**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 math
from dataclasses import dataclass
from typing import Tuple, List

class SegmentTree:
    def __init__(self, arr):
        self.n = len(arr)
        self.tree = [0] * (4 * self.n)
        self.build(arr, 0, 0, self.n - 1)

    def build(self, arr, v, tl, tr):
        if tl == tr:
            self.tree[v] = arr[tl]
        else:
            tm = (tl + tr) // 2
            self.build(arr, v*2+1, tl, tm)
            self.build(arr, v*2+2, tm+1, tr)
            self.tree[v] = self.tree[v*2+1] + self.tree[v*2+2]

    def query(self, v, tl, tr, l, r):
        if l > r:
            return 0
        if l == tl and r == tr:
            return self.tree[v]
        tm = (tl + tr) // 2
        return (self.query(v*2+1, tl, tm, l, min(r, tm))
                + self.query(v*2+2, tm+1, tr, max(l, tm+1), r))

class LCA:
    def __init__(self, n, edges):
        self.n = n
        self.log = int(math.log2(n)) + 1
        self.parent = [[-1] * n for _ in range(self.log)]
        self.depth = [0] * n
        self.adj = [[] for _ in range(n)]
        for u, v in edges:
            self.adj[u].append(v)
            self.adj[v].append(u)
        self.dfs(0, -1)
        self.build()

    def dfs(self, v, p):
        self.parent[0][v] = p
        for u in self.adj[v]:
            if u != p:
                self.depth[u] = self.depth[v] + 1
                self.dfs(u, v)

    def build(self):
        for i in range(1, self.log):
            for j in range(self.n):
                if self.parent[i-1][j] != -1:
                    self.parent[i][j] = self.parent[i-1][self.parent[i-1][j]]

    def lca(self, u, v):
        if self.depth[u] < self.depth[v]:
            u, v = v, u
        for i in range(self.log-1, -1, -1):
            if self.depth[u] - (1 << i) >= self.depth[v]:
                u = self.parent[i][u]
        if u == v:
            return u
        for i in range(self.log-1, -1, -1):
            if self.parent[i][u] != self.parent[i][v]:
                u = self.parent[i][u]
                v = self.parent[i][v]
        return self.parent[0][u]

@dataclass
class State:
    n: int
    edges: List[Tuple[int, int]]
    values: List[int]
    queries: List[Tuple[int, int]]

@dataclass
class Parameters:
    pass

def simulate_fast(state: State, parameters: Parameters, n_steps: int) -> Tuple[State, int]:
    # This function is not needed for this problem
    return state, n_steps

def is_in_cycle(state: State, parameters: Parameters) -> bool:
    # This function is not needed for this problem
    return False

def compute_cycle(state: State, parameters: Parameters, n_steps: int) -> State:
    # This function is not needed for this problem
    return state

def read_test_case() -> Tuple[State, Parameters, int]:
    n = int(input())
    edges = [tuple(map(lambda x: int(x)-1, input().split())) for _ in range(n-1)]
    values = list(map(int, input().split()))
    q = int(input())
    queries = [tuple(map(lambda x: int(x)-1, input().split())) for _ in range(q)]
    return State(n, edges, values, queries), Parameters(), 1

def print_result(results: List[int], parameters: Parameters):
    for result in results:
        print(result)

def run_dynamic(state: State, parameters: Parameters, n_steps: int) -> List[int]:
    n, edges, values, queries = state.n, state.edges, state.values, state.queries

    # Euler tour
    euler_tour = []
    first_occurrence = [-1] * n
    def dfs(v, p):
        first_occurrence[v] = len(euler_tour)
        euler_tour.append(v)
        for u in adj[v]:
            if u != p:
                dfs(u, v)
                euler_tour.append(v)

    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)
    dfs(0, -1)

    # Prepare arrays for segment tree
    m = len(euler_tour)
    a = [0] * m
    for i in range(m):
        a[i] = values[euler_tour[i]] ^ i

    # Build segment tree
    seg_tree = SegmentTree(a)

    # Prepare LCA
    lca = LCA(n, edges)

    # Process queries
    results = []
    for x, y in queries:
        if first_occurrence[x] > first_occurrence[y]:
            x, y = y, x
        l, r = first_occurrence[x], first_occurrence[y]
        z = lca.lca(x, y)
        if z == x:
            result = seg_tree.query(0, 0, m-1, l, r)
        else:
            result = (seg_tree.query(0, 0, m-1, l, first_occurrence[z]) +
                      seg_tree.query(0, 0, m-1, first_occurrence[z], r) -
                      a[first_occurrence[z]])
        results.append(result)

    return results

def main():
    t = int(input())
    for _ in range(t):
        state, parameters, n_steps = read_test_case()
        results = run_dynamic(state, parameters, n_steps)
        print_result(results, parameters)

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 potentially cause issues in the algorithm. Here are some considerations:

1. **Small Tree with Complex Structure**: A small tree with a non-trivial structure can help test the correctness of the LCA and segment tree implementations.
2. **Queries Involving Root and Leaf Nodes**: Queries that involve the root and leaf nodes can test the edge cases of the path traversal.
3. **Queries Involving Nodes with High Depth Difference**: Queries between nodes with a significant depth difference can test the segment tree's ability to handle large ranges efficiently.
4. **Queries Involving the Same Node**: Queries where the start and end nodes are the same can test the correctness of the path sum calculation for a single node.

Given these considerations, here is a small-scale test input that meets the requirements:

```plaintext
1
7
1 2
1 3
2 4
2 5
3 6
3 7
2 3 5 7 9 11 13
4
1 4
2 7
3 3
5 6
```

### Explanation:
- **Tree Structure**:
  - Node 1 is the root.
  - Node 1 has children 2 and 3.
  - Node 2 has children 4 and 5.
  - Node 3 has children 6 and 7.
- **Node Values**:
  - Values are [2, 3, 5, 7, 9, 11, 13].
- **Queries**:
  - Query (1, 4): Path is 1 -> 2 -> 4.
  - Query (2, 7): Path is 2 -> 1 -> 3 -> 7.
  - Query (3, 3): Path is just node 3.
  - Query (5, 6): Path is 5 -> 2 -> 1 -> 3 -> 6.

This test case is designed to cover various edge cases and ensure that the LCA and segment tree implementations are working 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:
import sys
input = sys.stdin.read
data = input().split()

def dfs(node, parent, depth):
    euler.append((node, depth))
    level[node] = depth
    par[node][0] = parent
    for child in adj[node]:
        if child != parent:
            dfs(child, node, depth + 1)
            euler.append((node, depth))

def preprocess_lca(n):
    for j in range(1, LOG):
        for i in range(1, n + 1):
            if par[i][j - 1] != -1:
                par[i][j] = par[par[i][j - 1]][j - 1]

def lca(u, v):
    if level[u] < level[v]:
        u, v = v, u
    diff = level[u] - level[v]
    for i in range(LOG):
        if diff & (1 << i):
            u = par[u][i]
    if u == v:
        return u
    for i in range(LOG - 1, -1, -1):
        if par[u][i] != par[v][i]:
            u = par[u][i]
            v = par[v][i]
    return par[u][0]

def xor_sum(path):
    result = 0
    for i, node in enumerate(path):
        result += a[node] ^ i
    return result

index = 0
results = []
while index < len(data):
    t = int(data[index])
    index += 1
    for _ in range(t):
        n = int(data[index])
        index += 1
        adj = [[] for _ in range(n + 1)]
        for _ in range(n - 1):
            u = int(data[index])
            v = int(data[index + 1])
            index += 2
            adj[u].append(v)
            adj[v].append(u)

        a = [0] + [int(data[index + i]) for i in range(n)]
        index += n

        LOG = 20
        level = [0] * (n + 1)
        par = [[-1] * LOG for _ in range(n + 1)]
        euler = []

        dfs(1, -1, 0)
        preprocess_lca(n)

        q = int(data[index])
        index += 1
        for _ in range(q):
            x = int(data[index])
            y = int(data[index + 1])
            index += 2

            l = lca(x, y)

            # Path from x to l
            path_x_to_l = []
            while x != l:
                path_x_to_l.append(x)
                x = par[x][0]
            path_x_to_l.append(l)

            # Path from y to l
            path_y_to_l = []
            while y != l:
                path_y_to_l.append(y)
                y = par[y][0]
            path_y_to_l.append(l)

            # Reverse one of the paths to get the full path
            path_y_to_l.reverse()
            path_x_to_l.extend(path_y_to_l[1:])

            results.append(str(xor_sum(path_x_to_l)))

print("\n".join(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:
To create a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are the steps to construct such a test input:

1. **Small-Scale Test Case**: Ensure the test case is small enough to be manageable but complex enough to challenge the algorithm.
2. **Tree Structure**: Create a tree structure that can lead to non-trivial paths and potentially expose issues in the path reconstruction.
3. **Query Selection**: Choose queries that involve nodes with different depths and paths that are not straightforward.
4. **Node Values**: Assign node values that can create interesting XOR results.

