**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

    static List<List<Integer>> adjList;
    static int[] numAnts;
    static boolean[] visited;

    public static void main(String[] args) {
        FastReader reader = new FastReader();
        int n = reader.nextInt();
        adjList = new ArrayList<>();
        numAnts = new int[n + 1];
        visited = new boolean[n + 1];

        for (int i = 0; i <= n; i++) {
            adjList.add(new ArrayList<>());
        }

        for (int i = 0; i < n - 1; i++) {
            int u = reader.nextInt();
            int v = reader.nextInt();
            adjList.get(u).add(v);
            adjList.get(v).add(u);
        }

        dfs(1, -1);

        boolean possible = true;
        for (int i = 1; i <= n; i++) {
            if (numAnts[i] == n && visited[i]) {
                System.out.println("YES");
                return;
            }
            if (numAnts[i] > 0 && numAnts[i] < n) {
                possible = false;
            }
        }

        if (possible) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }

    static void dfs(int node, int parent) {
        visited[node] = true;
        numAnts[node] = 1;

        for (int child : adjList.get(node)) {
            if (child != parent) {
                dfs(child, node);
                numAnts[node] += numAnts[child];
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can cause the algorithm to fail. The key issue in the code is that it does not correctly handle the condition where the ants can only move to a house with fewer ants. The code assumes that if a node has `n` ants, it is a valid solution, but it does not consider the movement constraints.

### Step-by-Step Analysis:

1. **Tree Structure**: The tree should be structured in a way that makes it difficult to gather all ants in a single house.
2. **Movement Constraints**: The ants can only move to a house with fewer ants. This constraint should be exploited to create a scenario where no single house can gather all the ants.
3. **Edge Cases**: Consider a tree where the root has multiple children, and each child has a subtree that makes it impossible to gather all ants in the root.

### Challenging Test Input:

Let's create a tree with 5 nodes where the ants cannot be gathered in a single house due to the movement constraints.

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

### Explanation:

- **Node 1** is connected to nodes 2, 3, and 4.
- **Node 2** is connected to node 5.

### Why This Input is Challenging:

- **Node 1** has three children: 2, 3, and 4.
- **Node 2** has one child: 5.
- **Initial State**: Each node has one ant.
- **Movement Constraints**:
  - Ants from nodes 3 and 4 cannot move to node 1 because node 1 already has an ant.
  - Ants from node 5 cannot move to node 2 because node 2 already has an ant.
  - Node 1 cannot gather all the ants because it cannot move the ants from nodes 3 and 4.

### Expected Output:

The expected output for this test input should be `NO` because it is impossible to gather all the ants in a single house given the movement constraints.

### Final Test Input:

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        List<Integer>[] tree = new ArrayList[n + 1];
        for (int i = 1; i <= n; i++) {
            tree[i] = new ArrayList<>();
        }
        for (int i = 0; i < n - 1; i++) {
            int u = scanner.nextInt();
            int v = scanner.nextInt();
            tree[u].add(v);
            tree[v].add(u);
        }
        boolean[] explored = new boolean[n + 1];
        boolean result = dfs( tree, explored, 1);

        System.out.println(result ? "YES" : "NO");
        scanner.close();
    }

    private static boolean dfs( List<Integer>[] tree, boolean[] explored, int node) {
        explored[node] = true;

        boolean canGather = true;
        for (int neighbor : tree[node]) {
            if (!explored[neighbor]) {
                canGather &= dfs( tree, explored, neighbor);
            }
        }
        return canGather;
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on scenarios where the DFS approach might fail. The code provided does not consider the condition where ants ignore commands if there are fewer ants in the target house than in the source house. Therefore, we need a test case where this condition leads to a failure in gathering all ants.

A good scenario to consider is a tree where the ants are initially distributed in such a way that the DFS traversal will not be able to move all ants to a single house due to the ants' behavior.

Let's construct a small-scale test input:

1. **Tree Structure**: Create a tree with a structure that makes it difficult to gather all ants in one house.
2. **Initial Distribution**: Ensure that the ants are distributed in a way that the DFS traversal will fail to gather them all due to the ants' behavior.

Consider the following tree structure:
- 1 - 2 - 3 - 4
- 1 - 5

In this structure, if we start the DFS from node 1, we can see that it is impossible to gather all ants in a single house because the ants in nodes 2, 3, and 4 will not move to node 1 if there are fewer ants in node 1 than in their current nodes.

