**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sort
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$ of length $n$. Then, you are (forced against your will) to answer $q$ queries.

For each query, you are given a range bounded by $l$ and $r$. In one operation, you can choose an integer $i$ ($l \leq i \leq r$) and set $a_i = x$ where $x$ is any character you desire. Output the minimum number of operations you must perform such that $\texttt{sorted(a[l..r])} = \texttt{sorted(b[l..r])}$. The operations you perform on one query does not affect other queries.

For an arbitrary string $c$, $\texttt{sorted(c[l..r])}$ denotes the substring consisting of characters $c_l, c_{l+1}, ... , c_r$ sorted in lexicographical order.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) – the number of test cases.

The first line of each test case contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^5$) – the length of both strings and the number of queries.

The following line contains $a$ of length $n$. It is guaranteed $a$ only contains lowercase latin letters.

The following line contains $b$ of length $n$. It is guaranteed $b$ only contains lowercase latin letters.

The following $q$ lines contain two integers $l$ and $r$ ($1 \leq l \leq r \leq n$) – the range of the query.

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

Output: For each query, output an integer, the minimum number of operations you need to perform in a new line.

Examples:
input:
3
5 3
abcde
edcba
1 5
1 4
3 3
4 2
zzde
azbe
1 3
1 4
6 3
uwuwuw
wuwuwu
2 4
1 3
1 6
output:
0
1
0
2
2
1
1
0

Note: For the first query, $\texttt{sorted(a[1..5])} =$ abcde and $\texttt{sorted(b[1..5])} =$ abcde, so no operations are necessary.

For the second query, you need to set $a_1 = $ e. Then, $\texttt{sorted(a[1..4])} = \texttt{sorted(b[1..4])} = $ bcde.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll int
using pl=long long;
int main()
{
	 ios_base::sync_with_stdio(false);
	 cin.tie(NULL);
	 cout.tie(NULL);
	 pl t;cin>>t;
	 while(t--){
	    ll n,k;cin>>n>>k;
	    string a,b;cin>>a>>b;
	    while(k--)
	    {
	        int p,q;cin>>p>>q;
	        string n1=a.substr(p-1,q-p+1);
	        string n2=b.substr(p-1,q-p+1);
	        int n=q-p+1;
	        unordered_map<char,int> mp;
	        for(int i=0;i<n;i++)
	        {
	            mp[n1[i]]++;
	        }
	        for(int i=0;i<n;i++)
	        {
	            mp[n2[i]]--;
	        }
	        int l=0;
	        for(auto i:mp)
	        {
	            l+=abs(i.second);
	        }
	        cout<<l/2<<endl;
	    }
	 }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
abcde
edcba
1 5
1 4
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sort
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$ of length $n$. Then, you are (forced against your will) to answer $q$ queries.

For each query, you are given a range bounded by $l$ and $r$. In one operation, you can choose an integer $i$ ($l \leq i \leq r$) and set $a_i = x$ where $x$ is any character you desire. Output the minimum number of operations you must perform such that $\texttt{sorted(a[l..r])} = \texttt{sorted(b[l..r])}$. The operations you perform on one query does not affect other queries.

For an arbitrary string $c$, $\texttt{sorted(c[l..r])}$ denotes the substring consisting of characters $c_l, c_{l+1}, ... , c_r$ sorted in lexicographical order.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) – the number of test cases.

The first line of each test case contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^5$) – the length of both strings and the number of queries.

The following line contains $a$ of length $n$. It is guaranteed $a$ only contains lowercase latin letters.

The following line contains $b$ of length $n$. It is guaranteed $b$ only contains lowercase latin letters.

The following $q$ lines contain two integers $l$ and $r$ ($1 \leq l \leq r \leq n$) – the range of the query.

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

Output: For each query, output an integer, the minimum number of operations you need to perform in a new line.

Examples:
input:
3
5 3
abcde
edcba
1 5
1 4
3 3
4 2
zzde
azbe
1 3
1 4
6 3
uwuwuw
wuwuwu
2 4
1 3
1 6
output:
0
1
0
2
2
1
1
0

Note: For the first query, $\texttt{sorted(a[1..5])} =$ abcde and $\texttt{sorted(b[1..5])} =$ abcde, so no operations are necessary.

For the second query, you need to set $a_1 = $ e. Then, $\texttt{sorted(a[1..4])} = \texttt{sorted(b[1..4])} = $ bcde.

Buggy Code:
from math import floor
for _ in range(int(input())):
    n,q = list(map(int,input().split()))
    a = input()
    b = input()
    pa= []
    pa.insert(0,[0]*26)
    pb= []
    pb.insert(0,[0]*26)
    for i in range(1,n+1):
        pa.append(list(pa[-1]))
        pb.append(list(pb[-1]))
        pa[i][ord(a[i-1])-ord('a')]+=1
        pb[i][ord(b[i-1])-ord('a')]+=1
    # print(pa,pb)
    for i in range(q):
        x,y = list(map(int,input().split()))
        ans = 0
        for i in range(26):
            ca = pa[y][i]-pa[x-1][i]
            cb = pb[y][i]-pb[x-1][i]
            ans+=abs(ca-cb)
        print(floor(ans/2))

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

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

Think step by step.

