**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Unfair Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob gathered in the evening to play an exciting game on a sequence of $$$n$$$ integers, each integer of the sequence doesn't exceed $$$4$$$. The rules of the game are too complex to describe, so let's just describe the winning condition — Alice wins if the bitwise XOR of all the numbers in the sequence is non-zero; otherwise, Bob wins.

The guys invited Eve to act as a judge. Initially, Alice and Bob play with $$$n$$$ numbers. After one game, Eve removes one of the numbers from the sequence, then Alice and Bob play with $$$n-1$$$ numbers. Eve removes one number again, after which Alice and Bob play with $$$n - 2$$$ numbers. This continues until the sequence of numbers is empty.

Eve seems to think that in such a game, Alice almost always wins, so she wants Bob to win as many times as possible. Determine the maximum number of times Bob can win against Alice if Eve removes the numbers optimally.

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

The first and only line of each test case contains four integers $$$p_i$$$ ($$$0 \le p_i \le 200$$$) — the number of ones, twos, threes, and fours in the sequence at the beginning of the game.

Output: For each test case, print the maximum number of times Bob will win in a separate line, if Eve removes the numbers optimally.

Examples:
input:
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
output:
1
1
3
3
12

Note: In the first example, Bob wins when Eve has not removed any numbers yet.

In the second example, Bob wins if Eve removes one one and one three.

Buggy Code:
import sys
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
lii = lambda: list(map(int, input().split()))

def solve():
    a, b, c, d = lii()
    ans = a//2 + b//2 + c//2 + d//2
    if a%2 == b%2 == c%2 == 1 or (a%2 == b%2 == c%2 == 0 and d):
        ans += 1
    return ans

for _ in range(ii()):
    print(solve())

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic of the `solve` function. The function calculates the maximum number of times Bob can win by considering the number of pairs of each type of number (1, 2, 3, 4) and then adding an additional win if certain conditions are met. The conditions are based on the parity (odd or even) of the counts of these numbers.

The buggy code has a logical flaw in the condition for adding the additional win. Specifically, the condition:
```python
if a%2 == b%2 == c%2 == 1 or (a%2 == b%2 == c%2 == 0 and d):
    ans += 1
```
is not entirely correct. This condition checks if all counts are odd or all counts are even and there is at least one 4. However, it doesn't correctly handle all edge cases.

To expose this bug, we can create a test case where the counts of the numbers are such that the condition should not trigger an additional win, but the current logic incorrectly does. Let's consider the following test case:

```plaintext
1
1 1 0 1
```

### Explanation:
- `a = 1` (one 1)
- `b = 1` (one 2)
- `c = 0` (zero 3)
- `d = 1` (one 4)

### Expected Output:
- The total number of pairs is `0` (since there are no pairs of any number).
- The additional win condition should not be triggered because `c` is 0, which means not all counts are odd.
- Therefore, the expected output should be `0`.

### Actual Output:
- The current code will incorrectly trigger the additional win condition because `a%2 == b%2 == 0%2 == 1` is false, but `a%2 == b%2 == 0%2 == 0 and d` is true.
- Therefore, the actual output will be `1`.

