**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

Buggy Code:
import java.util.*;

public class C_Ticket_Hoarding{
    public static  void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            long m = sc.nextLong();
            long k = sc.nextLong();

            long[][] arr = new long[n][2];
            for(int i=0; i<n; i++){
                arr[i][0] = sc.nextLong();
                arr[i][1] = i;
            }

            Arrays.sort(arr,(a,b)->(a[0]>=b[0]?1:-1));

            int days = (int)((k+m-1)/m);
            PriorityQueue<long[]> pq = new PriorityQueue<>((a,b)->(a[1]>=b[1]?1:-1));
            long kk = k;
            long rate = 0;
            long sum = 0;
            try{
                for(int i=0; i<days; i++){
                    pq.add(new long[]{arr[i][0],arr[i][1]});
                }

                // long len = m;
                while(pq.size()>1){
                    long[] rem = pq.remove();
                    sum += (rem[0]+rate)*m;
                    k -= m;
                    rate += m;
                    // System.out.println("Sum=" + sum + rem[0]);
                }
                if(pq.size()>0)
                    sum += (pq.remove()[0]+rate)*k;
            }catch(RuntimeException re){
                System.out.println("error in prev pq section ");
            }

            // pq.clear();
            long x = kk%m;
            try{
                for(int i=0; i<days; i++){
                    pq.add(new long[]{arr[i][0],arr[i][1]});
                }
                if(kk%m==0){
                    x = m;
                }
                long sum1 = 0;
                if(pq.size()>0){
                    sum1 = (pq.remove()[0])*x;
                }
                rate = x;
                while(pq.size()>0){
                    long[] rem = pq.remove();
                    sum1 += (rem[0]+rate)*m;
                    rate += m;
                    // System.out.println("summ=" + sum1) ;
                }
                // System.out.println(sum + " " + sum1);
                long ans = (long)Math.min(sum,sum1);
                System.out.println(ans);
            }catch( RuntimeException re){
                System.out.println("Runtime error in above section");
            }
        }
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

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

public class A {
    static final long mod = 1000000007;

    private static Map.Entry<Long, Long> pair(long key, long value) {
        return new AbstractMap.SimpleEntry<>(key, value);
    }

    private static boolean isPrime(int n) {
        return n > 1 && IntStream.range(2, (int) Math.sqrt(n) + 1).noneMatch(i -> n % i == 0);
    }

    private static int getGCD(long a, long b) {
        return BigInteger.valueOf(a).gcd(BigInteger.valueOf(b)).intValue();
    }

    private static int getLCM(int a, int b) {
        return BigInteger.valueOf(Math.abs(a)).multiply(BigInteger.valueOf(Math.abs(b))).divide(BigInteger.valueOf(a).gcd(BigInteger.valueOf(b))).intValue();
    }

    private static long binaryExponential(long a, long b, long m) {
        return b > 0 ? ((b & 1) != 0 ? (a * binaryExponential((a * a) % m, b >> 1, m)) % m : binaryExponential((a * a) % m, b >> 1, m)) : 1;
    }

    static Scanner Sc = new Scanner(System.in);

    public static void main(String[] args) {
        Test_Cases();
    }

    static void Test_Cases() {
        long tc = Sc.nextLong();
        while (tc-- > 0) {
            int n = Sc.nextInt();
            solve(n);
        }
    }

    private static void solve(int n) {
        int m = Sc.nextInt();
        int k = Sc.nextInt();
        int[] arr = new int[n];
        List<Pair> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            arr[i] = Sc.nextInt();
            list.add(new Pair(arr[i], i));
        }
        int[] qu = new int[n];
        calculate2(list,n,k,m,qu);
        calculateAns(n,qu,arr);
    }

    static void calculate2(List<Pair> list,int n,int k,int m,int[]qu) {
        list.sort(new Comparator<Pair>() {
            @Override
            public int compare(Pair p1, Pair p2) {
                return Integer.compare(p1.first, p2.first);
            }
        });

        for (Pair pair : list) {
            if (k > 0) {
                qu[pair.second] = Math.min(m, k);
                k -= Math.min(m, k);
            }
        }
    }
    static void calculateAns(int n,int[]q,int []arr){
        int ans = 0;
        int extra = 0;
        for (int i = 0; i < n; i++) {
            ans += (q[i] * (arr[i] + extra));
            extra += q[i];
        }
        System.out.println(ans);
    }
}

 class Pair {
    int first, second;

    Pair(int first, int second) {
        this.first = first;
        this.second = second;
    }
}

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

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

Think step by step.

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

**Explanation:**
- This test case is designed to highlight a potential issue in the buggy code where it might not correctly handle the scenario of purchasing tickets with a limit of 1 ticket per day.
- The input consists of:
  - `t = 1` (one test case)
  - `n = 3`, `m = 1`, `k = 3` (3 days, 1 ticket per day, need to buy 3 tickets)
  - `a = [1, 2, 3]` (prices for each day)
- The optimal way to buy 3 tickets is:
  - Day 1: Buy 1 ticket for 1 unit.
  - Day 2: Buy 1 ticket for 3 units (since the price increases by 1 due to the previous purchase).
  - Day 3: Buy 1 ticket for 5 units (since the price increases by 1 due to the previous purchase).
- The total cost should be 1 + 3 + 5 = 9.
- If the buggy code incorrectly calculates the total cost, this test case will expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int long long
void solve()
{
    int n, m, k;
    cin >> n >> m >> k;
    vector<pair<ll, ll>> a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].first;
        a[i].second = i;
    }
    sort(a.begin(), a.end());
    int ans = 0;
    int need = (k + m - 1) / m;
    int fir = (k % m == 0 ? m : k % m);
    vector<pair<ll, ll>> res;
    for (int i = 0; i < need; i++)
    {
        res.push_back(a[i]);
    }
    auto cmp = [&](pair<ll, ll> a, pair<ll, ll> b)
    {
        return a.second < b.second;
    };
    sort(res.begin(), res.end(), cmp);
    // cout << fir << ' ';
    int len = res.size();
    for (int i = 0; i < len; i++)
    {
        if (i)
            ans += res[i].first * m;
        else
            ans += res[i].first * fir;
        if (i)
        {
            ans += (fir + (i - 1) * m) * m;
        }
    }
    cout << ans << '\n';
}
signed main()
{
    int _;
    cin >> _;
    while (_--)
        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
1
3 2 5
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define pb push_back
#define pob pop_back
#define ff first
#define ss second
#define ump unordered_map
#define pll pair<ll, ll>
#define tpll tuple<ll, ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vpll vector<pair<ll, ll>>
#define all(x) x.begin(), x.end()
#define gl greater<ll>
#define pq priority_queue<ll>
#define pqm priority_queue<ll, vl, gl>
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define neg cout << "-1\n"
#define inf 1e18
#define md 1e9 + 7
#define f0(i, n) for (ll i = 0; i < (n); i++)
#define f(i, a, n) for (ll i = a; i <= (n); i++)
#define rf(i, n, a) for (ll i = n; i >= (a); i--)
#define memo(x, y) memset((x), (y), sizeof(x));
#define lower(s) transform(all(s), s.begin(), ::tolower)
#define upper(s) transform(all(s), s.begin(), ::toupper)

bool sec(const pair<int, int> &a, const pair<int, int> &b){return (a.second < b.second);}