### Test Input

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

### Explanation

- **Tree Structure**:
  - Node 1 is the root.
  - 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**:
  - Node 1: 2
  - Node 2: 3
  - Node 3: 6
  - Node 4: 5
  - Node 5: 7
- **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

- **Query 1 (1 to 5)**:
  - Path: 1, 2, 3, 4, 5
  - XOR Sum: \( (2 \oplus 0) + (3 \oplus 1) + (6 \oplus 2) + (5 \oplus 3) + (7 \oplus 4) = 2 + 2 + 4 + 6 + 3 = 17 \)
- **Query 2 (2 to 4)**:
  - Path: 2, 3, 4
  - XOR Sum: \( (3 \oplus 0) + (6 \oplus 1) + (5 \oplus 2) = 3 + 7 + 7 = 17 \)
- **Query 3 (1 to 3)**:
  - Path: 1, 2, 3
  - XOR Sum: \( (2 \oplus 0) + (3 \oplus 1) + (6 \oplus 2) = 2 + 2 + 4 = 8 \)

### Final Test Input

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

This test input is designed to challenge the algorithm by requiring it to handle non-trivial paths and compute the XOR sums 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:
// @idr1ska
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll N = 5e5 + 5;
const ll M = 20;
ll a[N];
pair<ll, ll> dp_[M][N];
ll dp[M][N];
vector<ll> g[N];
ll h[N];

void dfs1(ll v, ll p) {
    h[v] = h[p] + 1;
    dp_[0][v] = {p, a[v]};
    for (int i = 1; i < M; ++i) {
        dp_[i][v] = {dp_[i - 1][dp_[i - 1][v].first].first,
                     dp_[i - 1][v].second + dp_[i - 1][dp_[i - 1][v].first].second};
    }
    for (ll u : g[v]) {
        if (u != p) dfs1(u, v);
    }
}

void dfs(ll v, ll p, ll j) {
    dp[0][v] = (a[v] >> j) & 1;
    for (ll i = 1; i <= j; ++i) {
        dp[i][v] = dp[i - 1][v] + dp[i - 1][dp_[i - 1][v].first];
    }
    dp[j + 1][v] = dp[j][v];
    for (ll i = j + 2; i < M; ++i) {
        dp[i][v] = dp[i - 1][v] + dp[i - 1][dp_[i - 1][v].first];
    }
    for (ll u : g[v]) {
        if (u != p) dfs(u, v, j);
    }
}

pair<ll, ll> e[(ll) 1e5 + 5];

pair<ll, ll> lca(ll x, ll y) {
    if (h[y] > h[x]) {
        swap(x, y);
    }
    ll sum = 0;
    for (int i = M - 1; i >= 0; --i) {
        if (h[dp_[i][x].first] >= h[y]) {
            sum += dp_[i][x].second;
            x = dp_[i][x].first;
        }
    }
    if (x == y) {
        return {x, sum};
    }
    for (int i = M - 1; i >= 0; --i) {
        if (dp_[i][x] != dp_[i][y]) {
            sum += dp_[i][x].second;
            x = dp_[i][x].first;
            sum += dp_[i][y].second;
            y = dp_[i][y].first;
        }
    }
    return {x, sum};
}

ll get_up(ll v, ll cnt) {
    for (int i = M - 1; i >= 0; --i) {
        if ((cnt >> i) & 1) {
            v = dp_[i][v].first;
        }
    }
    return v;
}

ll ans_for(ll v, ll j, ll p) {
    ll sum = 0;
    for (int i = M - 1; i >= 0; --i) {
        if (h[p] <= h[dp_[i][v].first]) {
            if (i >= j + 1) {
                sum += ((1 << (i - 1)) - dp[i][v] - dp[i][v]) * (1 << j);
            } else {
                sum += ((1 << (i)) - dp[i][v] - dp[i][v]) * (1 << j);
            }
            v = dp_[i][v].first;
        }
    }
    return sum;
}

pair<ll, ll> f[(ll) 1e5 + 5];

void solve() {
    ll n;
    cin >> n;
    for (int i = 0; i <= n; ++i) {
        g[i].clear();
    }
    for (int i = 0; i < n - 1; ++i) {
        ll v, u;
        cin >> v >> u;
        g[v].push_back(u);
        g[u].push_back(v);
    }
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }
    dfs1(1, 0);
    ll q;
    cin >> q;
    for (int i = 0; i < q; ++i) {
        ll v, u;
        cin >> v >> u;
        f[i] = {v, u};
        e[i] = lca(v, u);
        e[i].second += a[e[i].first] ^ (h[v] - h[e[i].first]);
    }
    for (int i = 0; i < M - 1; ++i) {
        dfs(1, 0, i);
        for (int j = 0; j < q; ++j) {
            auto [v, u] = f[j];
            e[j].second += ans_for(get_up(v, (1 << i)), i, e[j].first);
            ll sz = h[v] + h[u] - h[e[j].first] * 2;
            if (sz >= (1 << i)) {
                ll o = (sz - (1 << i)) / (1 << (i + 1)) * (1 << (i + 1)) + (1 << i);
                e[j].second += ans_for(get_up(u, sz - (o - (1 << i) - 1)), i, e[j].first);
                e[j].second += ans_for(get_up(u, max(sz - (o + (1 << i) - 1), (ll) 0)), i, get_up(u, sz - (o - 1)));
            }
        }
    }
    for (int i = 0; i < q; ++i) {
        cout << e[i].second << '\n';
    }
}

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

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

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

Think step by step.

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


**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 u32 = unsigned;
using i64 = long long;
using u64 = unsigned long long;

constexpr int N = 500005;
int tt, n, q;
int a[N];
int ah[N], ax[2 * N], am, at[2 * N];
inline void alink(int u, int v) { ax[++am] = ah[u]; ah[u] = am; at[am] = v; }
int dep[N], fa[N];
int tp[19][N], cnt[19][N];
i64 f[19][N], g[19][N];

void dfs(int u, int w) {
  for (int j = 0; j <= 18; j++) cnt[j][u] += (a[u] >> j & 1);
  for (int e = ah[u]; e; e = ax[e]) {
    int v = at[e];
    if (v != w) {
      dep[v] = dep[u] + 1;
      fa[v] = u;
      tp[0][v] = v;
      for (int j = 0; j <= 18; j++) cnt[j][v] += cnt[j][u];
      dfs(v, u);
    }
  }
}
inline int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int j = 18; j >= 0; j--) if ((dep[u] - dep[v]) >> j & 1) u = fa[tp[j][u]];
  if (u == v) return u;
  for (int j = 18; j >= 0; j--) if (fa[tp[j][u]] != fa[tp[j][v]]) u = fa[tp[j][u]], v = fa[tp[j][v]];
  return fa[u];
}
inline int getanc(int u, int d) {
  for (int j = 18; j >= 0; j--) if (d >> j & 1) u = fa[tp[j][u]];
  return u;
}
inline i64 cost(int j, int u, int v, int w) {
  int cnt1 = cnt[j][u] + cnt[j][v] - cnt[j][w] - cnt[j][fa[w]];
  int cnt0 = dep[u] + dep[v] - 2 * dep[w] + 1 - cnt1;
  return (i64)(cnt0 - cnt1) * (1 << j);
}

i64 qry(int u, int w, int v) {
  if (v == w) {
    i64 ans = 0;
    for (int j = 18; j >= 0; j--) {
      if ((dep[u] - dep[w] + 1) >> j & 1) {
        ans += f[j][u];
        if (dep[f[j][u]] >= w) ans += cost(j, f[j][u], w, w);
        u = fa[tp[j][u]];
      }
    }
    return ans;
  } else if (u == w) {
    i64 ans = 0;
    int start = v, last = 0;
    for (int j = 0; j <= 18; j++) {
      if ((dep[v] - dep[w] + 1) >> j & 1) {
        ans += g[j][v];
        if (v != start) ans += cost(j, start, last, last);
        last = tp[j][v];
        v = fa[tp[j][v]];
      }
    }
    return ans;
  } else {
    int len = dep[u] + dep[v] - 2 * dep[w];
    int h = __lg(len);
    if (dep[u] - dep[w] >= (1 << h)) {
      return f[h][u] + cost(h, fa[tp[h][u]], w, v) + qry(fa[tp[h][u]], w, v);
    } else {
      int x = getanc(v, len - (1 << h));
      return cost(h, v, x, x) + qry(x, x, v) + qry(u, w, fa[x]);
    }
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  // cerr << sizeof(tp) * 6 << "\n" << flush;
  cin >> tt;
  while (tt--) {
    cin >> n;
    am = 0; for (int i = 1; i <= n; i++) ah[i] = 0;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      alink(u, v), alink(v, u);
    }
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      f[0][i] = g[0][i] = a[i];
    }
    dfs(1, 0);

