**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;

public class Main{
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    FastReader in = new FastReader();
    int mod = 1000000007;

    void solve(int t) throws Exception{
        int n = in();
        String[][] a = new String[n][2];
        HashMap<String,List<Integer>> h = new HashMap<>();
        for(int i = 0; i < n; i++){
            String s = ins();
            String[] r = s.split(" ");
            a[i] = r;
            if(h.containsKey(r[0])==false) h.put(r[0], new ArrayList<>());
            if(h.containsKey(r[1])==false) h.put(r[1], new ArrayList<>());
            h.get(r[0]).add(i);
            h.get(r[1]).add(i);
        }
        int x = 1;
        int ans = 1;
        for(int i = 0; i < n; i++){
            ans += x;
            x = x * 2;
        }
        int[][] dp = new int[n+1][ans+10];
        for(int[] r: dp){
            Arrays.fill(r, -1);
        }
        int max = 0;
        for(int i = 0; i < n; i++){
            max = Math.max(max, 1 + cc(a, i, (1 << i), dp, h));
        }
        out.println(n - max);
    }

    public int cc(String[][] s, int prev, int mask, int[][] dp, HashMap<String,List<Integer>> h){
        if(dp[prev][mask] != -1){
            return dp[prev][mask];
        }
        List<Integer> left = h.get(s[prev][0]);
        List<Integer> right = h.get(s[prev][1]);
        int x = 0;
        for(int j = 0; j < left.size(); j++){
            if(((1 << left.get(j)) & mask) == 0){
                x = Math.max(x, 1 + cc(s, left.get(j), ((1 << left.get(j)) | mask), dp, h));
            }
        }
        for(int j = 0; j < right.size(); j++){
            if(((1 << right.get(j)) & mask) == 0){
                x = Math.max(x, 1 + cc(s, right.get(j), ((1 << right.get(j)) | mask), dp, h));
            }
        }
        return dp[prev][mask] = x;
    }

    void run() throws Exception{
        int tc = in(), t=1;
        while(t<=tc){solve(t);t++;}
        out.flush();out.close();System.gc();
    }

    public static void main(String[] args) throws Exception{
        try {
            new Main().run();
        } catch (Exception e) {
            e.printStackTrace();
            return;
        }
    }

    public int search(int[] a, int i, int j, int t) {
        while(i <= j){int m = i +(j - i)/2;
            if (t > a[m]) {i = m + 1;
            }else if (t < a[m]) {j = m - 1;
            } else {return m;}
        }return i;
    }

    int floor(int arr[], int lo, int hi, int key){
        while(lo<=hi){
            int mid = lo + (hi - lo)/2;
            if(arr[mid]>key){
                hi = mid - 1;
            }else{
                lo = mid + 1;
            }
        }return lo;
    }

    int ceil(int arr[], int lo, int hi, int key){
        while(lo<=hi){
            int mid = lo + (hi - lo)/2;
            if(arr[mid]<key){
                lo=mid+1;
            }else{
                hi=mid-1;
            }
        }return lo;
    }

    String sort(String s){
        char[] arr = s.toCharArray();
        ArrayList<Character> ls = new ArrayList<Character>();
        for(char x: arr)ls.add(x);
        Collections.sort(ls);
        for(int i=0; i < arr.length; i++)arr[i] = ls.get(i);
        return String.valueOf(arr);
    }

    //////////////////////////////////////////////////////////////////////////////////////////////////////
    ////////////////////////////////////////////Lazy work/////////////////////////////////////////////////

    long fib(long n) {    // Fn =  ( [(1 + sqrt(5))/2] ^ n - [(1 - sqrt(5))/2] ^ n )/ sqrt(5)   (binet's formular for fib)
        return (long) Math.round((Math.pow((1 + Math.sqrt(5)) / 2, n) - Math.pow((1 - Math.sqrt(5)) / 2, n)) / Math.sqrt(5));
    }
    boolean isperfectsquare(double number){double sqrt=Math.sqrt(number);return ((sqrt - Math.floor(sqrt)) == 0);}

    int pow(int i,int j,int p){
        if(j==0){
            return 1;
        }
        if(j==1){
            return i;
        }
        long ans=pow(i,j/2,p);
        ans=(ans*ans)%p;
        if(j%2==1){
            ans=(ans*i)%p;
        }
        return (int)ans;
    }

    int in(){ return in.nextInt();}
    long inl(){ return in.nextLong();}
    String ins(){ return in.nextLine();}
    double ind() { return Double.parseDouble(in.next());}
    long log2(long N){return (long)(Math.log(N) / Math.log(2));}
    int binaryToInt(String s){return Integer.parseInt(s,2);}
    long digitsum(long n){long sum = 0;while(n>0){sum += n%10;n = n/10;}return sum;}
    String toBinaryString(int s){return Integer.toBinaryString(s);}
    int lcm(int a,int b){return a*b/gcd(a,b);}
    long lcm(long a,long b){return a*b/gcd(a,b);}
    int gcd(int a, int b){ if (a==0) return b; return gcd(b%a,a);}
    long gcd(long a, long b){ if (a==0) return b; return gcd(b%a,a);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    int ctz(int f){int x = 0;for (int i=5;i<=f;i=i*5)x += f/i;return x;}
    void fill(int[] a){ for(int i=0;i<a.length;i++) a[i] = in.nextInt(); }
    void fill(long[] a){ for(int i=0;i<a.length;i++) a[i] = in.nextLong(); }
    void fill(String[] a){ for(int i=0;i<a.length;i++) a[i] = in.next(); }
    void fill(ArrayList<ArrayList<Integer>> a){for (int i=0;i<a.size()-1;i++){int u = in()-1;int v = in()-1;addEdge(a,u,v);}}
    void addEdge(ArrayList<ArrayList<Integer>> a,int u,int v){a.get(u).add(v);a.get(v).add(u);}
    void fill(int[][] m){for (int i=0;i<m.length;i++){for(int j=0;j<m[i].length;j++){m[i][j] = in();}}}
    void swap(char[] c,int i,int j){ char t = c[i]; c[i] = c[j]; c[j] = t;}
    void swap(int[] c,int i,int j){int t = c[i]; c[i] = c[j]; c[j] = t;}
    void swap(long[] c,int i,int j){long t = c[i]; c[i] = c[j]; c[j] = t;}
    void p(int[] a){ for(int i=0;i<a.length;i++) out.printf(a[i]+" "); out.println(); }
    void p(int[][] a){ for(int i=0;i<a.length;i++){for(int j=0;j<a[i].length;j++)out.printf(a[i][j]+" ");out.println();}}
    void p(long[] a){for(int i=0;i<a.length;i++) out.printf(a[i]+" "); out.println();}
    void p(char[] a){for(int i=0;i<a.length;i++) out.printf(a[i]+" "); out.println();}
    void p(String s){for(int i=0;i<s.length();i++) out.printf(s.charAt(i)+" "); out.println();}
    void p(ArrayList<Integer> a){a.forEach(e -> out.printf(e+" "));out.println();}
    void p(LinkedList<Integer> a){a.forEach(e -> out.printf(e+" "));out.println();}
    void p(HashSet<Integer> a){a.forEach(e -> out.printf(e+" "));out.println();}
    void p(HashMap<Integer,Integer> a){for(Map.Entry<Integer, Integer> mp : a.entrySet())out.println(mp.getKey()+" "+mp.getValue());}
    void reverse(int[] a,int i,int j){while(i<j){int t = a[i];a[i] = a[j];a[j]= t;i++;j--;}}
    String reverse(String s,int i,int j){char[] a = s.toCharArray();while(i<j){char t = a[i];a[i] = a[j];a[j]= t;i++;j--;}return String.valueOf(a);}
    boolean isok(int i,int j,int n,int m){return i >=0 && j>= 0 && i<n && j<m;}
    Pair[] alldir(){ return new Pair[]{new Pair(0,1),new Pair(1,0),new Pair(0,-1),new Pair(-1,0),
    new Pair(-1,1),new Pair(-1,-1),new Pair(1,-1),new Pair(1,1)};}
    Pair[] sides(){ return new Pair[]{new Pair(0,1),new Pair(1,0),new Pair(0,-1),new Pair(-1,0)};}
    Pair[] corners(){ return new Pair[]{new Pair(-1,1),new Pair(-1,-1),new Pair(1,-1),new Pair(1,1)};}