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

// ll ans=0;
// void dfs(ll node,vector<vector<pair<int,int>>>& tree, ll par, ll itr,ll ind ){
//     ans=max(ans,itr);
//     for(auto& [x,y]:tree[node]){
//         if(x==par) continue;
//         dfs(x,tree, node ,itr+(y<ind),y);
//     }
// }

// ll factorial(int n)
// {
//     ll f = 1;
//     for (int i = 1; i <= n; i++)
//         f = (f * i) % M;
//     return f;
// }

void __solve()
{

	int n,m,k; cin>>n>>m>>k;
	vector<int> a(n);

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

	}
	sort(all(a));
	ll ans=0;
	ll dd=(k+m-1)/m;
	ll sm=0;
	for(int i=0;i<dd;i++){
		if(i==dd-1 and k%m){
			//baase;;
			int cc=k%m;
			ans+=cc*(a[i]+ m*i);

			break;
		}
		ans += m*(a[i]+ m*i);

	}
	cout<<ans<<endl;

}

int main()
{

    cin.tie(0)->sync_with_stdio(0);
    int _ts;
    cin >> _ts;
    while (_ts--)
    {
        __solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code logic seems to be flawed in how it handles the day-by-day ticket purchases and the subsequent price increases. Specifically, it might not correctly account for the dynamic price changes over multiple days.

Here are the steps to generate a challenging test input:

1. **Small-scale Test Case**: Ensure the test case is small enough to fit within the output length limit.
2. **Edge Cases**:
   - **Single Day**: Test the scenario where tickets are available for only one day.
   - **Maximum Tickets per Day**: Test the scenario where the number of tickets to be bought is exactly the maximum allowed per day.
   - **Exact Ticket Count**: Test the scenario where the number of tickets to be bought is exactly a multiple of the maximum tickets per day.
   - **Price Increase**: Test the scenario where the price increase due to ticket purchases significantly affects the total cost.
3. **Complex Scenario**: Test a scenario where the optimal strategy involves buying tickets over multiple days with different purchase amounts.

Let's generate a test input that covers these aspects:

```plaintext
4
1 10 1
100
4 2 4
1 2 3 4
5 2 5
1 2 3 4 5
6 3 9
1 2 3 4 5 6
```

Explanation:
1. **Single Day**:
   - `n = 1`, `m = 10`, `k = 1`, `a = [100]`
   - The only way to buy 1 ticket is to buy it on the only day available, which costs 100.

2. **Maximum Tickets per Day**:
   - `n = 4`, `m = 2`, `k = 4`, `a = [1, 2, 3, 4]`
   - The optimal strategy is to buy 2 tickets each on the first two days, costing 1 + 2 + 3 + 4 = 10.

3. **Exact Ticket Count**:
   - `n = 5`, `m = 2`, `k = 5`, `a = [1, 2, 3, 4, 5]`
   - The optimal strategy is to buy 2 tickets on the first two days and 1 ticket on the third day, costing 1 + 2 + 3 + 4 + 5 = 15.

4. **Complex Scenario**:
   - `n = 6`, `m = 3`, `k = 9`, `a = [1, 2, 3, 4, 5, 6]`
   - The optimal strategy is to buy 3 tickets each on the first three days, costing 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.

This test input should expose any issues in the buggy code, particularly in how it handles the price increases and the dynamic nature of the ticket purchases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

Buggy Code:
T = int(input())

from heapq import *

while T:
    '''
    k张票
    每天不能买超过m张票
    '''
    n,m,k = map(int,input().strip().split())
    arr = list(map(int,input().strip().split()))
    for i,num in enumerate(arr):
        arr[i] = (num,i)

    arr.sort(key = lambda x:(x[0],-x[1]))

    t = 0
    tmp = [[0,0] for _ in range(n)]
    for num,i in arr:
        t += m
        tmp[i][0] = num
        if t >= k:
            break

    res = 0
    t = 0
    for i in range(n-1,-1,-1):
        if not tmp[i][0]:
            continue

        if t + m >= k:
            tmp[i][1] = k - t
            break
        else:
            tmp[i][1] = m
            t += m

#     print(tmp)
    off = 0
    for num,t in tmp:
        res += (num + off) * t
        off += t

    print(res)
    T -= 1

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

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

Think step by step.

