**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mem(a, b) memset(a, b, sizeof(a))
#define read(x)        \
    for (auto &it : x) \
        cin >> it;
#define write(x)      \
    for (auto it : x) \
        cout << it << " ";
const int MOD = 1e9 + 7;
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        vector<int> v(n);

        map<int, int> m;

        for (int i = 0; i < n; i++)
        {
            cin >> v[i];
            m[v[i]]++;
        }
        vector<int> ans;
        for (auto it = m.begin(); it != m.end(); it++)
        {
            auto z = next(it);

            if (z != m.end() && it->first == ((z->first) - 1))
            {
                int p = (it->first * it->second);
                int q = (z->first * z->second);
                int x = p + q;

                if (x <= k)

                    ans.push_back(x);
                else
                {
                if(p>k){
                    int f=(k%it->first)-min(k%it->first,z->second);
                    ans.push_back(k-f);
                }
                else if(p==k)
                ans.push_back(k);
                else{
                    int f=k-p;
                    int  c=f/z->first,d=f%z->first;
                    int a=z->second-c;
                    d-=min(d,a);

                    ans.push_back(k-d);
                }

                }
            }
            else
            {
                int z = k % it->first;
                int p = (it->first * it->second);
                if (p <= k)
                {
                    ans.push_back(p);
                }
                else
                {

                    ans.push_back(k - z);
                }
            }
        }

        sort(all(ans));

        cout << ans[ans.size() - 1] << 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
3
5 10
1 1 2 2 3
8 100000
100000 100000 100000 100000 100000 100000 100000 100001
8 10
1 2 3 4 5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
p = int(input())

for _ in range(p):

    n, m = list(map(int, input().split()))
    arr = list(map(int, input().split()))

    arr.sort(reverse = True)

    while arr and arr[0] > m:
        arr.pop(0)

    if not arr:
        print(0)

    elif arr[0] == m:
        print(arr[0])

    else:

        ans = arr[0]
        left = 0
        temp = arr[0]
        right = 1

        while right < n:

            while arr[right]+1 < arr[left]:
                temp -= arr[left]
                left += 1

            if temp + arr[right] > m:
                if arr[left] == arr[right]:
                    temp -= (right-left)*arr[left]
                    left = right
                else:
                    temp -= arr[left]
                    left += 1

            temp += arr[right]
            ans = max(ans, temp)
            right += 1

        print(ans)

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

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

Think step by step.

Response:
```plaintext
3
1 10
10
2 10
10 9
5 10
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
def rectnt(arr,nl,i,n,cap):
    global maxi
    if(i>=n):
        x = sum(nl)
        if(x<=cap):
            maxi = max(maxi,x)
        return
    nl.append(arr[i])
    rectnt(arr,nl,i+1,n,cap)
    nl = nl[:-1]
    rectnt(arr,nl,i+1,n,cap)

def maxPetal(n,cap,petals):
    for i in range(len(petals)):
        petals[i] = int(petals[i])
    petals.sort()
    start = 0
    end = 0
    l=[]
    curr = 1e10
    while(start<len(petals)):
        if(curr!=1e10 and curr<len(petals) and petals[curr]==petals[start]):
            start+=1
            continue
        curr = petals[start]
        end = start
        while(end<len(petals)):
            if(petals[end]-petals[start]==0 or petals[end]-petals[start]==1):
                end+=1
            else:
                break
        l.append(petals[start:end])
        start+=1

    global maxi
    for i in range(len(l)):
        summ = sum(l[i])
        if(summ<=cap):
            maxi = max(maxi,summ)
        else:
            rectnt(l[i],[],0,len(l[i]),cap)

t = int(input())
for i in range(t):
    n,cap = input().split()
    petals = input().split()
    n = int(n)
    cap = int(cap)
    maxi = 0
    maxPetal(n,cap,petals)
    print(maxi)

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

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

Think step by step.

Response:
```plaintext
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
import java.util.HashMap;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCases = sc.nextInt();
        while (testCases-- > 0) {
            int n = sc.nextInt();
            long k = sc.nextLong();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) arr[i] = sc.nextInt();
            long ans = 0;
            HashMap<Integer, Integer> map = new HashMap<>();
            for (int i = 0; i < n; i++) {
                map.put(arr[i], map.getOrDefault(arr[i], 0) + 1);
            }
            for (int key : map.keySet()) {
                if (map.containsKey(key + 1)) {
                    long sum = (long) map.get(key + 1) * (key + 1) + (long) map.get(key) * key;
                    if (sum <= k) {
                        ans = Math.max(ans, sum);
                    } else {
                        long a = map.get(key);
                        long b = map.get(key + 1);
                        long div = k / key;
                        long req = k % key;
                        if (b >= req) {
                            ans = k;
                        } else {
                            ans = Math.max(ans, k - (req - b));
                        }
                    }
                } else {
                    if ((long) key * map.get(key) > k) {
                        ans = Math.max((k / key) * key, ans);
                    } else {
                        ans = Math.max(ans, (long) map.get(key) * key);
                    }
                }
                if (ans == k) break;
            }
            System.out.println(ans);
        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
import java.io.*;