Here is the test input:

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

This test input should expose the bug in the provided code. The correct output should be "NO" because it is impossible to gather all the ants in a single house given the constraints and the behavior of the ants.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define all(x) x.begin(), x.end()
#define pb push_back
#define ll long long
#define repp(n) for (int i = 0; i < n; i++)
// typedef __int128 lll;
const ll MOD = 998244353;
const ll mod = 1e9 + 7;
const ll inf = 1e17;
#define rep(i, a, b) for (long long i = a; i < b; i++)

ll n;
vector<ll> g[200010];
map<ll,ll> p;

void dfs(ll node,ll par){

    p[node]=par;

    for(auto x:g[node]){

        if(x!=par)dfs(x,node);
    }
}
void solve(){

    cin>>n;
    vector<ll> degree(n+2,0);
    repp(n-1){
        ll u,v;
        cin>>u>>v;
        g[u].pb(v);
        g[v].pb(u);
        degree[u]++;
        degree[v]++;
    }

    map<ll,ll> sz,vis;
    queue<ll> q;
    ll cnt = 0;
    for(int i=1;i<=n;i++){
        sz[i]=1;
        vis[i]=0;
        if(degree[i]==1){
            q.push(i);
            vis[i]=1;
            cnt++;
        }
    }

    while(q.size()){

        ll s = q.size();
        set<ll> temp;
        vector<pair<ll,ll>> v;
        while(s--){

            auto node = q.front();
            q.pop();
            v.pb({sz[node],node});

         }

         sort(all(v));

        for(auto p:v){

            for(auto c:g[p.second]){

                if(vis[c]==0){

                    if(sz[c]>=p.first){
                        temp.insert(c);
                        sz[c]+=p.first;
                    }
                    else{
                        cout<<"NO"<<endl;
                        return;
                    }
                }
            }
        }

        for(ll x:temp){
            vis[x]=1;
            q.push(x);
        }
    }

    cout<<"YES"<<endl;
}
int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    // 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
4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

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

#define int long long
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define FOR(i, n) for (int i = 0; (i) < (n); (i)++)
const int N=200004;
const int p = 1e9 + 7;
#define all(v) v.begin(), v.end()
#define sz(x) ((int)(x).size())
#define mx(a) *max_element(a.begin(), a.end())
#define mxi(a) max_element(a.begin(), a.end()) - a.begin()
#define mn(a) *min_element(a.begin(), a.end())
#define mni(a) min_element(a.begin(), a.end()) - a.begin()
#define checksetbit __builtin_popcountll
#define prezero __builtin_clz
#define postzero __builtin_ctzll
#define input int n,x;cin>>n>>x;  \
                vector<int>v(n),b(x);\
                for(int i=0;i<n;i++)cin>>v[i];\
                for(int i=0;i<x;i++)cin>>b[i];
int power(int x, int y)
{
    int res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

void solve() {
 int n;cin>>n;
 vector<int>ph[n+1];
 vector<int>v[n+1];
 vector<int>rr(n+1,0);
 for(int i=0;i<n-1;i++){
    int x,y;cin>>x>>y;
    ph[x].pb(y);
    ph[y].pb(x);
 }
 ph[0].pb(1);ph[1].pb(0);
function<void(int,int)>rrr=[&](int p,int pp){

    for(int tt:ph[p]){
        if(tt==pp)continue;
        rrr(tt,p);
        v[p].pb(rr[tt]);

        rr[p]+=rr[tt];
    }
    rr[p]+=1;
    if(sz(ph[p])==1){
        v[p].pb(0);
    }
    //cout<<p<<' '<<rr[p]<<' ';
};

function<void(int,int)>r_r=[&](int p,int pp){
    if(p==1) v[p].pb(0);
    for(int tt:ph[p]){
      if(tt==pp) continue;
      v[tt].pb(v[p].back()+1);
      r_r(tt,p);
    }
};
rrr(1,0);

r_r(1,0);

// for(int i=1;i<=n;i++){
//     for(int j=0;j<sz(v[i]);j++){
//         cout<<v[i][j]<<' ';
//     }
//     cout<<endl;
// }
for(int i=1;i<=n;i++){
    sort(all(v[i]));
    bool temp=1;
    int cnt=1;
    for(int j=0;j<sz(v[i]);j++){
        if(v[i][j]<=cnt){
            cnt+=v[i][j];
        }
        else{
            temp=0;break;
        }
    }
    if(temp){
        cout<<"YES\n";return;
    }
}
cout<<"NO\n";
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code seems to be trying to determine if all ants can be gathered in a single house by moving them according to the given rules. The key points to consider are:

1. **Tree Structure**: The tree should be structured in a way that maximizes the complexity of the movement.
2. **Ant Distribution**: The initial distribution of ants should be such that it tests the logic of the movement commands.
3. **Edge Cases**: Consider edge cases where the tree is a simple path, a star, or a balanced tree.

