**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Password Cracking
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dimash learned that Mansur wrote something very unpleasant about him to a friend, so he decided to find out his password at all costs and discover what exactly he wrote.

Believing in the strength of his password, Mansur stated that his password — is a binary string of length $$$n$$$. He is also ready to answer Dimash's questions of the following type:

Dimash says a binary string $$$t$$$, and Mansur replies whether it is true that $$$t$$$ is a substring of his password.

Help Dimash find out the password in no more than $$$2n$$$ operations; otherwise, Mansur will understand the trick and stop communicating with him.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

Output: None

Examples:
input:
4
3

0

0

1

4

4

2
output:
? 00

? 000

? 010

! 010

! 1100

! 0110

! 10

Note: In the first example, the string $$$010$$$ is given. Therefore, the answers to the queries are as follows:

"? 00" $$$00$$$ is not a substring of $$$010$$$, so the answer is $$$0$$$.

"? 000" $$$000$$$ is not a substring, so the answer is $$$0$$$.

"? 010" $$$010$$$ is a substring, so the answer is $$$1$$$.

In the second example, the string is $$$1100$$$, in the third $$$0110$$$, and in the fourth $$$10$$$.

Buggy Code:
# ﷽
import sys
import heapq
from heapq import *
from math import log2,ceil,floor,gcd,sqrt,log,perm,comb,factorial
from random import randrange
from bisect import bisect_left
from collections import defaultdict, Counter, deque, OrderedDict
from functools import lru_cache, reduce
from bisect import bisect_left, bisect_right
from types import GeneratorType

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

class dict_(dict):
    RANDOM = randrange(2 ** 62)

    def __init__(self, dd):
        super().__init__({k ^ self.RANDOM: v for k, v in dd.items()})

    def __missing__(self, key):
        return 0

    def __setitem__(self, key, value):
        super().__setitem__(key ^ self.RANDOM, value)

    def __getitem__(self, item):
        return super().__getitem__(item ^ self.RANDOM)

    def __contains__(self, item):
        return super().__contains__(item ^ self.RANDOM)

    def items(self):
        for k, v in super().items():
            yield (k ^ self.RANDOM, v)

    def keys(self):
        for k in super().keys():
            yield k ^ self.RANDOM

    def __repr__(self):
        return '{0}'.format({k ^ self.RANDOM: v for k, v in super().items()})

###############################################################################
# GRAPHS :

def add_edge(node1,node2,graph):

    if node1 not in graph:
        graph[node1] = []
    if node2 not in graph:
        graph[node2] = []
    graph[node1].append(node2)
    graph[node2].append(node1)

class disjointset:
    def __init__(self, n):
        self.rank, self.parent = array('i', [0] * (n + 1)), array('i', [i for i in range(n + 1)])
        self.n, self.even = n, array('b', [0] * (n + 1))

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

    def union(self, x, y):
        xpar, ypar = self.find(x), self.find(y)

        if xpar == ypar: return 1

        par, child = xpar, ypar
        if self.rank[xpar] < self.rank[ypar]:
            par, child = ypar, xpar

        elif self.rank[xpar] == self.rank[ypar]:
            self.rank[xpar] += 1

        self.parent[child] = par
        self.even[par] |= self.even[child]
        self.n -= 1
        return 0

def dijkstra(graph, start1):
    distances = {node: float('inf') for node in graph}  # Initialize distances to infinity
    distances[start1] = 0  # Distance from start1 to start1 is 0
    heap = [(0, start1,[start1])]  # Priority queue to store nodes and their distances
    paths = [-1 for i in range(n+1)]
    while heap:
        current_distance, current_node , path = heapq.heappop(heap)  # Get node with smallest distance

        if current_distance > distances[current_node]:
            continue  # Skip if distance is already greater than the known distance

        for neighbor, weight in graph[current_node]:
            distance = current_distance + weight  # Calculate tentative distance

            if distance < distances[neighbor]:
                lst = path
                lst.append(neighbor)
                paths[neighbor] = current_node
                distances[neighbor] = distance  # Update distance if shorter
                heapq.heappush(heap, (distance, neighbor, lst))  # Add neighbor to the heap

    return distances , paths

###############################################################################
def mergeSort(arr, n):
    # A temp_arr is created to store
    # sorted array in merge function
    temp_arr = [0]*n
    return _mergeSort(arr, temp_arr, 0, n-1)

# This Function will use MergeSort to count inversions

def _mergeSort(arr, temp_arr, left, right):

    # A variable inv_count is used to store
    # inversion counts in each recursive call

    inv_count = 0

    # We will make a recursive call if and only if
    # we have more than one elements

    if left < right:

        # mid is calculated to divide the array into two subarrays
        # Floor division is must in case of python

        mid = (left + right)//2

        # It will calculate inversion
        # counts in the left subarray

        inv_count += _mergeSort(arr, temp_arr,
                                left, mid)

        # It will calculate inversion
        # counts in right subarray

        inv_count += _mergeSort(arr, temp_arr,
                                mid + 1, right)

        # It will merge two subarrays in
        # a sorted subarray

        inv_count += merge(arr, temp_arr, left, mid, right)
    return inv_count

# This function will merge two subarrays
# in a single sorted subarray

def merge(arr, temp_arr, left, mid, right):
    i = left     # Starting index of left subarray
    j = mid + 1  # Starting index of right subarray
    k = left     # Starting index of to be sorted subarray
    inv_count = 0

    # Conditions are checked to make sure that
    # i and j don't exceed their
    # subarray limits.

    while i <= mid and j <= right:

        # There will be no inversion if arr[i] <= arr[j]

        if arr[i] <= arr[j]:
            temp_arr[k] = arr[i]
            k += 1
            i += 1
        else:
            # Inversion will occur.
            temp_arr[k] = arr[j]
            inv_count += (mid-i + 1)
            k += 1
            j += 1

    # Copy the remaining elements of left
    # subarray into temporary array
    while i <= mid:
        temp_arr[k] = arr[i]
        k += 1
        i += 1

    # Copy the remaining elements of right
    # subarray into temporary array
    while j <= right:
        temp_arr[k] = arr[j]
        k += 1
        j += 1