import java.util.*;

public class Main {
    static int zu;

    public static void main(String[] args) {
        zu= in.nextInt();
        for (int i = 0; i < zu; i++) {
            solve();
        }
        out.flush();
    }

    static void solve() {
        int n= in.nextInt();
        long m= in.nextLong();
        long []nums=new long[n];
        HashMap<Long,Long>hashMap=new HashMap<>();
        for (int i = 0; i <n ; i++) {
            nums[i]= in.nextLong();
            hashMap.put(nums[i],hashMap.getOrDefault(nums[i],0l)+1);
        }
        int k=hashMap.size(),count=0;
        long [][]dp=new long[k][2];
        for(long key:hashMap.keySet()){
            dp[count][0]=key;
            dp[count][1]=hashMap.get(key);
            count++;
        }
        Arrays.sort(dp,(e1,e2)-> Math.toIntExact(e1[0] - e2[0]));
        long ans=0;
        for (int i = 0; i <k ; i++) {
            if(i>0&&dp[i-1][0]+1==dp[i][0]){
                long value=dp[i-1][0]*dp[i-1][1]+dp[i][0]*dp[i][1];
                if(value<=m){
                    ans=Math.max(ans,value);
                }
                else{
                    long exceed=value-m;
                    long a=exceed/dp[i-1][0];
                    long b=exceed%dp[i-1][0];
                    if(a>=b&&dp[i-1][1]>=a&&dp[i][1]>=b){
                        ans=m;
                    }
                    else{
                        if(dp[i-1][1]>a){
                            ans=Math.max(ans,m+b-dp[i-1][0]);
                        }
                        else{
                            ans=Math.max(ans,m+b-dp[i-1][0]+dp[i-1][1]-a-1);
                        }
                    }
                }
            }
            else{
                long value=dp[i][0]*dp[i][1];
                if(value<=m) ans=Math.max(ans,value);
                else{
                    ans=Math.max(ans,m/dp[i][0]*dp[i][0]);
                }
            }
            if(ans==m)break;
        }
        out.println(ans);
    }

    static long pow(long n){
        long ans=1;
        for (int i = 0; i < n; i++) {
            ans*=2;
        }
        return ans;
    }
    static void reverse(char[]chars){
        int n=chars.length;
        for (int i = 0; i <n/2 ; i++) {
            char a=chars[i];
            chars[i]=chars[n-1-i];
            chars[n-1-i]=a;
        }
    }
    static long gcd(long a,long b){
        if(a<b) return gcd(b,a);
        if(b==0) return a;
        return gcd(b,a%b);
    }
    static FastReader in = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

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

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

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

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

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

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

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

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

Think step by step.

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

This test case is designed to expose the bug in the given code by presenting a scenario where the difference in the number of petals between any two flowers is exactly one, and the total cost of the flowers is just within the budget. The correct output should be 7, as the girl can buy all three flowers (1, 2, and 3) which total 6 petals, and the cost is 6 coins, which is within the budget of 7 coins. However, the buggy code might incorrectly handle the budget and the calculation of the maximum number of petals, leading to a wrong output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
// Problem: B1. Bouquet (Easy Version)
// Contest: Codeforces - Codeforces Round 961 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1995/B1
// Memory Limit: 512 MB
// Time Limit: 1500 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <assert.h>
#include <bits/stdc++.h>
#include <math.h>
#include <stdint.h>
#include <stdio.h>
#include <string.h>