Response:
```plaintext
1
5 3
abcde
edcba
1 5
1 4
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sort
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$ of length $n$. Then, you are (forced against your will) to answer $q$ queries.

For each query, you are given a range bounded by $l$ and $r$. In one operation, you can choose an integer $i$ ($l \leq i \leq r$) and set $a_i = x$ where $x$ is any character you desire. Output the minimum number of operations you must perform such that $\texttt{sorted(a[l..r])} = \texttt{sorted(b[l..r])}$. The operations you perform on one query does not affect other queries.

For an arbitrary string $c$, $\texttt{sorted(c[l..r])}$ denotes the substring consisting of characters $c_l, c_{l+1}, ... , c_r$ sorted in lexicographical order.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) – the number of test cases.

The first line of each test case contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^5$) – the length of both strings and the number of queries.

The following line contains $a$ of length $n$. It is guaranteed $a$ only contains lowercase latin letters.

The following line contains $b$ of length $n$. It is guaranteed $b$ only contains lowercase latin letters.

The following $q$ lines contain two integers $l$ and $r$ ($1 \leq l \leq r \leq n$) – the range of the query.

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

Output: For each query, output an integer, the minimum number of operations you need to perform in a new line.

Examples:
input:
3
5 3
abcde
edcba
1 5
1 4
3 3
4 2
zzde
azbe
1 3
1 4
6 3
uwuwuw
wuwuwu
2 4
1 3
1 6
output:
0
1
0
2
2
1
1
0

Note: For the first query, $\texttt{sorted(a[1..5])} =$ abcde and $\texttt{sorted(b[1..5])} =$ abcde, so no operations are necessary.

For the second query, you need to set $a_1 = $ e. Then, $\texttt{sorted(a[1..4])} = \texttt{sorted(b[1..4])} = $ bcde.

Buggy Code:
def min_operations_to_match_sorted_substrings(t, test_cases):
    results = []

    for n, q, a, b, queries in test_cases:
        # Prefix frequency arrays for 'a' and 'b'
        prefix_freq_a = [[0] * 26 for _ in range(n + 1)]
        prefix_freq_b = [[0] * 26 for _ in range(n + 1)]

        # Build prefix frequency arrays
        for i in range(1, n + 1):
            for j in range(26):
                prefix_freq_a[i][j] = prefix_freq_a[i - 1][j]
                prefix_freq_b[i][j] = prefix_freq_b[i - 1][j]
            prefix_freq_a[i][ord(a[i - 1]) - ord('a')] += 1
            prefix_freq_b[i][ord(b[i - 1]) - ord('a')] += 1

        # Process each query
        for l, r in queries:
            freq_a = [prefix_freq_a[r][j] - prefix_freq_a[l - 1][j] for j in range(26)]
            freq_b = [prefix_freq_b[r][j] - prefix_freq_b[l - 1][j] for j in range(26)]

            operations = sum(abs(freq_a[j] - freq_b[j]) for j in range(26)) // 2
            results.append(operations)

    return results

# Reading input
import sys
input = sys.stdin.read
data = input().split()

index = 0
t = int(data[index])
index += 1
test_cases = []

for _ in range(t):
    n = int(data[index])
    q = int(data[index + 1])
    index += 2
    a = data[index]
    b = data[index + 1]
    index += 2
    queries = []
    for __ in range(q):
        l = int(data[index])
        r = int(data[index + 1])
        index += 2
        queries.append((l, r))
    test_cases.append((n, q, a, b, queries))

# Get results
results = min_operations_to_match_sorted_substrings(t, test_cases)

# Print results
for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
1
5 3
abcde
edcba
1 5
1 4
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sort
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$ of length $n$. Then, you are (forced against your will) to answer $q$ queries.

For each query, you are given a range bounded by $l$ and $r$. In one operation, you can choose an integer $i$ ($l \leq i \leq r$) and set $a_i = x$ where $x$ is any character you desire. Output the minimum number of operations you must perform such that $\texttt{sorted(a[l..r])} = \texttt{sorted(b[l..r])}$. The operations you perform on one query does not affect other queries.

For an arbitrary string $c$, $\texttt{sorted(c[l..r])}$ denotes the substring consisting of characters $c_l, c_{l+1}, ... , c_r$ sorted in lexicographical order.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) – the number of test cases.

The first line of each test case contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^5$) – the length of both strings and the number of queries.

The following line contains $a$ of length $n$. It is guaranteed $a$ only contains lowercase latin letters.

The following line contains $b$ of length $n$. It is guaranteed $b$ only contains lowercase latin letters.

The following $q$ lines contain two integers $l$ and $r$ ($1 \leq l \leq r \leq n$) – the range of the query.

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

Output: For each query, output an integer, the minimum number of operations you need to perform in a new line.

Examples:
input:
3
5 3
abcde
edcba
1 5
1 4
3 3
4 2
zzde
azbe
1 3
1 4
6 3
uwuwuw
wuwuwu
2 4
1 3
1 6
output:
0
1
0
2
2
1
1
0

Note: For the first query, $\texttt{sorted(a[1..5])} =$ abcde and $\texttt{sorted(b[1..5])} =$ abcde, so no operations are necessary.

For the second query, you need to set $a_1 = $ e. Then, $\texttt{sorted(a[1..4])} = \texttt{sorted(b[1..4])} = $ bcde.

Buggy Code:
import java.util.Scanner;

public class C_Sort {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- > 0) {
            int num = sc.nextInt();
            int queries = sc.nextInt();
            sc.nextLine();
            String a = sc.nextLine();
            String b = sc.nextLine();

            for (int query = 0; query < queries; query++) {
                int left = sc.nextInt() - 1;
                int right = sc.nextInt() - 1;

                int[] frequencyA = new int[26];
                int[] frequencyB = new int[26];

                for (int i = left; i <= right; i++) {
                    frequencyA[a.charAt(i) - 'a']++;
                }

                for (int i = left; i <= right; i++) {
                    frequencyB[b.charAt(i) - 'a']++;
                }

                int minChanges = 0;
                for (int i = 0; i < 26; i++) {
                    minChanges += Math.abs(frequencyA[i] - frequencyB[i]);
                }

                System.out.println(minChanges / 2);
            }
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
abcde
edcba
1 5
1 4
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sort
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$ of length $n$. Then, you are (forced against your will) to answer $q$ queries.

For each query, you are given a range bounded by $l$ and $r$. In one operation, you can choose an integer $i$ ($l \leq i \leq r$) and set $a_i = x$ where $x$ is any character you desire. Output the minimum number of operations you must perform such that $\texttt{sorted(a[l..r])} = \texttt{sorted(b[l..r])}$. The operations you perform on one query does not affect other queries.

For an arbitrary string $c$, $\texttt{sorted(c[l..r])}$ denotes the substring consisting of characters $c_l, c_{l+1}, ... , c_r$ sorted in lexicographical order.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) – the number of test cases.

The first line of each test case contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^5$) – the length of both strings and the number of queries.

The following line contains $a$ of length $n$. It is guaranteed $a$ only contains lowercase latin letters.

The following line contains $b$ of length $n$. It is guaranteed $b$ only contains lowercase latin letters.

The following $q$ lines contain two integers $l$ and $r$ ($1 \leq l \leq r \leq n$) – the range of the query.

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

Output: For each query, output an integer, the minimum number of operations you need to perform in a new line.

Examples:
input:
3
5 3
abcde
edcba
1 5
1 4
3 3
4 2
zzde
azbe
1 3
1 4
6 3
uwuwuw
wuwuwu
2 4
1 3
1 6
output:
0
1
0
2
2
1
1
0

Note: For the first query, $\texttt{sorted(a[1..5])} =$ abcde and $\texttt{sorted(b[1..5])} =$ abcde, so no operations are necessary.

For the second query, you need to set $a_1 = $ e. Then, $\texttt{sorted(a[1..4])} = \texttt{sorted(b[1..4])} = $ bcde.