    # Copy the sorted subarray into Original array
    for loop_var in range(left, right + 1):
        arr[loop_var] = temp_arr[loop_var]

    return inv_count
###############################################################################

def Divisors(n) :
    divisors = {}
    for num in [n]:
        lst = set()
        i = 1
        while i <= sqrt(num):

            if (num % i == 0) :
                # If divisors are equal, print only one
                if ( num / i == i) :
                    lst.add(i)
                else :
                    # Otherwise print both
                    lst.update([i,num//i])

            i = i + 1
        divisors[num] = list(lst)
    return divisors[num]

def binpow(a, b):
    if b==0:
        return 1
    res = binpow(a,b//2)
    res = pow(res,2,MOD)
    if b%2:
        return (res*a)%MOD
    return res

def mod_inverse(a):
    return binpow(a,MOD-2)

def factors(n):
    if n==0:
        return set()
    return set(reduce(list.__add__,
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))

# factors = [factors(i) for i in range(MAX)]

# factorial and inverse factorial

# fact = [1]*MAX
# invfact = [1]*MAX
# for i in range(1,MAX):
#     fact[i] = (fact[i-1]*i)%MOD
#     invfact[i] = (invfact[i-1]*mod_inverse(i))%MOD

def inverse_mod(num):
    a = mod-2
    result = 1
    result=pow(num,a,mod)
    return result

def mex(arr, N):
    # Sort the array
    arr1 = arr.copy()
    arr1.sort()
    mex = 0
    for idx in range(N):
        if arr1[idx] == mex:
            # Increment mex
            mex += 1
    # return mex as answer
    return mex
###############################################################################

def mergeDictionary(dict_1, dict_2):
   dict_3 = {**dict_1, **dict_2}
   for key, value in dict_3.items():
       if key in dict_1 and key in dict_2:
               dict_3[key] = value + dict_1[key]
   return dict_3

def rindex(lst, value):
    lst.reverse()
    i = lst.index(value)
    lst.reverse()
    return len(lst) - i - 1

def prefix_sum(arr, length_arr, prefix_sum):
    prefix_sum[0] = arr[0]

    for i in range(1, length_arr): # 1, 2, 3, 4
        prefix_sum[i] =  prefix_sum[i-1]+ arr[i]

def suffix_sum(arr, length_arr, suffix_sum):
    suffix_sum[length_arr-1] = arr[length_arr-1]

    for i in reversed(range(length_arr-1)): # 3, 2, 1, 0
        suffix_sum[i] = suffix_sum[i+1] + arr[i]

letters = [
    'a','b','c','d','e','f',
    'g','h','i','j','k','l',
    'm','n','o','p','q','r',
    's','t','u','v','w','x',
    'y','z']

# recursion limit fix decorator, change 'return' to 'yield' and add 'yield' before calling the function
def bootstrap(f):
    stack = []
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

###############################################################################
###############################################################################
###############################################################################
mod = 10**9 + 7
from random import randint
def ask(s) :
    print("?",s,flush=1)
    return int(input())
def answer(s) :
    print("!",s,flush=1)
def solve():
    n = int(input())
    # arr = list(map(int,input().split()))
    cur = ""
    while len(cur) != n :
        r = randint(0,1)
        s1 = cur + str(r)

        if ask(s1) :
            cur = s1
        elif ask(cur+str(r^1)) :
            cur = cur+str(r^1)
        else:
            break
    while len(cur) != n :
        r = randint(0,1)
        s1 = str(r) + cur

        if ask(s1) :
            cur = s1
        elif ask(str(r^1) + cur) :
            cur = str(r^1) + cur
        else:
            break

    answer(cur)

###############################################################################
###############################################################################
###############################################################################

for t in range(int(input())):
    # case(t+1)
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input for the given problem, we need to consider edge cases and scenarios that can expose bugs in the provided buggy code. Here are some steps to think through the process:

1. **Edge Cases**:
   - **Short Passwords**: Test with very short passwords (e.g., length 1 or 2) to see if the code handles them correctly.
   - **Long Passwords**: Test with the maximum allowed length (e.g., 2000) to ensure the code can handle large inputs.
   - **All Zeros or All Ones**: Test with passwords consisting entirely of zeros or ones to check if the code can handle uniform strings.
   - **Alternating Patterns**: Test with alternating patterns (e.g., "010101...") to ensure the code can handle repetitive patterns.
   - **Random Strings**: Test with random binary strings to ensure the code can handle arbitrary inputs.

2. **Specific Scenarios**:
   - **Initial Guesses**: Test scenarios where the initial guesses are incorrect to see how the code handles incorrect guesses.
   - **Correct Substrings**: Test scenarios where the correct substrings are guessed early to ensure the code can confirm the correct password quickly.
   - **Incorrect Substrings**: Test scenarios where the incorrect substrings are guessed early to ensure the code can handle incorrect guesses efficiently.

Based on these considerations, let's generate a small-scale test input that includes a mix of these edge cases and specific scenarios:

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

### Explanation:
1. **Test Case 1**: A password of length 3 with the correct sequence "010".
   - The input will be "0", "000", and "010".
   - The expected output is "010".

2. **Test Case 2**: A password of length 4 with the correct sequence "0110".
   - The input will be "0", "0000", "0110".
   - The expected output is "0110".

3. **Test Case 3**: A password of length 5 with the correct sequence "10101".
   - The input will be "1", "0", "1", "0", "1".
   - The expected output is "10101".

4. **Test Case 4**: A password of length 2 with the correct sequence "10".
   - The input will be "1", "0".
   - The expected output is "10".

This test input is designed to cover a range of scenarios, including short and long passwords, alternating patterns, and specific guesses to challenge the code's ability to handle various inputs efficiently.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Password Cracking
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dimash learned that Mansur wrote something very unpleasant about him to a friend, so he decided to find out his password at all costs and discover what exactly he wrote.

Believing in the strength of his password, Mansur stated that his password — is a binary string of length $$$n$$$. He is also ready to answer Dimash's questions of the following type:

Dimash says a binary string $$$t$$$, and Mansur replies whether it is true that $$$t$$$ is a substring of his password.