#include <algorithm>
#include <array>
#include <complex>
#include <iomanip>
#include <iostream>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#define IOS()              \
  ios::sync_with_stdio(0); \
  cin.tie(0);              \
  cout.tie(0)
#define YES() cout << "YES" << endl
#define NO() cout << "NO" << endl
#define OK() cout << "OK" << endl
#define range(x) x.begin(), x.end()
#define lowbit(x) (x & (-x))
typedef long long LL;
typedef unsigned long long ULL;
// typedef pair<int, int> pii;
using namespace std;
const LL INF64 = 1e18;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
vector<vector<int>> adj;
vector<bool> vis;
LL n, m, p;
void solve() {
  cin >> n >> m;
  LL a[N]{};
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  LL num = 0, i = 0, j = 0, ans = 0;
  while (i < n) {
    num += a[i];
    while (num > m || a[i] - a[j] > 1) {
      num -= a[j];
      j++;
    }
    ans = max(ans, num);
    i++;
  }
  cout << ans << endl;
}
signed main() {
  IOS();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose a potential issue in the buggy code. The girl has 5 flowers with petals 1, 2, 3, 4, and 5, and she has 10 coins. The correct solution should be to pick the flowers with 2, 3, and 4 petals, which sum up to 9 petals and cost 9 coins. However, the buggy code might not correctly handle the condition where the difference in the number of petals between any two flowers should not exceed one, and might incorrectly pick a different combination of flowers. This test case aims to catch such a mistake.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
//#pragma GCC ("O3,unroll-loops")
//#pragma target("avx2,bmi,bmi2,lzcnt,popcnt")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e5;

ll a[N];
inline void solve() {
	ll n, k;
	cin >> n >> k;
	map<ll, ll>mp;
	for (int i = 1; i <= n; i++) {
		cin >> a[i], mp[a[i]]++;
	}
	ll ans = 0;
	for (auto &[x, y] : mp) {
		ans = max(ans, min(k / x, y) * x);
		if (mp.find(x + 1) != mp.end()) {
			ans = max(ans, min(k / (x + 1), mp[x + 1]) * (x + 1));
			ll xi = k - min(k / x, y) * x;
			ll yi = min(xi / (x + 1), mp[x + 1]);
			ll yu = xi - yi*(x + 1);
			ans = max(ans, yi*(x + 1) - xi + k);
			if (mp[x + 1] > xi / (x + 1)) {
				ll temp = min({k / x, y, mp[x + 1] - yi, yu});
				ans = max(k - yu + temp, ans);
			}
		}
	}
	cout << ans << '\n';
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	std::cin >> T;
	while (T--) {
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
3 4
1 2 3
4 10
1 2 3 4
5 15
1 2 3 4 5
6 20
1 2 3 4 5 6
7 25
1 2 3 4 5 6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
t=int(input())

for PPP in range(t):
    n,m=map(int, input().split(" "))
    T=sorted(list(map(int,input().split(" "))))

    max=0
    i=0
    BOO=True
    while i <=len(T)-1 and BOO:
        Res=T[i]
        boo=True
        j=i+1
        while j<=len(T)-1 and boo:
            if abs(T[j]-T[i])<=1:
                Res+=T[j]
            else:
                break

            if Res>m:
                break
            if Res>=max and Res<=m:
                max=Res

            j+=1

        if Res>=max and Res<=m:
            max=Res
        if(max==m):
            BOO=False
            break
        i+=1
    print(max)

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

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

Think step by step.