    for (int j = 1; j <= 18; j++) {
      for (int i = 1; i <= n; i++) {
        tp[j][i] = tp[j - 1][fa[tp[j - 1][i]]];
        f[j][i] = f[j - 1][i] + f[j - 1][fa[tp[j - 1][i]]] + cost(j - 1, fa[tp[j - 1][i]], tp[j][i], tp[j][i]);
        g[j][i] = g[j - 1][i] + g[j - 1][fa[tp[j - 1][i]]] + cost(j - 1, i, tp[j - 1][i], tp[j - 1][i]);
      }
    }

    cin >> q;
    for (int i = 1; i <= q; i++) {
      int u, v;
      cin >> u >> v;
      int w = getlca(u, v);
      cout << qry(u, w, v) << "\n";
    }
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
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:
def find(nodes):
    edges=set()
    for edge in range(nodes-1):
        i,j=map(int,input().split())
        edges.add((i,j))
    arr=[0]
    arr.extend(map(int,input().split()))
    memo={}
    def dfs(start,end,n):
        if (start,end,n) in memo:
            return memo[(start,end,n)]
        visited[start]=True
        if start==end:
            return n^arr[start]
        result=-1
        for i in range(1,nodes+1):
            if not visited[i] and ((start,i) in edges or (i,start) in edges):
                result=dfs(i,end,n+1)
                if result!=-1:
                    result+=(n^arr[start])
        memo[(start,end,n)]=result
        return result
    queries=int(input())
    for query in range(queries):
        start,end=map(int,input().split())
        visited=[False for i in range(nodes+1)]
        print(dfs(start,end,0))
numTests=int(input())
for i in range(numTests):
    nodes=int(input())
    find(nodes)

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

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

Think step by step.

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

1. **Edge Cases:**
   - **Single Node Tree:** The simplest case with only one node.
   - **Linear Tree:** A tree where nodes form a straight line.
   - **Complete Tree:** A tree where every node has exactly one or two children.
   - **Cyclic Structure:** Although the input guarantees a tree, we should still consider how the code handles large and deep trees.

2. **Query Cases:**
   - **Same Node Queries:** Queries where the start and end nodes are the same.
   - **Long Path Queries:** Queries that cover a large number of nodes.
   - **Short Path Queries:** Queries that cover only a few nodes.

3. **Value Distribution:**
   - **Uniform Values:** All nodes have the same value.
   - **Random Values:** Nodes have different values.

Given these considerations, let's create a small-scale test input that includes a linear tree with a few nodes and a mix of query types.

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

**Explanation:**
- **Tree Structure:**
  - The tree is a linear path: 1 -> 2 -> 3 -> 4 -> 5.
- **Node Values:**
  - The values of the nodes are 1, 2, 3, 4, 5 respectively.
- **Queries:**
  - **Query 1 (1, 5):** Path is 1 -> 2 -> 3 -> 4 -> 5.
  - **Query 2 (3, 3):** Path is just node 3.
  - **Query 3 (1, 2):** Path is 1 -> 2.

This test input includes:
- A simple linear tree to ensure the path is straightforward.
- Queries that cover different path lengths and include a self-query.

This should help expose any issues in the buggy code, particularly related to the path traversal and memoization.


**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
MAXN = 500005
LOG = 20
tree = [[] for _ in range(MAXN)]
a = [0] * MAXN
depth = [0] * MAXN
up = [[-1] * LOG for _ in range(MAXN)]
prefixXor = [0] * MAXN
prefixSum = [0] * MAXN

def dfs(node, parent):
    up[node][0] = parent
    for i in range(1, LOG):
        if up[node][i - 1] != -1:
            up[node][i] = up[up[node][i - 1]][i - 1]
        else:
            up[node][i] = -1

    prefixXor[node] = prefixXor[parent] ^ a[node]
    prefixSum[node] = prefixSum[parent] + (a[node] ^ depth[node])

    for child in tree[node]:
        if child != parent:
            depth[child] = depth[node] + 1
            dfs(child, node)

def lca(u, v):
    if depth[u] < depth[v]:
        u, v = v, u
    diff = depth[u] - depth[v]
    for i in range(LOG):
        if (diff >> i) & 1:
            u = up[u][i]
    if u == v:
        return u
    for i in range(LOG - 1, -1, -1):
        if up[u][i] != up[v][i]:
            u = up[u][i]
            v = up[v][i]
    return up[u][0]

def query(x, y):
    lcaNode = lca(x, y)
    return prefixSum[x] + prefixSum[y] - 2 * prefixSum[lcaNode] + (a[lcaNode] ^ depth[lcaNode])
sys.stdin = open(0)
sys.stdout = open(1, 'w')
t = int(input())
for _ in range(t):
    n = int(input())
    for i in range(1, n + 1):
        tree[i].clear()
        depth[i] = 0
        prefixXor[i] = 0
        prefixSum[i] = 0

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

    a[1:n+1] = map(int, input().split())

    for i in range(MAXN):
        for j in range(LOG):
            up[i][j] = -1

    dfs(1, 0)

    q = int(input())
    for _ in range(q):
        x, y = map(int, input().split())
        print(query(x, y))

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

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

Think step by step.

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


**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:
def solve(t, test_cases):
    results = []

    for case in test_cases:
        n, edges, values, queries = case

        # Build the tree using adjacency list
        tree = [[] for _ in range(n + 1)]
        for u, v in edges:
            tree[u].append(v)
            tree[v].append(u)

        # Function to find path using DFS
        def find_path(u, v, parent):
            if u == v:
                return [u]
            for neighbor in tree[u]:
                if neighbor != parent:
                    path = find_path(neighbor, v, u)
                    if path:
                        return [u] + path
            return []

        for x, y in queries:
            path = find_path(x, y, -1)
            sum_xor = 0
            for i, node in enumerate(path):
                sum_xor += values[node - 1] ^ i
            results.append(sum_xor)

    return results

# Read input
import sys
input = sys.stdin.read
data = input().split()

idx = 0
t = int(data[idx])
idx += 1
test_cases = []

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

# Solve the problem
results = solve(t, test_cases)

# Print the results
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:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following scenarios:

1. **Edge Cases:**
   - A tree with a single node.
   - A tree with a linear structure (a chain of nodes).
   - A tree with a star structure (one central node connected to all others).

2. **Complex Cases:**
   - A tree with a balanced structure.
   - A tree with multiple levels and branches.

3. **Queries:**
   - Queries involving the root node.
   - Queries involving leaf nodes.
   - Queries involving nodes in the middle of the tree.
   - Queries involving the same node (e.g., (1, 1)).

4. **Values:**
   - Values that are powers of 2 to maximize the effect of XOR operations.
   - Values that are large and small to test the range of the values array.

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

```plaintext
1
5
1 2
2 3
3 4
4 5
1 2 4 8 16
5
1 5
2 4
3 3
1 3
5 1
```

### Explanation:
- **Tree Structure:**
  - The tree is a simple linear chain: 1 -> 2 -> 3 -> 4 -> 5.
- **Values:**
  - The values are powers of 2: [1, 2, 4, 8, 16].
- **Queries:**
  - **1 5:** Path is 1, 2, 3, 4, 5.
  - **2 4:** Path is 2, 3, 4.
  - **3 3:** Path is 3 (single node).
  - **1 3:** Path is 1, 2, 3.
  - **5 1:** Path is 5, 4, 3, 2, 1 (reverse of the first query).

This test input should expose any issues in the path finding or the XOR computation 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>
using namespace std;
using LL = long long;

const int M = 20;

void solve() {
    int n;
    cin >> n;

    vector<vector<int>> e(n + 1);
    vector<int> a(n + 1);

    for (int i =0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }

    for (int i = 0; i < n; ++i) {
        cin >> a[i + 1];
    }

    vector<int> dep(n + 1);
    vector fa(n + 1, vector<int>(M));
    vector f(n + 1, array<array<int, 2>, M>());

    auto dfs = [&](auto &&self, int u) -> void {
        for (int i = 0; i < M; ++i) {
            f[u][i][(a[u] >> i) & 1]++;
        }
        for (int v : e[u]) {
            if (v == fa[u][0]) {
                continue;
            }
            dep[v] = dep[u] + 1;
            fa[v][0] = u;
            f[v] = f[u];
            self(self, v);
        }
    };

    dep[1] = 1;
    dfs(dfs, 1);

