**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <cstdint>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <unordered_map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <unordered_set>
#include <vector>
#include <complex>
#include <stack>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9; x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

typedef long long ll;
#define int ll
typedef vector<int> vi;
typedef pair<int,int> pi;
#define pb push_back
#define all(v) v.begin(), v.end()
#define MAX 200000
#define MOD 1000000007
#define nl '\n'
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
const int inf = 1e9;

void takeinput(vector<int> &a, int start, int end){for(int i=start;i<=end;i++)cin>>a[i];}
void yes(){cout<<"YES"<<nl;}
void no(){cout<<"NO"<<nl;}
void testcase(int t){cout<<"Case #"<<t<<" : ";}

int exp(int ,int);
int ceildiv(int, int);
int modinv(int);

const int K = 21;
class Tree{
    private :
    vector<vector<int>> adj;
    int n;
    vector<int> sz;
    vector<int> subor;
    vector<map<int, int>> orCounts;

    public :
    Tree(int n) : n(n){
        adj.assign(n + 1, vector<int>());
        sz.assign(n + 1, 1); subor.assign(n + 1, 0);
        orCounts.assign(n + 1, map<int, int>());
    }

    void addEdge(int u, int v){
        adj[u].pb(v); adj[v].pb(u);
    }
    void mergeCounts(map<int, int> &target, map<int, int> &source){
        for(int bit = 0; bit <= K; bit++){
            target[bit] += source[bit];
        }
    }

    void dfs(int u, int p){
        for(auto ng : adj[u]){
            if(ng == p) continue;
            dfs(ng, u);
            sz[u] += sz[ng];
            subor[u] |= subor[ng];
            // mergeCounts(orCounts[u], orCounts[ng]);
        }
        subor[u] |= sz[u];
        // for(int bit = 0; bit <= K; bit++){
        //     if(sz[u] & (1ll << bit)){
        //         orCounts[u][bit]++;
        //     }
        // }
    }

    int getSubor(){
        return sz[1];
    }
};

void solve(int t){
    int k; cin>>k;
    vector<Tree> trees;
    for(int i = 0; i < k; i++){
        int n; cin>>n;
        Tree tree(n);
        for(int j = 2; j <= n; j++){
            int parent; cin>>parent; tree.addEdge(parent, j);
        }
        trees.pb(tree);
    }

    vector<int> s;
    for(auto tree : trees){
        tree.dfs(1, -1);
        s.push_back(tree.getSubor());
    }

    vector<multiset<pair<int, int>>> counts(K + 1);
    auto getMxBit = [](int x)->int{
        int bit = K;
        while((x & (1ll << bit)) == 0) bit--;
        return bit;
    };

    auto getAllSet = [&getMxBit](int x)->int{
        int bit = getMxBit(x);
        return (1ll << bit) - 1;
    };

    for(auto x : s){
        bool on = false;
        int mxBit = getMxBit(x);
        int sm = getAllSet(x);
        counts[mxBit].insert({x, 1});
        for(int bit = mxBit - 1; bit >= 0; bit--){
            counts[bit].insert({sm, 0});
        }
    }
    int ans = 0;
    for(int bit = K; bit >= 0; bit--){
        if(counts[bit].empty()) continue;
        auto [top, type] = *counts[bit].begin();
        ans |= top;

        counts[bit].erase(counts[bit].find({top, type}));
        if(type){
            top = getAllSet(top); type = 0;
        }
        for(int lowerbit = bit - 1; lowerbit >= 0; lowerbit--){
            counts[lowerbit].erase(counts[lowerbit].find({top, type}));
        }
    }
    cout<<ans<<nl;
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
    freopen("output.out","w",stdout);
    freopen("error.out","w",stderr);
    #endif
    // int t = 1;
    int t; cin>>t;
    for(int i=1;i<=t;i++){
        solve(i);
    }
}

int exp(int a, int b){
    int res = 1;
    while(b > 0){
        if( b % 2 == 1){
            res = (res * a) % MOD;
        }
        a = (a*a)%MOD;
        b /= 2;
    }
    return res;
}

int ceildiv(int a, int b){
    return (a + b - 1)/b;
}

int modinv(int a){
    return exp(a, MOD - 2);
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight issues in the algorithm. Here are the steps to create such a test input:

1. **Single Tree with a Single Vertex:**
   - This is a very simple case and should be handled correctly, but it's good to include it for completeness.

2. **Single Tree with a Chain Structure:**
   - A chain structure can expose issues in how the algorithm handles linear trees.

3. **Single Tree with a Balanced Binary Tree:**
   - This can expose issues in how the algorithm handles more complex tree structures.

4. **Multiple Trees with Different Sizes:**
   - This can expose issues in how the algorithm handles multiple trees and combines their results.

5. **Tree with a High Degree Root:**
   - A tree with a high-degree root can expose issues in how the algorithm handles large subtrees.

6. **Tree with Many Small Subtrees:**
   - This can expose issues in how the algorithm handles many small subtrees.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a combination of these scenarios:

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

### Explanation:
1. **Test Case 1:**
   - A single tree with a single vertex.
   - Expected output: 1

2. **Test Case 2:**
   - A single tree with two vertices forming a chain.
   - Expected output: 3 (1 | 2)

3. **Test Case 3:**
   - A single tree with four vertices forming a balanced binary tree.
   - Expected output: 7 (1 | 2 | 4)

4. **Test Case 4:**
   - A single tree with six vertices forming a more complex structure.
   - Expected output: 7 (1 | 2 | 4)

This test input should be sufficient to expose any issues in the buggy code, especially in how it handles different tree structures and combines their results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
# define ll long long
# define ld long double
# define pb push_back
# define pll pair<long long,long long>
# define vll vector<long long>
# define vld vector<long double>
# define vpll vector<pair<long long,long long>>
const int mod = 1e9+7;
const int N=1e6+10;

vll g[N];
vll sz;
vll sx;
vpll s;

void dfs(ll r,ll p,ll b){
    // ll x=0;
    for(auto i:g[r]){
        if(i==p) continue;
        dfs(i,r,b);
        sz[r]+=sz[i];
        sx[r]|=sx[i];
    }
    sz[r]++;
    if(sz[r]&(1<<b))
        s.pb({sx[r],r});
    sx[r]|=sz[r];
}

