**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <climits>
#include <cstdlib>
#include <algorithm>     // transform
#include <array>         // array
#include <forward_list>  // forward_list
#include <iterator>      // inserter, front_inserter, end
#include <map>           // map
#include <set>           // set
#include <string>        // string
#include <tuple>         // tuple, make_tuple
#include <type_traits>   // is_arithmetic, is_same, is_enum, underlying_type, is_convertible
#include <unordered_map> // unordered_map
#include <utility>       // pair, declval
#include <valarray>      // valarray
#include <unordered_set>
#include <queue>
#include <sstream>
#include <bitset>
using namespace std;
#define int long long
#define double long double
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template <typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v)
{
    string sep;
    for (const T &x : v)
        os << sep << x, sep = " ";
    return os;
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T)
{
    cerr << ' ' << H;
    dbg_out(T...);
}
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif
const int MOD = 1e9 + 7;
const int INF = 1e18;
const double EPS = 1e-9;
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

void solve();
int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

void solve(){
    int n;
    cin >> n;
    vector<int> a(n);
    vector<pair<int, int>> b;
    map<int,int> mp;
    for(int i=0; i<n; i++) {
        cin >> a[i];
        mp[a[i]]++;
    }
    // sort(a.begin(), a.end());
    // for(auto x: a) cout << x << " ";
    // cout << endl;
    for(auto x: mp) b.push_back({x.first, x.second});
    int curr = 0, m = b.size();
    int ans = 0;
    for(int i=0; i<m;){
        while(i<m and !b[i].second) i++;
        if(i>=m) break;
        if(b[i].second) ans++;
        i++;
        if(i>=m) break;
        int minn = INT_MAX, min_idx = i;
        int cnt = 0;
        for(int j=i; j<m; j++){
            if(b[j].second and (j-i+1)-cnt >= b[j].second and b[j].second < minn){
                minn = b[j].second;
                min_idx = j;
            }
        }
        if(b[min_idx].second) {
            // cout << b[min_idx].first << " ";
            b[min_idx].second--;
            if(b[min_idx].second == 0) cnt++;
        }
    }
    // cout << endl;
    cout << ans << endl;

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import sys
import math
import bisect
import collections
from collections import defaultdict, Counter
inf = float('inf')
mod = 10**9 + 7
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    if n == 1:
        print(1)
        continue
    a.sort()
    prev = -1
    l = 0
    r = n - 1
    res = 0
    while l < r:
        if a[l] > prev:
            prev = a[l]
            res += 1
        l += 1
        if r -1 > l:
            r -= 1
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
k=int(input())
if k==500:
    for t in range(k):
        n=int(input())
        arr=list(map(int,input().split()))
        if t==55 or t==56 or t==57:
            print(n,arr)
else:
    for t in range(k):
        n=int(input())
        arr=list(map(int,input().split()))
        freq=[0]*(max(arr)+1)
        for i in arr:
            freq[i]+=1
        arr=[i for i in freq if i>0]
        p=len(arr)
        dp=[float("INF") for _ in range(p+1)]
        dp[1]=1
        for i in range(1,p):
            for j in range(p,0,-1):
                if dp[j]==float("INF"):
                    continue
                if j-dp[j]<=1-arr[i]:
                    dp[j+arr[i]]=dp[j]
                dp[j]+=1
            #print(dp)
        print(min(dp))

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
from heapq import heappop, heappush
from collections import Counter, defaultdict

def solve():
    n = int(input())
    arr = list(map(int, input().split()))

    d_arr = []
    c = Counter(arr)
    for key in sorted(c):
        d_arr.append([key, c[key]])

    # print(d_arr)
    done = [-1] * len(d_arr)

    cur_val = 0
    for i in range(len(d_arr)):
        if cur_val >= d_arr[i][1]:
            done[i] = 0
            cur_val -= d_arr[i][1]
        else:
            cur_val += 1
    # print(done)
    rem = 0
    for i in range(len(d_arr)):
        if done[i] != -1:
            happen = False
            cur_val = d_arr[i][1] + 1 + rem
            for j in range(i+1, len(d_arr)):
                if done[j] == -1 and d_arr[j][1] <= d_arr[i][1] and cur_val >= d_arr[j][1]:
                    happen = True
                    done[j] = 0
                    cur_val -= d_arr[j][1]
            if happen:
                rem = cur_val
                done[i] = -1
            else:
                rem = 0
        # print(done, i)
    print(done.count(-1))

for t in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
public class Main
{
     private static List<Integer> findAllPrimes(int limit) {
        boolean[] isPrime = new boolean[limit + 1];
        for (int i = 2; i <= limit; i++) {
            isPrime[i] = true;
        }

        for (int p = 2; p * p <= limit; p++) {
            if (isPrime[p]) {
                for (int i = p * p; i <= limit; i += p) {
                    isPrime[i] = false;
                }
            }
        }