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j < M; ++j) {
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
        }
    }

    auto getLca = [&](int x, int y) {
        if (dep[x] < dep[y]) {
            swap(x, y);
        }

        for (int i = M - 1; i >= 0; --i) {
            if (dep[fa[x][i]] >= dep[y]) {
                x = fa[x][i];
            }
        }

        if (x == y) {
            return x;
        }

        while (x != y) {
            for (int i = M - 1; i >= 0; --i) {
                if (fa[x][i] != fa[y][i]) {
                    x = fa[x][i];
                    y = fa[y][i];
                }
            }
        }

        return fa[x][0];
    };

    int q;
    cin >> q;

    vector<array<int, 3>> que(q);
    for (int i = 0; i < q; ++i) {
        int u, v;
        cin >> u >> v;
        int lca = getLca(u, v);
        que[i] = {u, v, lca};
    }

    vector<LL> ans(q);

    auto jump = [&](int x, int y) {
        for (int i = M - 1; i >= 0; --i) {
            if ((y >> i) & 1) {
                x = fa[x][i];
            }
        }

        return x;
    };

    for (int i = 0; i < M; ++i) {
        vector<int> g(n + 1);

        auto dfs = [&](auto &&self, int u) -> void {
            int t = fa[u][i];

            g[u] = f[u][i][1] - f[t][i][1] + (dep[t] - g[t]);
            for (int v : e[u]) {
                if (v == fa[u][0]) {
                    continue;
                }

                self(self, v);
            }
        };

        dfs(dfs, 1);

        for (int j = 0; j < q; ++j) {
            int res = 0;

            auto [x, y, lca] = que[j];

            int len = dep[x] - dep[lca] + 1;

            int len1 = (1 << i);
            int t1 = len / len1;
            int t2 = len % len1;

            int x1 = jump(x, t1 * len1);

            t1 %= 2;

            if (t1 == 0) {
                res += g[x] - g[x1];

                if (t2) {
                    res += (f[x1][i][1] - f[fa[lca][0]][i][1]);
                }
            } else {
                res += g[x] - (dep[x1] - g[x1]);

                if (t2) {
                    res += (f[x1][i][0] - f[fa[lca][0]][i][0]);
                }
            }

            len = dep[y] - dep[lca];

            if (t2 != 0) {
                t2 = len1 - t2;
                t1 = (t1 + 1) % 2;
            }

            if (len <= t2) {
                if (t1 == 0) {
                    res += (f[y][i][0] - f[lca][i][0]);
                } else {
                    res += (f[y][i][1] - f[lca][i][1]);
                }
            } else {
                len -= t2;

                int t3 = len / len1;
                int t4 = len % len1;

                int tt = (t1 + t3) % 2;

                int y1 = jump(y, t4);

                if (tt == 0) {
                    res += f[y][i][1] - f[y1][i][1];
                } else {
                    res += f[y][i][0] - f[y1][i][0];
                }

                int y2 = jump(y1, t3 * len1);

                if (tt == 0) {
                    res += len1 * t3 - (g[y1] - ((t3 % 2 == 0) ? g[y2] : (dep[y2] - g[y2])));
                } else {
                    res += g[y1] - ((t3 % 2 == 0) ? g[y2] : (dep[y2] - g[y2]));
                }

                if (t1 == 0) {
                    res += f[y2][i][0] - f[lca][i][0];
                } else {
                    res += f[y2][i][1] - f[lca][i][1];
                }
            }

            ans[j] += 1LL * (1 << i) * res;
        }
    }

    for (int i = 0; i < q; ++i) {
        cout << ans[i] << "\n";
    }
}

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

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the code to fail. Here are the steps to think through this:

1. **Tree Structure**: The tree should be designed to have a specific structure that might cause issues in the depth and ancestor calculations.
2. **Node Values**: The values of the nodes should be chosen to highlight any issues in the XOR and sum calculations.
3. **Queries**: The queries should be designed to test the path calculations and the XOR operations along the paths.

Given the constraints and the nature of the problem, a good test case would be a tree with a linear structure (a chain) and specific node values that might cause issues in the depth and ancestor calculations. Here is a small-scale test input:

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

### Explanation:
- **Tree Structure**: The tree is a simple linear chain from node 1 to node 5.
- **Node Values**: The values are 1, 2, 3, 4, 5 respectively.
- **Queries**:
  - `1 5`: This query tests the path from node 1 to node 5.
  - `2 4`: This query tests the path from node 2 to node 4.
  - `3 3`: This query tests the path from node 3 to itself.

This test case is designed to be small and manageable, but it should be sufficient to expose any issues in the depth calculations, ancestor relationships, and the XOR sum computations. The linear structure and simple values make it easy to manually verify the expected results.


**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 ff first
#define ss second
#define MAX 500001
#define INF LLONG_MAX
#define pb push_back
#define pii pair<int,int>
//#define int long long
using namespace std;
struct fenwick{
    int N;
    vector<int> sums;
    void init(int n){
        sums.assign(n+1,0);
        N=n;
    }
    void upd(int x,int v){
        while(x<=N){
            sums[x]+=v;
            x+=(x&(-x));
        }
    }
    int query(int x){
        int res=0;
        while(x>0){
            res+=sums[x];
            x-=(x&(-x));
        }
        return res;
    }
};
long long n,x,y,z,pre[MAX][20],dp[MAX][20];
int st[MAX][20],dep[MAX],par[MAX],a[MAX],b[MAX],t=1,cnt1=0;
vector<vector<int>> g;
void dfs(int v,int prev){
    par[v]=prev;
    for(int i=0;i<20;i++){
        if(a[v]%2){
            pre[v][i]++;
        }
        a[v]/=2;
    }
    for(auto x:g[v]){
        if(x==prev){
            continue;
        }
        dep[x]=dep[v]+1;
        for(int i=0;i<20;i++){
            pre[x][i]=pre[v][i];
        }
        dfs(x,v);
    }
}
int lca(int x,int y){
    if(dep[x]>dep[y]){
        swap(x,y);
    }
    for(int i=19;i>=0;i--){
        if(st[y][i]!=-1 && dep[st[y][i]]>=dep[x]){
            y=st[y][i];
        }
    }
    if(x==y){
        return x;
    }
    for(int i=19;i>=0;i--){
        if(st[x][i]!=st[y][i]){
            x=st[x][i];
            y=st[y][i];
        }
    }
    return st[x][0];
}
int up(int x,int v){
    for(int i=0;i<20;i++){
        if((v&(1<<i))){
            x=st[x][i];
        }
    }
    return x;
}
long long calc(int x,int y,int bit){
    int h=dep[x]-dep[y]+1;
    h-=h%(1<<bit);
    long long res=dp[x][bit];
    x=up(x,h);
    h/=(1<<bit);
    if(x!=-1){
        if(h%2){
            res-=(dep[x]+1-dp[x][bit]);
        }
        else{
            res-=dp[x][bit];
        }
        int h1=pre[x][bit];
        if(par[y]!=-1){
            h1-=pre[par[y]][bit];
        }
        if(h%2){
            res+=(dep[x]-dep[y]+1-h1);
        }
        else{
            res+=h1;
        }
    }
    return res;
}
void solve(){
    cin >> n;
    g.clear();
    g.resize(n);
    for(int i=0;i<n-1;i++){
        cin >> x >> y;
        x--;
        y--;
        g[x].pb(y);
        g[y].pb(x);
    }
    for(int i=0;i<n;i++){
        cin >> a[i];
        b[i]=a[i];
    }
    for(int i=0;i<20;i++){
        pre[0][i]=0;
    }
    dep[0]=0;
    dfs(0,-1);
    for(int i=0;i<n;i++){
        st[i][0]=par[i];
    }
    for(int i=1;i<20;i++){
        for(int j=0;j<n;j++){
            if(st[j][i-1]==-1){
                st[j][i]=-1;
                continue;
            }
            st[j][i]=st[st[j][i-1]][i-1];
        }
    }
    for(int i=0;i<20;i++){
        for(int j=0;j<n;j++){
            dp[j][i]=pre[j][i];
            if(st[j][i]!=-1){
                dp[j][i]-=(pre[st[j][i]][i]+dp[st[j][i]][i]);
                dp[j][i]+=dep[st[j][i]]+1;
            }
        }
    }
    int q;
    cin >> q;
    for(int i=0;i<q;i++){
        cin >> x >> y;
        x--;
        y--;
        z=lca(x,y);
        long long res=-(b[z]^(dep[x]-dep[z]));
        int x1,y1;
        x1=x;
        y1=y;
        for(int j=0;j<20;j++){
            x=x1;
            y=y1;
            long long res1=0;
            res1+=calc(x,z,j);
            int h=(dep[x]+dep[y]-2*dep[z]+1);
            int h1=h/(1<<j);
            h%=(1<<j);
            long long h2=pre[y][j];
            int y1=y;
            y=up(y,h);
            if(y!=-1 && dep[y]>=dep[z]){
                if(h1%2){
                    res1+=calc(y,z,j);
                }
                else{
                    res1+=(dep[y]-dep[z]+1-calc(y,z,j));
                }
                h2-=pre[y][j];
            }
            else if(par[z]!=-1){
                h2-=pre[par[z]][j];
            }
            if(h1%2){
                res1+=(min(h,dep[y1]-dep[z]+1)-h2);
            }
            else{
                res1+=h2;
            }
            res+=res1*(1<<j);
        }
        cnt1++;
        if(t<10){
            cout << res << "\n";
        }
        if(t>10 && cnt1==177){
            cout << n << "\n";
            for(int z=0;z<n;z++){
                for(auto zor:g[z]){
                    cout << zor << " ";
                }
                cout << "\n";
            }
            for(int z=0;z<n;z++){
                cout << b[z] << " ";
            }
            cout << "\n";
            cout << x << " " << y << "\n";
        }
    }
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;
    while(t--){
        solve();
    }
}

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. Here are the steps to consider:

1. **Tree Structure**: Create a tree structure that can expose potential issues in the DFS and LCA calculations.
2. **Node Values**: Assign node values that can lead to unexpected behavior in the XOR and sum calculations.
3. **Queries**: Design queries that target specific paths or nodes to highlight the bug.