Let's create a test case that is small but challenging:

### Test Case: Simple Path
A simple path where all nodes are connected in a linear sequence. This will test the logic of moving ants in a straightforward manner.

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

### Test Case: Star Tree
A star tree where one node is connected to all other nodes. This will test the logic of moving ants from multiple branches to a central node.

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

### Test Case: Balanced Tree
A balanced tree where the structure is more complex, testing the logic of moving ants in a more intricate way.

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

### Chosen Test Case: Balanced Tree
Let's use the balanced tree test case as it is more complex and likely to expose any bugs in the code.

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

This test case will ensure that the code handles a more intricate tree structure and the logic for moving ants is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
n = int(input())
d = dict()
m=0
c=0
for _ in range(n-1):
    u,v = map(int, input().split())
    if u not in d:
        d[u]=0
    if v not in d:
        d[v]=0
    d[u]+=1
    d[v]+=1
    m=max(m, d[u], d[v])
for i in d.values():
    if i>2:
        c+=1
# print(c)
if c>0:
    print("YES")
elif n<5:
    print("YES")
else:
    print("NO")

**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 where the faulty logic might fail. The buggy code counts the number of branches connected to each house and checks if there is any house with more than two branches connected. If there is, it assumes it is possible to gather all the ants in a single house. This logic is flawed because it does not consider the structure of the tree or the movement rules of the ants.

### Step-by-Step Reasoning:
1. **Identify the Bug**: The bug is in the logic that counts the number of branches connected to each house and assumes that if any house has more than two branches, it is possible to gather all ants. This is incorrect because it does not account for the movement rules and the structure of the tree.

2. **Create a Tree with a Central Node**: To expose the bug, we can create a tree where a central node has many branches connected to it, but the structure of the tree makes it impossible to gather all ants in a single house according to the movement rules.

3. **Ensure the Tree is Valid**: The tree must be a valid connected graph with \( n-1 \) edges.

4. **Test Input**: We will create a tree with a central node connected to multiple leaf nodes, but the structure will make it impossible to gather all ants in a single house.

### Challenging Test Input:
```plaintext
5
1 2
1 3
1 4
1 5
```

### Explanation:
- **Input**:
  - \( n = 5 \)
  - The tree has a central node 1 connected to four leaf nodes (2, 3, 4, 5).

- **Output**:
  - The buggy code will count the number of branches connected to each house. House 1 has 4 branches, so it will print "YES".
  - However, according to the movement rules, it is impossible to gather all the ants in a single house because each leaf node has only one ant, and moving all ants to the central node would violate the rule that ants will not move to a house with fewer ants.

This test input should expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
n = int(input())
d = dict()
m=0
c=0
cm=0
for _ in range(n-1):
    u,v = map(int, input().split())
    if u not in d:
        d[u]=0
    if v not in d:
        d[v]=0
    d[u]+=1
    d[v]+=1
    m=max(m, d[u], d[v])
for i in d.values():
    if m>2 and i==m:
        cm+=1
    if i>=2:
        c+=1

if cm==1 and m==(n-1)//2:
    print("YES")
if cm>=n//2:
    print("YES")
elif cm>=c//2:
    print("YES")
elif n<5:
    print("YES")
else:
    print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
#!/usr/bin/env python3
from collections import defaultdict

n = int(input())
if n <= 4:
    print("YES")
    exit()
tree = [[] for _ in range(n + 1)]
for _ in range(n-1):
    u, v = map(int, input().split())
    tree[u].append(v)
    tree[v].append(u)
if any(len(n) > 2 for n in tree[1:]):
    print("YES")
else:
    print("NO")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
import java.util.*;

public class Annual_Ants {
    static List<Integer>[]g;
    static boolean ans;
    static int ind;
    static int min;
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=1;
        while(t-->0){
            int n=sc.nextInt();
            g=new List[n+1];
            ans=true;
            ind=1;
            min=(int)1e9;
            Arrays.setAll(g,i->new ArrayList<>());
            for (int i = 1; i <n ; i++) {
                int x=sc.nextInt();
                int y=sc.nextInt();
                g[x].add(y);
                g[y].add(x);
            }
            find(1,0,n);
            //System.out.println(ind);
            dfs(ind,0);
            System.out.println(ans?"YES":"NO");
        }
    }
    static int find(int x,int fa,int n){
        int max=0;
        int cnt=0;
        for(int y:g[x]){
            if(y!=fa){
                int z=find(y,x,n);
                max=Math.max(z,max);
                cnt+=z;
            }
        }
        if(Math.max(max,n-1-cnt)<min){
            min=Math.max(cnt,n-1-cnt);
            ind=x;
        }
        return cnt+1;
    }
    static int dfs(int x,int fa){
        int cnt=1;
        ArrayList<Integer>a=new ArrayList<>();
        for(int y:g[x]){
            if(y!=fa){
                int z=dfs(y,x);
                a.add(z);
            }
        }
        Collections.sort(a);
        for(int e:a){
            if(cnt<e){
                ans=false;
            }
            cnt+=e;
        }
        return cnt;
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
// time-limit: 2000
// problem-url: https://codeforces.com/contest/1949/problem/C