        List<Integer> primes = new ArrayList<>();
        for (int i = 2; i <= limit; i++) {
            if (isPrime[i]) {
                primes.add(i);
            }
        }

        return primes;
    }
     public static int[] getIntArr(int n){
        int arr[] = new int[n];
        for(int i=0; i<n; i++){
            arr[i] = sc.nextInt();
        }
        return arr;
    }
    public static BigInteger[] getBigIntegerArr(int n){
        BigInteger arr[] = new BigInteger[n];
        for(int i=0;i<n;i++){
            arr[i] = sc.nextBigInteger();
        }
        return arr;
    }
    public static MyScanner sc;
    public static boolean isStablished(BigInteger ele[][], int i, int j, int n, int m){
        boolean t = false;
        if(i-1 >=0){
            if(ele[i][j].compareTo(ele[i-1][j]) <=0 ) return true;
        }
        if(i+1 < n){
            if(ele[i][j].compareTo(ele[i+1][j])<=0) return true;
        }
        if(j-1 >=0){
            if(ele[i][j].compareTo(ele[i][j-1])<=0) return true;
        }
        if(j+1 < m){
            if(ele[i][j].compareTo(ele[i][j+1])<=0) return true;
        }

        return false;
    }
    public static BigInteger biggestNum(BigInteger ele[][], int i, int j, int n, int m){
        BigInteger q = BigInteger.ZERO;
        BigInteger w = BigInteger.ZERO;
        BigInteger e = BigInteger.ZERO;
        BigInteger r = BigInteger.ZERO;
        if(i-1 >=0){
            q = ele[i-1][j];
        }
        if(i+1 < n){
            w = ele[i+1][j];
        }
        if(j-1 >=0){
            e = ele[i][j-1];
        }
        if(j+1 < m){
            r = ele[i][j+1];
        }
        return q.max(r).max(w).max(e);
    }

    public static  BigInteger  people(String s, int p , int c, BigInteger dp[][]){
        if(c==0) return new BigInteger(s.substring(p));
        if(dp[p][c].compareTo(new BigInteger("-1"))!=0) return dp[p][c];
        BigInteger res = new BigInteger("10000000000000000000000000000000000000000000000000000000000");
        for(int  i=p;i<s.length()-c;i++){
            String ert = s.substring(p,  i +1);
            if(ert.length()==0){
                ert = "0";
            }
            BigInteger a = new BigInteger(ert);
            BigInteger b = people(s, i+1, c-1, dp);
            res = res.min(a.add(b));
            res = res.min(a.multiply(b));
        }
         dp[p][c] = res;
        return res;
    }

    public static int memo[][];
    public static int values[];
    public static int evaluate(int limit, int index) {
        if (limit < 0) {
            return -2;
        }
        if (index == 0) {
            return limit != 0 ? -2 : 0;
        }
        if (memo[index][limit] != -1) {
            // System.out.println("fnifnei");
            return memo[index][limit];
        }
        if (values[index] > 0) {
            memo[index][limit] = evaluate( limit - 1, index - 1);
        } else {
            memo[index][limit] = evaluate( limit, index - 1);
        }
        if (evaluate( limit, index - 1) != -2 && ( evaluate( limit, index - 1) + values[index] ) <= limit) {
            if (memo[index][limit] == -2 || (evaluate( limit, index - 1) + values[index]) < memo[index][limit]) {
                memo[index][limit] = evaluate( limit, index - 1) + values[index];
            }
        }
        return memo[index][limit];
    }
    public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
  sc = new MyScanner();
       out = new PrintWriter(new BufferedOutputStream(System.out));

       // Start writing your solution here. -------------------------------------

       /*
       int n      = sc.nextInt();        // read input as integer
       long k     = sc.nextLong();       // read input as long
       double d   = sc.nextDouble();     // read input as double
       String str = sc.next();           // read input as String
       String s   = sc.nextLine();       // read whole line as String

       int result = 3*n;
       out.println(result);                    // print via PrintWriter
       */

       /*
            String str[] = sc.nextLine().split(" ");
            BigInteger ele[] = new BigInteger[n];
            int k = 0;
            for(int i=0;i<str.length;i++){
                if(str[i]==" " || str[i]==""){
                    continue;
                }
                ele[k] = new BigInteger(str[i]);
                k++;
            }
        */
        /* for sorting array with 2D
         *
         *
         *
         *   Arrays.sort(array, Comparator.comparingInt(paid -> paid.key));
         *
         */
        int t = sc.nextInt();

        ArrayList<Integer> arr = new ArrayList<>();

        for(int v=0;v<t;v++){
            int n = sc.nextInt();
            int ele[] = getIntArr(n);
            memo = new int[5001][5001];
            values = new int[5001];
            for (int i = 0; i <= n; i++){
                Arrays.fill(memo[i], -1);
            }
            Arrays.fill(values, 0);
            for(int i=0;i<n;i++){
                values[ele[i]]++;
                //= nums[ele[i]] + 1;
            }
            int val1 = n;
            for(int i=n;i>=0;i--){
                int val2 = evaluate(i , n);
                if(val2!=-2){
                    val1 = i;
                }
            }

            arr.add(val1);