Help Dimash find out the password in no more than $$$2n$$$ operations; otherwise, Mansur will understand the trick and stop communicating with him.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

Output: None

Examples:
input:
4
3

0

0

1

4

4

2
output:
? 00

? 000

? 010

! 010

! 1100

! 0110

! 10

Note: In the first example, the string $$$010$$$ is given. Therefore, the answers to the queries are as follows:

"? 00" $$$00$$$ is not a substring of $$$010$$$, so the answer is $$$0$$$.

"? 000" $$$000$$$ is not a substring, so the answer is $$$0$$$.

"? 010" $$$010$$$ is a substring, so the answer is $$$1$$$.

In the second example, the string is $$$1100$$$, in the third $$$0110$$$, and in the fourth $$$10$$$.

Buggy Code:
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x2013 {
    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        int T = in.nextInt();
        StringBuilder sb;
        while(T-->0) {
            int N = in.nextInt();
            // Count how many bits
            int low = 0;
            int high = N;
            while(low != high) {
                int mid = (low+high+1)>>1;
                sb = new StringBuilder("? ");
                for(int i=0; i < mid; i++) {
                    sb.append(1);
                }
                System.out.println(sb);
                int response = in.nextInt();
                if(response == 1) {
                    low = mid;
                }
                else {
                    high = mid-1;
                }
            }
            int ones = low;
            if(ones == 0 || ones == N) {
                sb = new StringBuilder("! ");
                for(int i=0; i < N; i++) {
                    sb.append(min(ones, 1));
                }
                System.out.println(sb);
                continue;
            }
            char[] res = new char[N];
            Arrays.fill(res, '?');
            int zeros = 1;
            for(int t=0; t < N; t++) {
                res[t] = '1';
                sb = new StringBuilder("? ");
                for(int i=0; i < zeros; i++) {
                    sb.append(0);
                }
                for(int i=0; i < ones; i++) {
                    sb.append(1);
                }
                System.out.println(sb);
                if(in.nextInt() == 1) {
                    res[t] = '0';
                }
                ones -= (res[t]-'0');
                if(res[t] == '0') {
                    zeros++;
                }
                if(ones == 0) {
                    for(int a=t+1; a < N; a++) {
                        res[a] = '0';
                    }
                    break;
                }
            }
            sb = new StringBuilder("! ");
            for(char c: res) {
                sb.append(c);
            }
            System.out.println(sb);
        }
    }
}
class FastScanner {
    //I don't understand how this works lmao
    private int BS = 1 << 16;
    private char NC = (char) 0;
    private byte[] buf = new byte[BS];
    private int bId = 0, size = 0;
    private char c = NC;
    private double cnt = 1;
    private BufferedInputStream in;

    public FastScanner() {
        in = new BufferedInputStream(System.in, BS);
    }

    public FastScanner(String s) {
        try {
            in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
        } catch (Exception e) {
            in = new BufferedInputStream(System.in, BS);
        }
    }

    private char getChar() {
        while (bId == size) {
            try {
                size = in.read(buf);
            } catch (Exception e) {
                return NC;
            }
            if (size == -1) return NC;
            bId = 0;
        }
        return (char) buf[bId++];
    }

    public int nextInt() {
        return (int) nextLong();
    }

    public int[] nextInts(int N) {
        int[] res = new int[N];
        for (int i = 0; i < N; i++) {
            res[i] = (int) nextLong();
        }
        return res;
    }

    public long[] nextLongs(int N) {
        long[] res = new long[N];
        for (int i = 0; i < N; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public long nextLong() {
        cnt = 1;
        boolean neg = false;
        if (c == NC) c = getChar();
        for (; (c < '0' || c > '9'); c = getChar()) {
            if (c == '-') neg = true;
        }
        long res = 0;
        for (; c >= '0' && c <= '9'; c = getChar()) {
            res = (res << 3) + (res << 1) + c - '0';
            cnt *= 10;
        }
        return neg ? -res : res;
    }

    public double nextDouble() {
        double cur = nextLong();
        return c != '.' ? cur : cur + nextLong() / cnt;
    }

    public double[] nextDoubles(int N) {
        double[] res = new double[N];
        for (int i = 0; i < N; i++) {
            res[i] = nextDouble();
        }
        return res;
    }

    public String next() {
        StringBuilder res = new StringBuilder();
        while (c <= 32) c = getChar();
        while (c > 32) {
            res.append(c);
            c = getChar();
        }
        return res.toString();
    }

    public String nextLine() {
        StringBuilder res = new StringBuilder();
        while (c <= 32) c = getChar();
        while (c != '\n') {
            res.append(c);
            c = getChar();
        }
        return res.toString();
    }

    public boolean hasNext() {
        if (c > 32) return true;
        while (true) {
            c = getChar();
            if (c == NC) return false;
            else if (c > 32) return true;
        }
    }
}

**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 that can cause the algorithm to fail. Here are the steps to identify such cases:

1. **Edge Case with All Zeros or All Ones**:
   - The code seems to handle these cases correctly, but it's still worth testing.
   - Example: `1\n3\n000`

2. **Edge Case with Mixed Zeros and Ones**:
   - The code might fail if the pattern of zeros and ones is not straightforward.
   - Example: `1\n3\n101`

3. **Edge Case with Repeated Patterns**:
   - The code might not handle repeated patterns correctly.
   - Example: `1\n4\n1010`

4. **Edge Case with Single Bit**:
   - The code might fail if the password length is 1.
   - Example: `1\n1\n1`

5. **Edge Case with Long Password**:
   - The code might fail if the password length is close to the maximum allowed (2000).
   - Example: `1\n20\n10101010101010101010`

6. **Multiple Test Cases**:
   - The code might fail if it has to handle multiple test cases.
   - Example: `2\n3\n101\n4\n1010`

Given these considerations, let's create a small-scale test input that includes a mix of these edge cases:

```plaintext
3
3
101
4
1010
1
1
```

This test input includes:
- A password of length 3 with mixed zeros and ones.
- A password of length 4 with a repeated pattern.
- A password of length 1.

This 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: Password Cracking
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dimash learned that Mansur wrote something very unpleasant about him to a friend, so he decided to find out his password at all costs and discover what exactly he wrote.

Believing in the strength of his password, Mansur stated that his password — is a binary string of length $$$n$$$. He is also ready to answer Dimash's questions of the following type:

Dimash says a binary string $$$t$$$, and Mansur replies whether it is true that $$$t$$$ is a substring of his password.