Response:
```plaintext
3
5 1 5
1 2 3 4 5
5 10 10
1 2 3 4 5
5 2 9
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: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

Buggy Code:
for _ in range(int(input())):

    def solve():
        n, m, k= map(int, input().split(' '))
        li = list(map(int, input().split(' ')))

        dp = {}

        def solve2(idx, tkt):
            if tkt ==0:
                return 0
            if idx==n:
                return 10**10
            if (idx, tkt) in dp: return dp[(idx, tkt)]
            res = 10**10
            for i in range(min(tkt, m)+1):
                res = min(res, i*(li[idx]+k-tkt)+solve2(idx+1, tkt-i))
            dp[(idx, tkt)] = res
            return res
        print(solve2(0, k))
    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
1
3 2 4
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

Buggy Code:
import java.util.*;
public class CP {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        while (tc-- > 0) {
            int n = sc.nextInt();
            long m = sc.nextLong();
            long k = sc.nextLong();
            long inp[][] = new long[n][2];
            for(int i=0;i<n;i++){
                inp[i][0] = i;
                inp[i][1] = sc.nextLong();
            }
            Arrays.sort(inp,Comparator.comparingDouble(o -> o[1]));
            int nn = (int)(k + (m-1))/(int)m;
            long q = m;
            long rem = k%m;

            // System.out.println(nn);
            // System.out.println(q);
            // System.out.println(rem);
            long req[][] = new long[nn][2];
            for(int i=0;i<nn;i++){
                   req[i][0] = inp[i][0];
                req[i][1] = inp[i][1];
            }
              Arrays.sort(req,Comparator.comparingDouble(o -> o[0]));
              long sum = 0;
              long max  =  Long.MIN_VALUE;
              int idx = -1;
              for(int i = 0; i < nn; i++){
                if((sum+req[i][1]) > max){max = req[i][1];idx = i;}
                sum+= q;
              }
              sum = 0;
              long ans = 0;
              for(int i=0;i<nn;i++){
                if(i != idx || rem == 0){
                   ans+= (q*(sum+req[i][1]));
                   sum+= q;
                }
                else{
                    ans+= (rem*(sum+req[i][1]));
                    sum+= rem;
                }
              }
              System.out.println(ans);
        }
        sc.close();
    }
}
// 4
// 4 2 3
// 8 6 4 2
// 4 2 8
// 8 6 4 2
// 5 100 1
// 10000 1 100 10 1000
// 6 3 9
// 5 5 5 5 5 5
// OutputCopy
// 10
// 64
// 1
// 72

**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, let's consider the following points:

1. **Edge Cases**:
   - When `k` is very small.
   - When `k` is very large.
   - When `m` is very small.
   - When `m` is very large.
   - When `n` is very small.
   - When `n` is very large.

2. **Behavior of the Code**:
   - The code sorts the days based on the ticket prices and then tries to distribute the tickets.
   - It calculates the number of full days (`nn`) and the remainder (`rem`).
   - It then tries to distribute the tickets in a way that minimizes the cost.

3. **Potential Bugs**:
   - The code might not correctly handle the distribution of tickets when `k` is not a multiple of `m`.
   - The code might not correctly handle the sorting and distribution of tickets when the prices are very close to each other.

Given these points, let's create a test case that can expose the bug:

- **Small `k` and `m`**:
  - This will test the code's ability to handle small values and ensure it correctly distributes the tickets.
- **Large `k` and `m`**:
  - This will test the code's ability to handle large values and ensure it correctly calculates the total cost.
- **Mixed `a` values**:
  - This will test the code's ability to handle a mix of low and high ticket prices.

Here is a small-scale test input:

```plaintext
4
3 2 3
1 2 3
4 2 8
1 2 3 4
5 100 1
10000 1 100 10 1000
6 3 9
1 2 3 4 5 6
```

This test input includes:
1. A small `k` and `m` with mixed prices.
2. A large `k` with a small `m` and mixed prices.
3. A large `k` with a large `m` and mixed prices.
4. A large `k` with a small `m` and mixed prices.

This should help in exposing any potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ticket Hoarding
time_limit_ms: 2000
memory_limit_mb: 256
Description: As the CEO of a startup company, you want to reward each of your $$$k$$$ employees with a ticket to the upcoming concert. The tickets will be on sale for $$$n$$$ days, and by some time travelling, you have predicted that the price per ticket at day $$$i$$$ will be $$$a_i$$$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

- A person may purchase no more than $$$m$$$ tickets per day.
- If a person purchases $$$x$$$ tickets on day $$$i$$$, all subsequent days (i.e. from day $$$i+1$$$ onwards) will have their prices per ticket increased by $$$x$$$.

For example, if $$$a = [1, 3, 8, 4, 5]$$$ and you purchase $$$2$$$ tickets on day $$$1$$$, they will cost $$$2$$$ in total, and the prices from day $$$2$$$ onwards will become $$$[5, 10, 6, 7]$$$. If you then purchase $$$3$$$ more tickets on day $$$2$$$, they will cost in total an additional $$$15$$$, and the prices from day $$$3$$$ onwards will become $$$[13, 9, 10]$$$.

Find the minimum spending to purchase $$$k$$$ tickets.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$$$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price per ticket for each of the upcoming $$$n$$$ days.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print one integer: the minimum amount of money needed to purchase exactly $$$k$$$ tickets.

Examples:
input:
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
output:
10
64
1
72

Note: In the first test case, one optimal way to buy $$$3$$$ tickets is as follows:

- Buy $$$0$$$ tickets on the first day. The prices per ticket for the remaining days are $$$[6, 4, 2]$$$.
- Buy $$$0$$$ tickets on the second day. The prices per ticket for the remaining days are $$$[4, 2]$$$.
- Buy $$$1$$$ ticket on the third day with cost $$$4$$$. The price per ticket for the remaining day is $$$[3]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$6$$$.

In the second test case, there is only one way to buy $$$8$$$ tickets:

- Buy $$$2$$$ tickets on the first day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6, 4]$$$.
- Buy $$$2$$$ tickets on the second day with cost $$$16$$$. The prices per ticket for the remaining days are $$$[8, 6]$$$.
- Buy $$$2$$$ tickets on the third day with cost $$$16$$$. The price per ticket for the remaining day is $$$[8]$$$.
- Buy $$$2$$$ tickets on the fourth day with cost $$$16$$$.

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

import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.sqrt;
import static java.lang.Math.abs;
import static java.lang.Math.ceil;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.sort;
import static java.util.Arrays.toString;
public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;
    static StringBuilder result = new StringBuilder();
    static String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    static int nextInt() { return Integer.parseInt(next()); }
    static long nextLong() { return Long.parseLong(next()); }
    static double nextDouble() { return Double.parseDouble(next()); }

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

    static long[] nextLongArray(int n){
        long[] a = new long[n];
        for(int i=0; i <n; i++){
            a[i] = nextLong();
        }
        return a;
    }

    static int[][] nextIntMatrix(int n, int m){
        int[][] a = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = nextInt();
            }
        }
        return a;
    }

    static long[][] nextLongMatrix(int n, int m){
        long[][] a = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = nextLong();
            }
        }
        return a;
    }

    static double[][] nextDoubleMatrix(int n, int m){
        double[][] a = new double[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = nextDouble();
            }
        }
        return a;
    }

    static void add(Object obj){
        result.append(obj).append(' ');
    }
    static void addln(Object obj){
        result.append(obj).append('\n');
    }
    static void print(){
        System.out.println(result);
    }
    static void print(Object obj){
        System.out.print(obj);
    }
    static void println(Object obj){
        System.out.println(obj);
    }

    static <T extends Comparable<T>> void s(T[] array, Comparator<? super T> c) {
        Arrays.sort(array, c);
    }
    static <T extends Comparable<T>> void insertInMap(Map<T, Integer> map, T a){
        if(map.containsKey(a)){
            map.put(a, map.get(a)+1);
        }
        else{
            map.put(a, 1);
        }
    }
    static <T extends Comparable<T>> void deleteInMap(Map<T, Integer> map, T a){
        if(map.containsKey(a)){
            if(map.get(a)==1)
                map.remove(a);
            else
                map.put(a, map.get(a)-1);
        }
    }

    static int mod = 1000_000_007;
    public static List<Integer> manacherOdd(String s) {
        int n = s.length();
        s = "$" + s + "^";
        List<Integer> p = new ArrayList<>(n + 2);
        for (int i = 0; i < n + 2; i++) {
            p.add(0);
        }
        addln(p);
        int l = 1, r = 1;
        for (int i = 1; i <= n; i++) {
            p.set(i, Math.max(0, Math.min(r - i, p.get(l + (r - i)))));
            add(p);
            while (s.charAt(i - p.get(i)) == s.charAt(i + p.get(i))) {
                p.set(i, p.get(i) + 1);
            }
            addln(p);
            if (i + p.get(i) > r) {
                l = i - p.get(i);
                r = i + p.get(i);
            }
            addln(l+" "+r);
        }
        return p.subList(1, n + 1);
    }
    static long kadane(long a[])
    {
        int size = a.length;
        long max_so_far = 0;
        long max_ending_here = 0;

        for (int i = 0; i < size; i++) {
            max_ending_here = max_ending_here + a[i];
            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;
            if (max_ending_here < 0)
                max_ending_here = 0;
        }
        return max_so_far;
    }
    static long power(long x, long y)
    {
        long res = 1;
        while (y > 0) {
            if ((y & 1) != 0)
                res = (res * x)%mod;
            y = y >> 1;
            x = (x * x)%mod;
        }
        return ((res%mod)+mod)%mod;
    }
    static void fillbinary(long a, int[] A){
        int i=0;
        while(a>0){
            A[i++]+=a%2;
            a/=2;
        }
    }

    public static void main(String[] args) {
        int t = nextInt();
        while(t-->0){
            int n = nextInt();
            long m = nextLong();
            long k = nextLong();
            long[] a = nextLongArray(n);

            long x = k/m;
            if(k%m!=0)
                x++;
            TreeMap<Long, List<Integer>> map = new TreeMap<>();
            for (int i = 0; i < n; i++) {
                List<Integer> lis;
                if(map.containsKey(a[i])){
                    lis = map.get(a[i]);
                }
                else{
                    lis = new ArrayList<>();
                }
                lis.add(i);
                map.put(a[i], lis);
            }
            TreeMap<Integer, Long> rev = new TreeMap<>();
            long count = x;
            for(Map.Entry<Long, List<Integer>> e: map.entrySet()){
                List<Integer> temp = e.getValue();
                for(int i=0; i<temp.size() && count>0; i++){
                // for(int i=temp.size()-1; i>=0 && count>0; i--){
                    rev.put(temp.get(i), e.getKey());
                    count--;
                }
            }
            List<Long> nList = new ArrayList<>();
            for(Map.Entry<Integer, Long> e: rev.entrySet()){
                nList.add(e.getValue());
            }
            long max=0;
            long l = nList.size();
            count = 0;
            for(int i=0; i<l; i++){
                // long p = nList.get(i)+(((long)(l-1-i))*((long)(l-i)))/2L;
                long p = (m-(k%m))*(nList.get(i)+(k-count-(k%m)));
                if(p>=max){
                    max = p;
                }
                count+=m;
            }
            long ans = 0;
            count = 0;
            for(int i=0; i<l; i++){
                nList.set(i, nList.get(i)+count);
                // if(i==max_ind && k%m!=0){
                //     ans += nList.get(i)*(k%m);
                //     count+=k%m;
                // }
                // else{
                    ans += nList.get(i)*m;
                    count+=m;
                // }
            }
            // addln(ans+" "+max);
            if(k%m!=0)
                ans-=max;
            addln(ans);
        }
        print();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause the algorithm to fail. Here are the steps to think through this:

1. **Edge Cases**:
   - **Small Values**: Test with the smallest possible values for `n`, `m`, and `k`.
   - **Large Values**: Test with the largest possible values for `n`, `m`, and `k`.
   - **Boundary Values**: Test with values that are at the boundary of the constraints.

2. **Specific Scenarios**:
   - **Few Days with High Limits**: Test with a small number of days but a high limit on tickets per day.
   - **Many Days with Low Limits**: Test with a large number of days but a low limit on tickets per day.
   - **Price Increase Impact**: Test with a scenario where the price increase significantly affects the total cost.

3. **Optimal and Suboptimal Solutions**:
   - Test with a scenario where the optimal solution is not straightforward and requires careful consideration of the price increase.

Given these considerations, let's generate a test input that is small-scale but challenging:

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

### Explanation:
- **Test Case 1**:
  - **n = 3**: 3 days of ticket sales.
  - **m = 2**: A person can purchase up to 2 tickets per day.
  - **k = 5**: The total number of tickets to be bought is 5.
  - **a = [1, 2, 3]**: The prices per ticket for each of the 3 days.

### Why This Test Case is Challenging:
- **Few Days with High Limits**: With only 3 days, the algorithm needs to carefully manage the ticket purchases to minimize the total cost, especially considering the price increase.
- **Price Increase Impact**: Buying 2 tickets on the first day will increase the prices for the subsequent days, making the decision on how many tickets to buy each day crucial.
- **Optimal Solution**: The optimal solution involves buying tickets in a way that minimizes the total cost, which is not straightforward and requires careful calculation.

This test case should help expose any bugs in the algorithm that might not handle the price increase correctly or fail to find the optimal solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
#include <bits/stdc++.h>
#include <algorithm>
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pb push_back
#define all(x) x.begin(),x.end()
#define vec vector<ll>
#define ma map<ll, ll>
//string abcd = "xabcdefghijklmnopqrstuvwxyz";

typedef long long int ll;
using namespace std;

ll mydiv(ll a, ll b){
    if( a % b == 0)
        return a / b;
    else
        return (a / b) + 1;
}

long long int gcd(long long int a, long long int b)
{
  if (b == 0)
    return a;
  return gcd(b, a % b);
}

// Function to return LCM of two numbers
long long int lcm(ll a, ll b)
{
    return (a / gcd(a, b)) * b;
}

void display(vector<ll> v){
    for(auto & e : v)
        cout<<e<<" ";
    cout<<endl;
}

ll digits(ll x){
    ll count = 0;
    while(x != 0){
        count++;
        x = x / 10;
    }
    return count;
}

ll sum_of_n(ll k){
    ll l = 0, r = 1e9;
    ll ans = 0;
    while(l <= r){
        ll mid = (l + r) / 2;
        ll x = (mid * (mid + 1)) / 2;

        if(x == k)
            return mid;
        if(x > k){
            r = mid - 1;
        }
        else{
            ans = mid;
            l = mid + 1;
        }
    }
    return ans;
}

ll bin_exp(ll a, ll b, ll M){
    if(b == 0)
        return 1;

    ll res = bin_exp(a, b / 2, M);

    if(b & 1)
        return (a * (( res * res) % M)) % M;
    else
        return (res * res) % M;
}

void solve(){

    ll n;
    cin>>n;

    map <ll, priority_queue <ll, vector <ll> , greater <ll> > > pq;
    vector <ll> a(n), b(n);
    map <ll, vector <ll> > m;

    for(ll i = 0; i < n; i++){
        cin>>a[i];
        m[a[i]].pb(i);
        pq[a[i]].push(i);
    }

    for(auto & itr : m){
        ll e = itr.first;

        for(ll i = m[e].size() - 1; i >= 0; i--){
            ll x = m[e][i];
            ll val = -1;
            for(ll j = e + 1; (e ^ j) < 4; j++){
               // cout<<e ^ j<<endl;
                if(!pq[j].empty() && pq[j].top() < x){
                    x = pq[j].top();
                    val = j;
                }
            }

            if(val == -1)
                break;
            //cout<<val<<endl;
            a[x] = e;
            a[m[e][i]] = val;
            pq[val].pop();
            pq[val].push(m[e][i]);
        }
    }

    display(a);
}

int main(){

    ios::sync_with_stdio(false);
	cin.tie(nullptr);

    ll n;
    cin>>n;

    for(ll i = 0; i < n; i++)
        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 buggy code, we need to focus on edge cases where the XOR condition \(a_i \mathsf{XOR} a_j < 4\) might lead to unexpected behavior. Here are some considerations:

1. **Small Array with Similar Values**: This can test the logic of swapping and ensuring the lexicographically smallest array.
2. **Array with Values that XOR to Less Than 4**: This can test the swapping logic and ensure that all possible swaps are considered.
3. **Array with Values that XOR to More Than 4**: This can test the boundary conditions and ensure that invalid swaps are not performed.

Let's construct a small-scale test input that combines these elements:

```plaintext
3
4
1 2 3 0
5
3 1 2 0 4
3
4 5 6
```

### Explanation:
1. **Test Case 1:**
   - Array: `1 2 3 0`
   - Possible swaps:
     - \(1 \mathsf{XOR} 2 = 3\) (valid)
     - \(1 \mathsf{XOR} 3 = 2\) (valid)
     - \(1 \mathsf{XOR} 0 = 1\) (valid)
     - \(2 \mathsf{XOR} 3 = 1\) (valid)
     - \(2 \mathsf{XOR} 0 = 2\) (valid)
     - \(3 \mathsf{XOR} 0 = 3\) (valid)
   - Lexicographically smallest array: `0 1 2 3`

2. **Test Case 2:**
   - Array: `3 1 2 0 4`
   - Possible swaps:
     - \(3 \mathsf{XOR} 1 = 2\) (valid)
     - \(3 \mathsf{XOR} 2 = 1\) (valid)
     - \(3 \mathsf{XOR} 0 = 3\) (valid)
     - \(1 \mathsf{XOR} 2 = 3\) (valid)
     - \(1 \mathsf{XOR} 0 = 1\) (valid)
     - \(2 \mathsf{XOR} 0 = 2\) (valid)
     - \(3 \mathsf{XOR} 4 = 7\) (invalid)
     - \(1 \mathsf{XOR} 4 = 5\) (invalid)
     - \(2 \mathsf{XOR} 4 = 6\) (invalid)
     - \(0 \mathsf{XOR} 4 = 4\) (invalid)
   - Lexicographically smallest array: `0 1 2 3 4`

3. **Test Case 3:**
   - Array: `4 5 6`
   - Possible swaps:
     - \(4 \mathsf{XOR} 5 = 1\) (valid)
     - \(4 \mathsf{XOR} 6 = 2\) (valid)
     - \(5 \mathsf{XOR} 6 = 3\) (valid)
   - Lexicographically smallest array: `4 5 6` (no changes needed)

This test input should help expose any issues in the buggy code by testing various valid and invalid swaps and ensuring the lexicographically smallest array is produced correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

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

typedef vector<int> vi;
typedef vector<pair<int,int>> vp;
typedef vector<pair<long long, long long>> vpll;
typedef vector<long long int> vll;
typedef vector<vector<int>> vvi;
typedef long long int ll;
typedef vector<vector<long long int>> vvll;
typedef vector<string> vs;

void solve(){
    ll n;
    cin>>n;
    vll arr(n);
    map<ll,deque<ll>> mp;
    for(ll i=0; i<n; i++){
        cin>>arr[i];
        mp[arr[i]].push_back(i);
    }
    for(ll i=0; i<n; i++){
        ll temp = arr[i];
        if(temp<4){
            for(ll j=0; j<temp; j++){
                if(mp.find(j) == mp.end() || mp[j].empty()) continue;
                else if(mp[j].back() > i){
                    mp[temp].push_back(mp[j].back());
                    mp[temp].pop_front();
                    swap(arr[i],arr[mp[j].back()]);
                    mp[j].pop_back();
                    break;
                }
            }
        }else{
            ll val = temp;
            if(val & 1) val = val^1;
            if(val & (1<<1)) val = val^2;
            for(ll j=val; j<temp; j++){
                if(mp.find(j) == mp.end() || mp[j].empty()) continue;
                else if(mp[j].back() > i){
                    mp[temp].push_back(mp[j].back());
                    mp[temp].pop_front();
                    swap(arr[i],arr[mp[j].back()]);
                    mp[j].pop_back();
                    break;
                }
            }
        }
    }
    for(auto it:arr) cout<<it<<" ";
    cout<<endl;
}

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

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

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

Think step by step.

Response:
```plaintext
1
5
3 0 1 2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define FOR(i,x,n) for(int i = x; i<=n; i++)
#define FORO(i,n) for(int i = 0; i<n; i++)
#define ROF(i,x,n) for(int i = x; i>=n; i--)
#define nl "\n"
#define sp " "
#define INF 1000000000000000000
#define pb push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define pi pair<int, int>
#define pl pair<ll, ll>
#define pp pair<pair<int,int>,int>
#define vi vector<int>
#define vl vector<ll>
#define vvi vector<vector<int> >
#define vvl vector<vector<ll> >
#define vpi vector<pair<int, int> >
#define vpl vector<pair<ll, ll> >
#define PQ priority_queue
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define CASE int t;cin>>t; FOR(i,1,t)
#define FASTIO ios_base::sync_with_stdio(false);cin.tie(NULL);
#define MOD 1000000007
#define MOD2 765576143

ll logab(ll a, ll b){
    ll ans = 0;
    while (a > 1) {
        a /= b;
        ans++;
    }
    return ans;
}

unsigned long long power(unsigned long long x, ll y, ll p){
    unsigned long long res = 1;

    x = x % p;
    while (y > 0) {
        if (y & 1)
            res = (res * x) % p;