Buggy Code:
//option 2

import java.util.*;
import java.lang.*;
import java.io.*;

public class sortQ {

    public static double calculateTriangleArea(double a, double b, double c) {
        // Calculate the semiperimeter
        double s = (a + b + c) / 2;

        // Calculate the area using Heron's formula
        double area = Math.sqrt(s * (s - a) * (s - b) * (s - c));

        return area;
    }
public static int search(int []nums, int target) {
       int n=nums.length;
       int low=0,high=n-1;
       int answer=-1;
           while(low<=high){
           int mid=(low+high)/2;
           if(nums[mid]==target){
               answer=mid;
               break;
           }
           else if(target>nums[mid])low=mid+1;
           else high=mid-1;
       }
       return answer;
       }
public static void printDivisors(int n)
    {   for (int i=1; i<=Math.sqrt(n); i++)
        { if (n%i==0)
            {   if (n/i == i) System.out.print(" "+ i);
                else  System.out.print(i+" " + n/i + " " );
            }
        }
    }
 public static long calculateModulus(long number) {
        // Modulus value
        long MOD = 1000000007; // 10^9 + 7

        // Calculate the modulus
        long modulus = number % MOD;

        // Ensure that the modulus is non-negative
        if (modulus < 0) {
            modulus += MOD;
        }

        return modulus;
    }

//function to calculate NcX
private static long factorial(int n) {
        if (n == 0 || n == 1) {
            return 1;
        }
        return n * factorial(n - 1);
    }

    // Function to calculate nCx
    public static long calculateBinomialCoefficient(int n, int x) {
        if (x < 0 || x > n) {
            return 0; // Invalid input, as x cannot be greater than n or negative
        }

        // Using the formula: nCx = n! / (x! * (n-x)!)
        long result = factorial(n) / (factorial(x) * factorial(n - x));

        return result;
    }

//
 public static int findLCM(int a, int b)
    {  int greater = Math.max(a, b);
        int smallest = Math.min(a, b);
        for (int i = greater;; i += greater) {
            if (i % smallest == 0)  return i;
        }
    }
public static boolean isPerfectSquare(int num) {
        if (num < 0) {
            return false;  // Negative numbers are not perfect squares
        }

        int sqrt = (int) Math.sqrt(num);
        return sqrt * sqrt == num;
    }
public static String reverseString(String s) {
        return new StringBuilder(s).reverse().toString();
    }
  private static void reverseArray(int[] arr) {
        int start = 0;
        int end = arr.length - 1;

        while (start < end) {
            // Swapping elements at start and end indices
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;

            // Updating indices
            start++;
            end--;
        }
    }
 static int findMostSignificantBit(int num) {
        if (num == 0) {
            return 0;
        }

        int msb = 0;

        // Find the position of the most significant bit
        while (num > 1) {
            msb++;
            num = num >> 1;
        }

        return msb;
    }
public static long factorialMod(int n) {
    long result = 1;
    for (int i = 1; i <= n; i++) {
        result = (result * i) % 1_000_000_007;
    }
    return result;
}
public static int binaryToDecimal(String binaryString) {
    int decimal = 0;
    int power = 0;

    // Iterate from the least significant bit (rightmost)
    for (int i = binaryString.length() - 1; i >= 0; i--) {
        char bit = binaryString.charAt(i);

        // Check if the current bit is '1'
        if (bit == '1') {
            // Multiply the current bit's value by its position value
            decimal += Math.pow(2, power);
        }

        power++;
    }

    return decimal;
}
 public int romanToInt(String s) {
        int n = s.length();
        int result =0;
        int curr=0;
        for(int i=n-1;i>=0;i--){
            switch(s.charAt(i)){
            case 'M' : curr= 1000; break;
            case 'D' : curr= 500; break;
            case 'C' : curr= 100; break;
            case 'L' : curr= 50; break;
            case 'X' : curr= 10; break;
            case 'V' : curr= 5; break;
            case 'I' : curr= 1; break;
            }
            if(4*curr<result) result-=curr;
            else result+=curr;
        }
        return result;
    }
public static ArrayList<String> generate(int index, String s, String current) {
        ArrayList<String> subsequences = new ArrayList<>();

        // Base case: when we reach the end of the string, add the current subsequence
        if (index == s.length()) {
            if (!current.isEmpty()) { // Ignore the empty subsequence if needed
                subsequences.add(current);
            }
            return subsequences;
        }

        // Recursive call: exclude the current character
        subsequences.addAll(generate(index + 1, s, current));

        // Recursive call: include the current character
        subsequences.addAll(generate(index + 1, s, current + s.charAt(index)));

        return subsequences;
    }
static ArrayList<String> AllPossibleStrings(String s) {
	int n = s.length();
	ArrayList<String>ans=new ArrayList<>();
	for (int num = 0; num < (1 << n); num++) {
		String sub = "";
		for (int i = 0; i < n; i++) {
			//check if the ith bit is set or not
			if ((num & (1 << i))!=0) {
				sub += s.charAt(i);
			}
		}
		if (sub.length() > 0) {
			ans.add(sub);
		}
	}
	Collections.sort(ans);
	return ans;
}

public boolean isSubsequence(String s, String t) {
        int i = 0, j = 0;
        while (i < s.length() && j < t.length()) {
            if (s.charAt(i) == t.charAt(j)) {
                j++;
            }
            i++;
        }
        return j == t.length();
    }

        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            int t = sc.nextInt();
            while (t-- > 0) {
                int n = sc.nextInt();
                int q = sc.nextInt();
                String a = sc.next();
                String b = sc.next();

                int[] al = new int[q];
                int[] ar = new int[q];
                for (int i = 0; i < q; i++) {
                    al[i] = sc.nextInt() - 1; // Convert to zero-indexed
                    ar[i] = sc.nextInt() - 1; // Convert to zero-indexed
                }

                for (int i = 0; i < q; i++) {
                    int l = al[i];
                    int r = ar[i];

                    HashMap<Character, Integer> arAcount = new HashMap<>();
                    HashMap<Character, Integer> arBcount = new HashMap<>();

                    for (int j = l; j <= r; j++) {
                        char ca = a.charAt(j);
                        char cb = b.charAt(j);
                        arAcount.put(ca, arAcount.getOrDefault(ca, 0) + 1);
                        arBcount.put(cb, arBcount.getOrDefault(cb, 0) + 1);
                    }

                    int ans = 0;
                    for (char ch = 'a'; ch <= 'z'; ch++) {

                        ans += Math.abs(arAcount.getOrDefault(ch, 0) - arBcount.getOrDefault(ch, 0));
                    }

                    System.out.println(ans / 2);
                }
            }

        }
    }

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code calculates the minimum number of operations needed to make the sorted substrings of `a` and `b` equal. The issue might be in the way it counts the differences between the characters in the substrings.

2. **Edge Cases**:
   - **Single Character Substrings**: Test with single character substrings to see if the code handles them correctly.
   - **Identical Substrings**: Test with identical substrings to ensure the code outputs 0 operations.
   - **Completely Different Substrings**: Test with substrings where all characters are different to see if the code correctly calculates the number of operations.
   - **Substrings with Repeated Characters**: Test with substrings that have repeated characters to ensure the code handles them correctly.

3. **Small-Scale Test Input**: Generate a small-scale test input that includes the above edge cases.

Here is a small-scale test input that covers these scenarios:

```plaintext
4
5 3
abcde
edcba
1 5
1 4
3 3
5 2
abcde
abcde
1 5
2 4
4 2
abcc
dbca
1 4
1 2
6 3
aaabbb
bbbaaa
1 6
2 4
3 5
```

This test input includes:
- A case with completely different substrings (`abcde` vs `edcba`).
- A case with identical substrings (`abcde` vs `abcde`).
- A case with repeated characters (`abcc` vs `dbca`).
- A case with a longer string and multiple queries to test different ranges (`aaabbb` vs `bbbaaa`).

This should help in exposing any issues in the buggy code.

4
5 3
abcde
edcba
1 5
1 4
3 3
5 2
abcde
abcde
1 5
2 4
4 2
abcc
dbca
1 4
1 2
6 3
aaabbb
bbbaaa
1 6
2 4
3 5


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
#pragma GCC optimized O(2)
#include<iostream>
#include<string>
#include<cstdio>
#include<queue>
#include<vector>
#include<map>
#include<math.h>
#include<algorithm>
#include<iomanip>
using namespace std;
#define inf 0x7fffffff
#define endl '\n'
typedef long long ll;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
ll a[N];
ll b[N];
vector<int>v;
void solve() {
    v.clear();
    priority_queue<ll, vector<ll>, greater<ll>>q;
    int num = 0;
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++)cin >> a[i];
    for (int i = 1; i <= n; i++) {
        cin >> b[i];
        if (b[i] > a[i]) {
            num++; v.push_back(i);
        }

    }
    ll ans = 0;
    if (k >= num) { cout << 0 << endl; return; }
    for (int i = 0; i < k; i++) {

        ans -= a[v[i]];
        q.push(b[v[i]]);
    }
    for (int i = k; i < v.size(); i++) {
        if(q.empty()||b[v[i]]<=q.top())ans+= b[v[i]] - a[v[i]];
        else {
            if (a[v[i]] < q.top()) {
                ans += - a[v[i]] + q.top();
                q.pop(); q.push(b[v[i]]);
            }
        }
    }
    if (ans <= 0)cout << 0 << endl;
    else cout << ans << endl;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t; cin >> t; while (t--)
        solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
#include <iostream>
#include<math.h>
#include<algorithm>
#include<vector>
#include<map>
#include<deque>
#include<queue>
#include <numeric>
#include <iomanip>
#include<set>
#include <unordered_map>
#include<stack>

#define ali ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define LONG_MAX 1e14

using namespace std;

pair<ll, ll> arr[200005];
int main() {
	ali
	int t = 1;
	cin >> t;
	while (t--) {
		ll n, k, sum1 = 0, sum2 = 0, taken, ans = 0;
		cin >> n >> k;
		taken = n;
		for (int i = 0; i < n; i++) {
			cin >> arr[i].first;
			sum1 += arr[i].first;
		}
		for (int i = 0; i < n; i++) {
			cin >> arr[i].second;
			if (arr[i].second < arr[i].first) {
				sum1 -= arr[i].first;
				arr[i] = { -1, -1 };
				taken--;
				continue;
			}

			sum2 += arr[i].second;
		}
		sort(arr, arr + n, [](const pair<ll, ll>& a, const pair<ll, ll>& b) {
			if (a.first == b.first) {
				return a.second > b.second;
			}
			return a.first > b.first;
		});

		if (taken == 0) {
			cout << 0 << '\n';
			continue;
		}

		if (k == 0) {

			for (int i = 0; i < taken; i++) {
				//cout << taken << "  rwef\n";
				ans += arr[i].second - arr[i].first;
			}
			cout << max(0ll, ans) << '\n';
			continue;
		}
		ll ks = 0, profit = 0;
		for (int i = 0; i < k; i++) {
			ks += arr[i].first;
		}
		for (int i = k; i < taken; i++) {
			profit += (arr[i].second - arr[i].first);
		}
		ll index = 0;
		while (taken > k) {

			if (profit - ks >= ans) {
				ans = profit - ks;
			}
			else {
				break;
			}
			//ans = max(ans, profit - ks);
			ks -= arr[index].first;
			ks += arr[index + k].first;
			profit -= (arr[index + k].second - arr[index + k].first);
			index++;
			taken--;
		}
		cout << ans << '\n';

	}
	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
4 2
1 2 3 4
4 3 2 1
4 2
1 2 3 4
1 2 3 4
4 2
1 2 3 4
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: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<deque>
#include<queue>
using ll = long long;
using namespace std;

void slove() {
	int n, k;
	cin >> n >> k;
	vector<ll>a(n);
	vector<ll>b(n);
	//multimap<ll, ll, greater<ll>>p;
	for (int i = 0;i < n;++i)cin >> a[i];
	for (int i = 0;i < n;++i)cin >> b[i];
	ll sum = 0;
	priority_queue<pair<ll,ll>>q;
		for (int i = 0;i < n;++i) {
			if (b[i] > a[i]) {
				sum += (b[i] - a[i]);
				q.push({a[i],i });
			}
		}
		if(k){
		auto kk=q.top();
		q.pop();
		sum-=b[kk.second];
		while(q.size()>k){
		    auto t=q.top();
		    q.pop();
		    if(kk.first>b[t.second]-a[t.second]){
		        sum+=kk.first;
		        sum-=b[t.second];
		        kk=t;
		    }
		   // else break;
		}
		k--;
		while(k--){
		    if(q.empty()){
		        sum=0;break;
		    }
		    auto t=q.top();
		    q.pop();
		    sum-=b[t.second];
		}}
	if(sum<0)sum=0;
	cout << sum << endl;
}
int main() {
	cin.tie(0), cout.tie(0);
	ios_base::sync_with_stdio(0);
	int t = 1;
	cin >> t;
	while (t--) {
		slove();
	}
}

**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 0
1
2
2 1
1 2
2 1
4 2
1 1 1 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: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
import heapq
import sys

putin = sys.stdin.readline

def inp():
    return int(putin())

def invr():
    return map(int, putin().split())

def starting_price(main_ar, sec_arr):
    sub_summ = 0
    for val_a in sec_arr:
        sub_summ += val_a[0]
    for val_b in main_ar:
        sub_summ += val_b[0] + val_b[1]
    return sub_summ

i = 0
if __name__ == "__main__":
    if i != 44 or i != 45 or i != 46:
        i += 1
        arr = []
        t = inp()
        for _ in range(t):
            n, k = invr()
            a = list(invr())
            a = [-x for x in a]
            b = list(invr())
            k_arr = []
            max_value = 0
            sub_sum = 0

            for i in range(n):
                arr.append([a[i], b[i]])
            arr = [x for x in arr if x[0] + x[1] >= 0]
            arr.sort(key=lambda x: x[1], reverse=True)

            for k_range in range(k):
                if arr and k_range < len(arr):
                    heapq.heappush(k_arr, arr[0])
                    arr.pop(0)

            if arr:
                sub_sum = starting_price(arr, k_arr)
            if sub_sum > max_value:
                max_value = sub_sum
            while arr:
                if k_arr and len(k_arr) == k:
                    sub_sum -= k_arr[0][0]
                    sub_sum += arr[0][0] - (arr[0][0] + arr[0][1])
                if arr:
                    heapq.heappushpop(k_arr, arr[0])
                    arr.pop(0)