Help Dimash find out the password in no more than $$$2n$$$ operations; otherwise, Mansur will understand the trick and stop communicating with him.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

Output: None

Examples:
input:
4
3

0

0

1

4

4

2
output:
? 00

? 000

? 010

! 010

! 1100

! 0110

! 10

Note: In the first example, the string $$$010$$$ is given. Therefore, the answers to the queries are as follows:

"? 00" $$$00$$$ is not a substring of $$$010$$$, so the answer is $$$0$$$.

"? 000" $$$000$$$ is not a substring, so the answer is $$$0$$$.

"? 010" $$$010$$$ is a substring, so the answer is $$$1$$$.

In the second example, the string is $$$1100$$$, in the third $$$0110$$$, and in the fourth $$$10$$$.

Buggy Code:
import java.util.Scanner;

public class P2013C {
	static Scanner xyz = new Scanner(System.in);

	public static void main(String[] args) {
		int t = xyz.nextInt();
		while (t-- > 0) {
			int n = xyz.nextInt();
			if (n == 1) {
				if (query("1") == 1)
					System.out.println("! 1");
				else
					System.out.println("! 0");

			} else {
				if (query("00") == 1)
					test("00", n);
				else if (query("01") == 1)
					test("01", n);
				else if (query("10") == 1)
					test("10", n);
				else {
					String s = "";
					for (int i = 0; i < n; i++) {
						s = s + "1";
					}
					System.out.println("!" + s);
				}
			}
			System.out.flush();
		}
	}

	static int query(String s) {
		System.out.println("? " + s);
		int res = xyz.nextInt();
		return res;
	}

	static void test(String s, int n) {
		String ans = s;
		boolean b = false;
		for (int i = 0; i < n - 2; i++) {
			String c1 = ans + "0";
			String c2 = ans + "1";
			if (!b) {
				if (query(c1) == 1)
					ans = c1;
				else if (query(c2) == 1)
					ans = c2;
				else
					b = true;

			}
			if (b) {
				c1 = "0" + ans;
				c2 = "1" + ans;
				if (query(c1) == 1)
					ans = c1;
				else
					ans = c2;
			}
		}
		System.out.println("! " + ans);
	}

}

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

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

Think step by step.

Response:
```plaintext
3
3
3
3
? 00
0
? 01
0
? 10
0
? 11
1
? 00
1
? 01
0
? 10
0
? 11
1
? 00
0
? 01
1
? 10
0
? 11
1
```


**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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <set>
#include <utility>

using namespace std;

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef vector<ll> vi;
typedef pair<ll, ll> pii;

ll mod = 1e9 + 7;

vi v;
ll K, m, n, rem;
unordered_map<ll, unordered_map<ll, ll>> memo;

ll dp(ll x, ll k) {
    if(k == 0) return 0;
    if(x == n) return 9223372036854000000;

    if(memo.find(k) != memo.end() && memo[k].find(x) != memo[k].end()) return memo[k][x];

    ll res = dp(x+1, k);
    if(k % m != 0) {
        res = min(res, rem*(v[x] + (K-k)) + dp(x+1, k-rem));
    }
    if(k >= m) {
        res = min(res, m*(v[x] + (K-k)) + dp(x+1, k-m));
    }

    return memo[k][x] = res;
}

ll solve() {
    cin>>n>>m>>K;
    rem = K%m;
    v.resize(n);
    for(int i=0; i<n; i++) cin>>v[i];

    return dp(0, K);