    class Pair implements Comparable<Pair> {
        int k; int v;
        public Pair(int k, int v) {this.k = k;this.v = v;}
        public String toString() {return "(" + k + "," + v + ")";}
        public int compareTo(Pair a){
            long dif = this.v - a.v;
            if (dif > 0) return 1;
            if (dif < 0) return -1;
            return 0;
        }
    }
    class compare implements Comparator<Pair>{
        public int compare(Pair a,Pair b){
            long diff = a.v - b.v;
            if(diff>0) return 1;
            if(diff<0) return -1;
            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 buggy code, let's consider the following steps:

1. **Understand the Bug**: The buggy code seems to be using a dynamic programming approach to find the longest sequence of songs that can be arranged to form an exciting playlist. However, it may not correctly handle the constraints or edge cases.

2. **Edge Cases**:
   - **Single Song**: A playlist with a single song is already exciting.
   - **All Songs the Same Genre/Writer**: A playlist where all songs have the same genre or writer is trivially exciting.
   - **Mixed Genres and Writers**: A playlist with mixed genres and writers, where it is not immediately obvious how to form an exciting playlist.
   - ** songs that cannot form an exciting playlist without removals**: A playlist where some songs must be removed to achieve an exciting playlist.

3. **Challenging Input**:
   - We will create a playlist where the songs have mixed genres and writers, and it is not immediately clear how to form an exciting playlist without removing some songs.
   - The input will have a small number of songs to ensure it is a valid small-scale test case.

Given these considerations, here is a challenging test input:

```plaintext
1
4
a b
b a
c d
d c
```

This input has:
- 4 songs.
- Each song has a unique genre and writer combination.
- No two songs share the same genre or writer, making it challenging to form an exciting playlist without removals.

This input should expose the bug in the buggy code by requiring the algorithm to correctly determine that at least one song must be removed to form an exciting playlist.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
// st.order_of_key(a); no of elements less than a in st
const int N = 1e5 + 5;
#define ll long long
#define e "\n"
vector<int> z_function(string s)
{
    int n = s.size();
    vector<int> z(n);
    int l = 0, r = 0;
    for (int i = 1; i < n; ++i)
    {
        if (i < r)
            z[i] = min(r - i, z[i - l]);
        while (i + z[i] < n && s[z[i]] == s[i + z[i]])
            ++z[i];
        if (i + z[i] > r)
            l = i, r = i + z[i];
    }
    return z;
}
class DisjointSets
{
private:
    vector<int> parents;
    vector<int> sizes;

public:
    DisjointSets(int size) : parents(size), sizes(size, 1)
    {
        for (int i = 0; i < size; i++)
        {
            parents[i] = i;
        }
    }
    /* @return the representative node in x's component */
    int find(int x)
    {
        return parents[x] == x ? x : (parents[x] = find(parents[x]));
    }
    bool unite(int x, int y)
    {
        int x_root = find(x);
        int y_root = find(y);
        if (x_root == y_root)
        {
            return false;
        }
        if (sizes[x_root] < sizes[y_root])
        {
            swap(x_root, y_root);
        }
        sizes[x_root] += sizes[y_root];
        parents[y_root] = x_root;
        return true;
    }
    bool connected(int x, int y)
    {
        return find(x) == find(y);
    }
};
void dfs(map<string, list<string>> &adj, map<string, list<string>> &adj1, string s, map<string, int> &vis, map<string, int>& vis1,int c)
{
    if (c % 2 == 0)
    {
        vis[s] = 1;
        for (auto h : adj[s])
        {
            if (vis.find(h) == vis.end())
            {
                dfs(adj, adj1, h, vis,vis1, c + 1);
            }
        }
    }
    else
    {
        vis1[s] = 1;
        for (auto h : adj1[s])
        {
            if (vis.find(h) == vis.end())
            {
                dfs(adj, adj1, h, vis,vis1,c + 1);
            }
        }
    }
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        map<string, list<string>> adj;
        map<string, list<string>> adj1;
        vector<pair<string, string>> as;
        int c = 0;
        for (int i = 0; i < n; i++)
        {
            string a, b;
            cin >> a >> b;
            as.push_back({a, b});
            adj[a].push_back(b);
            adj1[b].push_back(a);
        }
        int an = 20;

        int yu = 0;
        for (int i = 0; i < as.size(); i++)
        {
            map<string, int> vis;
            map<string, int> vis1;
            if (vis.find(as[i].first) == vis.end())
            {
                int ans = 0;
                dfs(adj, adj1, as[i].first, vis,vis1, 0);
            }
            if (vis1.find(as[i].second) == vis1.end())
            {
                int ans = 0;
                dfs(adj, adj1, as[i].second, vis,vis1 ,1);
            }
            int sd = 0;
            for (auto h : as)
            {
                if (vis.find(h.first)==vis.end()||vis1.find(h.second)==vis1.end())
                {
                    sd++;
                }
            }
            an = min(an, sd);
        }