//start-time: 2024-03-25 11:01:18
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

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

    vector<int> power(n, 1);
    vector<int> dist(n);
    auto dfs = [&](int u, int p, auto&& dfs) -> void {
        dist[u] = p == -1? 0 : dist[p] + 1;
        for(auto v : g[u]){
            if(v != p) {
                dfs(v, u, dfs);
                if(power[v] <= power[u]) {
                    power[u] += power[v];
                    power[v] = 0;
                }
            }
        }
    };

    for(int i = 0; i < n; i++){
        if(g[i].size() == 1){
            power[g[i][0]]++;
            power[i] = 0;
        }
    }

    for(int i = 0; i < n; i++){
        if(g[i].size() == 1) {
            dfs(i, -1, dfs);
            break;
        }
    }

    int times = 0;
    for(int i = 0; i < n; i++){
        int cnt = 0;
        if(power[i]){
            for(auto adj : g[i]){
                cnt += power[adj] > 0;
            }
        }
        if(cnt == 1 && times <= 10){
            int j = -1;
            for(auto adj : g[i]){
                if(power[adj] == 0){
                    j = adj;
                }
            }
            dfs(j, -1, dfs);
            times++;
        }
    }

    int x = count(power.begin(), power.end(), 0);

    if(x == power.size() - 1){
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    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
4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
#include<bits/stdc++.h>
#define lc p << 1
#define rc p << 1 | 1
using namespace std;
const int N = 2e5 + 5, mod = 1e6 + 7;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
int T; ll n, m, k, x, y;
int f[N], d[N], vis[N];
vector<int> e[N];
bool bfs() {
    queue<int> q;
    for(int i = 1; i <= n; i++) {
        if(f[i] == 1) q.push(i);
    }
    while(q.size()) {
        int u = q.front(); q.pop();
        vis[u] = 1;
        for(auto v : e[u]) {
            if(vis[v]) continue;
            if(d[v] < d[u]) return 0;
            d[v] += d[u]; f[v]--;
            if(f[v] == 1) q.push(v);
        }
    }
    //for(int i = 1; i <= n; i++) cout << d[i] << ' ';cout << endl;
    return true;
}
void solve() {
    cin >> n;
    for(int i = 1; i <= n; i++) d[i] = 1;
    for(int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
        f[u]++, f[v]++;
    }
    if(bfs()) {
        cout << "YES" << '\n';
    } else {
        cout << "NO" << '\n';
    }
}
int main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    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
4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
n=int(input())
if n==1:
    print("YES")
else:
    h = {}
    h2 = [1 for _ in range(n)]
    for i in range(n - 1):
        a, b = map(int, input().split())
        if a in h:
            h[a].append(b)
        else:
            h[a] = [b]
        if b in h:
            h[b].append(a)
        else:
            h[b] = [a]
    h=dict(sorted(h.items(), key=lambda item: len(item[1])))
    seen=set()
    for x in h:
        seen.add(x)
        if h[x]:
            for y in h[x]:
                if y in seen:
                    continue
                elif h2[x-1]<=h2[y-1]:
                    h2[y-1]+=h2[x-1]
                    h2[x-1]=0
                    break
    m=max(h2)
    if m==n:
        print("YES")
    else:
        print("NO")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
import sys
input = sys.stdin.readline
from collections import deque, defaultdict

MOD = 10**9 + 7

def ii():
    return int(input())

def lii():
    return [int(x) for x in input().split()]

def solve():
    n = ii()
    adj = defaultdict(lambda: set())
    ants = [1]*(n+1)
    deg = [0]*(n+1)
    leaves = deque()

    for i in range(n-1):
        u, v = lii()
        adj[u].add(v)
        adj[v].add(u)
        deg[u] += 1
        deg[v] += 1

    if n <= 3:
        return "YES"

    for i in range(1, n+1):
        if deg[i] == 1:
            leaves.append(i)

    while leaves:
        curr = leaves.popleft() # get leaf and parent
        if len(adj[curr])  == 0:
            if ants[curr] == n:
                return "YES"
            else:
                return "NO"
        neigh = list(adj[curr])[0]

        adj[neigh].remove(curr) # remove leaf from tree
        deg[neigh] -= 1

        if deg[neigh] == 1: # if neigh becomes a leaf add it
            leaves.append(neigh)
        if ants[neigh] < ants[curr]: #if more ants in curr than parent we mustve came to the end or its impossible
            if ants[neigh] + ants[curr] == n:
                return "YES"
            else:
                return "NO"
        ants[neigh] += ants[curr]

print(solve())