### Step-by-Step Reasoning:

1. **Tree Structure**:
   - A linear tree (a chain) can be a good starting point because it simplifies the path and LCA calculations.
   - For example, a tree with nodes connected in a linear fashion: 1 - 2 - 3 - 4 - 5.

2. **Node Values**:
   - Assign values to nodes that can cause unexpected behavior in the XOR and sum calculations.
   - For example, values that are powers of 2 or have specific bit patterns can be useful.

3. **Queries**:
   - Design queries that target paths that should be straightforward but can still expose bugs.
   - For example, queries that go from the start of the chain to the end, or from the middle to the end.

### Test Input:

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

### Explanation:

- **Tree Structure**: The tree is a simple linear chain with nodes 1, 2, 3, 4, and 5 connected in that order.
- **Node Values**: The values are 2, 3, 6, 5, and 4. These values are chosen to have different bit patterns to test the XOR and sum calculations.
- **Queries**:
  - The first query (1, 5) should compute the sum for the path 1 - 2 - 3 - 4 - 5.
  - The second query (2, 4) should compute the sum for the path 2 - 3 - 4.
  - The third query (3, 5) should compute the sum for the path 3 - 4 - 5.

This test input is designed to be small-scale and complete, focusing on a linear tree structure and specific node values to maximize the probability of exposing the bug in the faulty solution.


**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;
const int MAXN=5e5+5;
const int sqr=9;
int dis[MAXN],cp[MAXN][20],pc[MAXN][20],sp[MAXN][20],root[MAXN],A[MAXN];
vector<int> ds[MAXN];
int getlog(long long n) { return 63-__builtin_clzll(n); }
void dfs(int i,int pre)
{
	root[i]=sp[i][0]=pre,dis[i]=dis[pre]+1;
	for(int j=1;(1<<j)<=dis[i];j++) sp[i][j]=sp[sp[i][j-1]][j-1];
	for(int j=sqr+1;j<=18;j++)
	{
	    cp[i][j]=cp[pre][j]+((A[i]>>j)&1);
	    pc[i][j]=pc[pre][j]+1-((A[i]>>j)&1);
	}
	for(int j=0;j<=sqr;j++)
    {
        cp[i][j]=cp[sp[i][j]][j]+(((A[i]>>j)^((dis[i]-1)>>j))&1);
        pc[i][j]=pc[sp[i][j]][j]+1-(((A[i]>>j)^((dis[i]-1)>>j))&1);
    }
	for(auto v:ds[i]) if(v!=pre) dfs(v,i);
}
int shift(int v,int k)
{
	while(k) v=sp[v][getlog(k&-k)],k-=k&-k;
	return v;
}
int lca(int a,int b)
{
	int x=dis[a],y=dis[b],m=min(x,y);
	a=shift(a,x-m),b=shift(b,y-m);
	if(a==b) return a;
	for(int i=getlog(m);i+1;i--) if(sp[a][i]!=sp[b][i]) a=sp[a][i],b=sp[b][i];
	return root[a];
}
int getdis(int a,int b) { return dis[a]+dis[b]-dis[lca(a,b)]*2; }
long long getans1(int x,int y,int k,int v)
{
	long long ans=0;
	int ds=min((1<<k)-1,dis[x]-dis[y]+1),dl=(((dis[x]-dis[y])>>k)+1)<<k,res=shift(x,dl),d=dis[x]-dl;
	for(int i=0;i<=ds;i++)
	{
		if(((dis[x]-1)^(v+i))&(1<<k)) ans+=pc[x][k]-pc[res][k];
		else ans+=cp[x][k]-cp[res][k];
		x=root[x];
		if(d==dis[y]-(1<<k)) res=root[y],d=dis[y]-1;
		else res=root[res],d--;
	}
	return ans;
}
long long getans2(int x,int y,int k,int v)
{
	y=root[y];
	long long ans=0;
	int dx=dis[x],dy=dis[y];
	bool ck=((v>>k)&1);
	if((1<<k)-(v%(1<<k))>=dx-dy)
	{
		if(ck) ans+=pc[x][k]-pc[y][k];
		else ans+=cp[x][k]-cp[y][k];
	}
	else
	{
		int p=shift(x,(1<<k)-(v%(1<<k)));
		if(ck) ans+=pc[x][k]-pc[p][k];
		else ans+=cp[x][k]-cp[p][k];
		x=p,dx=dis[p],ck^=1;
		while(dx-(1<<k)>dy)
		{
			p=sp[x][k];
			if(ck) ans+=pc[x][k]-pc[p][k];
			else ans+=cp[x][k]-cp[p][k];
			x=p,dx-=(1<<k),ck^=1;
		}
		if(ck) ans+=pc[x][k]-pc[y][k];
		else ans+=cp[x][k]-cp[y][k];
	}
	return ans;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		int n,q;
		cin>>n;
		for(int i=1;i<n;i++)
		{
			int l,r;
			cin>>l>>r;
			ds[l].push_back(r),ds[r].push_back(l);
		}
		for(int i=1;i<=n;i++) cin>>A[i];
		dfs(1,0);
		cin>>q;
		while(q--)
		{
			int l,r,lc;
			cin>>l>>r;
			lc=lca(l,r);
			long long ans=0;
			int d=getdis(l,r);
			for(int i=0;i<=18;i++) if(i<=sqr) ans+=((getans1(l,lc,i,0)+getans1(r,lc,i,d^((1<<i)-1))-(((A[lc]>>i)^((dis[l]-dis[lc])>>i))&1))<<i);
			else ans+=((getans2(l,lc,i,0)+getans2(r,lc,i,d^((1<<i)-1))-(((A[lc]>>i)^((dis[l]-dis[lc])>>i))&1))<<i);
			cout<<ans<<"\n";
		}
		for(int i=1;i<=n;i++) ds[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
1
4
1 2
2 3
3 4
1 2 3 4
3
1 4
2 3
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Playing Quidditch (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This afternoon, you decided to enjoy the first days of Spring by taking a walk outside. As you come near the Quidditch field, you hear screams. Once again, there is a conflict about the score: the two teams are convinced that they won the game! To prevent this problem from happening one more time, you decide to get involved in the refereeing of the matches.

Now, you will stay in the stadium to watch the game and count the score. At the end of the game, you will decide the winner.

Today, two teams are competing: the red Gryffindor (R) and the blue Ravenclaw (B) team. Each team is composed of $$$P$$$ players ($$$1 \leq P \leq 10$$$).

The field is a rectangle of $$$N$$$ lines and $$$M$$$ columns ($$$3 \leq N, M \leq 99$$$, $$$N$$$ and $$$M$$$ are odd). All the positions are integers, and several entities are allowed to be at the same position in the field. At the beginning of the game, the field contains goals for the two teams (each team can own between one and five goals), the players, and exactly one Quaffle. In this version of the problem, one Bludger and a Golden Snitch can be present.

A game is composed of $$$T$$$ steps ($$$0 \leq T \leq 10000$$$). At each step, one entity on the field (a player or a ball) performs one action. All entities can move. A player can also catch a ball or throw the Quaffle that it is carrying. To catch a ball, a player must be located on the same cell as it. The Quaffle does not perform any action while it is being carried; it only follows the movements of the player. If a player carrying the Quaffle decides to throw it, the Quaffle is simply put at the current position of the player. If a player is on the same cell as a Bludger (either after a movement from the player or the Bludger), the player is eliminated. If the player is eliminated while it is carrying the Quaffle, the Quaffle remains on the cell containing both the player and the Bludger after the move. It is guaranteed that this never occurs while the player is in a cell containing a goal.

To win a point, a player must leave the Quaffle at a goal of the other team. When it does, the team of the player wins one point, and the Quaffle instantly moves to the middle of the field (the cell at the $$$(M+1)/2$$$-th column of the $$$(N+1)/2$$$-th line of the field, starting from 1). There is no goal in the middle of the field. If a player puts the ball in its own goal, the other team wins the point. If a player catches the Golden Snitch, their team wins 10 points and the game is over.

Input: On the first line, the integers $$$N$$$ and $$$M$$$.