        cout << an << e;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4
a b
c b
a d
c d
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
// 1<=t<=1000 1<=n<=16
#include<bits/stdc++.h>
using namespace std;
const int MAXN=21;

string a[MAXN],b[MAXN];
int fa[MAXN];
map<int,int> mp;

int fath(int id){
	if (fa[id]==id) return id;
	return fa[id]=fath(fa[id]);
}

void mer(int x,int y){
	fa[fath(x)]=fath(y);
}

int main(){
	int t;
	cin >> t;
	while (t--){
		int n;
		cin >> n;
		for (int i=1;i<=n;i++) fa[i]=i;
		for (int i=1;i<=n;i++)
			cin >> a[i] >> b[i];

		for (int i=1;i<=n;i++)
			for (int j=i+1;j<=n;j++)
				if (a[i]==a[j] or b[i]==b[j]){
					if (fath(i)==fath(j)) continue;
					mer(i,j);
				}

		int maxn=0;
		mp.clear();
		for (int i=1;i<=n;i++)
			mp[fath(i)]++;

		for (auto i:mp) maxn=max(maxn,i.second);

		cout << n-maxn << endl;
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3
pop taylorswift
rock taylorswift
pop johndoe
```

1
3
pop taylorswift
rock taylorswift
pop johndoe


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [1] * n

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
            elif self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        songs = [input().split() for _ in range(n)]

        uf = UnionFind(n)

        for i in range(n):
            for j in range(i + 1, n):
                gi, wi = songs[i]
                gj, wj = songs[j]

                if gi == gj or wi == wj:
                    uf.union(i, j)

        component_size = {}
        for i in range(n):
            root = uf.find(i)
            if root not in component_size:
                component_size[root] = 0
            component_size[root] += 1

        largest_component = max(component_size.values())

        print(n - largest_component)

solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios where the union-find structure might not correctly handle the connections between songs. The key points to focus on are:

1. **Small-scale input**: The input should be small enough to be manageable but still complex enough to expose the bug.
2. **Maximize failure probability**: Create a scenario where the union-find structure might not correctly identify the largest connected component.

Given the constraints and the nature of the problem, a good test case would involve a small number of songs where the union-find might incorrectly merge components or fail to identify the correct largest component.

Let's consider the following scenario:

- We have 4 songs.
- The songs are such that they form a connected component but the union-find might not correctly identify the largest component due to the way it handles the connections.

Here is the test input:

```plaintext
1
4
a b
b c
c d
d a
```

In this test case:
- Songs 1 and 2 share a writer (`b`).
- Songs 2 and 3 share a writer (`c`).
- Songs 3 and 4 share a writer (`d`).
- Songs 4 and 1 share a writer (`a`).

This forms a cycle, and the union-find should correctly identify all four songs as part of the same connected component. However, if the union-find implementation is faulty, it might not correctly handle the cycle and might incorrectly split the components.

The expected output for this test case should be:
```
0
```

This test case should help expose any issues in the union-find implementation, particularly in how it handles cycles and connected components.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
def solve():
    t = int(input())

    for _ in range(t):
        n = int(input())
        songs = [tuple(input().split()) for _ in range(n)]

        genre_map = {}
        writer_map = {}

        for i in range(n):
            genre, writer = songs[i]

            if genre not in genre_map:
                genre_map[genre] = []
            genre_map[genre].append(i)

            if writer not in writer_map:
                writer_map[writer] = []
            writer_map[writer].append(i)

        visited = [False] * n

        def dfs(song_idx):
            stack = [song_idx]
            visited[song_idx] = True
            component_size = 0
            while stack:
                u = stack.pop()
                component_size += 1
                genre, writer = songs[u]

                for v in genre_map[genre]:
                    if not visited[v]:
                        visited[v] = True
                        stack.append(v)

                for v in writer_map[writer]:
                    if not visited[v]:
                        visited[v] = True
                        stack.append(v)

            return component_size

        largest_component = 0
        for i in range(n):
            if not visited[i]:

                component_size = dfs(i)
                largest_component = max(largest_component, component_size)

        print(n - largest_component)

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
pop taylorswift
4
a a
b b
c c
d d
7
a b
b c
c d
d e
e f
f g
g h
```

3
1
pop taylorswift
4
a a
b b
c c
d d
7
a b
b c
c d
d e
e f
f g
g h


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

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

/*
   author : Multi-Thread
*/
public class C {
//public class Main {

	// static int INF = 998244353;
	static int INF = (int) 1e9 + 7;
	static int MAX = Integer.MAX_VALUE;
	static int MIN = Integer.MIN_VALUE;

	public static void main(String[] args) {

		int test = fs.nextInt();
//		int test = 1;

		for (int cases = 0; cases < test; cases++) {

//			solve();
			final_answer.append(solve() + "\n");
		}
		out.print(final_answer.toString());
		out.flush();
	}

	static StringBuilder final_answer = new StringBuilder();

	static String solve() {
		int n = fs.nextInt();
		int ar[][] = new int[n][2];
		HashMap<String, Integer> genreMap = new HashMap<String, Integer>();
		HashMap<String, Integer> artistMap = new HashMap<String, Integer>();
		for (int i = 0; i < n; i++) {
			String a = fs.next();
			if (genreMap.get(a) == null) {
				genreMap.put(a, genreMap.size() + 1);
			}
			String b = fs.next();
			if (artistMap.get(b) == null) {
				artistMap.put(b, artistMap.size() + 1);
			}
			ar[i][0] = genreMap.get(a);
			ar[i][1] = artistMap.get(b);
		}
		int ans = Integer.MAX_VALUE;
		int total = (1 << n);
		for (int i = 1; i < total; i++) {
			// select elements
			ArrayList<int[]> al = new ArrayList<int[]>();
			for (int j = 0; j < 32; j++) {
				int x = i >> j;
				if ((x & 1) == 1) {
					al.add(ar[j]);
				}
			}

			// generate graph
			ArrayList<ArrayList<Integer>> graph = new ArrayList<ArrayList<Integer>>();
			for (int j = 0; j < al.size(); j++) {
				graph.add(new ArrayList<>());
			}
			for (int j = 0; j < al.size(); j++) {
				for (int k = j + 1; k < al.size(); k++) {
					int[] f = al.get(j);
					int[] s = al.get(k);
					if (f[0] == s[0] || f[1] == s[1]) {
						graph.get(j).add(k);
						graph.get(k).add(j);
					}
				}
			}