Response:
```plaintext
3
5 10
1 2 2 3 4
8 20
1 2 3 4 5 6 7 8
11 15
1 2 2 3 3 3 4 4 5 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
def read_input():
    import sys
    input = sys.stdin.read
    data = input().split()
    index = 0
    t = int(data[index])
    index += 1
    test_cases = []
    for _ in range(t):
        n = int(data[index])
        m = int(data[index + 1])
        index += 2
        petals = list(map(int, data[index:index + n]))
        index += n
        test_cases.append((n, m, petals))
    return test_cases

def calculate_prefix_sums(petals):
    sorted_petals = sorted(petals)
    prefix_costs = [0] * len(petals)
    prefix_petals = [0] * len(petals)

    prefix_costs[0] = sorted_petals[0]
    prefix_petals[0] = sorted_petals[0]

    for i in range(1, len(petals)):
        prefix_costs[i] = prefix_costs[i - 1] + sorted_petals[i]
        prefix_petals[i] = prefix_petals[i - 1] + sorted_petals[i]

    return sorted_petals, prefix_costs, prefix_petals

def find_max_petals(n, m, petals):
    sorted_petals, prefix_costs, prefix_petals = calculate_prefix_sums(petals)

    max_petals = 0
    for i in range(n):
        for j in range(i, n):
            if sorted_petals[j] - sorted_petals[i] > 1:
                break
            total_cost = prefix_costs[j] - (prefix_costs[i - 1] if i > 0 else 0)
            if total_cost <= m:
                total_petals = prefix_petals[j] - (prefix_petals[i - 1] if i > 0 else 0)
                max_petals = max(max_petals, total_petals)

    return max_petals

def solve(test_cases):
    results = []
    for n, m, petals in test_cases:
        results.append(find_max_petals(n, m, petals))
    return results

if __name__ == "__main__":
    test_cases = read_input()
    results = solve(test_cases)
    for result in results:
        print(result)

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

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

Think step by step.

Response:
```plaintext
3
5 10
1 2 3 4 5
10 100
1 2 3 4 5 6 7 8 9 10
10 10
1 1 1 1 1 1 1 1 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
import java.util.*;
public class traa
{
    public int[] ssort(int n, int[] a)
    {
        int i=0,j=0,m=0,h=0;
        for(i=0;i<n-1;i++)
        {
            m=i;
            for(j=i+1;j<n;j++)
            {
                if(a[m]>a[j])
                m=j;
            }
            h=a[m];
            a[m] = a[i];
            a[i] = h;
        }
        return a;
    }
    public int finc(int n, int m, int[] a)
    {
        int i=0,j=0,max=0;
        for(i=0;i<n;i++)
        {
            int sum=0;
            for(j=i;j<n;j++)
            {
                if((a[j]-a[i])<=1)
                {
                    sum = sum + a[j];
                    if(sum>=m)
                    break;
                }
                else
                break;
            }

            if(sum>max && sum<=m)
            max=sum;
        }
        return max;
    }
    public static void main(String args[])
    {
        Scanner in = new Scanner(System.in);
        traa sef = new traa();
        int t=0,i=0,n1=0,m1=0,j=0;
        t = in.nextInt();
        int[] ans = new int[t];
        for(i=0;i<t;i++)
        {
            n1 = in.nextInt();
            int[] arr = new int[n1];
            m1 = in.nextInt();
            for(j=0;j<n1;j++)
            arr[j] = in.nextInt();
            arr = sef.ssort(n1, arr);
            ans[i] = sef.finc(n1, m1, arr);
        }
        for(i=0;i<t;i++)
        System.out.println(ans[i]);
    }
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases with extreme values or specific conditions that might cause the code to fail.
2. **Boundary Conditions**: Test cases where the number of flowers or the budget is at the limit of the constraints.
3. **Complex Cases**: Test cases with a mix of different petal counts and budgets that require careful handling of the logic.

Given the constraints and the nature of the problem, here are some potential edge cases and complex cases to consider:

1. **All Flowers with the Same Number of Petals**: This should be straightforward but can still expose issues in the sorting or summing logic.
2. **Flowers with a Small Range of Petal Counts**: This can test the logic for selecting the optimal subset of flowers.
3. **Budget Exactly Matching the Cost of a Valid Bouquet**: This can test the edge case where the budget is just enough to buy a specific set of flowers.
4. **Large Number of Flowers with a Small Budget**: This can test the performance and correctness of the algorithm when the budget is very tight.
5. **Flowers with a Large Range of Petal Counts**: This can test the logic for handling a wide range of petal counts.

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

```plaintext
5
3 10
3 3 3
4 10
1 2 3 4
5 15
1 2 3 4 5
10 20
1 2 3 4 5 6 7 8 9 10
10 100
1 2 3 4 5 6 7 8 9 10
```

This test input includes:
1. A case with all flowers having the same number of petals.
2. A case with a small range of petal counts.
3. A case with a budget that is just enough to buy a specific set of flowers.
4. A case with a large number of flowers and a small budget.
5. A case with a large range of petal counts and a large budget.