void solve(){
    ll n;cin>>n;
    ll ni[n];
    ll h[n];
    vll p;
    for(int i=0;i<n;i++){
        cin>>ni[i];
        if(!i)
            h[i]=0;
        else
            h[i]=h[i-1]+ni[i-1];
        p.pb(-1);
        for(int j=1;j<ni[i];j++){
            ll x;cin>>x;x--;
            p.pb(x+h[i]);
        }
    }
    ll tot=h[n-1]+ni[n-1];
    // cout<<tot<<"\n";
    ll ans=0;
    for(int b=19;b>=0;b--){
        // for(int j=)
        vll v;
        s.clear();
        for(int j=0;j<tot;j++){
            g[j].clear();
            // cout<<p[j]<<"\n";
            if(p[j]==-1)
                v.pb(j);
            else if(p[j]!=-2)
                g[p[j]].pb(j);
        }
        sz.assign(tot,0);
        sx.assign(tot,0);
        for(auto i:v)
            dfs(i,i,b);
        sort(s.begin(), s.end());
        // pll xx={1<<b,-1};
        // auto it=lower_bound(s.begin(), s.end(),xx)-s.begin();
        // for(auto it:sz)
        //     cout<<it<<"\n";
        if(s.size()){
            ans|=sz[s[0].second];
            p[s[0].second]=-2;
        }
    }
    cout<<ans<<"\n";
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout<<setprecision(30);
    int t=1;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
1
2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
t = int(input())
for i in range(t):
    k = int(input())
    sizes = []
    p = []
    for _ in range(k):
        n = int(input())
        sizes.append(n)
        p.append(list(map(int, input().split())))
    ans = 0
    while sizes:
        s = sizes.pop()
        if (1 << (s.bit_length()-1)) & ans != 0:
            ans |= (1 << s.bit_length()-1) - 1
        else:
            ans |= s
    print(ans)
    if t == 10000 and i == 25 and ans == 6:
        print("HOLA")
        print(k)
        print(p)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases that might cause the algorithm to produce incorrect results. The code seems to handle the bitwise OR operation in a way that might not always yield the maximum possible result. Specifically, the code might not correctly handle cases where the sizes of the subtrees have specific bit patterns that could lead to suboptimal results.

Let's break down the problem and think about a test case that could expose the bug:

1. **Small Test Case with Specific Bit Patterns:**
   - We need a test case with a small number of trees where the sizes of the subtrees have specific bit patterns that could cause the code to fail.
   - Consider a test case with two trees, where the sizes of the subtrees have bits that could cause the bitwise OR operation to be suboptimal.

2. **Example Test Case:**
   - Let's create a test case with two trees where the sizes are 4 and 7.
   - The binary representation of 4 is `100` and the binary representation of 7 is `111`.
   - The maximum possible bitwise OR result for these sizes is `111` (7 in decimal).

3. **Constructing the Input:**
   - The first tree has 4 nodes.
   - The second tree has 7 nodes.
   - The parent array for the first tree can be `1 1 1` (since it has 4 nodes).
   - The parent array for the second tree can be `1 1 1 1 1 1` (since it has 7 nodes).

Here is the test input:

```plaintext
1
2
4
1 1 1
7
1 1 1 1 1 1
```

This test case should expose the bug in the code if the bitwise OR operation is not handled correctly, leading to a suboptimal result. The expected output for this test case is `7`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
for i in range(int(input())):
    k = int(input())
    ans = 0
    for _ in range(k):
        n = int(input())
        l = list(map(int, input().split()))
        if (1 << (n.bit_length()-1)) & ans != 0:
            ans |= ((1 << (n.bit_length()-1)) - 1)
        else:
            ans |= n
    print(ans)

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

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

Think step by step.

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

**Explanation:**

1. **First Test Case:**
   - **Tree 1:**
     - 3 nodes: 1 (root), 2, 3
     - Edges: 2 -> 1, 3 -> 1
   - The size of the tree is 3. The maximum result is 3.

2. **Second Test Case:**
   - **Tree 1:**
     - 3 nodes: 1 (root), 2, 3
     - Edges: 2 -> 1, 3 -> 1
   - The size of the tree is 3. The maximum result is 3.

3. **Third Test Case:**
   - **Tree 1:**
     - 3 nodes: 1 (root), 2, 3
     - Edges: 2 -> 1, 3 -> 1
   - The size of the tree is 3. The maximum result is 3.

**Expected Output:**
```plaintext
3
3
3
```

**Rationale:**
- The buggy code incorrectly handles the bitwise OR operation, especially when the sizes of the trees are powers of 2 or close to powers of 2.
- By providing multiple trees of the same size (3 in this case), we can ensure that the bitwise OR operation is consistently applied, which can help expose the bug in the code.
- The test input is small-scale and complete, ensuring that the output is within the limit and valid for the problem constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using dl = long double;
using ll = long long;

void solve(){
	int k; cin >> k;
	vector<vector<int>> trs(k);
	vector<int> szs(k);
	for(int i=0; i<k ;++i){
		cin >> szs[i];
		trs[i] = vector<int>(szs[i], -1);
		for(int j=1; j<szs[i]; ++j){
			int val; cin >> val;
			trs[i][j] = val;
		}
	}

	sort(szs.rbegin(),szs.rend());
	int hbit = 1;
	while((1<<hbit) <= szs[0]){
		hbit++;
	}
	hbit--;
	int ans = szs[0];
	vector<bool> qset(hbit+1, 0);
	qset[hbit] = 1;
	int itr = 1;

	for(int i=hbit-1; i>=0; --i){
		if(itr >= k) break;
		else if((1<<(i+1))&szs[itr]){
			ans |= (1<<(i+1)) - 1;
			break;
		}
		else if((1<<i)&szs[itr]){
			ans |= szs[itr];
		}
		itr++;
	}

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

signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);cout.tie(nullptr);
	int tc; cin >> tc;
	while(tc--)
		solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
import java.util.*;

public class Main {
	private void solve() {
		int n = ni();
		int[] a = new int[n];
		for(int i = 0; i < n; i++) {
			a[i] = ni();
			for(int j = 1; j < a[i]; j++) {
				ni();
			}
		}
		List<Integer>[] f = new List[18];
		for(int i = 0; i < 18; i++) {
			f[i] = new ArrayList<>();
		}
		for(int i = 0; i < n; i++) {
			f[firstBit(a[i])].add(a[i]);
		}
		int res = 0;
		for(int i = 17; i >= 0; i--) {
			if(f[i].size() >= 2) {
				res |= (2 << i) - 1;
				break;
			}
			if(f[i].size() == 1) {
				if((res >> i & 1) == 1) {
					res |= (1 << i) - 1;
					break;
				}
				res |= 1 << i;
				int x = f[i].get(0) ^ (1 << i);
				if(x > 0) {
					f[firstBit(x)].add(x);
				}
			}
		}
		pr(res);
	}

	static Scanner sc = new Scanner(System.in);
	public static void main(String[] args) {
        var pg = new Main();
        int t = ni();
        while(t-- > 0) pg.solve();
    }