			// check if a single component is generated
			boolean vis[] = new boolean[graph.size()];
			int cc = 0;
			for (int j = 0; j < graph.size(); j++) {
				if (!vis[j]) {
					dfs(j, graph, vis);
					++cc;
				}
			}
			if (cc == 1) {
				ans = Integer.min(ans, n - al.size());
			}
		}
		return ans + "";
	}

	static void dfs(int src, ArrayList<ArrayList<Integer>> graph, boolean[] vis) {
		vis[src] = true;
		for (Integer z : graph.get(src)) {
			if (!vis[z]) {
				dfs(z, graph, vis);
			}
		}
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

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

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

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

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static class OutputWriter {
		private final PrintWriter writer;

		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}

		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}

		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0)
					writer.print(' ');
				writer.print(objects[i]);
			}
		}

		public void println() {
			writer.print("\n");
		}

		public void printLine(Object... objects) {
			print(objects);
			writer.println();
		}

		public void close() {
			writer.close();
		}

		public void flush() {
			writer.flush();
		}

	}

	private static final FastReader fs = new FastReader();
	private static final OutputWriter out = new OutputWriter(System.out);
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Identify the Bug**: The bug in the code seems to be related to the way the subsets of songs are being generated and checked. Specifically, the code uses a bit mask to select subsets of songs, but it has a fixed loop that iterates up to 32, which is not appropriate for the number of songs (n ≤ 16).

2. **Edge Cases**:
   - **Single Song**: A single song is always exciting, so this should return 0.
   - **All Songs with the Same Genre or Writer**: This should also return 0 because no removals are needed.
   - **Mixed Songs**: A mix of songs with different genres and writers that cannot be arranged in an exciting way should return the correct number of removals.

3. **Small-Scale Test Case**:
   - We need a test case with a small number of songs (n ≤ 16) that will expose the issue with the bit mask loop.

Let's create a test case with 4 songs where the bit mask loop will fail if it iterates up to 32 instead of 4.

```plaintext
1
4
pop taylorswift
rock taylorswift
pop themotans
rock themotans
```

**Explanation**:
- The first line indicates that there is 1 test case.
- The second line indicates that there are 4 songs.
- The next 4 lines describe the songs:
  - Song 1: Genre "pop", Writer "taylorswift"
  - Song 2: Genre "rock", Writer "taylorswift"
  - Song 3: Genre "pop", Writer "themotans"
  - Song 4: Genre "rock", Writer "themotans"

In this case, the correct answer should be 2, because we need to remove either the "rock" songs or the "pop" songs to make the playlist exciting. The buggy code, however, might fail due to the incorrect bit mask loop, which iterates up to 32 instead of 4. This will cause it to miss valid subsets and potentially return an incorrect answer.

1
4
pop taylorswift
rock taylorswift
pop themotans
rock themotans


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
import java.util.*;

public class G {

    static int P = 131;
   // static int P2 = 13331;
    static long mod = (long) 1e13 + 7;

    //获取一个字符串的 hash 值
    static long evaluate(String s) {
        long h = 0;
        for (char c : s.toCharArray()) h = (h * P + c) % mod;
        return h;
    }

    //获取一个字符串的 hash 值
    static long evaluate2(String s) {
        long h = 0;
        for (char c : s.toCharArray()) h = h * P + c;
        return h;
    }

    static class DSU {
        DSU(int n) {
            init(n);
        }

        private int[] p, sz;

        void init(int n) {
            p = new int[n];
            sz = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = i;
                sz[i] = 1;
            }
        }

        //用于连接两个连通分量
        void union(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return;
            sz[x] += sz[y];
            p[y] = x;
        }

        //用于查询某个元素所在连通分量的根节点
        int find(int x) {
            while (p[x] != x) {
                p[x] = p[p[x]];
                x = p[x];
            }
            return x;
        }

        //查询两个元素是否在同一连通分量中
        boolean same(int x, int y) {
            return find(x) == find(y);
        }

        //返回 x 所在连通分量的元素个数
        int size(int x) {
            return sz[find(x)];
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        StringBuilder sb = new StringBuilder();
        while (t-- > 0) {
            int n = sc.nextInt();
            DSU dsu = new DSU(n);
            long[][] a = new long[n][5];//idx genre writer
            for (int i = 0; i < n; i++) {
                a[i][0] = i;
                a[i][1] = evaluate(sc.next());
                a[i][2] = evaluate(sc.next());
            }
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (a[i][1] == a[j][1] || a[i][2] == a[j][2]) {
                        dsu.union(i, j);
                    }
                }
            }
            int max = 0;
            for (int i = 0; i < n; i++) {
                max = Math.max(dsu.size(i), max);
            }
            sb.append(n - max).append('\n');

        }
        System.out.println(sb);
    }
}

**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 code, we need to focus on edge cases that can highlight the issues in the algorithm. The primary issue in the code is the use of a single hash function, which can lead to hash collisions, especially with a small number of songs. To create a challenging test case, we can use a set of songs with similar or identical hashes for their genres and writers, which can confuse the algorithm.

Here's a step-by-step approach to generating a test case:

1. **Small Number of Songs**: Since the input size is limited to 16 songs, we can use a small number of songs to make the test case more manageable and easier to understand.

2. **Identical Hashes**: Use strings that produce the same hash value for both genres and writers. This will ensure that the algorithm incorrectly groups songs that should not be grouped together.

3. **Valid Input**: Ensure the input follows the format specified in the problem description.

Let's create a test case with 4 songs, where the hashes for some songs are identical, leading to incorrect grouping:

```plaintext
1
4
a a
b b
a a
b b
```

### Explanation:
- **Test Case 1**:
  - **Song 1**: Genre "a", Writer "a"
  - **Song 2**: Genre "b", Writer "b"
  - **Song 3**: Genre "a", Writer "a"
  - **Song 4**: Genre "b", Writer "b"

In this test case, the songs with genres "a" and writers "a" will have the same hash value, and the same for songs with genres "b" and writers "b". The algorithm might incorrectly group these songs, leading to an incorrect result.