These scenarios should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 512
Description: This is the easy version of the problem. The only difference is that in this version, the flowers are specified by enumeration.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ flowers in the store, each of which is characterized by the number of petals, and a flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use in her bouquet should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, output a single integer — the maximum possible number of petals in the bouquet that the girl can assemble while meeting all the conditions listed above.

Examples:
input:
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
output:
7
13
610
13
1033

Note: In the first test case, you can assemble a bouquet with $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the third test case, you can assemble a bouquet with only one flower of any type, so the answer is $610$. In the fourth test case, you can assemble a bouquet with $(4, 4, 5)$, which gives you $13$ petals, and it is the maximum amount of petals that the girl can buy.

Buggy Code:
import java.util.Scanner;
import java.util.Arrays;
import java.util.Collections;

public class B_1_Bouquet_Easy_Version {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        for(int i=0;i<t;i++){
            int n = scan.nextInt();
            long m = scan.nextLong();
            Long[] pe = new Long[n];
            for(int j=0;j<n;j++){
                pe[j]=scan.nextLong();
            }
            long count = 0;
            Arrays.sort(pe,Collections.reverseOrder());
            Arrays.toString(pe);
            long mcount =0;
            for(int j=0;j<n;j++){
                count=0;
                long high = pe[j];
                for(int s=j;s<n;s++){
                    if(pe[s]==high||pe[s]==high-1){
                        count = count+pe[s];
                        if(count<=m){
                            if(mcount<count){
                             mcount=count;
                            }
                        }
                    }
                }
            }
            System.out.println(mcount);
        }
        scan.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
#include<iostream>
using namespace std;

int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int a[n];
for(int i=0; i<n; i++){
    cin>>a[i];
}

int points =0;
int ones = 0, twos = 0;
bool visited[n] = {false};

for(int i=0; i<n; i++){
    if(visited[i]) continue;

    int count = 1;
    for(int j = i+1; j<n; j++){
        if(a[i] == a[j]){
            count ++;
            visited[j] = true;
        }
    }
if(count  == 1){
    ones = ones +1;
}
else{
    twos = twos +1;
}
}

if(ones!=0 && twos == 0){
    if(n%2 ==0){
        points = (n/2)*2;
    }
    else{
        points = (n/2 +1)*2;
    }
}

else if(ones==0 && twos!=0 ){
   points = twos*1;
}

else{
    points = ones*2 + twos*1;
}
cout<<points<<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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios where the code might fail. The code appears to have issues in correctly handling the distribution of marbles and calculating the points for Alice. Specifically, the code might not correctly handle cases where the number of unique colors and the distribution of those colors are not straightforward.

Here are the steps to think through the problem:

1. **Unique Colors and Distribution**:
   - The code should handle cases where there are multiple unique colors and different counts of each color.
   - The code might fail if there are an odd number of marbles or if the distribution of colors is uneven.

2. **Optimal Play**:
   - Alice and Bob play optimally, so the test case should reflect a situation where the optimal strategy is not trivial.
   - For example, if there are multiple colors with a small number of marbles, Alice and Bob might need to play strategically to maximize or minimize Alice's score.

3. **Edge Cases**:
   - Test cases with a single color.
   - Test cases with all unique colors.
   - Test cases with an odd number of marbles.
   - Test cases with a mix of unique and repeated colors.

Given these considerations, a challenging test input could be:

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

### Explanation:
1. **Test Case 1 (5 marbles, 3 colors):**
   - Marbles: `[1, 1, 2, 2, 3]`
   - Alice can take the first marble (1), then Bob takes the second marble (1), Alice takes the third marble (2), Bob takes the fourth marble (2), and Alice takes the fifth marble (3).
   - Alice's score: 3 points (1 point for having at least one marble of each color, 1 point for having all marbles of color 3).

2. **Test Case 2 (6 marbles, 2 colors):**
   - Marbles: `[1, 1, 1, 2, 2, 2]`
   - Alice can take the first marble (1), then Bob takes the second marble (1), Alice takes the third marble (1), Bob takes the fourth marble (2), Alice takes the fifth marble (2), and Bob takes the sixth marble (2).
   - Alice's score: 2 points (1 point for having at least one marble of each color, 1 point for having all marbles of color 1).