        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

unsigned long long modInverse(unsigned long long n, int p){
    return power(n, p - 2, p);
}
unsigned long long mul(unsigned long long x,
                       unsigned long long y, ll p){
    return x * 1ull * y % p;
}
unsigned long long divide(unsigned long long x,
                          unsigned long long y, ll p){
    return mul(x, modInverse(y, p), p);
}

unsigned long long nCrModPFermat(unsigned long long n,
                                 ll r, ll p){

    if (n < r)
        return 0;
    if (r == 0)
        return 1;
    if (n - r < r)
        return nCrModPFermat(n, n - r, p);

    unsigned long long res = 1;

    for (int i = r; i >= 1; i--)
        res = divide(mul(res, n - i + 1, p), i, p);
    return res;
}

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

     vl v(n);
     map<ll,vl> mp;
     FORO(i,n){
        cin>>v[i];
        mp[v[i]].pb(i+1);
     }

     vl vis(n,0);

     FORO(i,n){
        ll nm = v[i];
        if(vis[nm])continue;
        ll dv = nm/4;
        ll fst = dv*4;
       vl ind,yh;
       FORO(j,4){
        if(mp[fst].size()){
            for(int g=0;g<mp[fst].size();g++){
            ind.pb(mp[fst][g]-1);
            yh.pb(fst);
            vis[fst]=1;
            }

            }
            fst++;
        }
        sort(all(ind));
        sort(all(yh));

        for(int o=0;o<ind.size();o++)v[ind[o]]=yh[o];

     }

     FORO(i,n)cout<<v[i]<<" ";
     cout<<endl;
}

int main(){
    FASTIO;
    CASE{

        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 test input that exposes the bug in the given buggy code, we need to focus on scenarios where the XOR condition is particularly important. The code seems to have issues with handling the XOR condition and the grouping of elements that can be swapped. Here are the steps to create a challenging test input:

1. **Identify the XOR Condition**: The condition \(a_i \mathsf{XOR} a_j < 4\) is crucial. We need to create an array where this condition is frequently met but also where the lexicographically smallest array is not trivial to achieve.

2. **Create a Small-Scale Test Input**: Since we need a small-scale input, we can use an array of length \(n\) where \(n\) is relatively small but still challenging.

3. **Ensure Diverse XOR Values**: Include elements that have different values but can be swapped with each other based on the XOR condition.

4. **Test Edge Cases**: Include edge cases such as all elements being the same, or elements that are close to each other in value but have different XOR results.

Here is a test input that fits these criteria:

```plaintext
1
6
3 1 0 2 5 4
```

### Explanation:
- **Array**: `[3, 1, 0, 2, 5, 4]`
- **Length**: 6
- **XOR Condition**:
  - \(3 \mathsf{XOR} 1 = 2\)
  - \(3 \mathsf{XOR} 0 = 3\)
  - \(3 \mathsf{XOR} 2 = 1\)
  - \(1 \mathsf{XOR} 0 = 1\)
  - \(1 \mathsf{XOR} 2 = 3\)
  - \(5 \mathsf{XOR} 4 = 1\)

### Expected Output:
The lexicographically smallest array that can be made with the given swaps is:
```plaintext
0 1 2 3 4 5
```

### Why This Test Input is Challenging:
- **Diverse XOR Values**: The elements in the array have diverse values that can be swapped based on the XOR condition.
- **Non-trivial Sorting**: The elements need to be carefully swapped to achieve the lexicographically smallest order, which is not immediately obvious.
- **Edge Cases**: The input includes elements that are close to each other in value but have different XOR results, making it challenging for the buggy code to handle correctly.

This test input should help expose any issues in the buggy code, particularly in how it handles the XOR condition and the grouping of elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
import sys
from collections import deque

input = lambda: sys.stdin.readline().rstrip()

def print(*args, sep=' ', end='\n'):
    sys.stdout.write(sep.join(map(str, args)) + end)

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

def lii():  # list int input
    return list(map(int, input().split()))

def lsi():  # list string input
    return input().split()

def main():
    t = ii()
    for _ in range(t):
        n = ii()
        arr =  lii()
        for i in range(1, n):
            l = 0
            r = i
            while r - l > 0:
                m = (r + l) // 2
                if arr[m] > arr[i]:
                    r = m
                else:
                    l = m + 1
            if abs(arr[l] - arr[i]) < 4:
                arr[l], arr[i] = arr[i], arr[l]
        print(*arr)

if __name__ == '__main__':
    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:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on the conditions under which the code might fail. The code attempts to sort the array by swapping elements if their XOR is less than 4. The bug in the code is that it does not correctly handle the lexicographical order of the array after performing the swaps.

### Step-by-Step Analysis:
1. **Identify the Bug**: The code uses a binary search to find the position to swap, but it does not ensure that the array remains lexicographically smallest after the swap. It only checks if the absolute difference between the elements is less than 4, which is not sufficient to ensure the lexicographical order.

2. **Edge Cases**:
   - **Small Array**: Test with a small array to easily verify the output.
   - **Specific Values**: Use values that have a high probability of XOR being less than 4, such as 0, 1, 2, and 3.
   - **Multiple Test Cases**: Include multiple test cases to ensure the bug is consistently exposed.

3. **Test Case Construction**:
   - **Test Case 1**: An array with elements 0, 1, 2, 3. This array should be sortable to the lexicographically smallest order.
   - **Test Case 2**: An array with elements that have XOR less than 4 but are not in lexicographical order initially.
   - **Test Case 3**: An array with repeated elements that can be swapped to create a lexicographically smaller array.

### Challenging Test Input:
```plaintext
3
4
3 2 1 0
5
2 7 1 5 6
8
1 3 2 1 3 2 1 3
```

### Explanation:
1. **Test Case 1**:
   - Input: `3 2 1 0`
   - Expected Output: `0 1 2 3`
   - This test case checks if the code can sort an array that is initially in reverse lexicographical order.

2. **Test Case 2**:
   - Input: `2 7 1 5 6`
   - Expected Output: `1 5 2 6 7`
   - This test case checks if the code can handle elements that have an XOR less than 4 but are not initially in lexicographical order.

3. **Test Case 3**:
   - Input: `1 3 2 1 3 2 1 3`
   - Expected Output: `1 1 1 2 2 3 3 3`
   - This test case checks if the code can handle repeated elements and ensure they are in lexicographical order.