This test case will expose the bug in the current logic of the `solve` function.

```plaintext
1
1 1 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Unfair Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob gathered in the evening to play an exciting game on a sequence of $$$n$$$ integers, each integer of the sequence doesn't exceed $$$4$$$. The rules of the game are too complex to describe, so let's just describe the winning condition — Alice wins if the bitwise XOR of all the numbers in the sequence is non-zero; otherwise, Bob wins.

The guys invited Eve to act as a judge. Initially, Alice and Bob play with $$$n$$$ numbers. After one game, Eve removes one of the numbers from the sequence, then Alice and Bob play with $$$n-1$$$ numbers. Eve removes one number again, after which Alice and Bob play with $$$n - 2$$$ numbers. This continues until the sequence of numbers is empty.

Eve seems to think that in such a game, Alice almost always wins, so she wants Bob to win as many times as possible. Determine the maximum number of times Bob can win against Alice if Eve removes the numbers optimally.

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

The first and only line of each test case contains four integers $$$p_i$$$ ($$$0 \le p_i \le 200$$$) — the number of ones, twos, threes, and fours in the sequence at the beginning of the game.

Output: For each test case, print the maximum number of times Bob will win in a separate line, if Eve removes the numbers optimally.

Examples:
input:
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
output:
1
1
3
3
12

Note: In the first example, Bob wins when Eve has not removed any numbers yet.

In the second example, Bob wins if Eve removes one one and one three.

Buggy Code:
import sys
import heapq
import math
import functools

input = sys.stdin.readline
from collections import Counter, deque, defaultdict

def sieve_of_eratosthenes(n):
    sieve = [True] * (n + 1)
    sieve[0], sieve[1] = False, False  # 0 and 1 are not prime numbers
    primes = []

    for current in range(2, int(n ** 0.5) + 1):
        if sieve[current]:
            primes.append(current)
            for multiple in range(current * current, n + 1, current):
                sieve[multiple] = False

    for current in range(int(n ** 0.5) + 1, n + 1):
        if sieve[current]:
            primes.append(current)

    return primes

############ ---- Input Functions ---- ############
def inp():
    return (int(input()))

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

def insr():
    s = input()
    return (list(s[:len(s) - 1]))

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

dp = [[[0 for _ in range(201)] for _ in range(201)] for _ in range(201)]
for i in range(201):
    for j in range(201):
        for k in range(201):
            if k % 2 == 1:
                dp[i][j][k] = (k + 1) // 2 + (j - 1) // 2 + (i - 1) // 2
            else:
                dp[i][j][k] = k // 2 + i // 2 + j // 2

def solve():
    a, b, c, d = inlt()
    print(dp[a][b][c] + d // 2)

def main():
    # for _ in range(1):
    for _ in range(inp()):
        solve()

main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can highlight the incorrect logic in the code. The code's main issue seems to be in how it calculates the maximum number of times Bob can win. Specifically, the code uses a precomputed `dp` array, which might not handle all edge cases correctly.

### Step-by-Step Reasoning:

1. **Understand the Problem:**
   - The game involves a sequence of integers (1, 2, 3, 4).
   - Alice wins if the bitwise XOR of all numbers is non-zero.
   - Bob wins if the bitwise XOR is zero.
   - Eve removes numbers optimally to maximize Bob's wins.

2. **Analyze the Buggy Code:**
   - The `dp` array is precomputed to store the number of Bob's wins based on the counts of 1s, 2s, and 3s.
   - The code does not consider the 4s correctly, as it only adds `d // 2` to the result.

3. **Identify Edge Cases:**
   - **Zeroes and Even Distribution:** Test cases where the counts of 1s, 2s, and 3s are such that the XOR can easily become zero.
   - **Odd and Even Counts:** Test cases where the counts of 1s, 2s, and 3s are odd or even, which can affect the XOR result.
   - **Large Counts of 4s:** Test cases with a large number of 4s to see how the code handles them.

4. **Generate a Challenging Test Input:**
   - A small-scale test input that can expose the bug in the precomputed `dp` array and the handling of 4s.

### Chosen Test Input:
```plaintext
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
```

This test input includes:
- A mix of small and large counts of 1s, 2s, and 3s.
- Different combinations of 4s to see how the code handles them.

### Explanation:
- **Test Case 1 (1 1 1 0):** Bob wins when Eve has not removed any numbers yet.
- **Test Case 2 (1 0 1 2):** Bob wins if Eve removes one 1 and one 3.
- **Test Case 3 (2 2 2 0):** Bob wins 3 times.
- **Test Case 4 (3 3 2 0):** Bob wins 3 times.
- **Test Case 5 (0 9 9 9):** Bob wins 12 times.

These test cases should help expose any issues in the precomputed `dp` array and the handling of 4s in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Unfair Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob gathered in the evening to play an exciting game on a sequence of $$$n$$$ integers, each integer of the sequence doesn't exceed $$$4$$$. The rules of the game are too complex to describe, so let's just describe the winning condition — Alice wins if the bitwise XOR of all the numbers in the sequence is non-zero; otherwise, Bob wins.

The guys invited Eve to act as a judge. Initially, Alice and Bob play with $$$n$$$ numbers. After one game, Eve removes one of the numbers from the sequence, then Alice and Bob play with $$$n-1$$$ numbers. Eve removes one number again, after which Alice and Bob play with $$$n - 2$$$ numbers. This continues until the sequence of numbers is empty.

Eve seems to think that in such a game, Alice almost always wins, so she wants Bob to win as many times as possible. Determine the maximum number of times Bob can win against Alice if Eve removes the numbers optimally.

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

The first and only line of each test case contains four integers $$$p_i$$$ ($$$0 \le p_i \le 200$$$) — the number of ones, twos, threes, and fours in the sequence at the beginning of the game.

Output: For each test case, print the maximum number of times Bob will win in a separate line, if Eve removes the numbers optimally.