        }
         for(int i=0;i<arr.size();i++){
            System.out.println(arr.get(i));
         }
	}

  //-----------PrintWriter for faster output---------------------------------
  public static PrintWriter out;

  //-----------MyScanner class for faster input----------
  public static class MyScanner {
     BufferedReader br;
     StringTokenizer st;

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

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

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

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

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

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

  }
  //--------------------------------------------------------

}

**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 and scenarios that are likely to cause the code to fail. Here are some key considerations:

1. **Edge Cases:**
   - When all cakes have the same tastiness value.
   - When the number of cakes is very small (e.g., 1 or 2).
   - When the number of cakes is large and the tastiness values are distributed in a way that maximizes the complexity of the problem.

2. **Optimal Play:**
   - Alice should always try to eat the cake with the highest possible tastiness that is greater than any she has eaten before.
   - Bob should try to minimize the number of cakes Alice can eat by choosing the best possible cake for himself.

3. **Specific Scenarios:**
   - A sequence of tastiness values that forces Alice to make suboptimal choices if she doesn't play optimally.
   - A sequence of tastiness values that forces Bob to make suboptimal choices if he doesn't play optimally.

Given these considerations, let's create a small-scale test input that is complete and valid:

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `5 1 2 3 4 5`
   - **Output:** `3`
   - **Explanation:** Alice can eat cakes with tastiness values 1, 3, and 5. Bob will eat the remaining cakes.

2. **Test Case 2:**
   - **Input:** `4 5 4 3 2`
   - **Output:** `2`
   - **Explanation:** Alice can eat cakes with tastiness values 2 and 4. Bob will eat the remaining cakes.

3. **Test Case 3:**
   - **Input:** `6 1 3 2 4 6 5`
   - **Output:** `3`
   - **Explanation:** Alice can eat cakes with tastiness values 1, 4, and 6. Bob will eat the remaining cakes.

This test input covers a variety of scenarios, including increasing, decreasing, and mixed sequences of tastiness values, which should help expose any bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import java.util.*;

public class WorldIsMine {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        StringBuilder fin_ans = new StringBuilder();
        int t = in.nextInt();
        while (t-- > 0){
            int n = in.nextInt();
            int[] arr = new int[n];
            int[] freq = new int[n+1];
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
                freq[arr[i]]++;
            }
            int i = 0, ans = 0, curr = 0;
            while (i <= n){
                while (i <= n && freq[i] == 0) i++;
                if (i > n) break;
                ans++;
                curr++;
                for (int j = i+1; j <= n; j++){
                    if (freq[j] == curr){
                        curr = 0;
                        freq[j] = 0;
                        break;
                    }
                }
                i++;
            }
            fin_ans.append(ans).append("\n");
        }
        System.out.print(fin_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
5
4
1 2 3 4
3
3 2 1
5
2 2 3 3 4
6
4 3 2 1 2 3
7
1 2 3 2 1 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

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

#define vi vector<int>
#define fr(i,n) for(int i = 0; i < n; i++)
#define mod 1000000007
#define int long long int
#define endl "\n"

void solve()
{
  int n; cin>>n;
  vi v(n); fr(i,n) cin>>v[i];
  sort(v.begin(),v.end());
  map<int,int> freq;
  fr(i,n) freq[v[i]]++;
  vector<pair<int,int>> store;
  for(auto x:freq) {
    store.push_back({x.second,-x.first});
  }
  sort(store.begin(),store.end());

  int mn=INT_MAX;
  for(int i=0; i<store.size(); i++) {
    vi bobRemove(n+5,0);
    vi Vis(n+5,0);
    vi Alicetaken(n+5,0);
    for(int j=0; j<=i; j++) bobRemove[-store[j].second]=1;
    int val=0;
    int alice=0; int bob=0; int prev=0;
    for(int i=0; i<n; i++) {
      if(i%2==0) {
        while(alice<n && (v[alice]==prev or Vis[alice])) alice++;
        if(alice==n) break;
        val++;
        Alicetaken[v[alice]]++;
        Vis[alice]++;
        prev=v[alice];
      }
      else {
        while(bob<n  && (Alicetaken[v[bob]] or Vis[bob] or (!bobRemove[v[bob]]))) bob++;
        if(bob==n) continue;
        Vis[bob]++;
      }
    }
    mn=min(mn,val);
  }
  cout<<mn<<endl;
}

int32_t main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;

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

  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. The key points to focus on are:

1. **Cases where Alice can eat multiple cakes but the code might fail to recognize all possible valid choices.**
2. **Cases where the cakes have repeated tastiness values, which can confuse the logic.**
3. **Cases where the number of cakes is small, but the distribution of tastiness values is such that it tests the logic thoroughly.**

Let's construct a test case that fits these criteria:

1. **Multiple cakes with the same tastiness value:**
   - This can test how the code handles repeated values and whether it correctly identifies the optimal moves for Alice and Bob.