                if sub_sum > max_value:
                    max_value = sub_sum

            print(max_value)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
import heapq
def bin(a,n,x):
  l,r = 0,n-1
  while l<r:
    m = (l+r+1)//2
    if a[m]<=x: l=m
    else: r=m-1
  return r
def binr(a,n,x):
  l,r = 0,n
  while l<r:
    m = (l+r)//2
    if a[m]<=x: l=m+1
    else: r=m
  return r
def main():
  #n = int(input())
  n,k = (int(i) for i in input().split())
  a = [int(i) for i in input().split()]
  #k = int(input())
  b = [int(i) for i in input().split()]
  #j = [int(i) for i in input().split()]
  #c = [list(i) for i in input().split()]
  #b = [int(i) for i in input().split()]
  p = [b[i]-a[i] for i in range(n)]
  x = [[b[i],i] for i in range(n)]
  x.sort(reverse=1)
  pa = [0]
  sm=0
  for bx,i in x:
    if a[i]<bx: sm +=bx-a[i]
    pa.append(sm)
  h = []
  idx = []
  ans = m = loss = 0
  for i in range(k):
    bx,j = x[i]
    loss += a[j]
    m+=1
    heapq.heappush(h, -a[j])
    heapq.heappush(h, -j)
  #print(x)
  #print(pa)
  ans = max(ans, pa[n]-pa[k]-loss)
  #print(k-1,loss)
  for i in range(k, n-1):
    bx,j = x[i]
    loss += a[j]
    heapq.heappush(h, -a[j])
    loss += heapq.heappop(h)
    ans = max(ans, pa[n]-pa[i+1]-loss)
  return ans

t = int(input())
while t:
  print(main())
  t -= 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
# LUOGU_RID: 158619758
print(1)
print(1)
print(0)
print(7)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the algorithm's logic might fail. The buggy code seems to be providing a fixed output for each test case, which suggests it is not actually solving the problem but rather hardcoding the results. We need to create a test case that will make the hardcoded results incorrect.

### Key Points to Consider:
1. **Small Scale**: The test case should be small enough to be manageable and fit within the constraints.
2. **Edge Cases**: Consider scenarios where the number of items and the value of \( k \) are at their limits or in specific configurations that could expose the bug.
3. **Profit Calculation**: Ensure the test case requires a non-trivial calculation of Alice's profit to differentiate from the hardcoded results.

### Test Case Generation:
Let's create a test case with:
- 2 test cases to ensure we cover multiple scenarios.
- \( n \) and \( k \) values that are not trivial.
- Item prices that require careful selection to maximize profit.

### Test Case 1:
- \( n = 3 \)
- \( k = 1 \)
- \( a = [1, 2, 3] \)
- \( b = [2, 3, 4] \)

### Test Case 2:
- \( n = 4 \)
- \( k = 2 \)
- \( a = [1, 2, 3, 4] \)
- \( b = [5, 4, 3, 2] \)

### Explanation:
- In Test Case 1, Alice should buy all 3 items. Bob will take the first item for free and pay for the other two. Alice's profit is \((3 + 4) - (1 + 2 + 3) = 1\).
- In Test Case 2, Alice should buy the first two items. Bob will take both for free. Alice's profit is \(0\).

### Test Input:
```plaintext
2
3 1
1 2 3
2 3 4
4 2
1 2 3 4
5 4 3 2
```

This test input should expose the bug in the provided buggy code, which outputs a fixed set of results regardless of the input.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

public class Main {

    public static void main(String[] args) {
        AReader sc = new AReader();
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc);
        }
    }

    public static void solve(AReader sc) {
        int n = sc.nextInt(), k = sc.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        Integer[] ids = new Integer[n];
        for (int i = 0; i < n; i++) {
            ids[i] = i;
            b[i] = sc.nextInt();
        }
        Arrays.sort(ids, (o1, o2) -> b[o2] - b[o1]);
        PriorityQueue<Integer> pq = new PriorityQueue<>((o1, o2) -> a[o2] - a[o1]);
        long total = 0;
        long earn = 0;
        for (int i = 0; i < n; i++) {
            if (a[ids[i]] <= b[ids[i]]) {
                total += a[ids[i]];
                earn += b[ids[i]];
            }
        }
        for (int i = 0; i < k; i++) {
            pq.add(ids[i]);
            earn -= b[ids[i]];
        }
        long ans = Math.max(0, earn - total);
        for (int i = k; i < n; i++) {
            pq.add(ids[i]);
            earn -= b[ids[i]];
            if (pq.size() > k) {
                Integer j = pq.poll();
                total -= a[j];
            }
            ans = Math.max(ans, earn - total);
        }
        System.out.println(ans);
    }

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

        private String innerNextLine() {
            try {
                return reader.readLine();
            } catch (IOException ex) {
                return null;
            }
        }

        public boolean hasNext() {
            while (!tokenizer.hasMoreTokens()) {
                String nextLine = innerNextLine();
                if (nextLine == null) {
                    return false;
                }
                tokenizer = new StringTokenizer(nextLine);
            }
            return true;
        }

        public String nextLine() {
            tokenizer = new StringTokenizer("");
            return innerNextLine();
        }

        public String next() {
            hasNext();
            return tokenizer.nextToken();
        }

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

        public long nextLong() {
            return Long.parseLong(next());
        }
        // public BigInteger nextBigInt() {
        // return new BigInteger(next());
        // }
        // 若需要nextDouble等方法，请自行调用Double.parseDouble包装
    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The provided code seems to have an issue with how it handles the selection of items for Alice and Bob. Specifically, it might not correctly account for the optimal selection of items to maximize Alice's profit while considering the constraints of Bob's free items.

2. **Identify Edge Cases**:
   - **Case with \( k = 0 \)**: Bob can take all items for free, so Alice should not buy any items to avoid a loss.
   - **Case with \( k = n \)**: Bob takes all items for free, so Alice should not buy any items.
   - **Case with \( k \) such that \( 0 < k < n \)**: Alice needs to carefully choose items to maximize her profit while ensuring Bob takes the least profitable items for him.

3. **Construct a Test Case**:
   - **Case 1**: \( k = 0 \)
     - Alice should not buy any items, so the profit should be 0.
   - **Case 2**: \( k = n \)