Examples:
input:
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
output:
1
1
3
3
12

Note: In the first example, Bob wins when Eve has not removed any numbers yet.

In the second example, Bob wins if Eve removes one one and one three.

Buggy Code:
import java.util.*;
import java.io.*;
import java.util.StringTokenizer;
import java.lang.*;
public final class Codechef{

	static void doit()throws Exception{
		int [] a=ia(4);
		int ans=0;
		if(a[0]==1 && a[1]==1 && a[2]==1){
			pl(1);return;}
		for(int i:a){
			ans+=i/2;}
		pl(ans);

	}

	public static void main(String... args)throws Exception{
		int testcases=in.nextInt();
		while(testcases-->0){
			doit();
		}
		out.close();
	}

	static List<Integer> primes(int n){
		boolean prime[] = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++) {
            if (prime[p] == true) {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        List<Integer> ans=new ArrayList<>();
        for (int i = 2; i <= n; i++) {
            if (prime[i] == true)
                ans.add(i);
        }
        return ans;
	}

	public static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			if (System.getProperty("ONLINE_JUDGE")== null) {
				try {
					br = new BufferedReader(
						new InputStreamReader(
							new FileInputStream(
								"input.txt")));
				}
				catch (Exception e) {
				}
			}
		}
		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;

		}
	}
	public static class FastWriter {
		BufferedWriter bwr;
		public FastWriter() {
			bwr = new BufferedWriter(new OutputStreamWriter(System.out));
			if (System.getProperty("ONLINE_JUDGE")== null) {
				try {
					bwr = new BufferedWriter(
						new OutputStreamWriter(
							new FileOutputStream(
								"output.txt")));
				}
				catch (Exception e) {
				}
			}
		}
		public void print(Object object) throws IOException {
			bwr.append("" + object);
		}
		public void println(Object object) throws IOException {
			print(object);
			bwr.append("\n");
		}
		public void close() throws IOException {
			bwr.flush();
		}
	}
	static FastReader in=new FastReader();
	static FastWriter out=new FastWriter();

	static int i() {return in.nextInt();}
	static String s() {return in.next();}
	static long l() {return in.nextLong();}
	static int[] ia(int n) {int A[]=new int[n]; for(int i=0;i<n;i++) { A[i]=in.nextInt();}return A;}
	static long[] la(int n) {long A[]=new long[n]; for(int i=0;i<n;i++) {A[i]=in.nextLong();}return A;}
	static int[][] ia(int n,int m) {int a[][] = new int[n][m]; for(int i=0;i<n;i++) a[i] = ia(m);return a;}
	static long[][] la(int n,int m){long a[][] = new long[n][m]; for(int i=0;i<n;i++) a[i] = la(m);return a;}

	static void p(Object o)throws IOException{ out.print(o);}
	static void pl(Object o)throws IOException{out.println(o);}
	static void pl()throws IOException{out.println("");}
	static void y()throws IOException { pl("yes"); }
	static void Y()throws IOException { pl("YES"); }
	static void n()throws IOException { pl("no"); }
	static void N()throws IOException { pl("NO"); }
	static <E> void p(List<E> a)throws IOException {for (E e : a) {p(e);p(" ");}pl();}
	static void p(int[] a)throws IOException { for (int e : a) { p(e);p(" "); } pl(); }
	static void p(long[] a)throws IOException { for (long e : a) { p(e);p(" "); } pl(); }
	static long modl = (long) (1e9 + 7);
	static int modi = (int) (1e9 + 7);
	static int MAXI=Integer.MAX_VALUE;
	static int MINI=Integer.MIN_VALUE;
	static long MAXL=Long.MAX_VALUE;
	static long MINL=Long.MIN_VALUE;
	static void swap(int[] nums,int a,int b){int tmp=nums[a];nums[a]=nums[b];nums[b]=tmp;}

	private static long sqrt(long num) {long l = 0;long h = Integer.MAX_VALUE;long mid;while (h - l > 1) {  mid = (h + l)/2;    if (mid*mid <= num) {l = mid;}else {h = mid;}}return l;}
	static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); }
	static long gcd(long a, long b) { if (a == 0) return b; return gcd(b % a, a); }
	static int lcm(int a, int b){ return (a / gcd(a, b)) * b;}
	static long lcm(long a, long b){return (a / gcd(a, b)) * b;}
	static String reverse(String s) {StringBuffer p=new StringBuffer(s); p.reverse();return p.toString();}
	static int min(int a,int b) {return Math.min(a, b);}
    static int min(int a,int b,int c) {return min(a, min(b, c));}
    static int min(int a,int b,int c,int d) {return min(a, min(b, min(c, d)));}
    static int max(int a,int b) {return Math.max(a, b);}
    static int max(int a,int b,int c) {return max(a, max(b, c));}
    static int max(int a,int b,int c,int d) {return Math.max(a, Math.max(b, Math.max(c, d)));}
	static long min(long a,long b) {return Math.min(a, b);}
	static long min(long a,long b,long c) {return Math.min(a, Math.min(b, c));}
	static long min(long a,long b,long c,long d) {return Math.min(a, Math.min(b, Math.min(c, d)));}
	static long max(long a,long b) {return Math.max(a, b);}
	static long max(long a,long b,long c) {return Math.max(a, Math.max(b, c));}
	static long max(long a,long b,long c,long d) {return Math.max(a, Math.max(b, Math.max(c, d)));}
	static long sum(int A[]) {long sum=0;for(int i : A) { sum+=i;}return sum;}
	static long sum(long A[]) {long sum=0;for(long i : A) { sum+=i;}return sum;}

	private static class Pair<K,V>{K first;V second;public Pair(K f, V s) {first = f;second = s;}}
}

**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 code, we need to focus on edge cases where the code might fail. The buggy code has a specific condition for a very narrow set of inputs, and then it uses a general approach that might not be correct for all cases. Let's break down the problem and identify potential edge cases:

1. **Specific Condition Handling:**
   - The code has a specific condition for when `a[0] == 1 && a[1] == 1 && a[2] == 1`. This condition is handled separately, but it might not cover all edge cases.

2. **General Condition:**
   - The general approach is to count the number of pairs in the sequence and use this to determine the number of times Bob can win. This might not be correct for all cases, especially when the sequence has a mix of numbers.

3. **Edge Cases to Consider:**
   - **Single Number:** When the sequence has only one number.
   - **Two Numbers:** When the sequence has only two numbers.
   - **All Zeros:** When the sequence has only zeros.
   - **Mixed Numbers:** When the sequence has a mix of numbers.

Given these considerations, let's create a test input that includes these edge cases:

```plaintext
4
1 0 0 0
2 0 0 0
0 0 0 0
1 1 1 1
```

### Explanation:
1. **Test Case 1: 1 0 0 0**
   - Sequence: [1]
   - XOR: 1
   - Bob wins 0 times because the sequence is already non-zero.

2. **Test Case 2: 2 0 0 0**
   - Sequence: [1, 1]
   - XOR: 0
   - Bob wins 1 time because removing one number makes the sequence non-zero.

3. **Test Case 3: 0 0 0 0**
   - Sequence: []
   - XOR: 0
   - Bob wins 0 times because the sequence is empty.

4. **Test Case 4: 1 1 1 1**
   - Sequence: [1, 2, 3, 4]
   - XOR: 0
   - Bob wins 2 times because removing any two numbers can make the sequence non-zero.

This test input should expose the bug in the code, especially in the general approach where it counts pairs without considering the actual XOR values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Unfair Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob gathered in the evening to play an exciting game on a sequence of $$$n$$$ integers, each integer of the sequence doesn't exceed $$$4$$$. The rules of the game are too complex to describe, so let's just describe the winning condition — Alice wins if the bitwise XOR of all the numbers in the sequence is non-zero; otherwise, Bob wins.

The guys invited Eve to act as a judge. Initially, Alice and Bob play with $$$n$$$ numbers. After one game, Eve removes one of the numbers from the sequence, then Alice and Bob play with $$$n-1$$$ numbers. Eve removes one number again, after which Alice and Bob play with $$$n - 2$$$ numbers. This continues until the sequence of numbers is empty.

Eve seems to think that in such a game, Alice almost always wins, so she wants Bob to win as many times as possible. Determine the maximum number of times Bob can win against Alice if Eve removes the numbers optimally.

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

The first and only line of each test case contains four integers $$$p_i$$$ ($$$0 \le p_i \le 200$$$) — the number of ones, twos, threes, and fours in the sequence at the beginning of the game.

Output: For each test case, print the maximum number of times Bob will win in a separate line, if Eve removes the numbers optimally.

Examples:
input:
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
output:
1
1
3
3
12

Note: In the first example, Bob wins when Eve has not removed any numbers yet.

In the second example, Bob wins if Eve removes one one and one three.

Buggy Code:
import java.io.BufferedReader;import java.io.DataInputStream;import java.io.File;import java.io.FileInputStream;import java.io.FileNotFoundException;import java.io.FileWriter;import java.io.IOException;import java.io.InputStreamReader;import java.util.*;import java.util.*;

public class Codeforces implements Runnable
{
   //static FastReader sc = new FastReader();
   static long modulo = (int)Math.pow(10,9)+7;

   public static void main(String[] args)  {

     new Thread(null, new Codeforces(), "Main", 1<<28).start();
    }
    public void run()
    {
      FastReader sc = new FastReader();

      int tests=sc.nextInt();

      for(int t=0;t<tests;t++)
      {
        int n1=sc.nextInt();
        int n2=sc.nextInt();
        int n3=sc.nextInt();
        int n4=sc.nextInt();

        int bit1=n4;
        int bit2=n2;
        int bit3=n1;
        bit2+=n3;
        bit3+=n3;
       // print(bit1+" "+bit2+" "+bit3);
        int ans=0;
        while(n1>0 || n2>0 || n3>0 || n4>0)
        {
          if(bit1%2==1){bit1--;n4--;}
          else if(bit2%2==1 && bit3%2==1 && n3>0){bit2--;bit1--;n3--;}
          else if(bit2%2==1 && n2>0){bit2--;n2--;}
          else if(bit2%2==1 && n1>0 && n3>0){bit2--;n1--;n3--;}
          else if(bit3%2==1 && n1>0){bit3--;n1--;}
          else if(bit3%2==1 && n3>0 && n2>0){bit3--;n2--;n3--;}
          else break;
         // if(bit1%2==0 && bit2%2==0 && bit3%2==0)ans++;

        }

        print( (n1+n2+n3+n4)/2 + " ");
      }
    }

  public static void weightedDFS(long ans[],ArrayList<ArrayList<Pair>> graph,boolean visited[],int cur,ArrayList<Integer> itr)
  {
    for(Pair p: graph.get(cur))
    {
      int node=(int)p.first;
      long weight=p.second;
      ans[0]&=weight;

      if(!visited[node])
      {
        visited[node]=true;
        weightedDFS(ans,graph,visited,node,itr);
      }
    }
    itr.add(cur);
  }

    public static long[] Djistra(ArrayList<ArrayList<Pair>> graph,int st,long flights[])
    {
      PriorityQueue<Pair> pq = new PriorityQueue<>(new PairComparator());
     // print(flights);
      pq.add(new Pair(0,st));
      long distances[] = new long[flights.length];
      boolean visited[] = new boolean[flights.length];
     // visited[st]=true;
      Arrays.fill(distances,Long.MAX_VALUE);
      distances[st]=0;
     // Arrays.sort(flights);

      for(int i=0;i<flights.length;i++)
      {
        if(i!=st)
        {
          pq.add(new Pair(flights[st]+flights[i],i));
        }
      }

      Set<Integer> set = new HashSet<>();
      for(int i=0;i<flights.length;i++)set.add(i);

      while(!pq.isEmpty())
      {
        Pair p=pq.poll();
        int node=(int)p.second;
        long weight = p.first;
        if(visited[node])continue;
     //   print(weight+" "+node);
     //  print("");

        visited[node]=true;
        set.remove(node);
        distances[node]=weight;

        for(Pair p1:graph.get(node))
        {
          int dest=(int)p1.first;
          long nextweight=p1.second;
       //   print(flights[1]+"");

         if(!visited[dest]) pq.add(new Pair(Math.min(weight + nextweight,weight+flights[node]+flights[dest]),dest));
        }
        for(Integer i:set)pq.add(new Pair(weight+flights[node]+flights[i],i));
      }
      return distances;

    }

    public static int diff(String str1,String str2)
    {
      int cnt=0;
       for(int i=0;i<str1.length();i++)
       {
         if(str1.charAt(i)!=str2.charAt(i))cnt++;
       }
       return cnt;
    }

    public static boolean poss(int i)
    {
      String temp = Integer.toString(i);
      //  boolean flag=true;
        for(char c:temp.toCharArray())
        {
          if(c=='0' || c=='1')continue;
          return false;
        }
       return true;
    }

  public static long dfs(int arr[][],int i,int j,boolean visited[][])
  {
    long ans=0;
    if(visited[i][j])return ans;
    ans=arr[i][j];
    visited[i][j]=true;
    if(arr[i][j]==0 )return ans;
     if(i-1>=0)ans+=dfs(arr,i-1,j,visited);
     if(i+1<arr.length)ans+=dfs(arr,i+1,j,visited);
     if(j-1>=0)ans+=dfs(arr,i,j-1,visited);
     if(j+1<arr[0].length)ans+=dfs(arr,i,j+1,visited);

   //  print(ans+" ");
     return ans;
  }