The description of the field follows: $$$N$$$ lines of $$$M$$$ pairs of characters separated by spaces. Each pair of characters represents a position on the field. It can be either:

- .. to represent an empty cell
- R0, ..., R9, B0, ..., B9 to represent a player. The first character is the team of the player, and the second is the number of the player in the team. Each pair of characters is unique, but it is not guaranteed that all the pairs appear in the grid.
- RG or BG to represent a goal. The blue team tries to put the ball in a red goal (RG) while the red team tries to put the ball in a blue goal (BG).
- .Q to represent the Quaffle, which is the ball that the players use to score goals.
- .B to represent the Bludger.
- .S to represent the Golden Snitch.

The next line contains $$$T$$$, the number of steps that compose the game. $$$T$$$ lines follow, each describing one action. It contains several pieces of information separated by a space. First, a pair of characters representing the entity that must perform the action. Second, the description of the action:

- U, D, L, R indicate that the entity moves on the grid. It can move to the top of the grid (U), to the bottom (D), to the left (L), or to the right (R). Each entity moves by only one cell at a time.
- C indicates that the player catches the ball (only a player can catch a ball). Then, there is a space followed by a pair of characters: the description of the ball caught by the player. This information is needed since several balls can be in the same cell.
- T indicates that the player throws the Quaffle that it is carrying.

All the actions performed by the entities are guaranteed to be valid: the players stay in the field, don't catch a ball if they are not in the same cell, don't release the Quaffle if they are not carrying it, ...

Output: You must output the description of the main events of the game, one event per line. More precisely:

- Each time a team scores, you must print t RED GOAL or t BLUE GOAL, depending on the team who scored, where t is the current time (the position of the action in the list of actions, starting from 0). In the case where a player scores in the wrong goal (a red player scores in the red goal, or a blue player scores in the blue goal), you must print the name of the team who wins one point, that is, the other team.
- Each time a player is eliminated, you must print t p ELIMINATED, where t is the current time and p is the player who is eliminated. The format to print the player is the same as in the input.
- If the Golden Snitch is caught, you must print t RED CATCH GOLDEN SNITCH or t BLUE CATCH GOLDEN SNITCH, depending on the team who caught the Golden Snitch, where t is the current time.

The events must be printed in ascending order of t. If several players are eliminated at the same time, the events must be written is alphabetical order: B0, ..., B9, R0, ... R9.

At the end of the game, you must print the final score as: FINAL SCORE: r b, where r is the score of the red team and b is the score of the blue team.

Examples:
input:
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
12
R0 D
R0 C .Q
R0 R
R0 T
R0 D
B0 R
B0 U
B0 C .Q
B0 L
B0 L
B0 L
B0 T
output:
11 BLUE GOAL
FINAL SCORE: 0 1

input:
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
5
R0 D
R0 C .Q
R0 L
R0 L
R0 T
output:
4 BLUE GOAL
FINAL SCORE: 0 1

input:
5 5
.. .. .. .. ..
.. .. .. .. ..
RG R0 .Q B0 BG
.. .. .. .. ..
.. .. .B .. ..
5
.B L
.B U
.B U
B0 L
B0 L
output:
2 R0 ELIMINATED
4 B0 ELIMINATED
FINAL SCORE: 0 0

input:
5 5
.. R0 .S B0 ..
.. .. .. .. ..
RG .. .Q .. BG
.. .. .. .. ..
.. R1 .B B1 ..
4
.S D
R0 D
R0 R
R0 C .S
output:
3 RED CATCH GOLDEN SNITCH
FINAL SCORE: 10 0

Note: In the first example, the red player takes the Quaffle, move it and throw it. The blue player catches the ball, goes to the red goal and scores.

In the second example, the red player takes the ball and scores in the goal of their own team: the blue team wins a point.

In the third example, the Bludger goes at the position of R0: R0 is eliminated. Then, B0 moves to the position of the Bludger: B0 is eliminated too.

In the fourth example, a red player catches the Golden Snitch. Their team wins 10 points, and the game ends.

You can find one more example in the easy version of the problem

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

#define ll long long
#define pll pair<long long, long long>
#define pb push_back
#define F first
#define S second
#define all(x) (x).begin(), (x).end()

const ll N = 122;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll block = 520;
const ll base = 35711;
ll n,m,q;
pll R[N], B[N];
ll RG[N][N], BG[N][N];
void to_nho_cau(){
    cin >> n >> m;
    ll ib,jb;
    for(int i = 1; i <= n;i++){
        for(int j = 1; j <= m;j++){
            char c1, c2; cin >> c1 >> c2;
            if(c1 == 'R' && c2 == 'G'){
                RG[i][j] = 1;
            }
            else if(c1 == 'B' && c2 == 'G'){
                BG[i][j] = 1;
            }
            else if(c1 == 'R'){
                R[c2 - '0'] = {i, j};
            }
            else if(c1 == 'B'){
                B[c2 - '0'] = {i, j};
            }
            else if(c1 == '.' && c2 == 'B'){
                ib = i, jb = j;
            }
        }
    }
    cin >> q;
    ll timer = 0, scoreR = 0, scoreB = 0;
    while(q--){
        string a; cin >> a;
        if(a[0] == 'R'){ // res
            ll idx = a[1] - '0', i = R[idx].F, j = R[idx].S;
            char b; cin >> b;
            if(b == 'R') R[idx] = {i, j + 1};
            else if(b == 'L') R[idx] = {i, j - 1};
            else if(b == 'U') R[idx] = {i - 1, j};
            else if(b == 'D') R[idx] = {i + 1, j};
            else if(b == 'C'){
                string dm; cin >> dm;
                if(dm[1] == 'S'){
                    cout << timer << ' ' << "RED CATCH GOLDEN SNITCH" << '\n';
                    scoreR += 10;
                }
            }
            else if(b == 'T'){
                if(BG[i][j]){
                    cout << timer << ' ' << "RED GOAL" << '\n';
                    scoreR++;
                }
            }
        }
        else if(a[0] == 'B'){ // blue
            ll idx = a[1] - '0', i = B[idx].F, j = B[idx].S;
            char b; cin >> b;
            if(b == 'R') B[idx] = {i, j + 1};
            else if(b == 'L') B[idx] = {i, j - 1};
            else if(b == 'U') B[idx] = {i - 1, j};
            else if(b == 'D') B[idx] = {i + 1, j};
            else if(b == 'C'){
                string dm; cin >> dm;
                if(dm[1] == 'S'){
                    cout << timer << ' ' << "BLUE CATCH GOLDEN SNITCH" << '\n';
                    scoreB += 10;
                }
            }
            else if(b == 'T'){
                if(RG[i][j]){
                    cout << timer << ' ' << "BLUE GOAL" << '\n';
                    scoreB++;
                }
            }
        }
        else if(a[1] == 'B'){ // quai vat
            char b; cin >> b;
            if(b == 'R') jb++;
            else if(b == 'L') jb--;
            else if(b == 'U') ib--;
            else if(b == 'D') ib++;
        }
        else if(a[1] == 'Q' || a[1] == 'S'){
            char b; cin >> b;
        }
        for(int i = 0; i <= 9;i++){
                if(B[i].F == ib && B[i].S == jb){
                    cout << timer << ' ' << 'B' << i << ' ' << "ELIMINATED" << '\n';
                    B[i] = {inf, inf};
                }
            }
            for(int i = 0; i <= 9;i++){
                if(R[i].F == ib && R[i].S == jb){
                    cout << timer << ' ' << 'R' << i << ' ' << "ELIMINATED" << '\n';
                    R[i] = {inf, inf};
                }
            }
        timer++;
    }
    cout << "FINAL SCORE: " << scoreR << ' ' << scoreB << '\n';
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    //cin >> t;
    while(t--) to_nho_cau();
}

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

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

Think step by step.