2. **A sequence of cakes where Alice can eat multiple cakes:**
   - This can test whether the code correctly tracks the maximum tastiness value Alice has eaten and ensures she only eats cakes with strictly greater tastiness.

3. **A sequence of cakes where Bob can strategically remove cakes to minimize the number of cakes Alice can eat:**
   - This can test the code's ability to handle Bob's optimal moves.

Here is a small-scale test input that meets these criteria:

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

Explanation:
- **Test Case 1 (1 2 3 4 5):**
  - Alice can eat cakes with tastiness 1, 3, and 5. Bob will eat cakes with tastiness 2 and 4.
  - Expected output: 3

- **Test Case 2 (1 1 2 2 3):**
  - Alice can eat cakes with tastiness 1 and 3. Bob will eat the other cake with tastiness 1 and one of the cakes with tastiness 2.
  - Expected output: 2

- **Test Case 3 (3 3 3 3 3 3):**
  - Alice can only eat one cake with tastiness 3 because all other cakes have the same tastiness.
  - Expected output: 1

This test input will help identify if the code correctly handles repeated values and the optimal moves for both players.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define multiordered_set tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
#define lop(i,n) for(ll i=0;i<n;i++)
#define lop1(i,n) for(ll i=1;i<=n;i++)
#define lopr(i,n) for(ll i=n-1;i>=0;i--)
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define ll long long int
#define pb push_back
#define all(v) v.begin(),v.end()
#define IOS  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define endl '\n'
#define ff first
#define ss second
#define mem(arr,val) memset(arr,val,sizeof(arr))
#define pii pair<int,int>
#define pll pair<ll,ll>
#define LCM(a,b) (a/__gcd(a,b))*b
#define mii map<int,int>
#define mll map<ll,ll>
#define ub upper_bound
#define lb lower_bound
#define sz(x) (ll)x.size()
#define ld long double
#define pcnt(x) __builtin_popcountll(x)
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
const long long I1=1e9;
const long long I2=1e18;
const int32_t M1=1e9+7;
const int32_t M2=998244353;
ll binpow(ll a, ll b)
{
  ll res=1;
  while(b>0)
  {
    if(b%2==1) res=res*a;
    a=a*a;
    b=b/2;
  }
  return res;
}
ll binexp(ll a, ll b, ll mod)
{
  ll res=1;
  while(b>0)
  {
    if(b%2==1) res=((res%mod)*(a%mod))%mod;
    a=((a%mod)*(a%mod))%mod;
    b=b/2;
  }
  return res;
}
ll fact(ll n,ll mod)
{
    ll ans=1;
    for(ll i=2;i<=n;i++)
      ans=((ans%mod)*(i%mod)+mod)%mod;
    ans%=mod;
    return ans;
}
ll nCr(ll n,ll r,ll mod)
{
    ll ri=binexp(fact(r,mod),mod-2,mod);
    ll nri=binexp(fact(n-r,mod),mod-2,mod);
    ll ans=(((fact(n,mod)%mod)*(ri%mod))%mod);
    ans=(((ans%mod)*(nri%mod))%mod);
    ans%=mod;
    ans=(ans+mod)%mod;
    return ans;
}

void solve()
{
  ll n;
  cin>>n;
  ll arr[n];
  lop(i,n) cin>>arr[i];
  ll freq[n+1]={};
  lop(i,n) freq[arr[i]]++;
  vector<pll> v;
  lop1(i,n) if(freq[i]!=0) v.pb({i,freq[i]});
  ll ans=1e18;
  ll m=sz(v);
  lop(i,m)
  {
    ll c=m;
    ll a=i;
    ll b=0;
    for(ll j=i;j<m;j++)
    {
      if(a>b+v[j].ss)
      {
        c--;
        b+=v[j].ss;
      }
      else if(j==m-1 && a==b+v[j].ss)
      {
        c--;
        b+=v[j].ss;
      }
      else a++;
    }
    ans=min(ans,c);
  }
  cout<<ans<<endl;
}

int main()
{
  IOS;
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
  ll t;
  t=1;
  cin>>t;
  while(t--)
  {
    solve();
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that maximize the failure probability. Here are the key points to consider:

1. **Edge Cases:**
   - Single cake (n=1)
   - All cakes have the same tastiness (n=1 to 5000)
   - Cakes with unique tastiness values (n=1 to 5000)
   - Cakes with repeated tastiness values (n=1 to 5000)

2. **Scenarios:**
   - Alice can only eat a few cakes, and Bob can eat many.
   - Alice can eat all the cakes.
   - Cakes are arranged in a way that Alice and Bob take turns optimally.

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug is:

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `3 1 1 1`
   - **Expected Output:** `1`
   - **Reason:** Alice can only eat one cake because all cakes have the same tastiness.

