlt^=(1<<i);
			res+=dp[i][u];
			u=f[i][u];
		}
	}
	if(dlt&(1<<id))
	{
		res+=sumt0[id][u];
		u=f[id][u];
		dlt^=(1<<id);
		for(int i=id-1;i>=0;i--)
		{
			if(dlt&(1<<i))
			{
				res+=sumt1[i][u];
				u=f[i][u];
			}
		}
	}
	else
	{
		for(int i=id-1;i>=0;i--)
		{
			if(dlt&(1<<i))
			{
				res+=sumt0[i][u];
				u=f[i][u];
			}
		}
	}
	int l=dep[u]-dep[lca],r=l+(dep[v]-dep[lca]),fl=(l/pw[id+1]+1)*pw[id+1]-1,fr;
//	cout<<l<<" "<<r<<"\n";
	if(fl>=r)
	{
		fl=fl-pw[id]+1;
		if(fl>r)
		{
			dlt=r-l+1;
			for(int i=18;i>=0;i--)
			{
				if(dlt&(1<<i)) res+=sumt0[i][v],v=f[i][v];
			}
		}
		else if(fl<=r&&fl>=l)
		{
			dlt=r-fl+1;
			if(dlt==pw[id+1])
			{
				res+=sumt1[id][v],v=f[id][v];
				res+=sumt0[id][v],v=f[id][v];
			}
			else
			{
				for(int i=18;i>=0;i--)
				{
					if(dlt&(1<<i)) res+=sumt1[i][v],v=f[i][v];
				}
				dlt=fl-l;
				for(int i=18;i>=0;i--)
				{
					if(dlt&(1<<i)) res+=sumt0[i][v],v=f[i][v];
				}
			}
		}
		else
		{
			dlt=r-l+1;
			for(int i=18;i>=0;i--)
			{
				if(dlt&(1<<i)) res+=sumt1[i][v],v=f[i][v];
			}
		}
	}
	else
	{
		ml=fl+1;
		fr=r/pw[id+1]*pw[id+1];
		mr=fr-1;
		dlt=r-fr+1;
		if(dlt<=pw[id])
		{
			for(int i=18;i>=0;i--)
			{
				if(dlt&(1<<i)) res+=sumt0[i][v],v=f[i][v],dlt^=(1<<i);
			}
		}
		else
		{
			dlt-=pw[id];
			for(int i=18;i>=0;i--)
			{
				if(dlt&(1<<i)) res+=sumt1[i][v],v=f[i][v],dlt^=(1<<i);
			}
			res+=sumt0[id][v];
			v=f[id][v];
		}
		dlt=mr-ml+1;
		for(int i=18;i>=id+1;i--)
		{
			if(dlt&(1<<i)) res+=dpr[i][v],v=f[i][v],dlt^=(1<<i);
		}
		int dl=fl-pw[id]+1;
		if(dl<=l)
		{
			dlt=fl-l+1;
			for(int i=18;i>=0;i--)
			{
				if(dlt&(1<<i)) res+=sumt1[i][v],v=f[i][v],dlt^=(1<<i);
			}
		}
		else
		{
			res+=sumt1[id][v];
			v=f[id][v];
			dlt=dl-l;
			for(int i=18;i>=0;i--)
			{
				if(dlt&(1<<i)) res+=sumt0[i][v],v=f[i][v],dlt^=(1<<i);
			}
		}
	}
	return res;
}
void init_clear(){
	for(int i=1;i<=n;i++) head[i]=0,ans[i]=0;
	tot=0;
	return ;
}
void cleart(){
	for(int i=1;i<=n;i++)
		for(int j=0;j<=19;j++) dp[j][i]=dpr[j][i]=sumt0[j][i]=sumt1[j][i]=0;
	return ;
}
void solve(){
	int u,v,q;
	cin>>n;
	for(int i=1;i<n;i++)
	{
		cin>>u>>v;
		add(u,v);
		add(v,u);
	}
	for(int i=1;i<=n;i++) cin>>w[i];
	cin>>q;
	for(int i=1;i<=q;i++)
	{
		cin>>tx[i]>>ty[i];
	}
	init0(1,0);
	for(int j=0;j<=18;j++)
	{
		init(j,1,0);
		for(int i=1;i<=q;i++)
		{
			u=tx[i],v=ty[i];
			int lca=LCA(u,v);
			ans[i]+=1ll*rep(j,u,v,lca)*(1<<j);
		}
//		cleart();
	}
	for(int i=1;i<=q;i++)
	{
		cout<<ans[i]<<'\n';
		ans[i]=0;
	}
	init_clear();
	return ;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	pw[0]=1;
	for(int i=1;i<=19;i++)
	{
		pw[i]=pw[i-1]*2;
	}
	int T;
	cin>>T;
	while(T--) solve();
	return 0;
}
/*
1
4
1 2
2 3
3 4
2 3 6 5
1
1 4
*/

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

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

Think step by step.