     - Alice should not buy any items, so the profit should be 0.
   - **Case 3**: \( 0 < k < n \)
     - Alice should buy items such that the profit is maximized. This can be tricky if the prices \( a_i \) and \( b_i \) are such that Bob can take the most profitable items for himself.

4. **Generate the Test Input**:
   - We will create a test case where \( n = 5 \) and \( k = 2 \).
   - The prices \( a_i \) and \( b_i \) will be such that Bob can take the most profitable items for himself, and Alice should carefully choose the remaining items.

Here is the test input:

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

### Explanation:
1. **First Test Case**:
   - \( n = 5 \), \( k = 0 \)
   - Alice should not buy any items, so the profit is 0.

2. **Second Test Case**:
   - \( n = 5 \), \( k = 5 \)
   - Alice should not buy any items, so the profit is 0.

3. **Third Test Case**:
   - \( n = 5 \), \( k = 2 \)
   - Alice should buy the items with prices \( a_i \) and \( b_i \) such that the profit is maximized.
   - Bob will take the two most profitable items for himself, which are the items with \( b_i = 5 \) and \( b_i = 4 \).
   - Alice should buy the items with \( b_i = 3 \), \( b_i = 2 \), and \( b_i = 1 \), and her profit will be \( 3 + 2 + 1 - (1 + 2 + 3) = 3 \).

This test input should expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

public class ShopGame {
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st;
        int t = Integer.parseInt(br.readLine());
        for(int q = 0; q < t; q++){
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int k = Integer.parseInt(st.nextToken());
            Pair[] pairs = new Pair[n];
            st = new StringTokenizer(br.readLine());
            for(int i = 0; i < n; i++){
                pairs[i] = new Pair(Integer.parseInt(st.nextToken()));
            }
            st = new StringTokenizer(br.readLine());
            for(int i = 0; i < n; i++){
                pairs[i].b = Integer.parseInt(st.nextToken());
            }
            Arrays.sort(pairs, (x, y) -> x.a - y.a);
            PriorityQueue<Integer> leastB = new PriorityQueue<Integer>();
            long amount = 0;
            for(int i = 0; i < n; i++){
                if(pairs[i].b <= pairs[i].a){
                    continue;
                }
                if(leastB.size() < k){
                    leastB.add(pairs[i].b);
                    amount -= pairs[i].a;
                }else{
                    if(leastB.size() == 0){
                        amount += pairs[i].b;
                        amount -= pairs[i].a;
                        continue;
                    }
                    if(leastB.peek() < pairs[i].b){
                        if(leastB.peek() > pairs[i].a){
                            amount += leastB.poll();
                            amount -= pairs[i].a;
                            leastB.add(pairs[i].b);
                        }
                    }else{
                        amount += pairs[i].b;
                        amount -= pairs[i].a;
                    }
                }
            }
            pw.println(Math.max(0, amount));
        }
        pw.close();
    }

    static class Pair{
        int b, a;
        Pair(int a){
            this.a = a;
        }
    }
}
/*
remove things with b[i] < a[i], that LOSES MONEY
bob will always choose the k things with the highest b cost

imagine we have x as the profit, and we want to consider buying item with values b and a
Call the cheapest b that bob is buying b2
replace b2, and we add b2 - a to x

 */

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace chrono;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define iter(i,n,incr) for(int i=0;i<n;i+=incr)
#define iterDec(i,n,decr) for(int i=n-1;i>=0;i-=decr)
#define fr(i,st,en,incr) for(int i=st;i<en;i+=incr)
#define frDec(i,st,en,decr) for(int i=st;i>en;i-=decr)
#define frAuto(i,a) for(auto i:a)
#define take(dt,i) dt i;cin>>i;
#define ll long long
#define pii pair<int,int>
#define pll pair<long long,long long>
#define pob pop_back
#define pub push_back
#define pof pop_front
#define puf push_front
#define vi vector<int>
#define vb vector<bool>
#define vll vector<long long>
#define vpii vector<pii>
#define vpll vector<pll>
#define invi(a) for(auto &i:a) cin>>i;
#define all(x) x.begin(),x.end()
#define ff first
#define ss second
#define yupp cout<<"YES"<<endl
#define nope cout<<"NO"<<endl
#define bugg cout<<"bug is here"<<endl

int MOD = 1e9 + 7;
const ll inf = 1e18;
const int N = 1e6 + 1;

ll gcd(ll a, ll b) {
	while (b != 0) {
		ll temp = b;
		b = a % b;
	   	a = temp;
	}
	return a;
}

ll pwr(ll num, int n){
	if(n==0) return 1;
	if(n==1) return num;

	ll tmp = pwr(num,n/2);
	if(n%2==0){
		return tmp*tmp;
	} else {
		return tmp*tmp*num;
	}
}

ll pwr_mod(ll num, int n){
	if(n==0) return 1;
	if(n==1) return num%MOD;

	ll tmp = pwr_mod(num, n/2);
	if(n%2==0){
		return (tmp*tmp)%MOD;
	} else {
		return (num*((tmp*tmp)%MOD))%MOD;
	}
}

int popct(ll num){
	int ct=0;
	while(num){
		if(num & 1) ct++;
		num >>= 1;
	}
	return ct;
}

int binLen(ll num){
	int ct=0;
	while(num){
		ct++;
		num /= 2;
	}
	return ct;
}

ll digitSum(ll num){
	ll sum = 0;
	while(num){
		sum += num%10;
		num /= 10;
	}
	return sum;
}

int numOfDivs(int num,int i){
	int ct = 0;
	while(num && num%i==0){
		num /= i;
		ct++;
	}
	return ct;
}

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

int md(ll i){
	if(i<0){
		return MOD + (i%MOD);
	}

	return i%MOD;
}

int mdd(int i, int mod){
	if(i<0){
		return mod + i%mod;
	}

	return i%mod;
}

//***************MAIN-BODY****************//

void solve(){
	take(int, n)
	take(int, k)

	vi a(n);invi(a);
	vi b(n);invi(b);

	vpii vec;
	iter(i,n,1){
		if(a[i] < b[i]){
			vec.pub({a[i], b[i]});
		}
	}

	sort(all(vec), [](const pii &a, const pii &b)->bool{
		if(b.ss == a.ss){
			return a.ff > b.ff;
		}

		return a.ss > b.ss;
	});

	ll profit = 0;

	n = vec.size();

	iter(i,n,1){
		if(i<k) profit -= vec[i].ss;
		profit += vec[i].ss - vec[i].ff;
	}

	if(k==0){
		cout<<profit<<endl;
		return;
	}

	int i=0,j=k;

	while(i<j && j<n){
		if(vec[i].ff > vec[j].ss){
			profit += vec[i].ff - vec[j].ss;
			j++;
		}
		i++;
	}

	cout<<max(profit, 0LL)<<endl;
}

//***************MAIN-BODY****************//

int main(){
	take(int, t)
	// int t=1;
	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:
```plaintext
1
4 2
1 2 3 4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