    public class DSU {
    	int[] p;
    	int[] size;
    	DSU(int n) {
    		p = new int[n];
    		size = new int[n];
    		for(int i = 0; i < n; i++) {
    			p[i] = i;
    			size[i] = 1;
    		}
    	}

    	int find(int x) {
    		return p[x] == x ? x : (p[x] = find(p[x]));
    	}

    	boolean same(int x, int y) {
    		return find(x) == find(y);
    	}

    	boolean merge(int x, int y) {
    		x = find(x);
    		y = find(y);
    		if(x == y) {
    			return false;
    		}
    		size[x] += size[y];
    		p[y] = x;
    		return true;
    	}

    	int size(int x) {
    		return size[find(x)];
    	}
    }

    public class Fenwick {
        public long[] count;
        public Fenwick(int n) {
            count = new long[n + 1];
        }
        public void add(int x, long val) {
            x += 1;
            while(x < count.length) {
                count[x] += val;
                x += x & -x;
            }
        }
        public long sum(int x) {
            long res = 0;
            while(x != 0) {
                res += count[x];
                x -= x & -x;
            }
            return res;
        }

        public long rangeSum(int i, int j) {
            return sum(j) - sum(i);
        }
    }

    private static long nl() {
        return sc.nextLong();
    }

    private static int ni() {
        return sc.nextInt();
    }

    private static String ns() {
        return sc.next();
    }

    private static void pr(Object obj) {
        System.out.println(obj);
    }

    private static void prn(Object obj) {
        System.out.print(obj);
    }

    private static void prf(String format, Object... objs) {
        System.out.printf(format, objs);
    }

    private static int firstBit(int n) {
    	return Integer.numberOfTrailingZeros(Integer.highestOneBit(n));
    }

    private long cases(long n) {
    	return n * (n + 1) / 2; // 1 ~ n的全部可能性
    }

    private long sum(long a, long b) {
        if(a > b) return 0;
        return (a + b) * (b - a + 1) / 2;
    }

    private int sum(int a, int b) {
        if(a > b) return 0;
        return (a + b) * (b - a + 1) / 2;
    }

    private int max(Integer... nums) {
        int res = Integer.MIN_VALUE;
        for(var x: nums) {
            if(res < x) res = x;
        }
        return res;
    }

    private int min(Integer... nums) {
        int res = Integer.MAX_VALUE;
        for(var x: nums) {
            if(res > x) res = x;
        }
        return res;
    }

    private long min(Long... nums) {
        long res = Long.MAX_VALUE;
        for(var x: nums) {
            if(res > x) res = x;
        }
        return res;
    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on scenarios that challenge the logic of the algorithm. The key issues in the code are:

1. **Incorrect handling of tree structures and their sizes**.
2. **Incorrect bitwise operations and the way the result is calculated**.

Given these points, we can construct a test case that:

1. **Has multiple trees with different structures**.
2. **Includes subtrees with sizes that have significant bitwise OR values**.

Let's construct a small-scale test input that meets these criteria:

### Test Case
- **Number of test cases**: 1
- **Number of trees**: 2
- **Tree 1**: 4 vertices
  - Root: 1
  - Parent of 2: 1
  - Parent of 3: 2
  - Parent of 4: 2
- **Tree 2**: 6 vertices
  - Root: 1
  - Parent of 2: 1
  - Parent of 3: 1
  - Parent of 4: 3
  - Parent of 5: 1
  - Parent of 6: 3

### Expected Output
- The maximum bitwise OR of the sizes of the subtrees removed should be 7.

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

### Explanation
- **Tree 1**:
  - Subtree of vertex 1: {1, 2, 3, 4} (size 4)
  - Subtree of vertex 2: {2, 3, 4} (size 3)
  - Subtree of vertex 3: {3} (size 1)
  - Subtree of vertex 4: {4} (size 1)

- **Tree 2**:
  - Subtree of vertex 1: {1, 2, 3, 4, 5, 6} (size 6)
  - Subtree of vertex 2: {2} (size 1)
  - Subtree of vertex 3: {3, 4, 6} (size 3)
  - Subtree of vertex 4: {4} (size 1)
  - Subtree of vertex 5: {5} (size 1)
  - Subtree of vertex 6: {6} (size 1)

By removing the entire tree 2 (size 6) and the subtree of vertex 2 in tree 1 (size 3), the bitwise OR of the sizes is \(6 | 3 = 7\).

This test case should expose the bug in the provided code by challenging its handling of tree structures and bitwise operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
void fast()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
#define int long long int
#define mod 1000000007
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define f(i, s, n) for (i = s; i < n; i++)
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define vi vector<ll>
#define vb vector<bool>
#define pii pair<ll, ll>
#define vpi vector<pii>
#define umi unordered_map<ll, ll>
#define mi map<ll, ll>
#define si set<ll>

void init_code()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif // ONLINE_JUDGE
}

void solve()
{
    int i, j, sum = 0;
    int numtrees = 0;
    cin >> numtrees;
    vector<int> fin;
    for (i = 0; i < numtrees; i++)
    {
        int x;
        cin >> x;
        for (j = 0; j < x - 1; j++)
        {
            int y;
            cin >> y;
        }
        fin.pb(x);
    }
    sort(all(fin));
    reverse(all(fin));
    int ans = 0;
    for (i = 0; i < fin.size(); i++)
    {
        for (j = 20; j >= 0; j--)
        {
            if (fin[i] & (1LL << j))
            {
                if (!(ans & (1LL << j)))
                {
                    ans |= (1LL << j);
                    fin[i]-=(1LL<<j);
                    fin.pb(fin[i]);
                    break;
                }
                else
                {
                    j--;
                    while (j >= 0)
                    {
                        ans |= (1LL << j);
                        j--;
                    }
                    cout << ans << endl;
                    return;
                }
            }
        }
    }
    cout << ans << endl;
    return;
}

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

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

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

Think step by step.

Response:
```plaintext
3
1
1
2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
```

This test input is designed to cover a few key scenarios:
1. A single tree with a single node.
2. A forest with two trees, where the second tree has a structure that can be removed in a way that maximizes the bitwise OR.
3. A forest with a single tree that has a more complex structure, which should be removed entirely to achieve the maximum bitwise OR.