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

1. **Simple Tree Structure**: Start with a simple tree structure to ensure the basic functionality is working.
2. **Long Path**: Consider a long path to test the LCA and path summation logic.
3. **Complex Tree**: Test with a more complex tree structure to ensure the LCA and path summation logic handle branching correctly.
4. **Edge Cases**: Test with edge cases such as a single node, a linear chain, and a star graph.
5. **Query Patterns**: Test with queries that span the entire tree, including queries with the same start and end node, and queries that involve deep paths.

Based on these considerations, here is a small-scale test input that covers a variety of scenarios:

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

### Explanation:
- **Tree Structure**: The tree is a simple linear chain from node 1 to node 5.
- **Node Values**: The values of the nodes are 1, 2, 3, 4, and 5 respectively.
- **Queries**:
  - `1 5`: Path from node 1 to node 5.
  - `2 4`: Path from node 2 to node 4.
  - `3 3`: Path from node 3 to node 3 (single node).
  - `1 1`: Path from node 1 to node 1 (single node).
  - `5 1`: Path from node 5 to node 1 (reverse direction).

This test input covers a simple but comprehensive set of scenarios to ensure the code handles various paths and queries correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Your Loss
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a tree with $n$ nodes numbered from $1$ to $n$, along with an array of size $n$. The value of $i$-th node is $a_{i}$. There are $q$ queries. In each query, you are given 2 nodes numbered as $x$ and $y$.

Consider the path from the node numbered as $x$ to the node numbered as $y$. Let the path be represented by $x = p_0, p_1, p_2, \ldots, p_r = y$, where $p_i$ are the intermediate nodes. Compute the sum of $a_{p_i}\oplus i$ for each $i$ such that $0 \le i \le r$ where $\oplus$ is the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) operator.

More formally, compute $$\sum_{i =0}^{r} a_{p_i}\oplus i$$.

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

The first line of each set of input data contains a single integer $n$ ($1 \le n \le 5 \cdot 10^5$) — the number of nodes.

The next $n-1$ lines of each set of input data contain $2$ integers, $u$ and $v$ representing an edge between the node numbered $u$ and the node numbered $v$. It is guaranteed that $u \ne v$ and that the edges form a tree.

The next line of each set of input data contains $n$ integers, $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 5 \cdot 10^5$) — values of the nodes.

The next line contains a single integer $q$ ($1 \le q \le 10^5$) — the number of queries.

The next $q$ lines describe the queries. The $i$-th query contains $2$ integers $x$ and $y$ ($1 \le x,y \le n$) denoting the starting and the ending node of the path.

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

Output: For each query, output a single number — the sum from the problem statement.