using namespace std;

using ll = long long;
using db = long double; // or double, if TL is tight
using str = string; // yay python!

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
using vvi = vector<vi>;
using vvl = vector<vl>;
using vvpi = vector<vpi>;
// pairs
#define mp make_pair
#define f first
#define s second

// vectors
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend()
#define sor(x) sort(all(x))
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back
#define pf push_front
#define er erase
#define ub upper_bound
#define lb lower_bound

// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const db PI = acos((db)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!=

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;

template<typename T>
using indexed_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define ook order_of_key
#define fbo find_by_order

void solve(){
    ll n, k, sum = 0;
    cin >> n >> k;
    vl a(n), b(n);
    F0R(i, n)cin >> a[i];
    F0R(i, n)cin >> b[i];
    vpl xy;
    F0R(i, n){
        if(b[i] - a[i] > 0){
            xy.pb({b[i], a[i]});
            sum -= a[i];
        }
    }
    if(k == 0){
        trav(el, xy)sum += el.f;
        cout << sum << "\n";
        return;
    }
    sort(rall(xy));
    if(sz(xy) <= k){
        cout << 0 << "\n";
        return;
    }
    ll j = k, m = sz(xy);
    set<ll> s;
    F0R(i, k)s.ins(xy[i].s);
    //trav(el, s)cout << el << " ";
    //cout << "\n";
    while(j < m){
        ll x = *s.rbegin();
        auto it = prev(s.end());
        if(xy[j].f < x){
            sum += x;
            s.erase(it);
            s.ins(xy[j].s);
            j++;
            //trav(el, s)cout << el << " ";
        //cout << "\n";
        }
        else break;
    }
    //cout << sum << " " << j << "\n";
    FOR(i, j, m){
        sum += xy[i].f;
        //cout << xy[i].f << " ";
    }
    cout << max(0LL, sum) << "\n";

}

int main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t; cin>>t;while(t--){
        solve();
        //cout << "\n";
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
from operator import itemgetter
t = int(input())
for o in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    c = [0] * n
    d = [[0, 0] for i in range(n)]
    sm = 0
    if k == 0:
        for i in range(n):
            sm += max(0, b[i] - a[i])
        print(sm)
        continue
    not_using = []
    for i in range(n):
        c[i] = b[i] - a[i]
        sm += c[i]
    for i in range(n):
        d[i][0] = c[i]
        d[i][1] = b[i]
    d.sort(key=itemgetter(1))
    for i in range(n - k, n):
        sm -= d[i][1]
    using = list(range(n - k, n))
    for i in range(n - 1, - 1, - 1):
        if i in using and i - k >= 0 and sm - d[i][0] + d[i][1] - d[i - k][1] >= sm:
            sm = sm - d[i][0] + d[i][1] - d[i - k][1]
            using.append(i - k)
            not_using.append(i)
        else:
            k -= 1
        if k <= 0:
            continue
    for i in range(n):
        if i not in not_using and d[i][0] < 0:
            sm -= d[i][0]
    if sm >= 0:
        print(sm)
    else:
        print(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
3 2
1 2 3
4 5 6
3 2
3 2 1
4 5 6
3 2
1 2 3
6 5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
import heapq
import sys

putin = sys.stdin.readline

def inp():
    return int(putin())

def invr():
    return map(int, putin().split())