Response:
```plaintext
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
10
R0 D
R0 C .Q
R0 R
R0 T
R0 D
B0 R
B0 U
B0 C .Q
B0 L
B0 L
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Playing Quidditch (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This afternoon, you decided to enjoy the first days of Spring by taking a walk outside. As you come near the Quidditch field, you hear screams. Once again, there is a conflict about the score: the two teams are convinced that they won the game! To prevent this problem from happening one more time, you decide to get involved in the refereeing of the matches.

Now, you will stay in the stadium to watch the game and count the score. At the end of the game, you will decide the winner.

Today, two teams are competing: the red Gryffindor (R) and the blue Ravenclaw (B) team. Each team is composed of $$$P$$$ players ($$$1 \leq P \leq 10$$$).

The field is a rectangle of $$$N$$$ lines and $$$M$$$ columns ($$$3 \leq N, M \leq 99$$$, $$$N$$$ and $$$M$$$ are odd). All the positions are integers, and several entities are allowed to be at the same position in the field. At the beginning of the game, the field contains goals for the two teams (each team can own between one and five goals), the players, and exactly one Quaffle. In this version of the problem, one Bludger and a Golden Snitch can be present.

A game is composed of $$$T$$$ steps ($$$0 \leq T \leq 10000$$$). At each step, one entity on the field (a player or a ball) performs one action. All entities can move. A player can also catch a ball or throw the Quaffle that it is carrying. To catch a ball, a player must be located on the same cell as it. The Quaffle does not perform any action while it is being carried; it only follows the movements of the player. If a player carrying the Quaffle decides to throw it, the Quaffle is simply put at the current position of the player. If a player is on the same cell as a Bludger (either after a movement from the player or the Bludger), the player is eliminated. If the player is eliminated while it is carrying the Quaffle, the Quaffle remains on the cell containing both the player and the Bludger after the move. It is guaranteed that this never occurs while the player is in a cell containing a goal.

To win a point, a player must leave the Quaffle at a goal of the other team. When it does, the team of the player wins one point, and the Quaffle instantly moves to the middle of the field (the cell at the $$$(M+1)/2$$$-th column of the $$$(N+1)/2$$$-th line of the field, starting from 1). There is no goal in the middle of the field. If a player puts the ball in its own goal, the other team wins the point. If a player catches the Golden Snitch, their team wins 10 points and the game is over.

Input: On the first line, the integers $$$N$$$ and $$$M$$$.

The description of the field follows: $$$N$$$ lines of $$$M$$$ pairs of characters separated by spaces. Each pair of characters represents a position on the field. It can be either:

- .. to represent an empty cell
- R0, ..., R9, B0, ..., B9 to represent a player. The first character is the team of the player, and the second is the number of the player in the team. Each pair of characters is unique, but it is not guaranteed that all the pairs appear in the grid.
- RG or BG to represent a goal. The blue team tries to put the ball in a red goal (RG) while the red team tries to put the ball in a blue goal (BG).
- .Q to represent the Quaffle, which is the ball that the players use to score goals.
- .B to represent the Bludger.
- .S to represent the Golden Snitch.

The next line contains $$$T$$$, the number of steps that compose the game. $$$T$$$ lines follow, each describing one action. It contains several pieces of information separated by a space. First, a pair of characters representing the entity that must perform the action. Second, the description of the action:

- U, D, L, R indicate that the entity moves on the grid. It can move to the top of the grid (U), to the bottom (D), to the left (L), or to the right (R). Each entity moves by only one cell at a time.
- C indicates that the player catches the ball (only a player can catch a ball). Then, there is a space followed by a pair of characters: the description of the ball caught by the player. This information is needed since several balls can be in the same cell.
- T indicates that the player throws the Quaffle that it is carrying.

All the actions performed by the entities are guaranteed to be valid: the players stay in the field, don't catch a ball if they are not in the same cell, don't release the Quaffle if they are not carrying it, ...

Output: You must output the description of the main events of the game, one event per line. More precisely:

- Each time a team scores, you must print t RED GOAL or t BLUE GOAL, depending on the team who scored, where t is the current time (the position of the action in the list of actions, starting from 0). In the case where a player scores in the wrong goal (a red player scores in the red goal, or a blue player scores in the blue goal), you must print the name of the team who wins one point, that is, the other team.
- Each time a player is eliminated, you must print t p ELIMINATED, where t is the current time and p is the player who is eliminated. The format to print the player is the same as in the input.
- If the Golden Snitch is caught, you must print t RED CATCH GOLDEN SNITCH or t BLUE CATCH GOLDEN SNITCH, depending on the team who caught the Golden Snitch, where t is the current time.

The events must be printed in ascending order of t. If several players are eliminated at the same time, the events must be written is alphabetical order: B0, ..., B9, R0, ... R9.

At the end of the game, you must print the final score as: FINAL SCORE: r b, where r is the score of the red team and b is the score of the blue team.

Examples:
input:
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
12
R0 D
R0 C .Q
R0 R
R0 T
R0 D
B0 R
B0 U
B0 C .Q
B0 L
B0 L
B0 L
B0 T
output:
11 BLUE GOAL
FINAL SCORE: 0 1

input:
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
5
R0 D
R0 C .Q
R0 L
R0 L
R0 T
output:
4 BLUE GOAL
FINAL SCORE: 0 1

input:
5 5
.. .. .. .. ..
.. .. .. .. ..
RG R0 .Q B0 BG
.. .. .. .. ..
.. .. .B .. ..
5
.B L
.B U
.B U
B0 L
B0 L
output:
2 R0 ELIMINATED
4 B0 ELIMINATED
FINAL SCORE: 0 0

input:
5 5
.. R0 .S B0 ..
.. .. .. .. ..
RG .. .Q .. BG
.. .. .. .. ..
.. R1 .B B1 ..
4
.S D
R0 D
R0 R
R0 C .S
output:
3 RED CATCH GOLDEN SNITCH
FINAL SCORE: 10 0

Note: In the first example, the red player takes the Quaffle, move it and throw it. The blue player catches the ball, goes to the red goal and scores.

In the second example, the red player takes the ball and scores in the goal of their own team: the blue team wins a point.

In the third example, the Bludger goes at the position of R0: R0 is eliminated. Then, B0 moves to the position of the Bludger: B0 is eliminated too.

In the fourth example, a red player catches the Golden Snitch. Their team wins 10 points, and the game ends.

You can find one more example in the easy version of the problem

Buggy Code:
#include <bits/stdc++.h>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define f first
#define s second
using namespace std;
using ll = long long;
using pii = pair <int, int>;
const int N = 103, inf = 2e9;
int n, m, goal[1000];
string s[N][N];
set <string> st[N][N];
int main() {
	ios :: sync_with_stdio(false);
	cin.tie(nullptr);
	cin >> n >> m;
	map <string, pii> pos;
	map <string, char> bro;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			cin >> s[i][j];
			pos[s[i][j]] = {i, j};
			st[i][j].insert(s[i][j]);
		}
	}
	int T;
	cin >> T;
	for (int t = 1; t <= T; ++t) {
		string e, act;
		cin >> e >> act;
		st[pos[e].f][pos[e].s].erase(e);
		if (act == "U")
			pos[e].f--;
		else if (act == "D")
			++pos[e].f;
		else if (act == "L")
			--pos[e].s;
		else if (act == "R")
			++pos[e].s;
		else if (act == "C") {
			string lol;
			cin >> lol;
			if (lol == ".S") {
				goal[e[0]] += 10;
				if (e[0] == 'R')
					cout << t - 1 << " RED CATCH GOLDEN SNITCH\n";
				else
					cout << t - 1 << " BLUE CATCH GOLDEN SNITCH\n";
				break;
			}
			else
				bro[e] = e[0];
		}
		else
			bro[e] = 0;
		if (bro[e])
			pos[".Q"] = pos[e];
		st[pos[e].f][pos[e].s].insert(e);
		vector <string> del;
		for (auto x : st[pos[".B"].f][pos[".B"].s]) {
			if (x[0] != '.') {
				cout << t - 1 << " " << x << " ELIMINATED\n";
				del.pb(x);
			}
		}
		for (auto x : del)
			st[pos[x].f][pos[x].s].erase(x);
		bool okb = 0, okr = 0;
		for (auto x : st[pos[".Q"].f][pos[".Q"].s]) {
			if (x == "BG")
				okb = 1;
			if (x == "RG")
				okr = 1;
		}
		if (t == T)
			break;
		if (okb && e[0] == 'R') {
			goal['R']++;
			cout << t << " RED GOAL\n";
			st[pos[".Q"].f][pos[".Q"].s].erase(".Q");
			pos[".Q"] = {(n + 1) / 2, (m + 1) / 2};
			st[(n + 1) / 2][(m + 1) / 2].insert(".Q");
		}
		else if (okr && e[0] == 'B') {
			goal['B']++;
			cout << t << " BLUE GOAL\n";
			st[pos[".Q"].f][pos[".Q"].s].erase(".Q");
			pos[".Q"] = {(n + 1) / 2, (m + 1) / 2};
			st[(n + 1) / 2][(m + 1) / 2].insert(".Q");
		}
	}
	cout << "FINAL SCORE: " << goal['R'] << ' ' << goal['B'];
	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
5 5
.. .. .. .. ..
.. .. .. .. ..
RG R0 .Q B0 BG
.. .. .. .. ..
.. .B .. .. ..
5
.B L
.B U
.B U
B0 L
B0 L
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Playing Quidditch (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This afternoon, you decided to enjoy the first days of Spring by taking a walk outside. As you come near the Quidditch field, you hear screams. Once again, there is a conflict about the score: the two teams are convinced that they won the game! To prevent this problem from happening one more time, you decide to get involved in the refereeing of the matches.

Now, you will stay in the stadium to watch the game and count the score. At the end of the game, you will decide the winner.

Today, two teams are competing: the red Gryffindor (R) and the blue Ravenclaw (B) team. Each team is composed of $$$P$$$ players ($$$1 \leq P \leq 10$$$).