3. **Test Case 3 (7 marbles, 3 colors):**
   - Marbles: `[1, 1, 1, 2, 2, 3, 3]`
   - Alice can take the first marble (1), then Bob takes the second marble (1), Alice takes the third marble (1), Bob takes the fourth marble (2), Alice takes the fifth marble (2), Bob takes the sixth marble (3), and Alice takes the seventh marble (3).
   - Alice's score: 4 points (1 point for having at least one marble of each color, 1 point for having all marbles of color 1 and color 3).

This test input should expose the bug in the given buggy code by challenging its ability to handle different distributions of marbles and colors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <map>
#include <queue>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> colors(n);
        map<int, int> freq;

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

        priority_queue<int> maxHeap;
        for (auto &entry : freq) {
            maxHeap.push(entry.second);
        }

        bool alice_turn = true;
        map<int, int> alice_count;

        while (!maxHeap.empty()) {
            int top = maxHeap.top();
            maxHeap.pop();

            if (alice_turn) {

                for (auto &entry : freq) {
                    if (entry.second == top) {
                        alice_count[entry.first]++;
                        entry.second--;
                        break;
                    }
                }
                if (top > 1) {
                    maxHeap.push(top - 1);
                }
            } else {

                if (top > 1) {
                    maxHeap.push(top - 1);
                }
            }

            alice_turn = !alice_turn;
        }

        int alice_score = 0;
        for (auto &entry : freq) {
            if (alice_count[entry.first] > 0) {
                alice_score++;
            }
            if (alice_count[entry.first] == entry.second + alice_count[entry.first]) {
                alice_score++;
            }
        }

        cout << alice_score << 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
6
1 1 1 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
using namespace std;

int calculate_score(const vector<int>& alice, const map<int, int>& total_counts) {
    set<int> alice_colors(alice.begin(), alice.end());
    int score = 0;
    for (int color : alice_colors) {
        score++; // At least one marble of this color
        if (total_counts.at(color) == count(alice.begin(), alice.end(), color)) {
            score++; // All marbles of this color
        }
    }
    return score;
}

int optimal_game(vector<int>& marbles) {
    map<int, int> freq;
    for (int color : marbles) {
        freq[color]++;
    }

    // Max-heap to store frequencies of colors
    priority_queue<pair<int, int>> pq;
    for (auto [color, count] : freq) {
        pq.push({count, color});
    }

    vector<int> alice_taken;
    bool alice_turn = true;

    while (!pq.empty()) {
        auto [count, color] = pq.top();
        pq.pop();

        if (alice_turn) {
            alice_taken.push_back(color);
        }

        count--; // One marble taken
        if (count > 0) {
            pq.push({count, color});
        }
        alice_turn = !alice_turn;
    }

    return calculate_score(alice_taken, freq);
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> marbles(n);
        for (int i = 0; i < n; i++) {
            cin >> marbles[i];
        }
        cout << optimal_game(marbles) << 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
2
4
1 1 2 2
6
1 1 1 2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
t = int(input())

for i in range(t):
    n = int(input())
    c_lis = tuple(map(int, input().split()))
    d = {1:[],2:[]}
    for e in c_lis:
        count = c_lis.count(e)
        if count not in d:
            d[count] = []
        if e not in d[count]:
            d[count].append(e)

    c1 = len(d[1]) #count
    p1 = ((c1+1)//2)*2
    t2 = (c1+1)%2                                         # 1 = Alice, 0 = Bob

    p2 = len(d[2])
    t3 = t2

    c3 = sum([len(lis) for lis in d.values() if (lis!=d[1] and lis!=d[2])])
    p3 = c3//2
    if p3%2 ==1: # odd
        if t3 == 1:
            p3+=1
        t4 = 1-t3
    else:
        t4 = t3

    c4 = sum([(key-3)*len(lis) for key, lis in d.items() if (lis!=d[1] and lis!=d[2])])
    if c4%2 == 0:
        t5 = t4
    else:
        t5 = 1-t4

    p5 = 0
    if not t5:
        p5 = c3

    print(p1+p2+p3+p5)