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
// C SJH
#include<bits/stdc++.h>
#define INT long long
#define pii pair<int, int>
#define x first
#define y second
#define inf INT_MAX
#define INF LONG_LONG_MAX
#define VI vector<int>
using namespace std;
const int NN = 2e5 + 100;
int a[NN], b[NN], chk[NN], n, m, k;
int dp[NN], cnt[NN];
int sz[NN];
VI adj[NN];
void dfs(int u, int fa = 0){
	sz[u] = 1;
	VI vec;
	dp[u] = 1;
	for(int v : adj[u]){
		if(v == fa) continue;
		dfs(v, u);
		if(!dp[v]) cnt[u] ++;
		dp[u] = min(dp[v], dp[u]);
		sz[u] += sz[v];
		vec.push_back(sz[v]);
	}
	if(dp[u]){
		sort(vec.begin(), vec.end());
		int sum = 1, fg = 0;
		for(int i = 0; i < vec.size(); i++){
			if(vec[i] > sum) fg = 1;
			sum += vec[i];
		}
		if(fg) dp[u] = 0;
	}
}
int suf[NN], sum[NN], pre[NN], id[NN];
int ans = 0;
void rdfs(int u, int up, int fa = 0){
	vector<pii> vec;
	vec.push_back({0, 0});
	for(int v : adj[u]){
		if(v == fa) continue;
		vec.push_back({sz[v], v});
	}
	vec.push_back({up, u});
	sort(vec.begin(), vec.end());
	for(int i = 1; i < vec.size(); i++) id[vec[i].y] = i;
	for(int i = 1; i < vec.size(); i++){
		sum[i] = sum[i - 1] + vec[i].x;
	}
	pre[0] = pre[vec.size()] = 0;
	for(int i = 1; i < vec.size(); i++){
		pre[i] = max(pre[i - 1], vec[i].x - sum[i - 1]);
	}
	for(int i = vec.size() - 1; i >= 1; i--){
		suf[i] = max(suf[i + 1], vec[i].x - sum[i - 1]);
	}

	for(int v : adj[u]){
		if(v == fa) continue;
		int tup = sz[u] - sz[v] + up;
		int kk = id[v];
//		if(u == 1 && v == 5){
//			cerr<<tup<<" "<<kk<<" "<<pre[kk - 1]<<" "<<suf[kk + 1]<<" "<<vec[kk].x<<" "<<sz[v] + 1<<endl;
//		}
		if(pre[kk - 1] <= (sz[v] + 1) && suf[kk + 1] <= 1){
			int tmp = cnt[u];
			if(dp[v] == 0) tmp --;
			if(!tmp) rdfs(v, tup, u);
		}
	}
//	if(u == 6){
//		for(pii v : vec) cerr<<v.x<<" ";
//		cerr<<endl;
//	}
	int tsum = 1, fg = 0;
	for(int i = 1; i < vec.size(); i++){
		if(vec[i].x > tsum) fg = 1;
		tsum += vec[i].x;
	}
	if(!fg && cnt[u] == 0) ans = 1;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif
	scanf("%d", &n);
	for(int i = 1; i <= n - 1; i++){
		int u, v;
		scanf("%d%d", &u, &v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1);
	rdfs(1, 0);
	if(ans) puts("YES");
	else puts("NO");
}

**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 buggy code, we need to focus on edge cases where the algorithm might fail. Specifically, we should look for scenarios where the ants cannot be gathered in a single house due to the constraints of the problem.

### Step-by-Step Analysis

1. **Tree Structure**: The tree structure should be such that it is difficult to move all ants to a single house. This can be achieved by creating a tree with a specific branching pattern that makes it impossible to gather all ants in one house.

2. **Ant Distribution**: The distribution of ants should be such that moving them to a single house is not possible due to the constraints of the problem.

### Example Test Case

Let's create a small-scale test case that exposes the bug. Consider a tree with 4 nodes connected in a linear fashion, but with a specific configuration that makes it impossible to gather all ants in one house.

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

### Explanation

- **Node 1** is connected to **Node 2**.
- **Node 2** is connected to **Node 3**.
- **Node 3** is connected to **Node 4**.

This tree structure is a simple linear chain. If we place one ant in each node, it becomes impossible to gather all ants in a single house because:

- Moving ants from Node 1 to Node 2 is not possible if there is already an ant in Node 2.
- Similarly, moving ants from Node 2 to Node 3 or Node 3 to Node 4 is not possible if there are already ants in the destination nodes.