The field is a rectangle of $$$N$$$ lines and $$$M$$$ columns ($$$3 \leq N, M \leq 99$$$, $$$N$$$ and $$$M$$$ are odd). All the positions are integers, and several entities are allowed to be at the same position in the field. At the beginning of the game, the field contains goals for the two teams (each team can own between one and five goals), the players, and exactly one Quaffle. In this version of the problem, one Bludger and a Golden Snitch can be present.

A game is composed of $$$T$$$ steps ($$$0 \leq T \leq 10000$$$). At each step, one entity on the field (a player or a ball) performs one action. All entities can move. A player can also catch a ball or throw the Quaffle that it is carrying. To catch a ball, a player must be located on the same cell as it. The Quaffle does not perform any action while it is being carried; it only follows the movements of the player. If a player carrying the Quaffle decides to throw it, the Quaffle is simply put at the current position of the player. If a player is on the same cell as a Bludger (either after a movement from the player or the Bludger), the player is eliminated. If the player is eliminated while it is carrying the Quaffle, the Quaffle remains on the cell containing both the player and the Bludger after the move. It is guaranteed that this never occurs while the player is in a cell containing a goal.

To win a point, a player must leave the Quaffle at a goal of the other team. When it does, the team of the player wins one point, and the Quaffle instantly moves to the middle of the field (the cell at the $$$(M+1)/2$$$-th column of the $$$(N+1)/2$$$-th line of the field, starting from 1). There is no goal in the middle of the field. If a player puts the ball in its own goal, the other team wins the point. If a player catches the Golden Snitch, their team wins 10 points and the game is over.

Input: On the first line, the integers $$$N$$$ and $$$M$$$.

The description of the field follows: $$$N$$$ lines of $$$M$$$ pairs of characters separated by spaces. Each pair of characters represents a position on the field. It can be either:

- .. to represent an empty cell
- R0, ..., R9, B0, ..., B9 to represent a player. The first character is the team of the player, and the second is the number of the player in the team. Each pair of characters is unique, but it is not guaranteed that all the pairs appear in the grid.
- RG or BG to represent a goal. The blue team tries to put the ball in a red goal (RG) while the red team tries to put the ball in a blue goal (BG).
- .Q to represent the Quaffle, which is the ball that the players use to score goals.
- .B to represent the Bludger.
- .S to represent the Golden Snitch.

The next line contains $$$T$$$, the number of steps that compose the game. $$$T$$$ lines follow, each describing one action. It contains several pieces of information separated by a space. First, a pair of characters representing the entity that must perform the action. Second, the description of the action:

- U, D, L, R indicate that the entity moves on the grid. It can move to the top of the grid (U), to the bottom (D), to the left (L), or to the right (R). Each entity moves by only one cell at a time.
- C indicates that the player catches the ball (only a player can catch a ball). Then, there is a space followed by a pair of characters: the description of the ball caught by the player. This information is needed since several balls can be in the same cell.
- T indicates that the player throws the Quaffle that it is carrying.

All the actions performed by the entities are guaranteed to be valid: the players stay in the field, don't catch a ball if they are not in the same cell, don't release the Quaffle if they are not carrying it, ...

Output: You must output the description of the main events of the game, one event per line. More precisely:

- Each time a team scores, you must print t RED GOAL or t BLUE GOAL, depending on the team who scored, where t is the current time (the position of the action in the list of actions, starting from 0). In the case where a player scores in the wrong goal (a red player scores in the red goal, or a blue player scores in the blue goal), you must print the name of the team who wins one point, that is, the other team.
- Each time a player is eliminated, you must print t p ELIMINATED, where t is the current time and p is the player who is eliminated. The format to print the player is the same as in the input.
- If the Golden Snitch is caught, you must print t RED CATCH GOLDEN SNITCH or t BLUE CATCH GOLDEN SNITCH, depending on the team who caught the Golden Snitch, where t is the current time.

The events must be printed in ascending order of t. If several players are eliminated at the same time, the events must be written is alphabetical order: B0, ..., B9, R0, ... R9.

At the end of the game, you must print the final score as: FINAL SCORE: r b, where r is the score of the red team and b is the score of the blue team.

Examples:
input:
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
12
R0 D
R0 C .Q
R0 R
R0 T
R0 D
B0 R
B0 U
B0 C .Q
B0 L
B0 L
B0 L
B0 T
output:
11 BLUE GOAL
FINAL SCORE: 0 1

input:
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
5
R0 D
R0 C .Q
R0 L
R0 L
R0 T
output:
4 BLUE GOAL
FINAL SCORE: 0 1

input:
5 5
.. .. .. .. ..
.. .. .. .. ..
RG R0 .Q B0 BG
.. .. .. .. ..
.. .. .B .. ..
5
.B L
.B U
.B U
B0 L
B0 L
output:
2 R0 ELIMINATED
4 B0 ELIMINATED
FINAL SCORE: 0 0

input:
5 5
.. R0 .S B0 ..
.. .. .. .. ..
RG .. .Q .. BG
.. .. .. .. ..
.. R1 .B B1 ..
4
.S D
R0 D
R0 R
R0 C .S
output:
3 RED CATCH GOLDEN SNITCH
FINAL SCORE: 10 0

Note: In the first example, the red player takes the Quaffle, move it and throw it. The blue player catches the ball, goes to the red goal and scores.

In the second example, the red player takes the ball and scores in the goal of their own team: the blue team wins a point.

In the third example, the Bludger goes at the position of R0: R0 is eliminated. Then, B0 moves to the position of the Bludger: B0 is eliminated too.

In the fourth example, a red player catches the Golden Snitch. Their team wins 10 points, and the game ends.

You can find one more example in the easy version of the problem

Buggy Code:
def parse_input():
    import sys
    input = sys.stdin.read
    data = input().splitlines()

    N, M = map(int, data[0].split())
    field = [line.split() for line in data[1:N+1]]
    T = int(data[N+1])
    actions = data[N+2:N+2+T]

    return N, M, field, T, actions

def find_entities(field):
    entities = {}
    for i, row in enumerate(field):
        for j, cell in enumerate(row):
            if cell != '..':
                entities[cell] = (i, j)
    return entities

def simulate_game(N, M, field, T, actions):
    entities = find_entities(field)
    scores = {'R': 0, 'B': 0}
    carrying_quaffle = {player: False for player in entities if player[0] in 'RB'}
    events = []

    for t, action in enumerate(actions):
        parts = action.split()
        entity = parts[0]
        action_type = parts[1]

        if action_type in 'UDLR':
            # Move the entity
            i, j = entities[entity]
            if action_type == 'U':
                i -= 1
            elif action_type == 'D':
                i += 1
            elif action_type == 'L':
                j -= 1
            elif action_type == 'R':
                j += 1
            entities[entity] = (i, j)

            # Check for Bludger collision
            if entity[0] in 'RB' and '.B' in entities and entities[entity] == entities['.B']:
                events.append(f"{t} {entity} ELIMINATED")
                del entities[entity]
                if carrying_quaffle[entity]:
                    carrying_quaffle[entity] = False
                    entities['.Q'] = (i, j)

        elif action_type == 'C':
            # Catch the ball
            ball = parts[2]
            if ball == '.Q':
                carrying_quaffle[entity] = True
                del entities['.Q']

        elif action_type == 'T':
            # Throw the Quaffle
            if carrying_quaffle[entity]:
                i, j = entities[entity]
                entities['.Q'] = (i, j)
                carrying_quaffle[entity] = False

        # Check for scoring
        if '.Q' in entities:
            i, j = entities['.Q']
            for goal in entities:
                if goal in ['RG', 'BG'] and entities[goal] == (i, j):
                    if goal == 'RG':
                        if entity[0] == 'B':
                            events.append(f"{t} BLUE GOAL")
                            scores['B'] += 1
                        else:
                            events.append(f"{t} RED GOAL")
                            scores['R'] += 1
                    elif goal == 'BG':
                        if entity[0] == 'R':
                            events.append(f"{t} RED GOAL")
                            scores['R'] += 1
                        else:
                            events.append(f"{t} BLUE GOAL")
                            scores['B'] += 1
                    # Reset Quaffle to center
                    entities['.Q'] = ((N+1)//2 - 1, (M+1)//2 - 1)

        # Check for Golden Snitch
        if '.S' in entities and entity[0] in 'RB' and entities[entity] == entities['.S']:
            if entity[0] == 'R':
                events.append(f"{t} RED CATCH GOLDEN SNITCH")
                scores['R'] += 10
            else:
                events.append(f"{t} BLUE CATCH GOLDEN SNITCH")
                scores['B'] += 10
            break  # Game ends immediately

    return events, scores

def main():
    N, M, field, T, actions = parse_input()
    events, scores = simulate_game(N, M, field, T, actions)

    for event in events:
        print(event)
    print(f"FINAL SCORE: {scores['R']} {scores['B']}")

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3 5
.. .. R0 .. ..
RG .. .Q .. BG
.. .. B0 .. ..
4
R0 D
R0 C .Q
R0 L
R0 T
```