}

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

    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif

    ll t; cin>>t;
    for(int i=1; i<=t; i++) {
        cout<<solve()<<endl;
        memo.clear();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1000000000 1
1
1 1 1000000000
1000000000
10 10 100
1 2 3 4 5 6 7 8 9 10
```

### Explanation:
1. **Test Case 1:**
   - **n = 1, m = 1000000000, k = 1**
   - **a = [1]**
   - This test case checks the scenario where the number of tickets to buy is 1, and the maximum tickets that can be bought in a day is very large. The price is 1, so the expected output is 1.

2. **Test Case 2:**
   - **n = 1, m = 1, k = 1000000000**
   - **a = [1000000000]**
   - This test case checks the scenario where the number of tickets to buy is very large, but the maximum tickets that can be bought in a day is only 1. The price is very high, so the expected output is 1000000000.

3. **Test Case 3:**
   - **n = 10, m = 10, k = 100**
   - **a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]**
   - This test case checks the scenario where the number of tickets to buy is 100, and the maximum tickets that can be bought in a day is 10. The prices are increasing linearly, so the expected output is 72.

These test cases are designed to expose potential issues in the buggy code, such as handling very large or very small values, and the interaction between the number of tickets to buy and the maximum tickets that can be bought in a day.


**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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

#define _test                                                                  \
  int _TEST;                                                                   \
  cin >> _TEST;                                                                \
  while (_TEST--)

int main() {
  _test {
    int n, m, k;
    cin >> n >> m >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    int bought = 0;
    int totalcost=0;

    for (int i = 0; i < n; i++) {
      int quantity = min(k - bought, m);
      int cost=quantity*(a[i]+bought);
      totalcost += cost;
      bought+=quantity;

    }
    cout<<totalcost<<endl;
  }
  return 0;
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose the bug in the provided buggy code. Here's why:

1. **Input Values:**
   - `n = 5`: There are 5 days.
   - `m = 2`: A maximum of 2 tickets can be bought per day.
   - `k = 5`: We need to buy 5 tickets.
   - `a = [1, 2, 3, 4, 5]`: The ticket prices for each day are 1, 2, 3, 4, and 5 respectively.

2. **Expected Behavior:**
   - The optimal strategy should be to buy 2 tickets on the first day, 2 tickets on the second day, and 1 ticket on the third day.
   - The total cost should be calculated as follows:
     - Day 1: 2 tickets at 1 each, cost = 2 * 1 = 2.
     - Day 2: 2 tickets at (2 + 2) each, cost = 2 * 4 = 8.
     - Day 3: 1 ticket at (3 + 4) each, cost = 1 * 7 = 7.
     - Total cost = 2 + 8 + 7 = 17.

3. **Buggy Code Behavior:**
   - The buggy code sorts the prices and then buys tickets in increasing order of the sorted prices.
   - It does not correctly account for the fact that the prices increase after buying tickets.
   - The code will incorrectly calculate the total cost by not considering the increased prices due to previous purchases.

4. **Expected Output:**
   - The correct output for this test case should be `17`.

This test case should clearly show the bug in the buggy code by highlighting the incorrect calculation of the total cost due to the increase in ticket prices after each purchase.


**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 <iostream>
#include <vector>
#include <string>
#include <climits>
#include <cassert>
#include <map>
#include <algorithm>
#include <queue>
#include <set>
#include <ctime>
#include <cstdlib>
#include <map>
#include <unordered_set>
#include <numeric>
#include <cmath>

using namespace std;
void solve()
{
  int n;
  long long m, k;

  cin >> n >> m >> k;

  vector<long long> arr(n);
  for (auto& x : arr)
  {
    cin >> x;
  }

  long long penalty = 0LL;
  long long take = k/m;

  penalty = (k - (k%m)) * (k%m);
  long long rem = k-(k%m);
  while (rem > 0)
  {
    rem -= m;
    penalty += rem*m;
  }

  vector<long long> suf(n);
  priority_queue<long long> pq;
  long long sum = 0LL;

  for (int i = n-1; i >= 0; i--)
  {
    sum += arr[i];
    pq.emplace(arr[i]);

    while ((long long) pq.size() > take)
    {
      sum -= pq.top();
      pq.pop();
    }
    suf[i] = (long long) pq.size() == take ? sum : LLONG_MAX;
  }

  long long ans = LLONG_MAX;

  for (int i = 0; i < n && suf[i] != LLONG_MAX; i++)
  {
    long long sub;
    if (k % m == 0)
    {
      sub = suf[i] * m;
    }
    else if (i +1 < n && suf[i+1] != LLONG_MAX)
    {
      sub = arr[i] * (k % m) + suf[i+1] * m;
    }

    ans = min(ans, sub);
  }

  cout << ans + penalty << '\n';
}

int main()
{
  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
3
5 2 5
1 2 3 4 5
5 1 5
1 2 3 4 5
5 5 5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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:
def solve():
    mp = {}
    p = []
    n, m, k = map(int, input().split())

    a = list(map(int, input().split()))
    b = a.copy()
    b.sort()
    x = k // m
    y = k % m

    id = x
    if y:
        id += 1

    for i in range(id):
        mp[b[i]] = mp.get(b[i], 0) + 1

    for i in range(n):
        if mp.get(a[i], 0):
            p.append(a[i])
            mp[a[i]] -= 1

    sum = 0
    ans = 0
    cnt = 0
    for i in range(len(p)):
        if i == 0:
            if y:
                cnt += y
            else:
                cnt += m
            ans += p[i] * cnt
        else:
            val = p[i] + cnt
            ans += (val * m)
            cnt += m

    print(ans)

def main():
    T = int(input())
    for _ in range(T):
        solve()

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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:
def merge_sort(arr,comp):
    if(len(arr)==1 or len(arr)==0):
        return arr
    arr1=[]
    n=len(arr)
    mid=int(n/2)
    L=merge_sort(arr[:mid],comp)
    R=merge_sort(arr[mid:],comp)
    i,j=0,0
    for k in range(n):
        if(j==len(R) or (i<len(L) and comp(L[i],R[j]))):
            arr1.append(L[i])
            i+=1
        else:
            arr1.append(R[j])
            j+=1
    return arr1
def comp(a,b):
    if(a[0]==b[0]):
        return a[1]<b[1]
    return a[0]<b[0]
def compb(a,b):
    return a[1]<b[1]
if __name__ == "__main__":
    t=int(input())
    while(t>0):
        t-=1
        [n,m,k]=input().split()
        n,m,k=int(n),int(m),int(k)
        arr=input().split()
        arr=[int(arr[i]) for i in range(n)]
        t1=int(k/m)
        m1=k-t1*m

        arr1=[[arr[i],i] for i in range(n)]
        arr1=merge_sort(arr1,comp)
        if(t1+1<len(arr1)):
            arr2=arr1[:t1+1]
        else:
            arr2=arr1[:t1]
        arr2=merge_sort(arr2,compb)
        ct,ans=0,0
        if(t1>1 or (t1>0 and m1==0)):
            for i in range(1,t1+1):
                ans+=m*(arr2[i-1][0]+ct)
                ct+=m
        if(t1<len(arr2) and t1>0 and m1>0):
            ans+=min((arr2[t1][0]+(ct+m))*m1+arr2[t1-1][0]*m,(arr2[t1][0]+(ct+m1))*m+arr2[t1-1][0]*m1)
        if(t1==0):
            ans=arr2[t1][0]*m1
        print(ans)

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

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

Think step by step.

Response:
```plaintext
1
5 1 5
10 9 8 7 6
```


**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 fileinput
import sys

def solve_case(prices, n_buy, max_day):
    if max_day > n_buy:
        max_day = n_buy

    MAX_VAR = float('inf')
    D = [MAX_VAR for i in range(n_buy + 1)]

    for t in range(max_day + 1):
        D[t] = t * prices[0] + (n_buy - t) * t

    for day in range(len(prices) - 1):
        new_D = [MAX_VAR for i in range(n_buy + 1)]
        for prev_tickets in range(n_buy + 1):
            if D[prev_tickets] == MAX_VAR:
                continue

            for newday_ticket in range(max_day + 1):
                if prev_tickets + newday_ticket > n_buy:
                    break

                total_tickets = prev_tickets + newday_ticket
                poss_price = D[prev_tickets] + newday_ticket * prices[day + 1] + (n_buy - total_tickets) * newday_ticket
                new_D[total_tickets] = min(poss_price, new_D[total_tickets])
        D, new_D = new_D, D
    return D[-1]

if __name__ == "__main__":
    datagen = fileinput.input([sys.argv[1]] if len(sys.argv) > 1 else None)
    n_cases = int(datagen.readline())
    for _ in range(n_cases):
        x, max_day, n_buy = list(map(int, datagen.readline().split()))
        prices = list(map(int, datagen.readline().split()))
        print(solve_case(prices, n_buy, max_day))

**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 2 5
1 1 1 1 1
```

**Explanation:**

This test case is designed to expose potential issues in the buggy code:

- **Small number of days and tickets:** The small scale of the test case (5 days, 5 tickets, 2 tickets per day) ensures that the output can be easily verified.
- **Uniform prices:** All ticket prices are set to 1, which simplifies the calculation and makes it easier to see if the algorithm is handling the constraints correctly.
- **Maximum tickets to be bought:** The number of tickets to be bought (5) is exactly equal to the number of days (5), which means the algorithm should consider purchasing tickets in a way that minimizes the impact of the price increase.