2. **Test Case 2:**
   - **Input:** `5 1 2 3 4 5`
   - **Expected Output:** `3`
   - **Reason:** Alice can eat cakes with tastiness values 1, 3, and 5. Bob will eat the remaining cakes.

3. **Test Case 3:**
   - **Input:** `5 5 4 3 2 1`
   - **Expected Output:** `3`
   - **Reason:** Alice can eat cakes with tastiness values 5, 3, and 1. Bob will eat the remaining cakes.

4. **Test Case 4:**
   - **Input:** `7 1 2 3 3 2 1 1`
   - **Expected Output:** `4`
   - **Reason:** Alice can eat cakes with tastiness values 1, 3, 3, and 1. Bob will eat the remaining cakes.

This test input is designed to cover a variety of scenarios, including edge cases and more complex arrangements of cake tastiness values. It should help in exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import sys
input = sys.stdin.readline
from math import pi, inf, atan2, isqrt, comb
from heapq import nsmallest, nlargest, heapify, heappop, heappush
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from itertools import accumulate, count, combinations, permutations, pairwise, repeat, cycle, zip_longest, product
from functools import cache

def main():
    M = int(input().strip())
    for _ in range(M):
        n = int(input().strip())
        nums = list(map(int, input().split()))
        c = Counter(nums)
        cs = sorted([(i, k) for i, k in c.items()])
        L = len(cs)

        dp = [0] * n
        for i in range(1, L):
            key, val = cs[i]
            if i >= val:
                dp[val] = max(dp[val], 1)
                for lst in range(i-val+1):
                    time = dp[lst] + lst
                    if i >= time + val:
                        dp[dp[lst]+val] = max(dp[dp[lst]+val], dp[lst]+1)
        # print(cs)
        # print(dp)
        print(L-max(dp))

main()

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

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

Think step by step.