The first and third test cases are straightforward, ensuring that the algorithm handles single-node and large trees correctly. The second test case is more complex, with a structure that requires careful handling to maximize the bitwise OR, which can help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a);i <= (b);++i)
#define per(i, a, b) for(int i = (a);i >= (b);--i)
#define lc (k << 1)
#define rc (k << 1 | 1)
#define lowbit(x) ((x) & -(x))
#define odd(x) ((x) & 1)
#define even(x) (!odd(x))
#define fir first
#define sec second
#define pb push_back
#define il inline
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define i128 __int128
#define f128 __float128
#define pii pair<int, int>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define SZ(x) ((int)(x).size())
#define debug(x) cerr << "In Line " << __LINE__ << " the " << #x << " = " << (x) << "\n"
using namespace std;
using ll = long long;
using ull = unsigned long long;
template<class T> using vc = vector<T>;
template<class Tx, class Ty>
il void chkmx(Tx& x, const Ty y) {x = max<common_type_t<Tx, Ty>>(x, y);}
template<class Tx, class Ty>
il void chkmn(Tx& x, const Ty y) {x = min<common_type_t<Tx, Ty>>(x, y);}
const int inf = 0x3f3f3f3f;
const ll llinf = 0x3f3f3f3f3f3f3f3fll;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
#define yes cout << "YES\n"
#define no cout << "No\n"
bool ms;

const int N = 1e6 + 5;
int head[N], cnt;
int sz[N];

struct node {
    int to, nxt;
} e[N];

void init(int n) {
    cnt = 0;
    memset(head, -1, sizeof(head[0]) * (n + 1));
}

void add(int u, int v) {
    e[cnt] = {v, head[u]};
    head[u] = cnt++;
}

void dfs(int u) {
    sz[u] = 1;
    for(int i = head[u];~i;i = e[i].nxt) {
        int v = e[i].to;
        dfs(v);
        sz[u] += sz[v];
    }
}

void solve() {
    int k;
    cin >> k;
    vc<vc<int>> vec(k + 1);
    rep(i, 1, k) {
        int n;
        cin >> n;
        vec[i].resize(n + 1);
        rep(j, 2, n)
            cin >> vec[i][j];
    }
    sort(vec.begin() + 1, vec.end(), [&](const vc<int>& a, const vc<int>& b) {
        return SZ(a) > SZ(b);
    });
    int ans = 0;
    rep(i, 1, k) {
        int n = SZ(vec[i]) - 1;
        init(n);
        rep(j, 2, n) {
            int f = vec[i][j];
            add(f, j);
        }
        int res = 0;
        dfs(1);
        rep(j, 1, n)
            chkmx(res, ans | sz[j] | (n - sz[j]));
        ans = res;
    }
    cout << ans << "\n";
}

bool mt;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cerr << fabs(&ms - &mt) / 1024 / 1024 << "\n";
    int t;
    cin >> t;
    while(t--)
        solve();
    cerr << 1e3 * clock() / CLOCKS_PER_SEC << " ms\n";
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
import sys, time, random
from collections import deque, Counter, defaultdict
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
inf = 2 ** 61 - 1
mod = 998244353

class dsu():
    n=1
    parent_or_size=[-1 for i in range(n)]
    def __init__(self,N):
        self.n=N
        self.num = N
        self.parent_or_size=[-1 for i in range(N)]
    def merge(self,a,b):
        assert 0<=a<self.n, "0<=a<n,a={0},n={1}".format(a,self.n)
        assert 0<=b<self.n, "0<=b<n,b={0},n={1}".format(b,self.n)
        x=self.leader(a)
        y=self.leader(b)
        if x==y:
            return x
        self.num -= 1
        if (-self.parent_or_size[x]<-self.parent_or_size[y]):
            x,y=y,x
        self.parent_or_size[x]+=self.parent_or_size[y]
        self.parent_or_size[y]=x
        return x
    def same(self,a,b):
        assert 0<=a<self.n, "0<=a<n,a={0},n={1}".format(a,self.n)
        assert 0<=b<self.n, "0<=b<n,b={0},n={1}".format(b,self.n)
        return self.leader(a)==self.leader(b)
    def leader(self,a):
        assert 0<=a<self.n, "0<=a<n,a={0},n={1}".format(a,self.n)
        if (self.parent_or_size[a]<0):
            return a
        self.parent_or_size[a]=self.leader(self.parent_or_size[a])
        return self.parent_or_size[a]
    def size(self,a):
        assert 0<=a<self.n, "0<=a<n,a={0},n={1}".format(a,self.n)
        return -self.parent_or_size[self.leader(a)]
    def groups(self):
        leader_buf=[0 for i in range(self.n)]
        group_size=[0 for i in range(self.n)]
        for i in range(self.n):
            leader_buf[i]=self.leader(i)
            group_size[leader_buf[i]]+=1
        result=[[] for i in range(self.n)]
        for i in range(self.n):
            result[leader_buf[i]].append(i)
        result2=[]
        for i in range(self.n):
            if len(result[i])>0:
                result2.append(result[i])
        return result2
    def count(self):
        return self.num

def solve():
    n = ii()
    k = []
    for _ in range(n):
        k.append(ii())
        p = li()

    k.sort(reverse = True)

    ans = k[0]

    for i in range(1, n):
        for bit in range(20, -1, -1):
            if k[i] & (1 << bit):
                if not ans & (1 << bit):
                    ans |= k[i]
                    k[i] = 0
                    break
                else:
                    ans |= (1 << bit) - 1
                    k[i] = 0
                    break
    print(ans)

for _ in range(ii()):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
1
2
1
2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

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

MOD = 998244353

ivar = lambda: int(input())
ivars = lambda: map(int, input().split())
ilist = lambda: list(map(int, input().split()))
stri = lambda: input().strip()

def solve():
    k = ivar()
    A = []
    for _ in range(k):
        n = ivar()
        A.append(n)
        ilist()

    bits = [False  for _ in range(21)]
    for x in reversed(sorted(A)):
        x = bin(x)[2:]
        if bits[len(x)-1]:
            for i in range(0, len(x)):
                bits[i] = True
        else:
            for i, c in enumerate(reversed(x)):
                if c == '1':
                    bits[i] = True

    ans = ''
    for x in reversed(bits):
        if x:
            ans += '1'
        else:
            ans += '0'

    return int(ans, 2)

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

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

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

public final class temp {
    // public static ArrayList<ArrayList<Integer>> arr = new
    // ArrayList<ArrayList<Integer>>();
    // public static ArrayList<ArrayList<Integer>> sol = new
    // ArrayList<ArrayList<Integer>>();

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nli() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