**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
7
1 1 2 2 3 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
def giaic():
    number=input()
    a=list(map(str,input().split()))
    b= list(set(a))
    c=[]
    temp =[]
    for number in b:
        temp.append(a.count(number))
    c.append(b)
    c.append(temp)
    d=[]

    if len(a) % 2 == 0:
        sobuocdi = len(a)/2
    else:
        sobuocdi = (len(a)+1)/2

    so1 = c[1].count(1)

    if sobuocdi < so1:
            return sobuocdi*2
    else:
        diem =so1*2
        conlai = sobuocdi - so1
        if len(c[1]) == 1:
            return (diem + 1)
        else:
            for i in range(2, len(c[1])+1):
                temp = c[1].count(i)
                if conlai - temp*i > 0:
                    conlai -= temp*i
                    diem += 2*temp

                else:
                    if conlai % i == 0:
                        diem = diem + 2*int(conlai/i)
                        break
                    else:
                        diem = diem + 2*int(conlai/i) + 1
                        break
    return diem

N = int(input())
for n in range(N):
    print(f"{giaic():.0f}")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
#from collections import defaultdict

for _ in range(int(input())):
    n = int(input())
    arr = list(map(int,input().split()))
    lst = set(arr)
    dic  = {}

    for i in arr:
        if i not in dic:
            dic[i] = 1
        else:
            dic[i] += 1

    assign = (n+1) // 2

    s = dict(sorted(dic.items(), key=lambda x:x[1]))

    ans = 0

    if len(s)>1:
        while assign > 0:
            for i in s:
                if assign == 0:
                    break
                if s[i] == 0:
                    continue
                if s[i] == 1:
                    ans += 2
                    assign -= 1
                else:
                    assign -= 1
                    ans += 1
                s[i] -= 1
    else:
        if n==1:
            ans = 2
        else:
            ans = 1
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
import java.util.*;
public class U1{
        public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t > 0){
            int n = sc.nextInt();
            int[] arr = new int[n+1];
            for(int i = 0 ; i < n ; i++){
                arr[sc.nextInt()]++;
            }
            Arrays.sort(arr);
            int i = 0;
            while(arr[i] == 0 && i <= n) i++;
            int score = 0;
            int x = (int) Math.ceil(n/2.0);
            while(x > 0 && arr[i] == 1 && i <= n){
                score += 2;
                i++;
                x--;
            }
            if(x <= n-i-1 && x > 0){
                score += x;
                i = n+1;
            }
            else if (x > 0){
                score += n-i+1;
                x -= n-i+1;
            }
            while(x > 0 && i <= n){
                if(arr[i] == 1){
                    score++;
                    arr[i]--;
                    i++;
                }
                else{
                    arr[i]--;
                }
                x--;
            }
            System.out.println(score);
            t--;
        }
    }
}

**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
6
1 1 2 2 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
import java.util.HashMap;
import java.util.Scanner;

public class CFGame_with_Colored {
    public static void solve(Scanner sc) {
        int n = sc.nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = sc.nextInt();

        HashMap<Integer, Integer> freqMap = new HashMap<>();

        for (int i = 0; i < n; i++) {
            if (freqMap.containsKey(a[i])) {
                freqMap.put(a[i], freqMap.get(a[i]) + 1);
            } else {
                freqMap.put(a[i], 1);
            }
        }

        // System.out.println(freqMap);
        int lim = 0;

        if (n % 2 == 0) {
            lim = n / 2;
        } else {
            lim = (n / 2) + 1;

        }
        int c = 0;
        int p = 0;
        for (Integer i : freqMap.keySet()) {
            Integer tar = freqMap.get(i);
            if (tar == 1) {
                p += 2;
                freqMap.put(i, 0);
                c++;
            } else if (tar > 1) {
                c++;
                p += 1;
                freqMap.put(i, tar - 1);
            }
            if (c == lim)
                break;

        }
        System.out.println(p);
    }

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider scenarios where the code's logic might fail. The key points to focus on are:

1. **Edge Cases**: Small values of `n` and specific color distributions.
2. **Optimal Play**: Ensuring that both players play optimally to maximize or minimize Alice's score.
3. **Frequency of Colors**: Cases where some colors have a frequency of 1, and others have a frequency greater than 1.

The buggy code has a flaw in how it handles the distribution of marbles. Specifically, it incorrectly handles the scenario where there are multiple colors with a frequency of 1. It also has issues with the logic for distributing marbles to Alice and Bob.

### Step-by-Step Analysis

1. **Small `n` with Mixed Frequencies**:
   - Test cases with a small number of marbles and a mix of colors with frequencies of 1 and greater than 1.

2. **Large `n` with Dominant Color**:
   - Test cases with a large number of marbles where one color dominates.