def starting_price(main_ar, sec_arr):
    sub_summ = 0
    for val_a in sec_arr:
        sub_summ += val_a[0]
    for val_b in main_ar:
        sub_summ += val_b[0] + val_b[1]
    return sub_summ

if __name__ == "__main__":
    arr = []
    t = inp()
    ueli = False
    for x in range(t):
        n, k = invr()
        a = list(invr())
        a = [-x for x in a]
        b = list(invr())
        k_arr = []
        max_value = 0
        sub_sum = 0

        for i in range(n):
            arr.append([a[i], b[i]])
        arr = [x for x in arr if x[0] + x[1] >= 0]
        arr.sort(key=lambda x: x[1], reverse=True)

        for k_range in range(k):
            if arr and k_range <= len(arr):
                heapq.heappush(k_arr, arr[0])
                arr.pop(0)

        if arr:
            sub_sum = starting_price(arr, k_arr)
        if sub_sum > max_value:
            max_value = sub_sum
        while arr:
            if k_arr and len(k_arr) == k:
                sub_sum -= k_arr[0][0]
                sub_sum += arr[0][0] - (arr[0][0] + arr[0][1])
            if arr:
                heapq.heappushpop(k_arr, arr[0])
                arr.pop(0)
            if sub_sum > max_value:
                max_value = sub_sum
        if x == 1 and max_value == 21:
            ueli = True
        if ueli == True and x == 847:
            #andrin = 16
            #print(-a[andrin] + b[andrin]*100)
            print(1)
        else:
            print(max_value)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

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

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        // try {
        int testCases = vt.nextInt();
        // int testCases = 1;
        while (testCases-- > 0) {
            // write code here
            solve();
        }
        out.flush();
        // } catch (Exception e) {
        // return;
        // }
    }

    static void solve() {
        int n = vt.nextInt();
        int k = vt.nextInt();
        long[] ar = new long[n];
        long[] br = new long[n];
        for (int i = 0; i < n; i++) {
            ar[i] = vt.nextLong();
        }
        for (int i = 0; i < n; i++) {
            br[i] = vt.nextLong();
        }
        long[][] diff = new long[n][3];
        int d = 0;
        for (int i = 0; i < n; i++) {
            diff[i][0] = br[i] - ar[i];
            if (diff[i][0] >= 0) {
                d++;
            }
            diff[i][1] = ar[i];
            diff[i][2] = br[i];
        }
        if (d <= k) {
            out.println(0);
            return;
        }
        sort2d(diff, 0, n - 1, 0);
        int l = 0;
        int r = 0;
        for (int i = 1; i < n; i++) {
            if (diff[i][0] == diff[i - 1][0]) {
                r = i;
            } else {
                sort2d(diff, l, r, 2);
                l = i;
                r = i;
            }
        }
        sort2d(diff, l, r, 2);
        long ans = 0;
        long val = 0;
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for (int i = n - 1; i >= 0; i--) {
            // out.println(diff[i][0] + " " + diff[i][1] + " " + diff[i][2]);
            ans -= diff[i][1];
            if (k == 0) {
                ans += diff[i][2];
                val = (long) Math.max(val, ans);
                continue;
            }
            if (pq.size() < k) {
                pq.add(diff[i][2]);
            } else {
                if (pq.peek() < diff[i][2]) {
                    ans += pq.poll();
                    pq.add(diff[i][2]);
                } else {
                    ans += diff[i][2];
                }
            }
            // out.println(ans);
            val = (long) Math.max(val, ans);
        }
        out.println(val);
    }

    static void sort2d(long arr[][], int l, int r, int index) {
        if (l < r) {
            int m = l + (r - l) / 2;
            sort2d(arr, l, m, index);
            sort2d(arr, m + 1, r, index);
            merge2d(arr, l, m, r, index);
        }
    }

    static void merge2d(long arr[][], int l, int m, int r, int index) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int m1 = arr[0].length;
        long L[][] = new long[n1][m1];
        long R[][] = new long[n2][m1];
        for (int i = 0; i < n1; ++i)
            for (int j = 0; j < m1; j++)
                L[i][j] = arr[l + i][j];

        for (int i = 0; i < n2; ++i)
            for (int j = 0; j < m1; j++)
                R[i][j] = arr[m + 1 + i][j];

        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i][index] <= R[j][index]) {
                for (int t = 0; t < m1; t++)
                    arr[k][t] = L[i][t];
                i++;
            } else {
                for (int t = 0; t < m1; t++)
                    arr[k][t] = R[j][t];
                j++;
            }
            k++;
        }
        while (i < n1) {
            for (int t = 0; t < m1; t++)
                arr[k][t] = L[i][t];
            i++;
            k++;
        }
        while (j < n2) {
            for (int t = 0; t < m1; t++)
                arr[k][t] = R[j][t];
            j++;
            k++;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

Examples:
input:
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
output:
1
1
0
7

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

public class CF_1969_D {

    private static Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) {
        int t = scanner.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    private static void solve() {
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        Integer[] idxes = new Integer[n];
        int[] profit = new int[n];
        int[] sort = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            idxes[i] = i;
        }
        //首先所有利润加起来
        long totalProfit = 0;
        for (int i = 0; i < n; i++) {
            b[i] = scanner.nextInt();
            sort[i] = -b[i];
            profit[i] = b[i] - a[i];
            if (profit[i] > 0) {
                totalProfit += profit[i];
            }
        }
        Arrays.sort(idxes, (id1, id2) -> Integer.compare(sort[id1], sort[id2]));
        LinkedList<Integer> linkedList = new LinkedList<>();
        //利润为负, 一定不选。
        long res = 0;
        for (int i = 0; i < n; i++) {
            int curIdx = idxes[i];
            if (profit[curIdx] > 0) {
                if (k > 0) {
                    totalProfit += sort[curIdx];
                    linkedList.addLast(curIdx);
                    k--;
                }else {
                    //选不选当前值呢
                    res = Math.max(res, totalProfit);
                    if (linkedList.size() > 0) {
                        Integer idx = linkedList.removeFirst();
                        totalProfit = totalProfit - sort[idx];
                        totalProfit = totalProfit - profit[idx];
                        totalProfit = totalProfit + sort[curIdx];
                    }
                }
            }
        }
        res = Math.max(res, totalProfit);

        System.out.println(res);
    }

}

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

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

Think step by step.

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