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

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

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

        public void c() throws IOException {
            bw.close();
        }
    }

   static List<Integer> lit;

    public static void main(String[] args) {

        try {

            FastReader fr = new FastReader();
            // Scanner fr = new Scanner(System.in);
            FastWriter fw = new FastWriter();
            Scanner sc = new Scanner(System.in);
            int t = sc.nextInt();

            while (t > 0) {
                int n;
                n=sc.nextInt();
                lit = new ArrayList<>();
                for(int i=0; i<n; i++) {
                    lit.add(sc.nextInt());
                    for(int j=1;j<lit.get(i);j++) sc.nextInt();
                }
                Collections.sort(lit, Collections.reverseOrder());

                int ans = 0;
                ans = ans | lit.get(0);

                for(int i = 1 ; i<lit.size(); ++i){

                    int temp  = ans;

                    int count = 0;
                    int val = lit.get(i);
                    while(val > 0){
                        val = val >> 1;
                        count++;
                    }
                    // val = 1 << (count - 1);
                    // val  = val -1;
                    // ans = ans | val;

                    temp = temp >> (count-1);
                    if((temp & 1) == 1){
                        val = 1 << (count-1);
                        val = val-1;
                        ans = ans | val;
                    }else{
                        ans = ans | lit.get(i);
                    }

                }
                if(ans == 61) System.out.println(lit.toString());
                System.out.println(ans);

                t--;
            }
            fw.c();
        } catch (Exception e) {
            System.out.println(e.toString());
        }
    }
}

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

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

Think step by step.

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

1. **Understand the Buggy Code**:
   - The code reads the number of trees and the structure of each tree.
   - It sorts the sizes of the subtrees in descending order.
   - It calculates the bitwise OR of the sizes of the subtrees to find the maximum result.
   - There is a potential issue in the bitwise OR calculation logic, especially in the loop where it processes the sorted list.

2. **Identify Edge Cases**:
   - **Single Tree with Multiple Nodes**: This can test the basic functionality of the code.
   - **Multiple Trees with Different Sizes**: This can test how the code handles multiple trees.
   - **Trees with Large Subtrees**: This can test the bitwise OR logic.
   - **Trees with Small Subtrees**: This can test the edge cases where the subtrees are very small.

3. **Create a Small-Scale Test Input**:
   - We will create a test case with a single tree that has a structure that can expose the bug.
   - The tree will have a root and multiple subtrees with sizes that can cause issues in the bitwise OR calculation.

Here is the test input:

```plaintext
1
4
7
1 1 3 3 3 3
```

**Explanation**:
- **Test Case 1**:
  - **Number of Trees**: 1
  - **Tree 1**:
    - **Size**: 7
    - **Parent Array**: [1, 1, 3, 3, 3, 3]
    - **Structure**:
      ```
        1
       / \
      2   3
          / \
         4   5
        / \
       6   7
      ```
    - **Subtree Sizes**:
      - Subtree of 1: 7
      - Subtree of 2: 1
      - Subtree of 3: 4
      - Subtree of 4: 2
      - Subtree of 5: 1
      - Subtree of 6: 1
      - Subtree of 7: 1

**Expected Output**:
- The maximum result should be the bitwise OR of the sizes of the subtrees: 7 | 4 | 2 | 1 = 7

This test input should expose any issues in the bitwise OR calculation logic and help identify the bug 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: Wooden Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a forest of $k$ rooted trees$^{\text{∗}}$. Lumberjack Timofey wants to cut down the entire forest by applying the following operation:

  * Select a subtree$^{\text{†}}$ of any vertex of one of the trees and remove it from the tree.

Timofey loves bitwise operations, so he wants the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of the sizes of the subtrees he removed to be maximum. Help him and find the maximum result he can obtain.

$^{\text{∗}}$ A tree is a connected graph without cycles, loops, or multiple edges. In a rooted tree, a selected vertex is called a root. A forest is a collection of one or more trees.

$^{\text{†}}$ The subtree of a vertex $v$ is the set of vertices for which $v$ lies on the shortest path from this vertex to the root, including $v$ itself.

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

The first line of each test case contains a single integer $k$ ($1 \leq k \leq 10^6$) — the number of trees in the forest.

This is followed by a description of each of the $k$ trees:

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$) — the size of the tree. The vertices of the tree are numbered with integers from $1$ to $n$. The root of the tree is vertex number $1$.

The second line contains $n - 1$ integers $p_2, p_3, \ldots p_n$ ($1 \leq p_i < i$), where $p_i$ — the parent of vertex $i$.

It is guaranteed that the sum of $k$ and $n$ for all sets of input data does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum result that can be obtained.

Examples:
input:
3
1
1

2
4
1 2 2
6
1 1 3 1 3
1
10
1 2 2 1 1 5 7 6 4
output:
1
7
10

Note: In the second test case, the trees look like this:

![](CDN_BASE_URL/149fbfb0b4a0bcf29a8a6b8b997b79b2)

The first operation removes the entire second tree.

![](CDN_BASE_URL/3a1651525e7c5087a7d88d51824c16d6)

The second operation removes vertex $4$ from the first tree.

![](CDN_BASE_URL/c5e219384aa686ba75274781503be59f)

The third operation removes the first tree. The result is $6|1|3 = 7$ ($|$ denotes bitwise OR).

In the third test case, the entire tree needs to be removed.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {

		int t = sc.nextint();
		while (t-- > 0) {

			sovle1();
		}

		out.close();

	}

	static void sovle1 () {
		k=sc.nextint();
		Queue<Integer>queue=new PriorityQueue<Integer>(Comparator.reverseOrder());
		for (int i = 0; i < k; i++) {
			int n=sc.nextint();
			queue.add(n);
			for (int j = 1; j < n; j++) {
				sc.nextint();
			}
		}
		int ans=0;
		while (!queue.isEmpty()) {

			int chu=queue.peek()|ans;

			int dang=queue.peek();

			for (int i = 0; i < 30; i++) {
				if (((1<<i)|dang)!=0) {
					chu=Math.max(chu, ans|(dang-1));
					dang-=(1<<i);
				}
			}
			ans=chu;queue.poll();
		}
		out.println(ans);
	}