Examples:
input:
1
4
1 2
2 3
3 4
2 3 6 5
3
1 4
3 4
1 1
output:
14
10
2

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=5e5+5;
const int M=1e5+5;
const int K=20;
int n,q,a[N];
inline int read(){
    int x=0,f=1;
    char c=getchar();
    while(!isdigit(c)){
        if(c=='-'){
            f=-1;
        }
        c=getchar();
    }
    while(isdigit(c)){
        x=(x<<3)+(x<<1)+c-'0';
        c=getchar();
    }
    return x*f;
}
void write(ll x){
    if(x<0){
        putchar('-');
    }
    if(x>9){
        write(x/10);
    }
    putchar(x%10+'0');
}
vector<int>E[N];
struct Query{
    int u,v,lca;
}Q[M];
int st[N][K],dep[N];
void dfs(int u,int fa){
    st[u][0]=fa;
    dep[u]=dep[fa]+1;
    for(auto v:E[u]){
        if(v==fa){
            continue;
        }
        dfs(v,u);
    }
}
void buildst(){
    for(int j=1;j<K;j++){
        for(int i=1;i<=n;i++){
            st[i][j]=st[st[i][j-1]][j-1];
        }
    }
}
int getlca(int u,int v){
    if(dep[u]>dep[v]){
        swap(u,v);
    }
    for(int i=K-1;~i;i--){
        if(dep[st[v][i]]>=dep[u]){
            v=st[v][i];
        }
    }
    if(u==v){
        return u;
    }
    for(int i=K-1;~i;i--){
        if(st[u][i]!=st[v][i]){
            u=st[u][i];
            v=st[v][i];
        }
    }
    return st[u][0];
}
int f[N][K][2];
void buildf(int pos){
    for(int i=1;i<=n;i++){
        for(int type=0;type<2;type++){
            f[i][0][type]=((a[i]>>pos)&1)^type;
        }
    }
    for(int j=1;j<K;j++){
        for(int i=1;i<=n;i++){
            for(int type=0;type<2;type++){
                if(j==pos+1){
                    f[i][j][type]=f[i][j-1][type]+f[st[i][j-1]][j-1][type^1];
                }else{
                    f[i][j][type]=f[i][j-1][type]+f[st[i][j-1]][j-1][type];
                }
            }
        }
    }
}
int getF(int u,int type,int pos){
    // printf("getF:u=%d type=%d pos=%d ",u,type,pos);
    int res=0;
    for(int i=0;i<K;i++){
        if(pos&(1<<i)){
            res=res+f[u][i][type];
            u=st[u][i];
        }
    }
    // printf("now u=%d res=%d\n",u,res);
    return res+f[u][K-1][type^1];
}
ll ans[M];
void solve(){
    n=read();
    for(int i=1;i<=n;i++){
        E[i].clear();
    }
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        E[u].push_back(v);
        E[v].push_back(u);
    }
    for(int i=1;i<=n;i++){
        a[i]=read();
    }
    dfs(1,0);
    buildst();
    q=read();
    for(int i=1;i<=q;i++){
        int u=read(),v=read();
        int lca=getlca(u,v);
        Q[i]=Query{u,v,lca};
    }
    for(int j=0;j<K-1;j++){
        buildf(j);
        for(int i=1;i<=q;i++){
            int u=Q[i].u,v=Q[i].v,lca=Q[i].lca,fa=st[Q[i].lca][0];
            int dis1=dep[u]-dep[lca],dis=dis1+dep[v]-dep[lca];
            ans[i]=ans[i]+(getF(u,1,0)<<j);
            // printf("i=%d j=%d ans=%lld\n",i,j,ans[i]);
            ans[i]=ans[i]-(getF(fa,((dis1+1)>>j)&1,(1<<j)-(dis1+1)%(1<<j))<<j);
            // printf("i=%d j=%d ans=%lld\n",i,j,ans[i]);
            ans[i]=ans[i]+(getF(v,(dis>>j)&1,dis%(1<<j)+1)<<j);
            // printf("i=%d j=%d ans=%lld\n",i,j,ans[i]);
            ans[i]=ans[i]-(getF(lca,(dis1>>j)&1,dis1%(1<<j)+1)<<j);
            // printf("i=%d j=%d ans=%lld\n",i,j,ans[i]);
        }
    }
    for(int i=1;i<=q;i++){
        printf("%lld\n",ans[i]);
    }
}
int main(){
    int T=read();
    while(T--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code involves tree traversal, LCA (Lowest Common Ancestor) computation, and bitwise operations, so we should consider the following scenarios:

1. **Small Tree with a Path of Maximum Length**: This can help in testing the correctness of the LCA and path computation.
2. **Tree with a Complex Structure**: This can help in testing the correctness of the tree traversal and the LCA computation.
3. **Queries with Repeated Nodes**: This can help in testing the correctness of the bitwise operations and the sum computation.
4. **Queries with Nodes at Different Depths**: This can help in testing the correctness of the depth calculations and the bitwise operations.

Let's create a small-scale test input that covers these scenarios:

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

Explanation:
- **Tree Structure**: The tree is a simple path from node 1 to node 5, which ensures that the LCA computation is straightforward.
- **Node Values**: The values of the nodes are 1, 2, 3, 4, 5, which are small and easy to handle.
- **Queries**:
  - The first query (1 5) tests the path from the root to a leaf, which should cover the entire path.
  - The second query (2 4) tests a path that is not from the root to a leaf, ensuring that the LCA computation is correct.
  - The third query (1 1) tests a path with a single node, which should return the value of the node itself.

This test input is small and complete, and it should help in exposing any bugs in the provided code.