### Test Input

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

This test case should expose the bug in the provided buggy code by making it impossible to gather all ants in a single house, leading to the correct output of "NO".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

input:
5
1 4
4 2
3 2
5 3
output:
NO

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
const int N=2e5+5;
vector<int>g[N];
int d[N],val[N];
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n;cin>>n;
	int u,v;
	for(int i=1;i<n;i++){
		cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
		d[u]++;
		d[v]++;
	}
	if(n==1){
		puts("YES");
		return 0;
	}
	queue<int>q;
	for(int i=1;i<=n;i++){
		if(d[i]==1){
			q.push(i);
		}
		val[i]=1;
	}
	while(!q.empty()){
		u=q.front();
		q.pop();
		for(auto v:g[u]){
			if(d[v]&&val[u]<=val[v]){
				val[v]+=val[u];
				val[u]=0;
				d[u]--;
				d[v]--;
				if(d[v]<=1)q.push(v);
				break;
			}
		}
	}
	int cnt=0;
	for(int i=1;i<=n;i++){
		if(val[i]){
			cnt++;
		}
	}
	if(cnt==1){
		puts("YES");
	}
	else{
		puts("NO");
	}
	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
4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;
const int MAX_NUM = 2e5 + 5;
const int N = 2e5 + 5;

int tree[4 * N];

void inv(int v, int tl, int tr, int pos) {
    if (tl == tr) {
        tree[v]++;
        return;
    }

    int tm = (tl + tr) / 2;

    if (pos <= tm) {
        inv(2 * v, tl, tm, pos);
    } else {
        inv(2 * v + 1, tm + 1, tr, pos);
    }

    tree[v] = tree[2 * v] + tree[2 * v + 1];
}

int sum(int v, int tl, int tr, int l, int r) {
    if (l <= tl && tr <= r) {
        return tree[v];
    }

    if (tl > r || tr < l) {
        return 0;
    }

    int tm = (tl + tr) / 2;

    return sum(2 * v, tl, tm, l, r) + sum(2 * v + 1, tm + 1, tr, l, r);
}

void clear() {
    for (int i = 0; i < 4 * N; i++) {
        tree[i] = 0;
    }
}

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

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

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

    vector<int> sorted_a = a;
    vector<int> sorted_b = b;

    sort(sorted_a.begin(), sorted_a.end());
    sort(sorted_b.begin(), sorted_b.end());

    if (sorted_a != sorted_b) {
        cout << "NO\n";
        return;
    }

    long long diff_a = 0;
    long long diff_b = 0;

    for (int i = n - 1; i >= 0; i--) {
        diff_a += sum(1, 0, N - 1, 0, a[i] - 1);
        inv(1, 0, N - 1, a[i]);
    }

    clear();

    for (int i = n - 1; i >= 0; i--) {
        diff_b += sum(1, 0, N - 1, 0, b[i] - 1);
        inv(1, 0, N - 1, b[i]);
    }

    clear();

    if (diff_a % 2 == diff_b % 2) {
        cout << "YES\n";
    } else {
        cout << "NO\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);

    int t;
    cin >> t;

    while (t--) {
        solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

using namespace std;
typedef long long ll;
void solve(){
     ll n;
     cin >> n;
     vector <ll> a(n);
     vector <ll> b(n);
     map <ll,ll> for_a;
     map <ll,ll> for_b;
     vector <ll> a3(n);
     map <ll,ll> must_a;
     map <ll,ll> must_b;
     for (int i = 0; i < n; i++){
        cin >> a[i];
        for_a[a[i]]++;
        a3[i] = a[i];
        must_a[a[i]]+=i;
     }
     for (int i = 0; i < n; i++){
        cin >> b[i];
        for_b[b[i]]++;
        must_b[b[i]]+= i;
     }
     for (auto u : for_a){
        if(u.second != for_b[u.first]){
            cout << "NO" << endl;
            return;
        }
     }
     sort(a3.begin(),a3.end());
     map <ll,ll> must;
     for (int i = 0; i < n; i++)
        must[a3[i]]+= i;
    ll kol1 = 0, kol2 = 0;
    for (auto u : must){
        kol1 += abs(u.second - must_a[u.first]);
        kol2 += abs(u.second - must_b[u.first]);
    }
    kol1/=2;
    kol2/=2;
    if(abs(kol2-kol1)%2==0)cout << "YES" << endl;
    else cout << "NO" << endl;

}
int main()
{
    ll 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
1
2
1 2
2 1
4
1 2 3 4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

// Aliases for convenience
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vb = vector<bool>;
// Constants
const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = 1e9 + 7;
const ld EPS = 1e-9;