  public static int DFS(ArrayList<ArrayList<Integer>> tree,int parent,int cur,int child[],int whites[],String str)
  {
      child[cur]=1;
      if(str.charAt(cur)=='W')whites[cur]++;
      int ans=0;
      for(Integer i: tree.get(cur))
      {
        if(i==parent)continue;
       ans+= DFS(tree,cur,i,child,whites,str);
        child[cur]+=child[i];
        whites[cur]+=whites[i];

      }
      if(whites[cur]*2==child[cur])ans++;

      return ans;

  }

     public static long query(int l , int r,long[] trees,long arr[])
    {
        long ans=query(0,arr.length-1,l,r,trees,0);
        return ans;
    }

    public static long query(int st,int end,int l,int r,long[]trees,int idx)
    {
        if(st>r || end<l)
        {

            return 0;
        }
        if(st>=l && end<=r)
        {

             long arr= trees[idx];
            return arr;
        }
        int mid=(st+end)/2;

        long arr1=query(st,mid,l,r,trees,2*idx+1);
        long arr2 =query(mid+1,end,l,r,trees,2*idx+2);
        long ans = (arr1+arr2);

        return ans%modulo;
    }

    public static void update(long [] trees, long diff, int idx,long arr[])
    {
        updateTrees(trees,diff,idx,0,arr.length-1,0);
        arr[idx]+=diff;
    }

    public static void updateTrees(long[] trees, long diff,int idx,int st,int end,int curidx)
    {
        if(st>idx || end<idx)
        {
            return;
        }
        if(st==idx && end==idx)
        {
            trees[curidx]+=diff;
            return;
        }

        int mid=(st+end)/2;

        updateTrees(trees, diff, idx, st, mid, 2*curidx+1);
        updateTrees(trees, diff, idx, mid+1, end, 2*curidx+2);

         trees[curidx]= (trees[2*curidx+1]+trees[2*curidx+2]);
      //   trees[curidx][1]=0;
    }

    public static long[] create(long arr[])
    {
        long trees[] = new long[4*arr.length];
        createTrees(trees,arr,0,arr.length-1,0);
        return trees;
    }

    public static void createTrees(long trees[],long arr[],int st,int end,int idx)
    {
        if(st==end)
        {
            trees[idx]=arr[st];
            trees[idx]%=modulo;
            return ;
        }

        int mid=(st+end)/2;

        createTrees(trees, arr, st, mid, 2*idx+1);
        createTrees(trees, arr, mid+1, end, 2*idx+2);

        trees[idx] = (trees[idx*2+1]+trees[2*idx+2])%modulo;
      }

    public static void solve()
    {
    }

    public static void print(String str) {System.out.println(str);}

    public static void print(int arr[])
    {StringBuilder sb = new StringBuilder("");for(int i=0;i<arr.length;i++) {    sb.append(arr[i]+" ");}print(sb.toString());}

    public static void print(long arr[])
    {StringBuilder sb = new StringBuilder("");for(int i=0;i<arr.length;i++) {    sb.append(arr[i]+" ");}print(sb.toString());}

    public static void print(boolean f,StringBuilder sb)
    {if(f)sb.append("YES\n");else sb.append("NO\n");}

    // public static int[] inputarr(int n) throws IOException
    // { int arr[] = new int[n]; for(int i=0;i<n;i++)arr[i]=sc.nextInt();return arr;}

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

    static long lcm(long a ,long b)
    { return ((a*b))/gcd(a,b);}

    static long sum(long n)
    {return (n*(n+1))/2;}

    public static void print(int arr[][])
    {for(int i=0;i<arr.length;i++) {for(int j=0;j<arr[0].length;j++) { System.out.print(arr[i][j]+" "); } System.out.println();}}

    public static int isPrime(int n)
    {if (n <= 1) return -1;if (n == 2 || n == 3)return -1; if (n % 2 == 0 )return 2;
     if(n%3==0)return 3; for (int i = 5; i <= Math.sqrt(n); i = i + 6){if (n % i == 0 )return i;if(n%(i+2)==0)return i+2;}return -1;}
   static class FastReader {
   BufferedReader br;  StringTokenizer st;public FastReader()
  { br = new BufferedReader(  new InputStreamReader(System.in));}
  String next() {  while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); }
  catch (IOException e) {  e.printStackTrace(); } } return st.nextToken(); }
  int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble()
{ return Double.parseDouble(next()); } String nextLine() { String str = ""; try { if(st.hasMoreTokens()){ str = st.nextToken("\n"); } else{ str = br.readLine(); } }
catch (IOException e) {  e.printStackTrace(); } return str;  }} }

class Pair {long first=-1;long second=-1;Pair(long f,long s) {first=f;second=s;}}

class Tuple {int first=-1;int second=-1;int third=-1;Tuple(int f,int s,int t) {first=f;second=s;third=t;}}

class PairComparator implements Comparator<Pair> {
  @Override
  public int compare(Pair p1, Pair p2) {
      // Compare integers in reverse order
      if(p1.first>p2.first)
      {
        return 1;
      }
      else if(p1.first==p2.first && p1.second<p2.second)
      {
        return 1;
      }
      else if(p1.first==p2.first && p1.second==p2.second)
      {
        return 0;
      }
      {
        return -1;
      }
  }
}
class TupleComparator implements Comparator<Tuple> {
  @Override
  public int compare(Tuple p1, Tuple p2) {
      // Compare integers in reverse order
      if(p1.third>p2.third)
      {
        return 1;
      }
      if(p1.third==p2.third)return 0;

      return -1;
  }

}

  /**   template to increase stack size
     * class Main implements Runnable {
 static void main(...){
  new Thread(null, new Main(), "Main", 1<<28).start();
 }

 public void run() {
  // your code here
 }
}
     */

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Bitwise XOR Calculation**:
   - The code attempts to manage the bitwise XOR of the sequence by keeping track of the counts of numbers contributing to specific bits (1, 2, 3, 4).
   - The logic for removing numbers to ensure Bob wins as many times as possible is complex and may have edge cases.

2. **Edge Cases**:
   - **Zeroes in the sequence**: The code does not handle zeroes explicitly, which can affect the XOR calculation.
   - **Balanced counts**: When the counts of numbers are balanced, the code might not handle the removal optimally.
   - **Odd and even counts**: The code uses modulo operations to check the parity of the counts, which can lead to incorrect decisions in certain cases.

3. **Test Input**:
   - We need a test input that forces the code to handle these edge cases.
   - A good candidate is a sequence where the counts of numbers are balanced and the XOR of the sequence is zero initially, but removing numbers should lead to a non-zero XOR.

Let's create a test input that fits these criteria:

- **Test Case 1**: Balanced counts with an even number of each type.
- **Test Case 2**: Balanced counts with an odd number of each type.
- **Test Case 3**: A mix of even and odd counts.

Here is the test input:

```plaintext
3
2 2 2 2
3 3 3 3
2 3 2 3
```

### Explanation:
1. **Test Case 1 (2 2 2 2)**:
   - The sequence has an even number of each type of number (1, 2, 3, 4).
   - The XOR of the sequence is zero initially.
   - Removing any number should lead to a non-zero XOR, but the code might not handle this correctly.

2. **Test Case 2 (3 3 3 3)**:
   - The sequence has an odd number of each type of number.
   - The XOR of the sequence is non-zero initially.
   - Removing numbers should be handled carefully to ensure Bob wins as many times as possible.

3. **Test Case 3 (2 3 2 3)**:
   - The sequence has a mix of even and odd counts.
   - This case tests the code's ability to handle different parities and ensure the correct number of wins for Bob.

This test input should expose any issues in the buggy code by forcing it to handle various edge cases and ensure the correct number of wins for Bob.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Unfair Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob gathered in the evening to play an exciting game on a sequence of $$$n$$$ integers, each integer of the sequence doesn't exceed $$$4$$$. The rules of the game are too complex to describe, so let's just describe the winning condition — Alice wins if the bitwise XOR of all the numbers in the sequence is non-zero; otherwise, Bob wins.

The guys invited Eve to act as a judge. Initially, Alice and Bob play with $$$n$$$ numbers. After one game, Eve removes one of the numbers from the sequence, then Alice and Bob play with $$$n-1$$$ numbers. Eve removes one number again, after which Alice and Bob play with $$$n - 2$$$ numbers. This continues until the sequence of numbers is empty.

Eve seems to think that in such a game, Alice almost always wins, so she wants Bob to win as many times as possible. Determine the maximum number of times Bob can win against Alice if Eve removes the numbers optimally.

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

The first and only line of each test case contains four integers $$$p_i$$$ ($$$0 \le p_i \le 200$$$) — the number of ones, twos, threes, and fours in the sequence at the beginning of the game.

Output: For each test case, print the maximum number of times Bob will win in a separate line, if Eve removes the numbers optimally.

Examples:
input:
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
output:
1
1
3
3
12

Note: In the first example, Bob wins when Eve has not removed any numbers yet.

In the second example, Bob wins if Eve removes one one and one three.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>

using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less_equal<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;

#define ll long long int
#define forn(i, n) for (int i = 0; i < int(n); i++)
#define cin(arr,n) for(int i=0; i<n;i++){cin>>arr[i];}
#define cin1(vec,n) forn(i,n){ll x;cin>>x;vec.push_back(x);}

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

  clock_t z = clock();

  ll t;
  cin>>t;
  while(t--){

    ll a,b,c,d;
    cin>>a>>b>>c>>d;

    ll n=a+b+c+d;

    ll bit1=a+c;
    ll bit2=b+c;
    ll bit3=d;

    ll ans=0;
    ll ct=0;
    while(n>0 && ct<20){
      ct++;
      //cout<<"a b c d are "<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
      if(bit1%2!=0 && bit2%2!=0){
        if(c>0){
          c--;
          n--;
        }
        else{
          a--;
          b--;
          n-=2;
        }
        bit1--;
        bit2--;
      }
      else if(bit1%2!=0){
        n--;
        bit1--;

        if(a>0){
          a--;
        }
        else{
          c--;
          bit2--;
        }
      }
      else if(bit2%2!=0){
        n--;
        bit2--;
        if(b>0){
          b--;
        }
        else{
          c--;
          bit1--;
        }
      }
      else if(bit3%2!=0){
          bit3--;;
          n--;
          d--;
      }
      else if(bit1%2==0 && bit2%2==0 && bit3%2==0){
        ans++;

        if(c>=2){
          c-=2;
          n-=2;
          bit1-=2;
          bit2-=2;
        }
        else if(a>=1 && b>=1 && c>=1){
          c--;
          a--;
          b--;
          n-=3;
          bit1-=2;
          bit2-=2;
        }
        else if(a>=2){
          a-=2;
          n-=2;
          bit1-=2;
        }
        else if(b>=2){
          b-=2;
          n-=2;
          bit2-=2;
        }
        else if(d>=2){
          d-=2;
          n-=2;
          bit3-=2;
        }
      }
    }

    cout<<ans<<endl;

  }
  //cerr << "Run Time : " << ((double)(clock() - z) / CLOCKS_PER_SEC);
  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
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Unfair Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob gathered in the evening to play an exciting game on a sequence of $$$n$$$ integers, each integer of the sequence doesn't exceed $$$4$$$. The rules of the game are too complex to describe, so let's just describe the winning condition — Alice wins if the bitwise XOR of all the numbers in the sequence is non-zero; otherwise, Bob wins.

The guys invited Eve to act as a judge. Initially, Alice and Bob play with $$$n$$$ numbers. After one game, Eve removes one of the numbers from the sequence, then Alice and Bob play with $$$n-1$$$ numbers. Eve removes one number again, after which Alice and Bob play with $$$n - 2$$$ numbers. This continues until the sequence of numbers is empty.

Eve seems to think that in such a game, Alice almost always wins, so she wants Bob to win as many times as possible. Determine the maximum number of times Bob can win against Alice if Eve removes the numbers optimally.

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

The first and only line of each test case contains four integers $$$p_i$$$ ($$$0 \le p_i \le 200$$$) — the number of ones, twos, threes, and fours in the sequence at the beginning of the game.

Output: For each test case, print the maximum number of times Bob will win in a separate line, if Eve removes the numbers optimally.

Examples:
input:
5
1 1 1 0
1 0 1 2
2 2 2 0
3 3 2 0
0 9 9 9
output:
1
1
3
3
12

Note: In the first example, Bob wins when Eve has not removed any numbers yet.

In the second example, Bob wins if Eve removes one one and one three.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
// typedef tree<int, null_type, less<int> rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;

typedef long long int ll;
typedef long double ld;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<ld> vd;
typedef vector<vd> vvd;
typedef vector<bool> vb;
typedef pair<ll, ll> pll;
typedef map<ll, ll> mll;

const double PI = 3.141592653589793;
const int Mod = 1e9+7;
const int mod = 998244353;
const int inf = INT_MAX;
const ll INF = LLONG_MAX;

#define f(i,a,b,d) for(ll i=a;i<b;i+=d)
#define rf(i,a,b,d) for(ll i=a;i>=b;i-=d)

ll solve()
{
	ll a,b,c,d;
	cin>>a>>b>>c>>d;

	vector<vector<vector<ll>>> dp(a+1, vector<vector<ll>>(b+1, vector<ll>(c+1,0)));

	f(i,0,a+1,1)
	{
		f(j,0,b+1,1)
		{
			f(k,0,c+1,1)
			{
				if(i%2==j%2 && i%2==k%2 && !(i==0 &&  j==0 && k==0))
				{
					dp[i][j][k]=1;
				}
				ll x=0;
				if(i>0)
					x=max(x,dp[i-1][j][k]);
				if(j>0)
					x=max(x,dp[i][j-1][k]);
				if(k>0)
					x=max(x,dp[i][j][k-1]);
				dp[i][j][k]+=x;
			}
		}
	}

	return dp[a][b][c]+(d/2);
}

int main() {
	fastio
    #ifndef ONLINE_JUDGE
    freopen("error.txt", "w", stderr);
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
	ll t=1;
	cin>>t;
	while(t--)
	{
		cout<<solve()<<endl;
		// solve();
	}

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

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

void solve(){
    int n,k;
    cin>>n>>k;
    int arr[n];
    for (int  i = 0; i < n; i++)
    {
        cin>>arr[i];

    }
    int count =0;
    int zero =0;
    for (int j = 0; j < n; j++)
    {
        if (arr[j]>=k){
            count+= arr[j];
        }
        else if(arr[j]==0){
            zero+=1;
        }
    }
    if(count>zero){
        cout<<zero<<"\n";
    }
    else if(count<zero){
        cout<<count<<"\n";
    }
    else{
        cout<<zero<<"\n";
    }

}
int32_t main(){
     int t;
     cin>>t;
     while(t--){
         solve();
     }
     return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the logic. The buggy code incorrectly handles the scenario where Robin Hood has enough gold to give to all or some of the people with zero gold.

### Key Points to Consider:
1. **Initial Gold**: Robin starts with 0 gold.
2. **Taking Gold**: Robin takes all gold from people with `a_i >= k`.
3. **Giving Gold**: Robin gives 1 gold to people with `a_i == 0` if he has any gold left.

### Potential Bugs:
1. **Incorrect Counting**: The code might not correctly count the number of people who get gold.
2. **Insufficient Gold Handling**: The code might not handle the case where Robin runs out of gold before giving to all people with zero gold.

### Challenging Test Input:
Let's create a test case where Robin encounters a mix of people with gold greater than or equal to `k` and people with zero gold, but the total gold taken is just enough to give to some of the people with zero gold, not all.

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

### Explanation:
- **n = 5**: There are 5 people.
- **k = 2**: The threshold for taking gold is 2.
- **a = [1, 0, 0, 3, 0]**:
  - The first person has 1 gold (less than 2, so Robin does nothing).
  - The second person has 0 gold (Robin can give 1 gold if he has any).
  - The third person has 0 gold (Robin can give 1 gold if he has any).
  - The fourth person has 3 gold (greater than or equal to 2, so Robin takes 3 gold).
  - The fifth person has 0 gold (Robin can give 1 gold if he has any).

### Expected Output:
- **Count of gold taken**: 3 gold from the fourth person.
- **Count of people getting gold**: 3 (the second, third, and fifth people).

### Correct Output:
```plaintext
3
```

### Why This Test Input is Challenging:
- The total gold taken (3) is exactly enough to give to 3 people with zero gold.
- This test case checks if the code correctly handles the situation where Robin has just enough gold to give to some, but not all, of the people with zero gold.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

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

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        vector<int> v(n);
        for(int i=0;i<n;i++)
        {
            cin>>v[i];
        }
        int ans=0;int cnt=0;
        for(int i=0;i<n;i++)
        {
            if(v[i]==0)
            {
                if(cnt>0)
                {
                    ans++;
                    cnt--;
                }
            }
            else
            {
                cnt=cnt+v[i];
            }
        }
        // cout<<"Ans:";
        cout<<ans<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 1
0 0 0 0 100
```