This test case should expose the bug in the hash function and the union-find logic, as the algorithm might incorrectly consider songs with identical hashes as part of the same group, even if they are not.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using i64=long long;
const int MAXN=(int)1e5+5;
void solve();
struct SONG{
    int x;
    string ge;
    string wr;
};
signed main(){
    // ios::sync_with_stdio(false);
    // cin.tie(nullptr);
    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
}
void solve(){
    int n;
    cin>>n;
    vector<SONG> song;
    vector<vector<int>>  adj(n+1);
    map<string,vector<int>> mp1,mp2;
    for (int i=1;i<=n;i++){
        string ge,wr;
        cin>>ge;
        cin>>wr;
        song.push_back({i,ge,wr});
        mp1[ge].push_back(i);
        mp2[wr].push_back(i);
        // cout<<ge<<'\n'<<wr<<'\n';
    }
    for (auto [x,ge,wr]:song){
        for (auto v:mp1[ge]){
            if (x==v) continue;
            adj[x].push_back(v);
        }
        for (auto v:mp2[wr]){
            if (x==v) continue;
            adj[x].push_back(v);
        }
    }
    vector<bool> p(n+1);
    auto dfs=[&](auto self,int x,int fa,int dis)->int {
        int mx=dis;
        p[x]=1;
        for (auto v:adj[x]){
            if (p[v]) continue;
            mx=max(mx,self(self,v,x,dis+1));
        }
        p[x]=0;
        return mx;
    };
    int ans=0;
    for (int i=1;i<=n;i++){
        ans=max(ans,dfs(dfs,i,0,1));
    }
    ans=n-ans;
    cout<<ans<<'\n';
}

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

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

Think step by step.

Response:
```plaintext
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
```

4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
import math
import sys
sys.setrecursionlimit(10**6)
mo = int(1e9 + 7)
genre = []
writter = []
best = mo

def dfs(g, w, completed):
    global best
    use = False

    for i in range(len(genre)):
        if not completed[i] and (genre[i] == g or writter[i] == w):
            completed[i] = True
            dfs(genre[i], writter[i], completed)
            completed[i] = False
            use = True

    if not use:
        best = min(best, len(completed) - sum(completed))
        return

tt = int(input())
for _ in range(tt):

    n = int(input())
    genre = []
    writter = []
    for i in range(n):
        g, w = map(str, input().split())
        genre.append(g)
        writter.append(w)

    best = mo
    completed = [False] * n
    for i in range(n):
        completed[i] = True
        dfs(genre[i], writter[i], completed)
        completed[i] = False
    print(best)

**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
pop taylorswift
2
pop taylorswift
pop justinbieber
3
pop taylorswift
pop justinbieber
rock taylorswift
```

3
1
pop taylorswift
2
pop taylorswift
pop justinbieber
3
pop taylorswift
pop justinbieber
rock taylorswift


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
from collections import deque

used = []
g = []

def gen(v, res):
    global used
    used[v] = True
    kol = 0
    mx = 0
    for u in g[v]:
        if not used[u]:
            mx = max(mx, gen(u, res + 1))
            kol += 1
    used[v] = False
    if kol == 0:
        return res
    else:
        return mx

for _ in range(int(input())):
    n = int(input())
    g = [[] for i in range(n)]
    zh, au = [], []
    for i in range(n):
        q, w = input().split()
        for j in range(len(zh)):
            if zh[j] == q:
                if not (j in g[i]):
                    g[i].append(j)
                if not (i in g[j]):
                    g[j].append(i)
        zh.append(q)
        for j in range(len(au)):
            if au[j] == w:
                if not (j in g[i]):
                    g[i].append(j)
                if not (i in g[j]):
                    g[j].append(i)
        au.append(w)
    mx = 0
    for i in range(n):
        used = [False for j in range(n)]
        mx = max(mx, gen(i, 1))
    print(n - mx)

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

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

Think step by step.

Response:
```plaintext
3
3
a b
c b
a c
4
a b
c d
e f
g h
5
a b
a c
b c
b d
c d
```

3
3
a b
c b
a c
4
a b
c d
e f
g h
5
a b
a c
b c
b d
c d


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

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

public class Main {

    static RealFastReader sc = new RealFastReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        //int n=sc.ni();
        //double d=sc.nd();
        //String s=sc.ns();
        int T=sc.ni();
        while(T-->0){
            solve();
        }

        out.close();
    }

    public static void solve() {
        int n = sc.ni();
        String arr[][]=new String[n][2];
        for(int i=0;i<n;i++){
            arr[i][0]=sc.ns();
            arr[i][1]=sc.ns();
        }

        System.out.println(n-bfs(arr));

    }
    public static int bfs(String arr[][]){
        Deque<Node> ad=new ArrayDeque<>();
        for(int i=0;i<arr.length;i++){
            ad.add(new Node(1<<i,1,i));
        }
        int max=0;
        while(!ad.isEmpty()){
            Node node=ad.poll();
            max=Math.max(node.sum,max);
            for(int i=0;i<arr.length;i++){
                int j=1<<i;
                if((j&node.val)==0){
                    String s=arr[node.la][0];
                    String s1=arr[node.la][1];
                    if(arr[i][0].equals(s)||arr[i][1].equals(s1)){
                        ad.add(new Node(node.val|j,node.sum+1,i));
                    }
                }
            }
        }
        return max;

    }

    //快读快写---------------------------------------------------------------------------------
    public static class RealFastReader {
        InputStream is;

        public RealFastReader(final InputStream is) {
            this.is = is;
        }

        private byte[] inbuf = new byte[8192];
        public int lenbuf = 0, ptrbuf = 0;

        public int readByte() {
            if (lenbuf == -1) {
                throw new InputMismatchException();
            }
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0) {
                    return -1;
                }
            }
            return inbuf[ptrbuf++];
        }

        private boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }

        private int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b)) ;
            return b;
        }

        public double nd() {
            return Double.parseDouble(ns());
        }

        public char nc() {
            return (char) skip();
        }

        public String ns() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        public char[] ns(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        public int[] na(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] nal(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public char[][] nm(int n, int m) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++) {
                map[i] = ns(m);
            }
            return map;
        }

        public int[][] nmi(int n, int m) {
            int[][] map = new int[n][];
            for (int i = 0; i < n; i++) {
                map[i] = na(m);
            }
            return map;
        }

        public int ni() {
            int num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        public long nl() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }
    }
}
class Node{
    int val;
    int sum;
    int la;
    public Node(int val,int l,int la){
        this.val=val;
        this.sum=l;
        this.la=la;
    }

}

**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 scenarios that might cause the algorithm to fail. The key points to consider are:

1. **Edge Cases**: Small playlists with specific patterns that might not be handled correctly.
2. **Maximal Removals**: Cases where the minimum number of removals is high.
3. **Complex Relationships**: Cases with multiple songs that have the same genre or writer but are not easily rearranged to form an exciting playlist.

Given the constraints and the need for a small-scale test input, let's create a test case that is likely to expose issues in the BFS approach used in the buggy code.

### Test Case
We will create a playlist with 8 songs where the relationships between songs are complex, and the minimum number of removals is not immediately obvious.