Response:
```plaintext
3
4
4 3 2 1
5
1 2 3 4 5
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: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
n = int(input())
for _ in range(n):
    k = int(input())
    a = list(map(int, input().split()))
    a = sorted(a)
    s = sorted(list(set(a)))
    c, d = [], []
    m = 0
    for i in s:
        count = a.count(i)
        c.append((i, count, m))
        d.append((count, i, m))
        m += 1
    d = sorted(d)
    pot = len(s)
    r = 0
    i = 0
    e = c + []
    f = [0]
    while min(f) < 5001:
        f = [d[i][0] if (r + d[i][0] < pot) and (d[i][2] >= e[r + d[i][0]][2]) else 5001 for i in range(pot)]
        if min(f) < 5001:
            i = f.index(min(f))
            r += d[i][0]
            e.remove((d[i][1], d[i][0], d[i][2]))
            d.pop(i)
            pot -= 1
    print(pot)

**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
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CFTemp {

    static int MOD = 998244353;
    static int INF = 0x3f3f3f3f;

    public static void main(String[] args) throws IOException {
        int t = Reader.nextInt();
        StringBuilder sb = new StringBuilder();
        while (t-- > 0) {
            int n = Reader.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = Reader.nextInt();
            }
            sb.append(solve(arr)).append("\n");
        }
        System.out.print(sb.toString());
    }

    static String solve(int[] arr) {
        StringBuilder sb = new StringBuilder();
        int n = arr.length;
        Arrays.sort(arr);

        List<int[]> list = new ArrayList<>();
        TreeMap<Integer, LinkedList<Integer>> map = new TreeMap<>();
        for (int i = 0; i < n; i++) {
            int j = i;
            while (j < n && arr[j] == arr[i]) {
                j++;
            }
            list.add(new int[] { i, j - i });
            map.computeIfAbsent(j - i, x -> new LinkedList<>()).add(list.size() - 1);
            i = j - 1;
        }

        int result = 0;
        int power = 0;
        for (int i = 0; i < list.size(); i++) {
            while (i < list.size() && list.get(i)[0] == -1) {
                i++;
            }
            if (i == list.size()) {
                break;
            }
            result++;
            power++;
            for (Map.Entry<Integer, LinkedList<Integer>> entry : map.entrySet()) {
                LinkedList<Integer> l = entry.getValue();
                while (!l.isEmpty() && l.getFirst() <= i) {
                    l.removeFirst();
                }
            }
            while (!map.isEmpty() && map.firstEntry().getValue().isEmpty()) {
                map.pollFirstEntry();
            }
            if (map.isEmpty()) {
                break;
            } else {
                Map.Entry<Integer, LinkedList<Integer>> next = map.firstEntry();
                if (next.getKey() <= power) {
                    power -= next.getKey();
                    int remove = next.getValue().pollFirst();
                    list.get(remove)[0] = -1;
                }
            }
        }

        return sb.append(result).toString();
    }

    static class Reader {
        static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        static StringTokenizer tokenizer = new StringTokenizer("");

        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        }

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

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

        static double nextDoublet() throws IOException {
            return Double.parseDouble(next());
        }
    }

}

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

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

Think step by step.

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

1. **All Cakes Have the Same Tastiness**: This scenario can test how the code handles cases where Alice cannot eat any more cakes after the first one.
2. **Cakes with Increasing Tastiness**: This scenario can test the code's ability to handle a sequence where Alice can always eat a cake on her turn.
3. **Cakes with Decreasing Tastiness**: This scenario can test the code's ability to handle a sequence where Bob's choices can significantly impact the game.
4. **Cakes with Mixed Tastiness Values**: This scenario can test the code's ability to handle a mix of increasing and decreasing tastiness values, which can be more complex.

Given the constraints and the need for a small-scale test input, let's create a test case that combines these elements:

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

### Explanation:
1. **Test Case 1: All Cakes Have the Same Tastiness**
   - Input: `3 1 1 1`
   - Expected Output: `1`
   - Explanation: Alice can only eat one cake because all cakes have the same tastiness.

2. **Test Case 2: Cakes with Increasing Tastiness**
   - Input: `5 1 2 3 4 5`
   - Expected Output: `3`
   - Explanation: Alice can eat cakes with tastiness values 1, 3, and 5.

3. **Test Case 3: Cakes with Decreasing Tastiness**
   - Input: `5 5 4 3 2 1`
   - Expected Output: `3`
   - Explanation: Alice can eat cakes with tastiness values 5, 3, and 1.

4. **Test Case 4: Cakes with Mixed Tastiness Values**
   - Input: `7 1 3 2 4 5 3 1`
   - Expected Output: `4`
   - Explanation: Alice can eat cakes with tastiness values 1, 3, 4, and 5.

This test input should help expose any bugs in the given buggy code by covering a variety of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import java.util.*;
import java.io.*;
public class CGreedy {

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        // your code goes here
        int t = sc.nextInt();
        int count=0;
        while (t-- > 0) {
            count++;
            int n = sc.nextInt();
            TreeMap<Integer, Integer> hp = new TreeMap<>();
            for (int i = 0; i < n; i++) {
                int x = sc.nextInt();
                hp.put(x, hp.getOrDefault(x, 0) + 1);
            }
            Pair[] arr = new Pair[hp.size()];
            int index = 0;
            for (Map.Entry<Integer, Integer> entry : hp.entrySet()) {
                arr[index] = new Pair(entry.getKey(), entry.getValue(),index);
                index++;
            }
            for(int i=0;i<arr.length-1;i++){
                if(arr[i].y<arr[i+1].y) arr[i].value=true;
            }
            if(count==237)System.out.println(Arrays.toString(arr));
            Comparator<Pair> c=new Comparator<Pair>() {
                @Override
                public int compare(Pair o1, Pair o2) {
                    if(o1.value!=o2.value){
                        if(o1.value)
                            return -1;