**Expected Output:**

The correct output for this test case should be `5`, as the optimal strategy is to buy 1 ticket each day, which results in a total cost of 5.

**Why This Test Case?**

- **Edge Case for Uniform Prices:** If the buggy code incorrectly handles the price increase or the number of tickets bought per day, it might produce a higher cost.
- **Simplicity for Verification:** The uniform prices and small numbers make it easy to verify the correctness of the solution.
- **Ensuring Correct Handling of Constraints:** The constraints of 2 tickets per day and 5 tickets in total require the algorithm to distribute the purchases correctly to avoid unnecessary price increases.


**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.*;

public class dragon {
    static Kattio io = new Kattio();

    public static void main(String[] args) throws IOException {

    	int t = io.nextInt();

    	while (t > 0) {
    		int n = io.nextInt();
    		int m = io.nextInt();
    		int k = io.nextInt();

    		int[] a = new int[n];
    		int[] b = new int[n];
    		for (int i = 0; i < n; i++) {
    			a[i] = io.nextInt();
    			b[i] = a[i];
    		}
    		Arrays.sort(b);
    		int num = (int)(Math.ceil((double)k/m));
    		HashSet<Integer> h = new HashSet<Integer>();
    		for (int i = 0; i < num; i++) {
    			h.add(b[i]);
    		}
    		long[] c = new long[num];
    		int j = 0;
    		for (int i = 0; i < n; i++) {
    			if (h.contains(a[i]) && j < num) {
    				c[j] = a[i];
    				j++;
    			}
    		}
    		long ans = 0;
    		long left = k;
    		long max = 0;
    		for (int i = 0; i < num; i++) {
    			if (c[i] > max) max = c[i];
    		}
    		long rem = k%m;

    		for (int i = 0; i < num; i++) {
    			if (c[i] == max && rem != 0) {
    				long w = rem*c[i] + rem*(left-rem);
    				long d = 0;
    				if (left <= m) {
    					d = left*c[i];
    				} else {
    					d = m*c[i] + m*(left-m);
    				}
    				if (w < d) {
    					ans += w;
    					left -= rem;
    				} else {
    					ans += d;
    					left -= m;
    				}
    			} else {
    				long d = 0;
    				if (left <= m) {
    					d = left*c[i];
    				} else {
    					d = m*c[i] + m*(left-m);
    				}
    				ans += d;
    				left -= Math.min(m, left);
    			}
    		}
    		io.println(ans);

    		t--;
    	}

		io.close();
    }

    static class Kattio extends PrintWriter {
        private BufferedReader r;
        private StringTokenizer st;

        public Kattio() {
            this(System.in, System.out);
        }

        public Kattio(InputStream i, OutputStream o) {
            super(o);
            r = new BufferedReader(new InputStreamReader(i));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(r.readLine());
                return st.nextToken();
            } catch (Exception e) {
            }
            return null;
        }

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Edge Cases:**
   - **Small `n` and `m` with large `k`:** This can test the algorithm's ability to handle cases where the number of tickets required is much larger than the daily limit.
   - **Large `n` and `m` with small `k`:** This can test the algorithm's ability to handle cases where the number of tickets required is much smaller than the daily limit.
   - **Equal `n` and `m`:** This can test the algorithm's ability to handle cases where the number of tickets required can be bought in a single day.
   - ** prices are in descending order:** This can test the algorithm's ability to handle cases where the prices decrease over time.

2. **Scenarios:**
   - **Multiple test cases with different configurations:** This can ensure that the algorithm is robust and can handle a variety of inputs.

Given these considerations, let's create a small-scale test input:

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

Explanation:
- **Test Case 1:**
  - `n = 3`, `m = 2`, `k = 5`
  - Prices: `[10, 5, 1]`
  - This case tests the algorithm's ability to handle a small number of days and a large number of tickets required, with prices in descending order.
- **Test Case 2:**
  - `n = 4`, `m = 3`, `k = 10`
  - Prices: `[1, 2, 3, 4]`
  - This case tests the algorithm's ability to handle a larger number of days and a larger number of tickets required, with prices in ascending order.
- **Test Case 3:**
  - `n = 5`, `m = 1`, `k = 5`
  - Prices: `[1, 2, 3, 4, 5]`
  - This case tests the algorithm's ability to handle a small daily limit and a number of tickets required equal to the number of days.

This test input should expose potential issues in the buggy code, particularly in how it handles the dynamic price changes and the optimal ticket purchase strategy.


**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 com.sun.source.tree.Tree;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.*;
import java.io.PrintWriter;

public class Codeforces {
    public static void main (String[] args) throws IOException {
        // your code goes here
        FastScanner fs = new FastScanner();
        //BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=fs.nextInt();
        PrintWriter out=new PrintWriter(System.out);
        while(t-->0) {
            int n=fs.nextInt();
            long m=fs.nextLong(),k= fs.nextLong();
            long[] a=fs.readArray1(n);
            PriorityQueue<Long> pq=new PriorityQueue<>();
            for(long i:a) pq.add(i);
            HashMap<Long,Integer> map=new HashMap<>();
            long req=k/m,curr=0,res=0,count=0,x=0;
            for(long i=0;i<req;i++){
                long val=pq.poll();
                map.put(val,map.getOrDefault(val,0)+1);
            }
            if (k%m!=0){
                x=pq.poll();
            }else count=1;
            for(long i:a){
                if (map.containsKey(i)){
                    res+=m*(curr+i);
                    curr+=m;
                    map.put(i,map.getOrDefault(i,0)-1);
                    if (map.get(i)==0) map.remove(i);
                    count++;
                }else if (i==x && count==0){
                    res+=k%m*(curr+i);
                    curr+=k%m;
                }
            }
            out.println(res);
        }
        out.close();
    }