This test input should expose the bug in the given buggy code by ensuring that the output is not the lexicographically smallest array possible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    mas = {}
    lit = []
    for i in range(n):
        pre = bin(a[i])[2:]
        if len(pre) < 3:
            lit.append(a[i])
        else:
            b = pre[:-3]
            if b in mas:
                mas[b].append(a[i])
            else:
                mas[b] = [a[i]]

    lit = sorted(lit)[::-1]
    for k,v in mas.items():
        mas[k] = sorted(v)[::-1]

    for i in range(n):
        pre = bin(a[i])[2:]
        if len(pre) < 3:
            print(lit.pop(), end=' ')
        else:
            b = pre[:-3]
            if b in mas:
                print(mas[b].pop(), end=' ')
            else:
                print(a[i], end=' ')
    print()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The key issues in the buggy code are:

1. **Incorrect Grouping of Elements**: The code groups elements based on the last three bits (i.e., the remainder when divided by 8). However, the problem requires elements to be swapped if their XOR is less than 4, which is not the same as grouping by the last three bits.

2. **Lexicographical Order**: The code attempts to sort and reverse the groups, which might not always produce the lexicographically smallest array.

To expose these issues, we can create a test case with elements that have the same last three bits but different values, and elements that can be swapped according to the XOR condition but are not correctly grouped.

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

```plaintext
1
8
1 3 2 6 4 7 5 0
```

### Explanation:
- **Test Case 1**:
  - **Array**: `[1, 3, 2, 6, 4, 7, 5, 0]`
  - **Expected Output**: The lexicographically smallest array that can be made with any number of swaps.

### Why This Test Case is Challenging:
1. **Elements with the Same Last Three Bits**:
   - `1 (001)`, `3 (011)`, `5 (101)`, `7 (111)` all have the last three bits as `001`, `011`, `101`, `111` respectively.
   - `2 (010)`, `6 (110)`, `4 (100)`, `0 (000)` all have the last three bits as `010`, `110`, `100`, `000` respectively.

2. **Valid Swaps**:
   - `1 (001)` can be swapped with `2 (010)` (XOR = 3).
   - `3 (011)` can be swapped with `6 (110)` (XOR = 3).
   - `5 (101)` can be swapped with `4 (100)` (XOR = 5, not valid).
   - `7 (111)` can be swapped with `0 (000)` (XOR = 7, not valid).

3. **Lexicographical Order**:
   - The lexicographically smallest array should be `[0, 1, 2, 3, 4, 5, 6, 7]`.

The buggy code will incorrectly group and sort these elements, leading to an incorrect output. This test case should help identify the issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
from collections import deque

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        n = int(input())
        nums = input().split(" ")
        nums = [int(num) for num in nums]

        idx_map = {}
        for idx in range(n):
            swap_idx, swap_target = n + 1, nums[idx]
            for x in range(1, 4):
                target = x ^ nums[idx]
                if target > nums[idx] and target in idx_map:
                    if idx_map[target][0] < swap_idx:
                        swap_idx = idx_map[target][0]
                        swap_target = target

            if nums[idx] not in idx_map:
                idx_map[nums[idx]] = deque()
            if swap_idx != n + 1:
                idx_map[swap_target].popleft()
                idx_map[swap_target].append(idx)
                idx_map[nums[idx]].append(swap_idx)

                temp = nums[idx]
                nums[idx] = nums[swap_idx]
                nums[swap_idx] = temp
            else:
                idx_map[nums[idx]].append(idx)

        print(*nums)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the logic of the code might fail. The code attempts to find and perform swaps based on the XOR condition, but it might not handle all edge cases correctly. Here are some considerations:

1. **Handling Duplicates**: The code should correctly handle cases where there are duplicate elements in the array.
2. **Swapping Logic**: The code might fail if it does not correctly identify and perform all possible swaps.
3. **Edge Cases with XOR Values**: The code should handle cases where the XOR values are close to the boundary conditions (0, 1, 2, 3).

Let's construct a small-scale test input that maximizes the failure probability:

```plaintext
3
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
```

### Explanation:
1. **Test Case 1**:
   - Array: `[1, 0, 3, 2]`
   - Expected Output: `[0, 1, 2, 3]`
   - This case tests the basic swap logic. The code should be able to sort the array since all elements can be swapped.

2. **Test Case 2**:
   - Array: `[2, 7, 1, 5, 6]`
   - Expected Output: `[1, 5, 2, 6, 7]`
   - This case tests the ability to handle multiple elements and perform multiple swaps. The code should correctly identify and perform the swaps to achieve the lexicographically smallest array.

3. **Test Case 3**:
   - Array: `[1, 2, 1, 2, 1, 2, 1, 2]`
   - Expected Output: `[1, 1, 1, 1, 2, 2, 2, 2]`
   - This case tests the ability to handle duplicate elements and ensure that all elements are correctly sorted.

These test cases are designed to cover a range of scenarios, including basic swaps, multiple swaps, and handling duplicates, which should help expose 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: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
import java.util.*;
import java.util.stream.Collectors;

/**
 * Approach 1: Using Trie
 * For each 'i' find number/s from trie such that i xor 'x' < 4
 * i.e first 6 bits should be same as i and last 2 bits can be anything
 * in this way each number would have 4 combinations:
 * 00,01,10,11 --> one of these combinations of last bit is the number itself
 *which means if number is n and has 01 as last 2 bits
 * then n-1,n,n+1,n+2 are required numbers
 * Case: If all numbers are less than 4 just sort the array and return
 */
public class Xour {

    static class Node{
        private final int number;
        private final int index;

        public Node(int number, int index){
            this.number = number;
            this.index = index;
        }

        public int getNumber(){
            return number;
        }

        public int getIndex(){
            return index;
        }
    }
    static class Group implements Iterable<Node>{

        private static List<Node> sortedNodes;
        private static int size = 0;

        //both should be of same length

        public Group(ArrayList<Integer> numbers, ArrayList<Integer> indices){
            Collections.sort(numbers);
            Collections.sort(indices);
            sortedNodes = new ArrayList<>();
            size = numbers.size();
            for(int i = 0; i < size; i++){
                sortedNodes.add(new Node(numbers.get(i), indices.get(i)));
            }
        }

        @Override
        public Iterator<Node> iterator() {
            return new GroupIterator();
        }

        static class GroupIterator implements Iterator<Node>{

           private int index = 0;

            @Override
            public boolean hasNext() {
                return index < size;
            }

            @Override
            public Node next() {
                return sortedNodes.get(index++);
            }
        }
    }

    static class DSU{

        Map<Integer, Integer> parentMap = new HashMap<>();

        public DSU(){

        }

        public void addEdge(int v1, int v2){
            int p1 = findParent(v1);
            int p2 = findParent(v2);

            if(p1 == p2){
                return;
            }

            parentMap.put(p1, p2);