```plaintext
1
8
pop taylorswift
pop themotans
electronic carlasdreams
electronic themotans
rock taylorswift
rock themotans
indie taylorswift
indie carlasdreams
```

### Explanation
- **Song 1**: pop taylorswift
- **Song 2**: pop themotans
- **Song 3**: electronic carlasdreams
- **Song 4**: electronic themotans
- **Song 5**: rock taylorswift
- **Song 6**: rock themotans
- **Song 7**: indie taylorswift
- **Song 8**: indie carlasdreams

### Why This Test Case?
1. **Mixed Genres and Writers**: Each genre and writer appears multiple times, but not in a simple pattern that can be easily rearranged.
2. **Complex Relationships**: Songs with the same writer or genre are not contiguous, requiring careful consideration of the minimum number of removals.
3. **Non-Obvious Solution**: The minimum number of removals is not immediately clear, which can expose bugs in the BFS approach.

### Expected Output
For this test case, the expected output should be the minimum number of removals necessary to make the playlist exciting. The correct output is:

```plaintext
2
```

This is because we can remove two songs to make the playlist exciting. For example, removing songs 3 and 8:

- Remaining songs: pop taylorswift, pop themotans, electronic themotans, rock taylorswift, rock themotans, indie taylorswift
- One possible exciting playlist: pop taylorswift, pop themotans, electronic themotans, rock taylorswift, rock themotans, indie taylorswift

This test case should help expose any issues in the BFS approach used in the buggy code.

1
8
pop taylorswift
pop themotans
electronic carlasdreams
electronic themotans
rock taylorswift
rock themotans
indie taylorswift
indie carlasdreams


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

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

// -------------------------Main class-------------------------

public class Main {

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

    static ArrayList<ArrayList<Integer>> al;

    // static boolean visited[];

    // -------------------------Main function-------------------------
    public static void main(String args[]) throws IOException {
        FastReader sc = new FastReader();
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i = 0; i <t; i++) {
            int n=sc.nextInt();
            al=new ArrayList<>();
            for(int j=0;j<n;j++)
            al.add(new ArrayList<>());
            String a[][]=new String[n][2];
            for(int j=0;j<n;j++)
            {
                a[j][0]=sc.next();
                a[j][1]=sc.next();
            }
            for(int j=0;j<n;j++)
            {
                for(int k=j+1;k<n;k++)
                {
                    if(a[j][0].equals(a[k][0])||a[j][1].equals(a[k][1]))
                    {
                        al.get(j).add(k);
                        al.get(k).add(j);
                    }
                }
            }
            int ans=n-1;
            for(int j=0;j<n;j++)
            {
                int currcount=dfs(j,new boolean[n]);
                ans=Math.min(ans,n-currcount);
            }
            pw.write(ans+"\n");
        }
        pw.flush();
        pw.close();
    }

    static int dfs(int curr, boolean[] visited)
    {
        visited[curr]=true;
        ArrayList<Integer> next=al.get(curr);
        int maxans=0;
        for(int j=0;j<next.size();j++)
        {
            if(!visited[next.get(j)])
            {
                boolean new_visited[]=new boolean[visited.length];
                for(int k=0;k<visited.length;k++)
                new_visited[k]=visited[k];
                maxans=Math.max(maxans,dfs(next.get(j),new_visited));
            }
        }
        return 1+maxans;
    }

    // -------------------------Other functions-------------------------
    // Time Complexity : log(n)
    static long fast_power_mod(long n, long mod) {
        long pow2 = 2;
        long res = 1;
        while (n > 0) {
            if (n % 2 == 1)
                res = (res % mod * pow2 % mod) % mod;
            pow2 = (pow2 % mod * pow2 % mod) % mod;
            n >>= 1;
        }
        return res;
    }

    // Time Complexity : O(n)
    // Best mod values for hashing = 1e9 + 9
    static void prefixHashAndModInverse(String s, long[] prefixhash, long[] modInversePowP)
    {
        long p = 31;
        long currhash = 0;
        long powP = 1;
        for(int j=0;j<s.length();j++) {
            char c = s.charAt(j);
            currhash = (currhash + (c - '0' + 1) * powP) % MOD;
            prefixhash[j] = currhash;
            powP = (powP * p) % MOD;
        }
        for(int j=0;j<s.length();j++)
        modInversePowP[j]=inverseMOD(power(p,j,MOD), MOD)%MOD;
    }

    // Use this for multiple nCr queries
    // Time Complexity : n*r
    static long nCrModpDP(long n, long r, long p) {
        long c[] = new long[(int) r + 1];
        c[0] = 1;
        for (long j = 1; j <= n; j++) {
            for (long k = Math.min(j, r); k > 0; k--)
                c[(int) k] = (c[(int) k] % p + c[(int) k - 1] % p) % p;
        }
        return c[(int) r];
    }

    // Use this for multiple nCr queries
    // Time Complexity : log(n) * n * r
    static long nCrModpLucas(long n, long r, long p) {
        if (r == 0)
            return 1;
        long ni = n % p;
        long ri = r % p;
        return (nCrModpLucas(n / p, r / p, p) % p * nCrModpDP(ni, ri, p) % p) % p;
    }

    // Use this for single nCr calculation when n <= 1e6
    // Time Complexity : n
    public static long nCrModpFermat(long n, long r, long p) {
        if (n < r)
            return 0;
        if (r == 0)
            return 1;
        long[] fac = new long[(int) (n + 1)];
        fac[0] = 1;
        for (long i = 1; i <= n; i++)
            fac[(int)i] = (fac[(int)i - 1] * i) % p;
        return (fac[(int) n] * inverseMOD(fac[(int)r], p) % p * inverseMOD(fac[(int) (n - r)], p) % p) % p;
    }

    // Time Complexity : log(mod)
    static long inverseMOD(long x, long mod)
    {
        return power(x,mod-2,mod)%mod;
    }

    // Time Complexity : log(exponent)
    static long power(long base, long exponent, long mod) {
        long ans = 1;
        base = base % mod;
        while (exponent > 0)
        {
            if (exponent%2!=0)
                ans = (ans * base) % mod;
            exponent/=2;
            base = (base * base) % mod;
        }
        return ans;
    }

    // Time Complexity : log(max(a,b))
    static long bitwiseAND(long a, long b) {
        long shiftcount = 0;
        while (a != b && a > 0) {
            shiftcount++;
            a = a >> 1;
            b = b >> 1;
        }
        return (long) (a << shiftcount);
    }

    // Time Complexity : n*m
    static void dfs(int j, ArrayList<ArrayList<Integer>> al, boolean[] visited) {
        visited[j] = true;
        for (int x = 0; x < al.get(j).size(); x++) {
            if (!visited[al.get(j).get(x)])
                dfs(al.get(j).get(x), al, visited);
        }
    }

    // Time Complexity : log(n) for composite numbers, n for prime numbers
    static long getPrimeFactors(long n) {
        int x = 2;
        long count = 0;