    public static boolean isPrime(int n){
        if (n%2==0) return false;
        for(int i=3;i<=Math.sqrt(n);i+=2){
            if (n%i==0) return false;
        }
        return true;
    }

    public static int sumOfDigits(long n){
        int sum=0;
        while(n!=0){
            sum+=n%10;
            n/=10;
        }
        return sum;
    }

    public static long gcd(long a, long b){
        if (a==0) return b;
        return gcd(b%a,a);
    }
    public static boolean larger(ArrayList<Integer> list1, ArrayList<Integer> list2){
        if (list2.size()==0) return true;
        int x=list1.size(),y=list2.size(),i=0,j=0;
        while(i<x && j<y){
            int first=list1.get(i++),second=list2.get(j++);
            if (first>second) return true;
            else if (first<second) return false;
        }
        return false;
    }
    public static String minString(String s1, String s2){
        if (s1.compareTo(s2)<=0) return s1;
        return s2;
    }
    public static String maxString(String s1, String s2){
        if (s1.compareTo(s2)<0) return s2;
        return s1;
    }
    public static int uppperBound(ArrayList<Integer> a, long x){
        int lo=0,hi=a.size()-1,res=a.size();
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            int curr=a.get(mid);
            if (curr>x){
                res=mid;
                hi=mid-1;
            }else lo=mid+1;
        }
        return res;
    }
    public static int lowerBound(ArrayList<Long> a, long x){
        int res=-1,lo=0,hi=a.size()-1;
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            long curr=a.get(mid);
            if (curr<=x){
                res=mid;
                lo=mid+1;
            }else hi=mid-1;
        }
        return res;
    }
    public static int uppperBound(long[] a, long x){
        int lo=0,hi=a.length-1,res=a.length;
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            long curr=a[mid];
            if (curr>x){
                res=mid;
                hi=mid-1;
            }else lo=mid+1;
        }
        return res;
    }
    public static long GCD(long a, long b){
        if (a==0) return b;
        return GCD(b%a,a);
    }

    public static long lcm_of_array(long[] arr)
    {
        long lcm = arr[0];
        for (int i = 1; i < arr.length; i++) {
            long num1 = lcm;
            long num2 = arr[i];
            long gcd_val = GCD(num1, num2);
            lcm = (lcm * arr[i]) / gcd_val;
        }
        return lcm;
    }
    static long power(long x, long y, long p) {
        long res = 1;
        x = x % p;
        if (x == 0)
            return 0;

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

            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        long[] readArray1(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }

}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the number of tickets to be bought is very close to the maximum possible or where the number of days is minimal.
2. **Complex Scenarios**: Scenarios where the prices change significantly due to the ticket purchase constraints.
3. **Large Ticket Purchases**: Cases where a large number of tickets are purchased on a single day to maximize the impact on subsequent days.

Given these considerations, let's create a test input that includes a mix of these scenarios:

```plaintext
3
5 1 5
1 2 3 4 5
4 2 8
1 2 3 4
1 100 100
1
```

### Explanation:
1. **Test Case 1**:
   - **n = 5**, **m = 1**, **k = 5**
   - **Prices**: [1, 2, 3, 4, 5]
   - This test case is designed to check the scenario where only one ticket can be bought per day, and the total number of tickets to be bought is equal to the number of days. This should force the algorithm to buy one ticket each day, and the prices should increase accordingly.

2. **Test Case 2**:
   - **n = 4**, **m = 2**, **k = 8**
   - **Prices**: [1, 2, 3, 4]
   - This test case is designed to check the scenario where the number of tickets to be bought is exactly double the maximum tickets that can be bought per day. This should force the algorithm to buy the maximum number of tickets each day, and the prices should increase significantly.

3. **Test Case 3**:
   - **n = 1**, **m = 100**, **k = 100**
   - **Prices**: [1]
   - This test case is designed to check the scenario where there is only one day to buy all the tickets, and the maximum number of tickets that can be bought per day is much larger than the number of tickets needed. This should force the algorithm to buy all tickets on the first day, and the price should remain the same.

These test cases are designed to cover a range of edge cases and complex scenarios to 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: 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 int long long
#define all(a) a.begin(),a.end()
#define loop(i,n) for(int i=0;i<n;i++)
#define nl cout<<endl;
#define pb push_back
#define bhalo ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

int char_int(char a){
    return a-'0';
}

string Toupper(string &s1){
    transform(s1.begin(), s1.end(), s1.begin(), ::toupper);
    return s1;
}

string Tolower(string &s1){
    transform(s1.begin(), s1.end(), s1.begin(), ::tolower);
    return s1;
}

int summation(vector<int> sum){
    return accumulate(sum.begin(), sum.end(), 0);
}

int summation(set<int> sum){
    return accumulate(sum.begin(), sum.end(), 0);
}

string Reverse(string &a){
    reverse(a.begin(), a.end());
    return a;
}

unordered_set<int> printDivisors(int n)
{
    // Note that this loop runs till square root
    unordered_set<int> s;
    for (int i=1; i<=sqrt(n); i++)
    {
        if (n%i == 0)
        {
            s.insert(i);
        }
    }
    return s;
}

void solve()
{
  int n,m,k;cin>>n>>m>>k;
  int arr[n];
  vector<pair<int,int>> a;
  for(auto &it:arr) cin>>it;
  for (int i = 0; i < n; i++)
  {
   a.push_back({arr[i],i});
  }
  int req=ceil((k*1.0)/m);
  sort(all(a));
  int brr[n];
  for (int i = 0; i < n; i++)
  {
    brr[a[i].second]=i;
  }
  vector<int> v;
  int mx=-1;
  for (int i = 0; i < n; i++)
  {
    if(brr[i]<req) v.push_back(arr[i]);
    if(brr[i]==req-1) mx=arr[i];
  }
  int ans=0,curr=0;
  for (int i = 0; i < v.size(); i++)
  {
    if(v[i]!=mx){
      ans+=(v[i]+curr)*m;
      curr+=m;
    }
    else{
    int x=k%m;
    if(x==0) x+=m;
    ans+=(v[i]+curr)*x;
    curr+=x;
    }
  }
  cout<<ans<<endl;
}
signed  main(){
	bhalo;
    int T ;
    T=1;
    cin>>T;

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

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases**:
   - **Small `n` and `m`**: To ensure the code handles small inputs correctly.
   - **Large `k`**: To test the code's ability to handle large ticket purchases.
   - **Uniform Prices**: To see how the code handles uniform ticket prices.
   - **Increasing Prices**: To test the code's ability to handle increasing prices.
   - **Decreasing Prices**: To test the code's ability to handle decreasing prices.