	static void sovle() {
		n=sc.nextint();
		arr=new int[n+1];
		for (int i = 1; i < arr.length; i++) {
			arr[i]=sc.nextint();
		}
		ls = new int[n + 1];Arrays.fill(ls, 0);
		rs = new int[n + 1];Arrays.fill(rs, n+1);
		Stack<Integer>stack=new Stack<Integer>();
		for (int i = 1; i < arr.length; i++) {
			while (!stack.isEmpty()&&arr[stack.peek()]>arr[i]) {
				stack.pop();
			}
			if (!stack.isEmpty()) {
				ls[i]=stack.peek();
			}
			stack.add(i);
		}
		stack.clear();
		for (int i = n; i >0; i--) {
			while (!stack.isEmpty()&&arr[stack.peek()]>arr[i]) {
				stack.pop();
			}
			if (!stack.isEmpty()) {
				rs[i]=stack.peek();
			}
			stack.add(i);
		}
		System.out.println(Arrays.toString(ls));
		System.out.println(Arrays.toString(rs));
		long sum=0;
		int [][]va=new int[n+1][2];
		int[]fu=new int[n+1];
		int gen=0;
		for (int i = 1; i < arr.length; i++) {
			va[i][0]=i-ls[i];
			va[i][1]=rs[i]-i;
			if (ls[i]!=0&&rs[i]!=n+1) {
				if (arr[ls[i]]<arr[rs[i]]) {
					fu[i]=rs[i];
				}else {
					fu[i]=ls[i];
				}
			}else if (ls[i]!=0) {
				fu[i]=ls[i];
			}else if (rs[i]!=n+1) {
				fu[i]=rs[i];
			}
			sum+=1l*va[i][0]*va[i][1]*arr[i];
			if (arr[i]==1) {
				gen=i;
			}
		}
		map=new HashMap<Integer, List<Integer>>();
		for (int i = 0; i < fu.length; i++) {
			map.put(i, new ArrayList<Integer>());
		}
		System.out.println(Arrays.toString(fu));
		//System.out.println(Arrays.toString(ls));
		//System.out.println(Arrays.toString(rs));
		for (int i = 1; i < fu.length; i++) {

			map.get(fu[i]).add(i);
		}
		ans=new long[n+1];
		ans2=new long[n+1];
		dfs(gen, 0, va);
		System.out.println(Arrays.toString(ans)+" "+sum);
		System.out.println(Arrays.toString(ans2));
		for (int i = 1; i < ans.length; i++) {
			out.print((sum-ans[i]+ans2[i])+" ");
		}
		out.println();

		System.out.println(map);
	}
	static int []ls,rs;
	static void dfs(int x,int fu,int[][]va) {
		//System.out.println(x+" "+Arrays.toString(va[x]));
		if (fu!=0) {
			if (x<fu) {
				ans[x]+=1l*arr[fu]*va[fu][1];
			}else {
				ans[x]+=1l*arr[fu]*va[fu][1];
			}
		}
		ans[x]+=ans[fu];
		int shang=-1;
		for (int a : map.get(x)) {
			dfs(a, x, va);
			ans2[x]+=ans2[a];
			if (shang!=-1) {
				if (shang<a) {
					if (arr[shang]>arr[a]) {
						ans2[x]+=1l*arr[a]*va[a][1]*(va[shang][0]+va[shang][1]-1);
						//System.out.println(arr[a]);
					}else {
						ans2[x]+=1l*arr[shang]*va[shang][0]*(va[a][0]+va[a][1]-1);
					}
				}else {
					if (arr[shang]>arr[a]) {
						ans2[x]+=1l*arr[a]*va[a][0]*(va[shang][0]+va[shang][1]-1);
					}else {
						ans2[x]+=1l*arr[shang]*va[shang][1]*(va[a][0]+va[a][1]-1);
					}
				}
			}
			shang=a;
		}
		ans[x]+=1l*va[x][0]*va[x][1]*arr[x];
	}

	static int n,m;
	static long [][]dp,min;
	static long k;
	static long[]ans,ans2;
	static int[] v,arr;
	static void sovle2() {

	}

	static Map<Integer, Integer> lMap, rMap;
	static Map<Integer, TreeSet<Integer>> vMap;

	static int[] dx = { 0, 0, 1, -1 };
	static int[] dy = { 1, -1, 0, 0 };
	static Map<Integer, List<Integer>> lpMap;

	static long pow(long x, long a) {
		if (a == 0) {
			return 1;
		} else {
			long c = pow(x, a / 2);
			if (a % 2 == 0) {
				return c * c % mod;
			} else {
				return c * c % mod * x % mod;
			}
		}
	}

	static long mod = (long) (1e9+7);

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

	static long[] c;

	static int lowbit(int x) {
		return x & -x;
	}

	static void add(int x, int v) {
		while (x < c.length) {
			c[x] += v;

			x += lowbit(x);
		}
	}

	static long sum(int x) {
		long ans = 0;
		while (x > 0) {
			// System.out.println(x);
			ans += c[x];
			x -= lowbit(x);
		}
		return ans;
	}

	static Map<Integer, List<Integer>> map;

	static void mapinit(int chu, int n) {
		map = new HashMap<Integer, List<Integer>>();
		for (int i = chu; i < n + chu; i++) {
			map.put(i, new ArrayList<Integer>());
		}
	}
}

class kattio extends PrintWriter {

	BufferedReader r;
	StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String out) throws IOException {
		super(out);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

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

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

	public double nextdouble() {
		return Double.parseDouble(next());
	}
}

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

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

Think step by step.

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

1. **Understand the Buggy Code**:
   - The code reads the input and processes each tree.
   - It uses a priority queue to keep track of the sizes of the subtrees.
   - The code then tries to maximize the bitwise OR of the sizes of the subtrees.

2. **Identify Potential Bugs**:
   - The code might not correctly handle the bitwise OR operation, especially when dealing with large numbers.
   - The code might have issues with the priority queue, such as incorrect handling of subtree sizes.
   - The code might not correctly process the input for trees with complex structures.

3. **Create a Test Input**:
   - We will create a test case with a small number of trees and a small number of vertices in each tree.
   - The test case should include a tree with a structure that might expose the bug, such as a tree with a deep hierarchy or a tree with a large number of leaves.

4. **Example Test Input**:
   - We will create a test case with 2 trees.
   - The first tree will have a simple structure with 3 vertices.
   - The second tree will have a more complex structure with 5 vertices.

Here is the test input:

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

Explanation:
- The first line `2` indicates that there are 2 test cases.
- The first test case:
  - `1` indicates that there is 1 tree.
  - `3` indicates that the tree has 3 vertices.
  - `1 1` indicates that the parent of vertex 2 is 1, and the parent of vertex 3 is 1.
- The second test case:
  - `1` indicates that there is 1 tree.
  - `5` indicates that the tree has 5 vertices.
  - `1 1 3 3` indicates that the parent of vertex 2 is 1, the parent of vertex 3 is 1, the parent of vertex 4 is 3, and the parent of vertex 5 is 3.

This test input is small and complete, and it includes a tree with a simple structure and a tree with a more complex structure, which should help in exposing any bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree of Life
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the heart of an ancient kingdom grows the legendary Tree of Life — the only one of its kind and the source of magical power for the entire world. The tree consists of $n$ nodes. Each node of this tree is a magical source, connected to other such sources through magical channels (edges). In total, there are $n-1$ channels in the tree, with the $i$-th channel connecting nodes $v_i$ and $u_i$. Moreover, there exists a unique simple path through the channels between any two nodes in the tree.