3. **Equal Distribution**:
   - Test cases where the number of marbles is even and the colors are distributed equally.

### Chosen Test Input

Let's consider a test case where `n = 6` and the colors are `[1, 1, 2, 2, 3, 3]`. This test case is designed to expose the bug in the code's handling of colors with frequencies greater than 1.

### Explanation

- **Initial State**: There are 6 marbles with colors `[1, 1, 2, 2, 3, 3]`.
- **Optimal Play**:
  - Alice can take a marble of color 1, reducing the count to `[1, 2, 2, 3, 3]`.
  - Bob can take a marble of color 2, reducing the count to `[1, 2, 3, 3]`.
  - Alice can take a marble of color 3, reducing the count to `[1, 2, 3]`.
  - Bob can take a marble of color 1, reducing the count to `[2, 3]`.
  - Alice can take a marble of color 2, reducing the count to `[3]`.
  - Bob can take the last marble of color 3.

- **Alice's Score**:
  - Alice has marbles of colors 1, 2, and 3.
  - She has taken all marbles of color 3.
  - Therefore, her score should be 4 (3 points for having at least one marble of each color and 1 point for having all marbles of color 3).

### Test Input

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

This test input will help expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Colored Marbles
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob play a game. There are $$$n$$$ marbles, the $$$i$$$-th of them has color $$$c_i$$$. The players take turns; Alice goes first, then Bob, then Alice again, then Bob again, and so on.

During their turn, a player must take one of the remaining marbles and remove it from the game. If there are no marbles left (all $$$n$$$ marbles have been taken), the game ends.

Alice's score at the end of the game is calculated as follows:

- she receives $$$1$$$ point for every color $$$x$$$ such that she has taken at least one marble of that color;
- additionally, she receives $$$1$$$ point for every color $$$x$$$ such that she has taken all marbles of that color (of course, only colors present in the game are considered).

For example, suppose there are $$$5$$$ marbles, their colors are $$$[1, 3, 1, 3, 4]$$$, and the game goes as follows: Alice takes the $$$1$$$-st marble, then Bob takes the $$$3$$$-rd marble, then Alice takes the $$$5$$$-th marble, then Bob takes the $$$2$$$-nd marble, and finally, Alice takes the $$$4$$$-th marble. Then, Alice receives $$$4$$$ points: $$$3$$$ points for having at least one marble for colors $$$1$$$, $$$3$$$ and $$$4$$$, and $$$1$$$ point for having all marbles of color $$$4$$$. Note that this strategy is not necessarily optimal for both players.

Alice wants to maximize her score at the end of the game. Bob wants to minimize it. Both players play optimally (i. e. Alice chooses a strategy which allows her to get as many points as possible, and Bob chooses a strategy which minimizes the amount of points Alice can get).

Calculate Alice's score at the end of the game.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains one integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of marbles;
- the second line contains $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$1 \le c_i \le n$$$) — the colors of the marbles.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output: For each test case, print one integer — Alice's score at the end of the game, assuming that both players play optimally.

Examples:
input:
3
5
1 3 1 3 4
3
1 2 3
4
4 4 4 4
output:
4
4
1

Note: In the second test case of the example, the colors of all marbles are distinct, so, no matter how the players act, Alice receives $$$4$$$ points for having all marbles of two colors, and no marbles of the third color.

In the third test case of the example, the colors of all marbles are the same, so, no matter how the players act, Alice receives $$$1$$$ point for having at least one (but not all) marble of color $$$4$$$.

Buggy Code:
#include <bits/stdc++.h>
#include <random>
#include<unordered_set>
using namespace std;
#define int long long
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
#define lowbit(x) ((x)&(-x))
#define endl '\n';

void solve() {

    int n;cin>>n;
    int a=0,b=0,c=0;

    map<int,int> ma;

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

    for(auto [_,x]:ma) {
        a+=x==1;
        b+=x%2;
        c+=x%2==0;
    }
    b-=a;

    int ans=b+c;

    while(a | b) {
        if(a) {
            a--;
            ans+=2;
        } else if(b) {
            b--;
            c++;
            ans++;
        }

        if(a) {
            a--;
        } else if(b) {
            b--;
            c++;
        }
    }
    cout<<ans<<endl;
}

main() {

	ios::sync_with_stdio(0); cout.tie(0);

    int _=1;
    cin>>_;
    while(_--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

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