                        return 1;
                    }
                    if(o1.y==o2.y) {
//                        if(o1.y==1)
                        return o1.x - o2.x;
//                        else return o2.x-o1.x;
                    }
                    return o1.y-o2.y;
                }
            };
            Pair[] bobArr=new Pair[arr.length];
            for(int i=0;i<arr.length;i++){
                bobArr[i]=arr[i];          //not shallow , not deep cloning
            }
            Arrays.sort(bobArr,c);
//            System.out.println(Arrays.toString(bobArr));
//            for(int i=0;i<bobArr.length;i++){
//                System.out.print(bobArr[i].value+" ");
//            }
//            System.out.println();
            int p1 = 0;
            int p2 = 0;
            int alice = 0;
            outerLoop:while(true) {
                //alice roll
                while(p1!=arr.length){
                    if( !arr[p1].alice && arr[p1].y!=0){
                        alice++;
                        arr[p1].alice=true;
                        break;
                    }
                    else {
                        p1++;
                    }
                }
                if(p1==arr.length) break;
                //
                //Bob turn
                while(p2!=bobArr.length){
                    if(bobArr[p2].alice){
                        p2++;
                    }
                    else if(bobArr[p2].x<=arr[p1].x){
                        p2++;
                    }
                    else if(bobArr[p2].y==0){
                        p2++;
                    }
                    else if(bobArr[p2].ind-p1<bobArr[p2].y){
                        p2++;
                    }
                    else{
                        bobArr[p2].y--;
                        break;
                    }
                }
            }
            System.out.println(alice);
//           outerLoop: while(true){
//                if(p2<=p1)
//                    p2=p1+1;
//                if(p1==arr.length || p2==arr.length)
//                    break;
//                while(arr[p2].y>p2-p1){
//                    p2++;
//                    if(p2==arr.length) break outerLoop;
//                }
//                while(arr[p2].y!=0){
//                    alice++;
//                    arr[p2].y--;
//                    p1++;
//                    if(p1==p2) p1++;
//                }
//            }
//            alice+=(arr.length-p1);
//            System.out.println(alice);
//        }

        }
            pw.flush();
    }
//    public static int AliceRoll(Pair[] arr,int p1){
//        int alice=0;
//        while(p1!=arr.length){
//            if(arr[p1].y!=0){
//                alice++;
//                arr[p1].alice=true;
//                break;
//            }
//            else {
//                p1++;
//            }
//        }
//        return alice;
//    }
//    public static boolean BobRoll(Pair[] bobArr,int p2){
//        while(p2!=bobArr.length){
//            if(bobArr[p2].alice){
//                p2++;
//            }
//            else if(bobArr[p2].x<=)
//        }
//    }
    static class Pair{
        int x,y;
        boolean alice;
        boolean bob;
        int ind;
        boolean value; //smaller than the next one(smaller in y)

        public Pair(int x,int y,int ind){
            this.x=x;
            this.y=y;
            this.ind=ind;
        }
        public String toString(){
            return "("+x+","+y+")";
        }
    }
    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

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

using namespace std;
#ifndef HARSHIT
    #define debug(...) cout << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__);
#else
    #define debug(...);
#endif
template <typename T> std::ostream &operator<<(std::ostream &stream, const vector<T> &vec) { for (size_t i = 0; i < vec.size(); i++) { stream << vec[i]; if (i != vec.size() - 1) stream << ' '; } return stream; } template <typename T> std::istream &operator>>(std::istream &stream, vector<T> &vec) { for (T &x : vec) stream >> x; return stream; } template <typename T, typename U> std::ostream &operator<<(std::ostream &stream, const pair<T, U> &pr) { stream << pr.first << ' ' << pr.second; return stream; } template <typename T, typename U> std::istream &operator>>(std::istream &stream, pair<T, U> &pr) { stream >> pr.first >> pr.second; return stream; } template <typename A, typename B> string to_string(pair<A, B> p); template <typename A, typename B, typename C> string to_string(tuple<A, B, C> p); template <typename A, typename B, typename C, typename D> string to_string(tuple<A, B, C, D> p); string to_string(const string &s) { return '"' + s + '"'; } string to_string(char c) { string s; s += c; return s; } string to_string(const char *s) { return to_string((string)s); } string to_string(bool b) { return (b ? "1" : "0"); } string to_string(vector<bool> v) { bool first = true; string res = "{ "; for (int i = 0; i < static_cast<int>(v.size()); i++) { if (!first) { res += ", "; } first = false; res += to_string(v[i]); } res += " }"; return res; } template <size_t N> string to_string(bitset<N> v) { string res = ""; for (size_t i = 0; i < N; i++) { res += static_cast<char>('0' + v[i]); } return res; } template <typename A> string to_string(A v) { bool first = true; string res = "{ "; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += " }"; return res; } template <typename A, typename B> string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; } template <typename A, typename B, typename C> string to_string(tuple<A, B, C> p) { return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")"; } template <typename A, typename B, typename C, typename D> string to_string(tuple<A, B, C, D> p) { return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")"; } void debug_out() { cout << endl; } template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) { cout << " " << to_string(H); debug_out(T...); }

#define int long long
#define endl '\n'
# define Harshit  cin.tie(nullptr)->sync_with_stdio(false);
#define X first
#define Y second
#define  V vector
#define pb push_back
  int n;
const bool multipleTestCases = 1;
int F(int idx , int ops , vector<int>& A){
    if(idx == n+1){
        return 0;
    }
    if(A[idx] <= ops){
        return min(1 + F(idx+1 , ops+1 , A) , F(idx+1 ,ops-A[idx] , A ));
    }
    return 1 + F(idx+1 , ops+1 , A);
}

void harshit() {

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

    cout << F(0, 0, A) << endl;
}

signed main()
{
    Harshit
    int t = 1;
    multipleTestCases and cin >> t;
    for (int i = 1; i <= t; i++)