However, the magical energy flowing through these channels must be balanced; otherwise, the power of the Tree of Life may disrupt the natural order and cause catastrophic consequences. The sages of the kingdom discovered that when two magical channels converge at a single node, a dangerous "magical resonance vibration" occurs between them. To protect the Tree of Life and maintain its balance, it is necessary to select several paths and perform special rituals along them. A path is a sequence of distinct nodes $v_1, v_2, \ldots, v_k$, where each pair of adjacent nodes $v_i$ and $v_{i+1}$ is connected by a channel. When the sages perform a ritual along such a path, the resonance vibration between the channels $(v_i, v_{i+1})$ and $(v_{i+1}, v_{i+2})$ is blocked for each $1 \leq i \leq k - 2$.

The sages' task is to select the minimum number of paths and perform rituals along them to block all resonance vibrations. This means that for every pair of channels emanating from a single node, there must exist at least one selected path that contains both of these channels.

Help the sages find the minimum number of such paths so that the magical balance of the Tree of Life is preserved, and its power continues to nourish the entire world!

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the number of nodes in the Tree of Life.

The $i$-th of the following $n - 1$ lines of each test case contains two integers $v_i$ and $u_i$ ($1 \leq v_i < u_i \leq n$) — the channel connecting nodes $v_i$ and $u_i$.

It is guaranteed that there exists a unique simple path through the channels between any two nodes.

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

Output: For each test case, output a single integer — the minimum number of paths that the sages need to select to prevent a catastrophe.

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

Note: In the first test case, there are two pairs of channels emanating from a single node: $(1, 2)$ and $(2, 3)$, $(2, 3)$ and $(3, 4)$. It is sufficient to perform the ritual along the path $1-2-3-4$. Thus, the answer is $1$.

In the second test case, there are no pairs of channels emanating from a single node, so the answer is $0$.

In the third test case, rituals can be performed along the paths $2-1-3$, $2-1-4$, and $3-1-4$.

Buggy Code:
#include <bits/stdc++.h>
int T,N,hd[500009],to[1000009],nxt[1000009],k,dp[500009];
long long ans;
void l(int u,int v) {
	to[++k]=v;nxt[k]=hd[u];hd[u]=k;
}
void dfs(int n,int f) {
	int dg=0;
	for(int i=hd[n];i;i=nxt[i]) {
		if(to[i]!=f) {
			dg++;
			dfs(to[i],n);
		}
	}
	dp[n]=0;
	if(!dg) return;
	if(n>1) dp[n]=dg;
	ans=ans+1ll*dg*(dg-1)/2;
	for(int i=hd[n];i;i=nxt[i]) {
		if(to[i]!=f) {
			dp[n]+=std::max(dp[to[i]]-dg+1,0);
		}
	}
}
signed main(void) {
	scanf("%d",&T);
	while(T--) {
		scanf("%d",&N);ans=0;
		for(int i=1;i<=N;i++) hd[i]=0;k=0;
		for(int i=1;i<N;i++) {
			int u,v;
			scanf("%d %d",&u,&v);
			l(u,v),l(v,u);
		}
		dfs(1,0);
		printf("%lld\n",ans+(dp[1]+1)/2);
	}
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree of Life
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the heart of an ancient kingdom grows the legendary Tree of Life — the only one of its kind and the source of magical power for the entire world. The tree consists of $n$ nodes. Each node of this tree is a magical source, connected to other such sources through magical channels (edges). In total, there are $n-1$ channels in the tree, with the $i$-th channel connecting nodes $v_i$ and $u_i$. Moreover, there exists a unique simple path through the channels between any two nodes in the tree.

However, the magical energy flowing through these channels must be balanced; otherwise, the power of the Tree of Life may disrupt the natural order and cause catastrophic consequences. The sages of the kingdom discovered that when two magical channels converge at a single node, a dangerous "magical resonance vibration" occurs between them. To protect the Tree of Life and maintain its balance, it is necessary to select several paths and perform special rituals along them. A path is a sequence of distinct nodes $v_1, v_2, \ldots, v_k$, where each pair of adjacent nodes $v_i$ and $v_{i+1}$ is connected by a channel. When the sages perform a ritual along such a path, the resonance vibration between the channels $(v_i, v_{i+1})$ and $(v_{i+1}, v_{i+2})$ is blocked for each $1 \leq i \leq k - 2$.

The sages' task is to select the minimum number of paths and perform rituals along them to block all resonance vibrations. This means that for every pair of channels emanating from a single node, there must exist at least one selected path that contains both of these channels.

Help the sages find the minimum number of such paths so that the magical balance of the Tree of Life is preserved, and its power continues to nourish the entire world!

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the number of nodes in the Tree of Life.

The $i$-th of the following $n - 1$ lines of each test case contains two integers $v_i$ and $u_i$ ($1 \leq v_i < u_i \leq n$) — the channel connecting nodes $v_i$ and $u_i$.

It is guaranteed that there exists a unique simple path through the channels between any two nodes.

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

Output: For each test case, output a single integer — the minimum number of paths that the sages need to select to prevent a catastrophe.

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

Note: In the first test case, there are two pairs of channels emanating from a single node: $(1, 2)$ and $(2, 3)$, $(2, 3)$ and $(3, 4)$. It is sufficient to perform the ritual along the path $1-2-3-4$. Thus, the answer is $1$.

In the second test case, there are no pairs of channels emanating from a single node, so the answer is $0$.

In the third test case, rituals can be performed along the paths $2-1-3$, $2-1-4$, and $3-1-4$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 500005;
long long dp[MAXN]; vector<int> G[MAXN];
long long ans = 0;
void dfs(int u, int f) {
	int d = 0; for (int v : G[u]) if (v != f) dfs(v, u), d++;
	dp[u] = 0; for (int v : G[u]) if (v != f) dp[u] += max((u == f ? 0ll : 1ll), dp[v] - (d - 1));
	ans += 1ll * d * (d - 1) / 2;
	// printf("?? %d %lld %lld?\n", u, dp[u], 1ll * d * (d - 1) / 2);
}
int main() {
	int t; scanf("%d", &t); while (t--) {
		int n; scanf("%d", &n);
		for (int i = 1; i <= n; i++) G[i].clear();
		for (int i = 1; i < n; i++) {
			int u, v; scanf("%d %d", &u, &v);
			G[u].push_back(v); G[v].push_back(u);
		}
		int rt = 1;
		ans = 0; dfs(rt, rt);
		printf("%lld\n", ans + (dp[rt] + 1) / 2);
	}
}
/*
1
7
1 2
1 3
2 4
2 5
3 6
3 7
*/

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases or specific scenarios that can maximize the failure probability. The buggy code seems to have issues with how it calculates the `dp` values and the final answer. Specifically, the code might not correctly handle cases where the tree has multiple branches and nodes with high degrees.