2. **Specific Scenarios**:
   - **`k` is a multiple of `m`**: To test the code's ability to handle exact multiples.
   - **`k` is not a multiple of `m`**: To test the code's ability to handle non-multiples.

3. **Input Format**:
   - The input should be formatted as specified in the problem description.

Given these considerations, let's create a test input that includes a combination of these edge cases and specific scenarios:

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

### Explanation:
1. **Test Case 1**:
   - **`n = 5`**, **`m = 2`**, **`k = 5`**
   - Prices: **`[1, 2, 3, 4, 5]`**
   - This case tests the code's ability to handle increasing prices and a `k` that is not a multiple of `m`.

2. **Test Case 2**:
   - **`n = 5`**, **`m = 2`**, **`k = 5`**
   - Prices: **`[5, 4, 3, 2, 1]`**
   - This case tests the code's ability to handle decreasing prices and a `k` that is not a multiple of `m`.

3. **Test Case 3**:
   - **`n = 3`**, **`m = 1`**, **`k = 3`**
   - Prices: **`[1, 1, 1]`**
   - This case tests the code's ability to handle uniform prices and a `k` that is a multiple of `m`.

4. **Test Case 4**:
   - **`n = 5`**, **`m = 3`**, **`k = 7`**
   - Prices: **`[1, 2, 3, 4, 5]`**
   - This case tests the code's ability to handle a larger `m` and a `k` that is not a multiple of `m`.

These test cases should help expose any bugs in the provided buggy code by covering a variety of scenarios and edge cases.


**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;
using vll = vector<ll>;
using vvll = vector<vll>;
using vi = vector<int>;

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

    int t; cin >> t;
    while (t--) {
        ll n, m, k; cin >> n >> m >> k;
        vll a(n);
        for (auto &i : a) cin >> i;

        vi idx(n);
        for (int i = 0; i < n; i++) idx[i] = i;
        sort(idx.begin(), idx.end(), [&] (int x, int y) { return a[x] < a[y]; });

        ll m2 = k % m;
        if (m2 == 0) m2 = m;
        ll n2 = (k+m-1) / m;

        ll rst1 = 0;
        auto idx1(idx);
        idx1.erase(idx1.begin()+n2, idx1.end());
        sort(idx1.begin(), idx1.end());
        ll sub1 = m2;
        rst1 = a[idx1[0]] * m2;
        for (int i = 1; i < n2; i++) {
            rst1 += (a[idx1[i]] + sub1) * m;
            sub1 += m;
        }

        if (n2 == 1) {
            cout << rst1 << "\n";
            continue;
        }

        auto idx2(idx);
        idx2.erase(idx2.begin()+n2-1, idx2.end());
        sort(idx2.begin(), idx2.end());
        ll subsum = 0;
        for (int i : idx2) subsum += a[i];
        for (int i = 0; i < idx2[0]; i++) {
            ll cur = a[i] * m2 + subsum * m + m2 * m * (n-1) + (n-1) * (n-2) / 2 * m * m;
            rst1 = min(rst1, cur);
        }
        cout << rst1 << "\n";
    }
}

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

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

Think step by step.

Response:
```plaintext
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:
t=int(input())

for _ in range(t):
    n,m,k=map(int,input().split())
    a=list(map(int,input().split()))
    k_cpy=k

    a_original=a

    a.sort()
    dic={}
    fac=0

    for ele in a:
        #if ele in dic:
        #    fac+=1
        #    ele=ele+dic[ele]*fac
        #else:
        #    fac=0
        if k>=m:
            dic[ele]=m
            k=k-m
        else:
            dic[ele]=k
            k=k-k
        if k==0:
            break

    x,price=0,0
    for elem in a_original:
        if x>=k_cpy:
            break
        if elem not in dic:
            continue
        else:
            val=elem+x
            x+=dic[elem]

            mapvalue=dic[elem]
            price+=val*mapvalue

    print(price)

**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 math
import sys
# sys.setrecursionlimit(int(1e6))

tt = int(input())
for _ in range(tt):
    n, m, k = map(int, input().split())
    l = list(map(int, input().split()))
    a = sorted(l)
    t = math.ceil(k / m)
    a = a[:t]
    isthere = {}
    for i in range(t):
        if a[i] in isthere:
            isthere[a[i]] += 1
        else:
            isthere[a[i]] = 1
    most = a[-1]
    ll = []
    for i in range(n):
        if l[i] in isthere and isthere[l[i]] > 0:
            ll.append(l[i])
            isthere[l[i]] -= 1
    # print(ll)
    ans = 0
    extra = 0
    for i in range(len(ll)):
        if ll[i] == most and k%m != 0:
            ans += (ll[i] + extra)*(k % m)
            extra += k % m
        else:
            ans += (ll[i] + extra)*m
            extra += m
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
1 100 100
1
1 100 101
1
2 100 200
1 1
```

### Explanation:
1. **First Test Case:**
   - **n = 1, m = 100, k = 100, a = [1]**
   - The price of tickets is always 1, and you need to buy 100 tickets. The optimal strategy is to buy all 100 tickets on the first day, which costs 100.

2. **Second Test Case:**
   - **n = 1, m = 100, k = 101, a = [1]**
   - The price of tickets is always 1, but you need to buy 101 tickets. Since you can only buy up to 100 tickets on the first day, you will need to buy 100 tickets on the first day and 1 ticket on the second day. The total cost will be 101.

3. **Third Test Case:**
   - **n = 2, m = 100, k = 200, a = [1, 1]**
   - The price of tickets is 1 on both days, and you need to buy 200 tickets. The optimal strategy is to buy 100 tickets on the first day and 100 tickets on the second day, which costs 200.

These test cases are designed to expose the following issues:
- The first test case checks if the code correctly handles a single day with a large number of tickets to be bought.
- The second test case checks if the code correctly handles a single day with a number of tickets slightly more than the maximum allowed per day.
- The third test case checks if the code correctly handles multiple days with a large number of tickets to be bought.