    {
        // cout << "Case #" << i << ": ";
        harshit();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define endl "\n"
#define JaldiKrtehaina       \
    ios::sync_with_stdio(0); \
    cin.tie(0);
// ;)
#define M 5010
ll dp[M][M];
ll ct[M];

bool isPrime(int n)
{
    if (n <= 1)
        return false;
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;
    return true;
}

ll highestPowerOf2(ll n)
{
    return (n & (~(n - 1)));
}
ll check(ll i, ll ctt)
{
    if (ctt < 0)
    {
        return -1;
    }
    if (i == 0)
    {
        if (ctt)
        {
            return -1;
        }
        else
        {
            return 0;
        }
    }
    if (dp[i][ctt] != -1)
    {
        return dp[i][ctt];
    }
    if (ct[i] != 0)
    {
        dp[i][ctt] = check(i - 1, ctt - 1);
    }
    else
    {
        dp[i][ctt] = check(i - 1, ctt);
    }
    if (check(i - 1, ctt) != -1 && check(i - 1, ctt) + ct[i] <= ctt)
    {
        if (dp[i][ctt] == -1 || check(i - 1, ctt) + ct[i] < dp[i][ctt])
        {
            dp[i][ctt] = check(i - 1, ctt) + ct[i];
        }
    }
    return dp[i][ctt];
}

void solve()
{
    ll n;
    cin >> n;
    vector<ll> a(n);
    for (ll i = 0; i < n + 1; i++)
    {
        fill_n(dp[i], n + 1, -1);
    }
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i];
        ct[a[i]]++;
    }
    ll ans = n;
    for (ll i = n - 1; i >= 0; i--)
    {
        if (check(n, i) != -1)
        {
            ans = i;
        }
    }
    cout << ans << endl;
    fill_n(ct, n + 1, 0LL);
}

int main()
{
    JaldiKrtehaina
        ll t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
q = int(input())
for it in range(q):
    n = int(input())
    a = [0 for i in range(n + 1)]
    for i in map(int, input().split()):
        a[i] += 1
    ans = [[i, []] for i in a if i > 0]
    ans[0][1] = [[1,0]]
    for i in range(1, len(ans)):
        for j in ans[i - 1][1]:
            # print(j)
            ans[i][1].append([j[0] + 1, j[1]])
            if j[0] >= ans[i][0]:
                ans[i][1].append([j[0] - ans[i][0], j[1] + 1])
    # print(ans)
    ma = 0
    for j in ans[len(ans) - 1][1]:
        if j[1] > ma:
            ma = j[1]
    print(len(ans) - ma)

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

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

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
3
1 1 2
5
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: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
from heapq import heappop, heappush
from collections import Counter, defaultdict

def solve():
    n = int(input())
    arr = list(map(int, input().split()))

    d_arr = []
    c = Counter(arr)
    for key in sorted(c):
        d_arr.append([key, c[key]])

    # print(d_arr)
    done = [-1] * len(d_arr)

    cur_val = 0
    for i in range(len(d_arr)):
        if cur_val >= d_arr[i][1]:
            done[i] = 0
            cur_val -= d_arr[i][1]
        else:
            cur_val += 1
    # print(done)
    for i in range(len(d_arr)):
        if done[i] != -1:
            happen = False
            stack = []
            cur_val = 1
            for j in range(len(d_arr)):
                if j == i:
                    continue
                if done[j] == -1:
                    if j > i and d_arr[j][1] <= d_arr[i][1] and cur_val >= d_arr[j][1]:
                        happen = True
                        done[j] = 0
                        cur_val -= (d_arr[j][1])
                        stack.append(j)
                    else:
                        cur_val += 1
                else:
                    while cur_val - d_arr[j][1] < 0:
                        temp = stack.pop(-1)
                        done[temp] = -1
                        cur_val += d_arr[temp][1] + 1
                    cur_val -= d_arr[j][1]
                if cur_val < 0:
                    raise Exception("")

            if happen:
                done[i] = -1
        # print(done, i)
    print(done.count(-1))

for t in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
4
4 3 2 1
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: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

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

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

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

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

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

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

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

        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class Pair
    {
        int val, i;
        Pair(int a, int b)
        {
            val = a;
            i = b;
        }
    }

    public static void main(String args[]) throws IOException
    {
        FastReader fr = new FastReader();
        int t = fr.nextInt();
        // int t = 1;
        StringBuilder _v = new StringBuilder();
        PrintWriter out = new PrintWriter(System.out);
        while(t-- > 0)
        {
            int n = fr.nextInt();
            int[] f = new int[n+1];
            for(int i = 0; i < n; i++)
                f[fr.nextInt()]++;
            List<Integer> ar = new ArrayList<>();
            for(int i = 1; i <= n; i++)
            {
                if(f[i] > 0)
                    ar.add(i);
            }
            // _v.append("Bob eats : ");
            int x = func(0, f, ar);
            // _v.append(x);
            // _v.append("\tAlice eats : ");
            _v.append(ar.size() - x);
            _v.append("\n");
        }
        out.print(_v);
        out.flush();
    }

    public static int func(int turn, int[] f, List<Integer> ar)
    {
        int n = ar.size();
        int ans = 0, smaller = 0;
        int startIndex = -1, a = turn;
        // Skip cakes eaten by alice
        for(int j = 0; j < n && a > 0; j++)
        {
            if(f[ar.get(j)] > 0)
            {
                a--;
            }
            startIndex = j;
        }
        for(int j = startIndex + 1; j < n; j++)
        {
            int temp = f[ar.get(j)];
            if(temp > 0)
            {
                if(smaller >= temp)
                {
                    // System.out.printf("Bob can eat %d with f = %d after turn = %d\n", ar.get(j), temp, turn);
                    // Bob can eat it out
                    f[ar.get(j)] = 0;
                    ans = Math.max(1 + func(turn + temp, f, ar), ans);
                    f[ar.get(j)] = temp;
                }
                smaller++;
            }
        }
        return 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
4
1 4 2 3
4
3 4 1 4
8
4 3 2 5 6 8 3 4
```