// Macros for convenience
#define FASTIO ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define fi first
#define se second
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i >= (b); --i)
#define trav(a, x) for (auto& a : x)
#define sz(x) (int)(x).size()
#define endl '\n'

// Debugging Tools
#ifdef LOCAL
#define debug(x) cerr << #x << " = " << (x) << endl
#else
#define debug(x)
#endif

// Utility Functions
template<typename T>
void read(T& x) { cin >> x; }
template<typename T, typename... Ts>
void read(T& x, Ts&... xs) { cin >> x; read(xs...); }

template<typename T>
void print(const T& x) { cout << x; }
template<typename T, typename... Ts>
void print(const T& x, const Ts&... xs) { cout << x << ' '; print(xs...); }
void println() { cout << '\n'; }
template<typename T, typename... Ts>
void println(const T& x, const Ts&... xs) { print(x, xs...); cout << '\n'; }

template<typename T>
void read_vector(vector<T>& v) { for (auto& el : v) read(el); }
template<typename T>
void print_vector(const vector<T>& v) { for (const auto& el : v) print(el, ' '); println(); }

class SequenceTransformer {
private:
    int n;
    vi seq1, seq2;

    bool checkCompatibility() {
        vi freq(200001, 0);
        FOR(i, 0, n) {
            freq[seq1[i]]++;
            freq[seq2[i]]--;
        }
        FOR(i, 1, 200001) {
            if (freq[i] != 0) return false;
        }
        return true;
    }

    int calculateParity() {
        vi pos2(200001);
        FOR(i, 0, n) pos2[seq2[i]] = i;
        vb visited(n, false);
        int cycleCount = 0;
        FOR(i, 0, n) {
            if (!visited[i]) {
                int current = i;
                while (!visited[current]) {
                    visited[current] = true;
                    current = pos2[seq1[current]];
                }
                cycleCount++;
            }
        }
        return n - cycleCount;
    }

public:
    void inputSequences() {
        read(n);
        seq1.resize(n); seq2.resize(n);
        read_vector(seq1);
        read_vector(seq2);
    }

    bool isTransformable() {
        if (!checkCompatibility()) return false;
        return calculateParity() % 2 == 0;
    }
};

void solve() {
    SequenceTransformer transformer;
    transformer.inputSequences();
    println(transformer.isTransformable() ? "YES" : "NO");
}

int main() {
    FASTIO;
    int t;
    read(t);
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases or scenarios that can make the faulty logic in the code more apparent. The code provided has a specific logic to check if the arrays can be transformed into each other by calculating the parity of the number of cycles. Let's break down the potential issues and create a test case that can expose them.

### Potential Issues:
1. **Cycle Detection**: The code uses a cycle detection algorithm to determine if the arrays can be transformed. It calculates the number of cycles and checks if the parity of the number of elements minus the number of cycles is even. This can be incorrect if the arrays have a specific structure that the cycle detection algorithm fails to handle correctly.

2. **Parity Calculation**: The parity calculation might be incorrect if the arrays have a specific permutation that the cycle detection algorithm does not handle correctly.

### Challenging Test Case:
We need to create a test case where the arrays have a permutation that the cycle detection algorithm might fail to handle correctly. One such case is when the arrays are permutations of each other but have a specific structure that makes the cycle detection algorithm return the wrong parity.

### Example Test Case:
Let's create a test case where the arrays are permutations of each other but the cycle detection algorithm might fail to handle correctly.

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

### Explanation:
- **Array `a`:** [1, 2, 3, 4]
- **Array `b`:** [4, 3, 2, 1]

This test case is designed to check if the code can correctly handle the scenario where the arrays are permutations of each other but the elements are in reverse order. The cycle detection algorithm should correctly identify that the arrays can be transformed, but if there is a bug in the parity calculation, it might return the wrong result.

### Expected Output:
- The expected output for this test case should be "YES" because the arrays can be transformed into each other by swapping elements.

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

This test input should help expose any issues in the cycle detection and parity calculation logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
t=int(input())
import bisect as b
for i in range(t):
    n=int(input())

    l=list(map(int,input().split()))
    l1=list(map(int,input().split()))

    d={}
    d1={}
    for i in range(len(l)):
        d[l[i]]=i
        d1[l1[i]]=i

    ret=0
    f=0
    push=0
    dd={}
    ll=[]
    for i in range(len(l)):

        ind=d.get(l1[i],-1)
        if ind==-1:
            f=-1
            break
        if l[i]==l1[i]:
            continue
        ii=b.bisect_left(ll,ind)
        #print(ll)
        push=len(ll)-ii
        dif=(ind+push)-i
        if dif==0:
            continue
        #print(l)
        el=l.pop(ind+push)
        ll.insert(ii,ind)

        l.insert(i,el)
        push+=1
        ret+=dif
        while len(ll)>0 and ll[0]<=i:
            ll.pop(0)
        #print(l,l1)
    if f==-1:
        print('no')
        continue
    if ret%2==1:
        print('no')
    if ret%2==0:
        print('yes')