        }

        private int findParent(int v){

            if(v == parentMap.get(v)){
                return v;
            }

            int last2Bits = v % 4;
            int num0 = v - last2Bits;
            int lastNum = num0+4;
            while(num0 <= lastNum){
                if(parentMap.containsKey(num0)){
                    parentMap.put(v, findParent(parentMap.get(num0)));
                    return parentMap.get(v);
                }
                num0++;
            }

            parentMap.put(v, v);
            return v;
        }

    }

    /**
     * Approach:
     * 1. Create groups after identifying numbers with xor < 4
     * 2. A vertex = number + index
     * 3. Sort numbers within each group and then populate index
     * @param args
     */
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int __ = 0; __ < t; __++){
            int n = sc.nextInt();
            int a[] = new int[n];

            for(int i = 0; i < n;  i++){
                a[i] = sc.nextInt();
            }

            createGroups(a);

        }
    }

    private static void createGroups(int[] a){

        Map<Integer, Set<Integer>> numIndex= convertToNumIndex(a);

        Map<Integer, List<Integer>> groupByNumbers = new HashMap<>();
        Map<Integer, List<Integer>> groupByIndices = new HashMap<>();
        Set<Integer> visited = new HashSet<>();
        int n = a.length;
        int totalSizeDone = 0;
        int[] sorted = new int[n];

        for(int i: a){
            int last2Bits = i % 4;
            int num0 = i - last2Bits;
            int lastNum = num0+3;
            List<Integer> group = new ArrayList<>();
            List<Integer> indices = new ArrayList<>();
            if(visited.contains(num0)){
                continue;
            }
            groupByNumbers.put(i, group);
            groupByIndices.put(i, indices);

            while(num0 <= lastNum){

                if(numIndex.containsKey(num0)){
                    visited.add(num0);
                    int times = numIndex.get(num0).size();
                    while(times-- > 0){
                        group.add(num0);
                    }
                    indices.addAll(numIndex.get(num0));

                }
                num0++;
            }

            totalSizeDone += group.size();
            Collections.sort(group);
            Collections.sort(indices);

            if(totalSizeDone == n){
                break;
            }

        }

        for(int num: groupByNumbers.keySet()){

            int i = 0;
            for(int index: groupByIndices.get(num)){
                sorted[index] = groupByNumbers.get(num).get(i);
                i++;
            }

        }

        print(sorted);
    }

    private static Map<Integer, Set<Integer>> convertToNumIndex(int[] a){

        Map<Integer, Set<Integer>> numIndexMap = new HashMap<>();
        for(int i = 0; i < a.length; i++){
            if(!numIndexMap.containsKey(a[i])){
                numIndexMap.put(a[i], new HashSet<>());
            }
            numIndexMap.get(a[i]).add(i);
        }

        return numIndexMap;
    }

    private static void print(int[] arr){
        for(int i: arr){
            System.out.print(i + " ");
        }
        System.out.println();
    }
}

**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
4
0 1 2 3
5
1 2 3 4 5
8
1 1 1 1 2 2 2 2
4
4 5 6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOUR
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ consisting of $n$ nonnegative integers.

You can swap the elements at positions $i$ and $j$ if $a_i~\mathsf{XOR}~a_j < 4$, where $\mathsf{XOR}$ is the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Find the lexicographically smallest array that can be made with any number of swaps.

An array $x$ is lexicographically smaller than an array $y$ if in the first position where $x$ and $y$ differ, $x_i < y_i$.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2\cdot10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_i$ ($0 \leq a_i \leq 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers — the lexicographically smallest array that can be made with any number of swaps.

Examples:
input:
4
4
1 0 3 2
5
2 7 1 5 6
8
1 2 1 2 1 2 1 2
4
16 4 1 64
output:
0 1 2 3
1 5 2 6 7
1 1 1 1 2 2 2 2
16 4 1 64

Note: For the first test case, you can swap any two elements, so we can produce the sorted array.

For the second test case, you can swap $2$ and $1$ (their $\mathsf{XOR}$ is $3$), $7$ and $5$ (their $\mathsf{XOR}$ is $2$), and $7$ and $6$ (their $\mathsf{XOR}$ is $1$) to get the lexicographically smallest array.

Buggy Code:
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class Mains {
    public static void main(String[] args)throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter wr = new BufferedWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(br.readLine());
        while(t-->0){
              int n=Integer.parseInt(br.readLine());
              String[] s = br.readLine().split(" ");
              List<Long> b = new ArrayList<>();
              for(String str:s)
              b.add(Long.parseLong(str));
              for(int i=0;i<n;i++){
                List<Long> c = new ArrayList<>(b.subList(i+1,n));
                for(int j=1;j<4;j++){
                    long v=b.get(i)^j;
                    if(c.contains(v))
                    if(v<b.get(i))
                    Collections.swap(b,c.indexOf(v)+i+1,i);
                }
              }
              for(long num:b)
              wr.write(Long.toString(num)+" ");
               wr.newLine();

        }
               wr.flush();
               wr.close();
               br.close();
    }
}

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

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

Think step by step.

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

1. The code only checks for swaps with XOR values 1, 2, and 3, but it does not consider the possibility of multiple swaps or more complex interactions.
2. The code does not handle cases where multiple elements can be swapped to achieve a lexicographically smaller array.

Let's construct a small-scale test input that can expose these issues:

### Test Input:
```plaintext
1
4
3 0 1 2
```

### Explanation:
- **Test Case:**
  - **Array:** [3, 0, 1, 2]
  - **Expected Output:** [0, 1, 2, 3]

### Why This Test Input is Challenging:
1. **Initial Configuration:**
   - The array [3, 0, 1, 2] is already close to being sorted, but not in the lexicographically smallest form.
2. **Possible Swaps:**
   - The code should consider the following swaps:
     - 3 XOR 0 = 3 (valid swap, but 3 > 0)
     - 3 XOR 1 = 2 (valid swap, but 3 > 1)
     - 3 XOR 2 = 1 (valid swap, and 3 > 1)
   - The correct sequence of swaps should result in [0, 1, 2, 3].

### Expected Behavior:
- The code should be able to swap 3 with 2 to get [2, 0, 1, 3].
- Then, it should swap 2 with 1 to get [1, 0, 2, 3].
- Finally, it should swap 1 with 0 to get [0, 1, 2, 3].

### Actual Behavior in Buggy Code:
- The code might only perform a single swap and not consider the sequence of swaps needed to achieve the lexicographically smallest array.
- This will expose the bug in the code's logic for handling multiple swaps and achieving the correct lexicographical order.

This test input is small and complete, making it easy to verify the behavior of the code and identify the bug.