        // ArrayList<Integer> al=new ArrayList<>();
        while (n > 1) {
            if (n % x == 0) {
                // if(!al.contains(x))
                // al.add(x);
                count++;
                n /= x;
            } else
                x++;
        }
        return count;
    }

    // Time Complexity : log(n)
    static ArrayList<Integer> primeFactorization(int x, int[] spf) {
        HashMap<Integer, Integer> map = new HashMap<>();
        ArrayList<Integer> al = new ArrayList<>();
        while (x != 1) {
            if (!al.contains(spf[x]))
                al.add(spf[x]);
            map.put(spf[x], map.getOrDefault(spf[x], 0) + 1);
            x /= spf[x];
        }
        return al;
        // return map;
    }

    // Time Complexity : n*10
    static long[][] getBitMap(long[] a) {
        int n = a.length;
        long[][] bit_map = new long[n][60];
        for (int j = 0; j < n; j++)
            Arrays.fill(bit_map[j], 0);
        long b[] = new long[n];
        for (int j = 0; j < n; j++)
            b[j] = a[j];
        for (int j = 0; j < n; j++) {
            int counter = 0;
            while (b[j] != 0) {
                bit_map[j][counter] = b[j] % 2;
                b[j] /= 2;
                counter++;
            }
        }
        return bit_map;
    }

    // Time Complexity : n*log(log(n))
    static ArrayList<Integer> sieveOfEratosthenes(int n) {
        boolean prime[] = new boolean[n + 1];
        for (int j = 0; j <= n; j++)
            prime[j] = true;
        for (long p = 2; p * p <= n; p++) {
            if (prime[(int) p]) {
                for (long j = p * p; j <= n; j += p)
                    prime[(int) j] = false;
            }
        }
        ArrayList<Integer> al = new ArrayList<>();
        for (long j = 2; j <= n; j++) {
            if (prime[(int) j])
                al.add((int) j);
        }
        return al;
    }

    // Time Complexity : n
    static boolean sortedIncreasing(int[] a) {
        int f = 0;
        for (int j = 1; j < a.length; j++) {
            if (a[j] < a[j - 1])
                f = 1;
        }
        return f == 0 ? true : false;
    }

    // Time Complexity : n
    static boolean sortedDecreasing(int[] a) {
        int f = 0;
        for (int j = 1; j < a.length; j++) {
            if (a[j] > a[j - 1])
                f = 1;
        }
        return f == 0 ? true : false;
    }

    // Time Complexity : sqrt(n)
    static ArrayList<Integer> getFactors(long n) {
        ArrayList<Integer> al = new ArrayList<>();
        // int count = 0;
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                al.add((int)i);
                // count++;
                if (n / i != i) {
                    al.add((int)(n / i));
                    // count++;
                }
            }
        }
        Collections.sort(al);
        return al;
        // return count;
    }

    // Time Complexity : n*log(n)
    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

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

    // Time Complexity : n*log(n)
    static void sort(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    // Time Complexity : log(min(a,b))
    static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    // Time Complexity : log(min(a,b))
    static long gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    // Time Complexity : log(min(a,b))
    static long lcm(long a, long b) {
        return ((a / gcd(a, b)) * b);
    }

    // Time Complexity : log(min(a,b))
    static long lcm(int a, int b) {
        return ((a / gcd(a, b)) * b);
    }

    // Time Complexity : log(n)
    static long floorSqrt(long x) {
        if (x == 0 || x == 1)
            return x;
        long l = 1;
        long r = (long) Math.sqrt(x) + 1;
        long ans = 0;
        while (l <= r) {
            long mid = l + (r - l) / 2;
            long curr = mid * mid;
            if (curr == x)
                return mid;
            else if (curr > 0 && curr <= x) {
                ans = mid;
                l = mid + 1;
            } else
                r = mid - 1;
        }
        return ans;
    }

    // Time Complexity : log(n*logn)
    static long getRemainderSum(long[] a, long totalsum, int x) {
        long k = 0;
        outer: for (int mul = x;; mul += x) {
            int l = 0;
            int r = a.length - 1;
            int index = -1;
            while (l <= r) {
                int mid = l + (r - l) / 2;
                if (a[mid] >= mul) {
                    index = mid;
                    r = mid - 1;
                } else
                    l = mid + 1;
            }
            if (index == -1)
                break outer;
            else
                k += a.length - index;
        }
        return (totalsum - k * x);
    }

    // -------------------------BIT-------------------------

    // private static class BIT {
    //     long[] bit;
    //     long[] arr;
    //     int len;

    //     public BIT(int len) {
    //         bit = new long[len + 1];
    //         arr = new long[len];
    //         this.len = len;
    //     }

    //     public void add(int ind, long val) {
    //         arr[ind] += val;
    //         ind++;
    //         for (; ind <= len; ind += ind & -ind) {
    //             bit[ind] += val;
    //         }
    //     }

    //     public long query(int ind) {
    //         ind++;
    //         long sum = 0;
    //         for (; ind > 0; ind -= ind & -ind) {
    //             sum += bit[ind];
    //         }
    //         return sum;
    //     }
    // }

    // -------------------------Input class-------------------------

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}

// -------------------------Other classes-------------------------

class Pair {

    // int first = 0;
    // int second = 0;
    // char first = 'a';
    // int second = -1;

    // Pair(int x, int y) {
    //     first = x;
    //     second = y;
    // }

    // Pair(char x, int y) {
    // first = x;
    // second = y;
    // }

}

class ArrayComparator implements Comparator<long[]> {
    public int compare(long[] a, long[] b) {
        return Long.compare(b[0], a[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 consider edge cases and scenarios that are likely to cause issues. The code's main logic involves creating an adjacency list to represent connections between songs that can be adjacent in an exciting playlist. The DFS function is used to find the largest connected component, and the result is the total number of songs minus the size of the largest connected component.

Here are some potential edge cases and challenging scenarios:

1. **Isolated Songs**: Songs that do not share a genre or writer with any other song.
2. **Large Connected Components**: Multiple songs that can form a large connected component.
3. **Mixed Genres and Writers**: A mix of songs with different genres and writers to test the connectivity logic.
4. **Small Test Cases**: Test cases with a small number of songs to ensure the logic is correct for small inputs.

Let's construct a test input that includes these elements:

```plaintext
4
3
pop taylorswift
rock themotans
pop themotans
4
pop taylorswift
rock themotans
pop themotans
jazz irinarimes
7
pop taylorswift
rock themotans
pop themotans
jazz irinarimes
pop irinarimes
rock theoffspring
pop taylorswift
5
a b
c d
e f
g h
i j
```

### Explanation:
1. **First Test Case**:
   - 3 songs: `pop taylorswift`, `rock themotans`, `pop themotans`.
   - The songs `pop taylorswift` and `pop themotans` can be adjacent, but `rock themotans` is isolated.
   - Expected output: 1 (remove `rock themotans`).

2. **Second Test Case**:
   - 4 songs: `pop taylorswift`, `rock themotans`, `pop themotans`, `jazz irinarimes`.
   - The songs `pop taylorswift` and `pop themotans` can be adjacent, but `rock themotans` and `jazz irinarimes` are isolated.
   - Expected output: 2 (remove `rock themotans` and `jazz irinarimes`).

3. **Third Test Case**:
   - 7 songs: `pop taylorswift`, `rock themotans`, `pop themotans`, `jazz irinarimes`, `pop irinarimes`, `rock theoffspring`, `pop taylorswift`.
   - The songs can form a large connected component, but some may be isolated.
   - Expected output: 3 (remove `rock themotans`, `jazz irinarimes`, and `rock theoffspring`).

4. **Fourth Test Case**:
   - 5 songs: `a b`, `c d`, `e f`, `g h`, `i j`.
   - All songs are isolated.
   - Expected output: 4 (remove all but one song).

This test input covers various scenarios and should help expose any issues in the provided code.

4
3
pop taylorswift
rock themotans
pop themotans
4
pop taylorswift
rock themotans
pop themotans
jazz irinarimes
7
pop taylorswift
rock themotans
pop themotans
jazz irinarimes
pop irinarimes
rock theoffspring
pop taylorswift
5
a b
c d
e f
g h
i j


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffling Songs
time_limit_ms: 3000
memory_limit_mb: 256
Description: Vladislav has a playlist consisting of $$$n$$$ songs, numbered from $$$1$$$ to $$$n$$$. Song $$$i$$$ has genre $$$g_i$$$ and writer $$$w_i$$$. He wants to make a playlist in such a way that every pair of adjacent songs either have the same writer or are from the same genre (or both). He calls such a playlist exciting. Both $$$g_i$$$ and $$$w_i$$$ are strings of length no more than $$$10^4$$$.

It might not always be possible to make an exciting playlist using all the songs, so the shuffling process occurs in two steps. First, some amount (possibly zero) of the songs are removed, and then the remaining songs in the playlist are rearranged to make it exciting.

Since Vladislav doesn't like when songs get removed from his playlist, he wants the making playlist to perform as few removals as possible. Help him find the minimum number of removals that need to be performed in order to be able to rearrange the rest of the songs to make the playlist exciting.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 16$$$) — the number of songs in the original playlist.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains two strings of lowercase letters $$$g_i$$$ and $$$w_i$$$ ($$$1 \leq |g_i|, |w_i| \leq 10^4$$$) — the genre and the writer of the $$$i$$$-th song. Where $$$|g_i|$$$ and $$$|w_i|$$$ are lengths of the strings.

The sum of $$$2^n$$$ over all test cases does not exceed $$$2^{16}$$$.

The sum of $$$|g_i| + |w_i|$$$ over all test cases does not exceed $$$4 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of removals necessary so that the resulting playlist can be made exciting.

Examples:
input:
4
1
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
output:
0
0
4
3

Note: In the first test case, the playlist is already exciting.

In the second test case, if you have the songs in the order $$$4, 3, 1, 2$$$, it is exciting, so you don't need to remove any songs.

In the third test case, you can remove songs $$$4, 5, 6, 7$$$. Then the playlist with songs in the order $$$1, 2, 3$$$ is exciting.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define io ios::sync_with_stdio(false); cin.tie(0);
#define ld long double
#define vi vector<int>
#define pii pair<int, int>
#define PB push_back
#define F first
#define S second
#define MP make_pair
#define all(v) v.begin(), v.end()
#define forn(a, n) for(int i = a ; i < n ; i++)
#define yesno(b) cout << ((b)? "YES" : "NO")
#define nl cout << "\n"
#define print(v,a) for(auto _ : v) cout << _ << a;
#define MOD 998244353ll
#define MOD2 1000000007ll

void possibles(vector<pair<char, char>>& vp, int& ans){
    sort(vp.begin(), vp.end());
    do {
        if(!ans) break;
        int t = vp.size() - 1;

        for(int i = 0; i < vp.size() - 1; ++i) {
            if ((vp[i].F == vp[i + 1].F) || (vp[i].S == vp[i + 1].S)) {
                t--;
                continue;
            }
            break;
        }
        ans = min(ans, t);
    } while (next_permutation(vp.begin(), vp.end()));
}

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

    vector<pair<string, string>> v(n);
    unordered_map<string, char> m1, m2;
    char ch = 'a', CH = 'A';

    for(auto &p : v){
        cin >> p.F >> p.S;
        m1[p.F] = ch++;
        m2[p.S] = CH++;
    }

    vector<pair<char, char>> vp;
    for(auto &p : v){
        vp.PB(MP(m1[p.F], m2[p.S]));
    }

    int ans = n - 1;
    possibles(vp, ans);

    cout << ans;
}

signed main(){
    io;
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int t = 1;
    cin >> t;
    // auto st = chrono::high_resolution_clock::now();
    forn(1, t + 1){
        // cout << "Case #" << i << ": ";
        solve();
        nl;
    }
    // cerr << "Time: " << (ld)(chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - st).count()) / 1000.0 << "s";

    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
pop taylorswift
4
electronic themotans
electronic carlasdreams
pop themotans
pop irinarimes
7
rap eminem
rap drdre
rap kanyewest
pop taylorswift
indierock arcticmonkeys
indierock arcticmonkeys
punkrock theoffspring
4
a b
c d
e f
g h
```