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
t = int(input())
for i in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))
    seta = {}
    setb = {}
    for i in range(n):
        seta[a[i]] = 1 + seta.get(a[i], 0)
        setb[b[i]] = 1 + setb.get(b[i], 0)
    if seta != setb:
        print("NO")
    else:

        c = a.copy()
        c.sort()
        hashsort = {}
        i = 0
        while i < n:
            hashsort[c[i]] = i
            i+=1
        i = 0
        hasha = {}
        while i < n:
            hasha[a[i]] = hasha.get(a[i], 0) + hashsort[a[i]] - i
            i+=1
        i = 0
        hashb = {}
        while i < n:
            hashb[b[i]] = hashb.get(b[i],0)+hashsort[b[i]] - i
            i+=1
        suma = 0
        sumb = 0
        i = 0
        while i < n:
            if hasha[a[i]] > 0:
                suma+=hasha[a[i]]
            if hashb[b[i]] > 0:
                sumb+=hashb[b[i]]
            i+=1
        suma = suma%2
        sumb = sumb%2
        if suma == sumb:
            print("YES")
        else:
            print("NO")

**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 3 4
4 3 2 1
5
1 3 5 7 9
9 7 5 3 1
6
1 2 3 4 5 6
6 5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
t=int(input())
for _ in range(t):
    n = int(input())
    a=list(map(int, input().split()))
    b=list(map(int, input().split()))
    if a==b:
        print("YES")
    elif set(a)!=set(b):
        print("NO")
    elif a==b[::-1]:
        print("YES")
    else:
        print("NO")

**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 create a test case where the arrays `a` and `b` are not identical but can be made the same through valid swaps. The buggy code incorrectly checks if `a` is equal to `b` or if `a` is equal to the reverse of `b`, which is not sufficient to determine if the arrays can be made the same.

A good test case to expose this bug is one where `a` and `b` have the same elements but are not in the same order or the reverse order. Specifically, we can create a test case where the elements are shuffled in a way that requires more than one swap to make the arrays the same.

Here is a small-scale test input that should expose the bug:

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

### Explanation:
- The arrays `a` and `b` have the same elements but are not in the same order or the reverse order.
- The buggy code will incorrectly output "NO" because it only checks if `a` is equal to `b` or if `a` is equal to the reverse of `b`.
- However, it is possible to make the arrays the same by performing the following swaps:
  1. Swap `a[1]` and `a[4]` (1-indexed) to get `a = [4, 2, 3, 1]`.
  2. Swap `a[2]` and `a[4]` (1-indexed) to get `a = [4, 1, 3, 2]`.
  3. Swap `a[3]` and `a[4]` (1-indexed) to get `a = [4, 1, 2, 3]`.

This test input should effectively expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
import java.util.ArrayList;
import java.util.Arrays;

import java.util.Scanner;

public class D1983 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-->0){
            int n = sc.nextInt();
            int a[] = new int[n];
            int b[] = new int[n];
            ArrayList<Integer> al = new ArrayList<>();
            ArrayList<Integer> bl = new ArrayList<>();
            for(int i=0;i<n;i++){
                a[i] = sc.nextInt();
                al.add(a[i]);
            }
            for(int i=0;i<n;i++){
                b[i] = sc.nextInt();
                bl.add(b[i]);
            }
            int boll=0;
            int[] sortedA = Arrays.copyOf(a, n);
            int[] sortedB = Arrays.copyOf(b, n);
            Arrays.sort(sortedA);
            Arrays.sort(sortedB);

            for (int i = 0; i < n; i++) {
                if (sortedA[i] != sortedB[i]) {
                    boll = 1;
                }
            }
            int count = 0;
            ArrayList c = new ArrayList();
            if(boll!=1){
                for(int i=0;i<n;i++){

                    if(!c.contains(i) && ! (al.get(i)==bl.get(i))){
                        if(al.get(bl.indexOf(al.get(i)))==bl.get(i)){
                            count++;
                            c.add(bl.indexOf(al.get(i)));

                        }
                        else{
                            boll = 1;
                            break;
                        }
                    }
                }
            }
//            System.out.println(count);
            if(boll==1 || count%2==1){
                System.out.println("NO");
            }
            else{
                System.out.println("YES");
            }

        }

    }

}

**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
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
```