Let's consider a small-scale test input that can expose the bug:

1. **Star Tree**: A tree where one central node is connected to all other nodes. This scenario can help check how the code handles a node with a high degree.
2. **Linear Tree**: A tree where nodes form a single path. This can help check the basic path handling.
3. **Balanced Binary Tree**: A tree where each node has exactly two children. This can help check the handling of balanced structures.

Let's create a test input that includes these scenarios:

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

Explanation:
- **Test Case 1**: A star tree with 5 nodes, where node 1 is the center. The expected output is 4 because there are 4 pairs of edges emanating from node 1.
- **Test Case 2**: A linear tree with 4 nodes. The expected output is 1 because there is only one path.
- **Test Case 3**: A balanced binary tree with 7 nodes. The expected output is 3 because there are 3 pairs of edges emanating from nodes 1, 2, and 3.

This test input should help expose any issues in the buggy code, particularly in how it handles nodes with multiple children and complex tree structures.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree of Life
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the heart of an ancient kingdom grows the legendary Tree of Life — the only one of its kind and the source of magical power for the entire world. The tree consists of $n$ nodes. Each node of this tree is a magical source, connected to other such sources through magical channels (edges). In total, there are $n-1$ channels in the tree, with the $i$-th channel connecting nodes $v_i$ and $u_i$. Moreover, there exists a unique simple path through the channels between any two nodes in the tree.

However, the magical energy flowing through these channels must be balanced; otherwise, the power of the Tree of Life may disrupt the natural order and cause catastrophic consequences. The sages of the kingdom discovered that when two magical channels converge at a single node, a dangerous "magical resonance vibration" occurs between them. To protect the Tree of Life and maintain its balance, it is necessary to select several paths and perform special rituals along them. A path is a sequence of distinct nodes $v_1, v_2, \ldots, v_k$, where each pair of adjacent nodes $v_i$ and $v_{i+1}$ is connected by a channel. When the sages perform a ritual along such a path, the resonance vibration between the channels $(v_i, v_{i+1})$ and $(v_{i+1}, v_{i+2})$ is blocked for each $1 \leq i \leq k - 2$.

The sages' task is to select the minimum number of paths and perform rituals along them to block all resonance vibrations. This means that for every pair of channels emanating from a single node, there must exist at least one selected path that contains both of these channels.

Help the sages find the minimum number of such paths so that the magical balance of the Tree of Life is preserved, and its power continues to nourish the entire world!

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the number of nodes in the Tree of Life.

The $i$-th of the following $n - 1$ lines of each test case contains two integers $v_i$ and $u_i$ ($1 \leq v_i < u_i \leq n$) — the channel connecting nodes $v_i$ and $u_i$.

It is guaranteed that there exists a unique simple path through the channels between any two nodes.

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

Output: For each test case, output a single integer — the minimum number of paths that the sages need to select to prevent a catastrophe.

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

Note: In the first test case, there are two pairs of channels emanating from a single node: $(1, 2)$ and $(2, 3)$, $(2, 3)$ and $(3, 4)$. It is sufficient to perform the ritual along the path $1-2-3-4$. Thus, the answer is $1$.

In the second test case, there are no pairs of channels emanating from a single node, so the answer is $0$.

In the third test case, rituals can be performed along the paths $2-1-3$, $2-1-4$, and $3-1-4$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
vector<int>D[500009];
int dp[500009];
long long wyn;
void DFS(int v, int last)
{
    vector<int>A;
    for(int som : D[v])
    {
        if(som==last){continue;}
        DFS(som,v);
        A.push_back(dp[som]);
    }
    if(A.size()==0)
    {
        dp[v]=1;
        return;
    }
    sort(A.begin(),A.end());
    wyn+=(long long)A.size()*(long long)(A.size()-1)/2;
    int s=0;
    for(int i=0;i<A.size();i++)
    {
        A[i]-=A.size()-1;
        A[i]=max(A[i],0);
        dp[v]+=max(A[i],1);
    }
    if(last==-1)
    {
        int a=0,b=A.back();
        for(int i=0;i+1<A.size();i++)
        {
            a+=A[i];
        }
        if(a>=b)
        {
            wyn+=(a+b+1)/2;
        }
        else
        {
            wyn+=a+(b-a+1)/2;
        }
    }
}
int main()
{ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int N;
    cin>>N;
for(int I=1;I<=N;I++)
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        D[i].clear();
        dp[i]=0;
    }
    for(int i=1;i<n;i++)
    {
        int a,b;
        cin>>a>>b;
        D[a].push_back(b);
        D[b].push_back(a);
    }
    if(n<=2){cout<<0<<'\n';continue;}
    wyn=0;
    int v=1;
    for(int i=1;i<=n;i++)
    {
        if(D[i].size()>1)
        {
            v=i;
            break;
        }
    }
    DFS(v,-1);
    cout<<wyn<<'\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:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The code seems to be attempting to solve the problem using a depth-first search (DFS) approach to calculate the minimum number of paths required to block all resonance vibrations. However, there are a few potential issues that could arise:

1. **Node with a single child**: The code might not handle nodes with only one child correctly, especially when it comes to updating the `dp` values and the `wyn` (result) value.
2. **Nodes with multiple children**: The code might have issues when a node has multiple children, particularly in how it calculates the contributions of these children to the `dp` value and the `wyn` value.
3. **Root node selection**: The code selects the root node as the first node with more than one child. This might not always be the optimal choice and could lead to incorrect results.

Given these points, let's construct a test case that can potentially expose these issues. A small-scale test case with a tree structure that has nodes with multiple children and a root node with more than one child should be effective.

Here is the test input:

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

### Explanation:
1. **First Test Case (n = 4)**:
   - Tree: 1 - 2, 1 - 3, 1 - 4
   - Node 1 has three children (2, 3, 4).
   - The correct answer should be 3, as we need to block all resonance vibrations between the three children.

2. **Second Test Case (n = 5)**:
   - Tree: 1 - 2, 1 - 3, 1 - 4, 1 - 5
   - Node 1 has four children (2, 3, 4, 5).
   - The correct answer should be 6, as we need to block all resonance vibrations between the four children.

3. **Third Test Case (n = 6)**:
   - Tree: 1 - 2, 1 - 3, 1 - 4, 1 - 5, 1 - 6
   - Node 1 has five children (2, 3, 4, 5, 6).
   - The correct answer should be 10, as we need to block all resonance vibrations between the five children.

This test input should help in identifying any issues with the handling of nodes with multiple children and the root node selection in the provided code.