**Explanation:**
- This test case is designed to expose the bug where Robin Hood should give gold to the first few people with 0 gold, but the buggy code might not handle the scenario correctly if Robin has enough gold from a single rich person later in the list.
- Here, Robin starts with 0 gold and encounters 5 people. The first four people have 0 gold, and the last person has 100 gold.
- According to the correct logic, Robin should give 1 gold to each of the first four people, but the buggy code might not correctly manage the gold count and might fail to give gold to all four people.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
#include <bits/stdc++.h>
#define endl '\n'

using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n,k;
        cin >> n >> k ;
        int gold = 0,poor=0;
        int temp;
        for (int i = 0; i < n; i++)
        {
            cin >> temp;
            if(temp>= k)gold+= temp;
            if(temp == 0) poor++;
        }
        (gold>poor) ? cout << poor << endl : cout << gold<<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 1
0 0 0 0 0
```

**Explanation:**
- This test case is designed to expose the bug where Robin starts with 0 gold and encounters only people with 0 gold.
- The input sets up a scenario where Robin has no gold to give to anyone, even though there are 5 people with 0 gold.
- The correct output should be 0, but the buggy code might incorrectly output 5 if it doesn't properly handle the condition where Robin has no gold to give.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
for _ in range(int(input())):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    print(min(a.count(0), max(a)))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
import sys
input = sys.stdin.read

def robin_hood(test_cases):
    results = []

    for t in test_cases:
        n, k, arr = t
        robin_gold = 0
        count = 0

        # Collect gold from those with gold >= k
        for i in range(n):
            if arr[i] >= k:
                robin_gold += arr[i]

        # Distribute gold to those with 0
        for i in range(n):
            if arr[i] == 0 and robin_gold > 0:
                robin_gold -= 1
                count += 1

        results.append(str(count))

    # Output all results at once
    sys.stdout.write("\n".join(results) + "\n")

def main():
    # Reading input
    data = input().splitlines()

    # Parse test cases
    t = int(data[0])
    test_cases = []
    idx = 1
    for _ in range(t):
        n, k = map(int, data[idx].split())
        arr = list(map(int, data[idx + 1].split()))
        test_cases.append((n, k, arr))
        idx += 2

    # Solve all test cases
    robin_hood(test_cases)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Edge Case: All People Have Gold >= k**: This will test if the code correctly handles the scenario where Robin doesn't have any gold to give away.
2. **Edge Case: All People Have 0 Gold**: This will test if the code correctly handles the scenario where Robin has a lot of gold to give away.
3. **Mixed Case: Some People Have Gold >= k, Some Have 0**: This will test if the code correctly handles the distribution of gold after collecting it.
4. **Edge Case: One Person with 0 Gold and No Gold to Give**: This will test if the code correctly handles a single person scenario.

Given these scenarios, let's construct a small-scale test input:

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

Explanation:
1. **Test Case 1 (3 1 1 1 1)**: All people have gold >= k. Robin will take all the gold and won't give any gold.
2. **Test Case 2 (3 1 0 0 0)**: All people have 0 gold. Robin will give 1 gold to each person.
3. **Test Case 3 (5 3 3 0 0 4 0)**: Mixed case with some people having gold >= k and some having 0 gold. Robin will take gold from the first and fourth persons and give gold to the second, third, and fifth persons.
4. **Test Case 4 (1 1 0)**: Single person with 0 gold and no gold to give. Robin will not give any gold.

This test input should help in identifying any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
test_num = int(input())
robbin_coins = 0
move = 0
for i in range(test_num):
    a, b = map(int, input().split())
    nums = list(map(int, input().split()))
    for x in range(a):
        if nums[x] > 0:
            robbin_coins += nums[x]
        elif nums[x] == 0:
            if robbin_coins > 0:
                robbin_coins -= 1
                nums[x] = 1
                move += 1

        if x+1 == a:
            print(move)
            move = 0
    robbin_coins = 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
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
import java.util.*;

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

    int t = sc.nextInt();

    for (int i = 0; i < t; i++) {
      int n = sc.nextInt();
      int k = sc.nextInt();

      int count = 0;
      int gold = 0;

      for (int j = 0; j < n; j++) {
        int a = sc.nextInt();

        if (a == 0) {
          count++;
        }
        if (a >= k) {
          gold = gold + a;
        }
      }

      System.out.println(count > gold ? gold : count);
    }
  }
}

**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 5
0 0 5 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
	    Scanner sc=new Scanner(System.in);
	    int t=sc.nextInt();
	    while(t-->0){
	        int n=sc.nextInt();
	        int k=sc.nextInt();
	        int[] a=new int[n];
	        int sum=0,zcnt=0;
	        for(int i=0;i<n;i++){
	             a[i]=sc.nextInt();
	             if(a[i]>=k){
	                 sum+=a[i];
	             }
	             if(a[i]==0){
	                 zcnt++;
	             }
	        }
	        if(sum<zcnt){
	            System.out.println(sum);
	        }else{
	            System.out.println(zcnt);
	        }
	    }
	}
}

**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 the following aspects:

1. **Robin's Gold Management**: The code incorrectly handles the situation where Robin runs out of gold before giving it to all the people who have 0 gold.
2. **Edge Cases**: We need to test scenarios where Robin has just enough gold to give to some people but not all, and where Robin runs out of gold before he can give to all the people who have 0 gold.

Let's construct a test case that will expose the bug:

- **Number of Test Cases**: 1 (to keep it simple)
- **Number of People (n)**: 5 (a small number to ensure the test case is manageable)
- **Threshold (k)**: 3 (a moderate threshold to ensure some people have more than k and some have 0)
- **Gold Distribution (a)**: [3, 0, 0, 0, 10] (This will ensure that Robin takes 13 gold but can only give to 3 people before running out of gold)

Here's the test input:

```plaintext
1
5 3
3 0 0 0 10
```

Explanation:
- **Test Case 1**:
  - **n = 5**: 5 people
  - **k = 3**: Threshold for taking gold
  - **a = [3, 0, 0, 0, 10]**: Gold distribution
    - Person 1: 3 gold (Robin takes 3 gold)
    - Person 2: 0 gold (Robin gives 1 gold)
    - Person 3: 0 gold (Robin gives 1 gold)
    - Person 4: 0 gold (Robin gives 1 gold)
    - Person 5: 10 gold (Robin takes 10 gold)

Robin starts with 0 gold, takes 3 gold from the first person, and then gives 1 gold to the next three people. After giving 1 gold to the third person, Robin has 1 gold left. Robin then takes 10 gold from the fifth person, ending with 11 gold. However, the buggy code will incorrectly print the number of people who received gold as 3 instead of 1 (since Robin can only give 1 gold to each of the first three people before running out).

This test input should expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cwchar>
#include <cwctype>
#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
//#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cuchar>
#endif
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <codecvt>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif
#if __cplusplus >= 201402L
#include <shared_mutex>
#endif
#if __cplusplus >= 201703L
#include <any>
#include <charconv>
#include <execution>
#include <filesystem>
#include <optional>
#include <string_view>
#include <variant>
#endif
#if __cplusplus > 201703L
#include <bit>
#include <compare>
#include <concepts>
#include <numbers>
#include <ranges>
#include <span>
#include <version>
#endif
#define ll long long
#define str string
#define np next_permutation
#define ch char
#define mp make_pair
#define pb push_back
#define fs fixed<<setprecision
#define pi pair<int,int>
#define pii pair<int,pi>
#define bistryi ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define F first
#define S second
#define rt return
#define du double
#define sz size()
#define all(x) x.begin(),x.end()
#define int ll
#define yes cout<<"Yes";
#define br break;
#define co continue;
#define push_back pb
#define no cout<<"No";
using namespace std;
const int N=1e6+9;
bool amir(int a, int b){
    return a>b;
}
signed main(){
    bistryi
    int tt;
    cin>>tt;
    while(tt--){
        int a,b,sum=0,sum1=0;
        cin>>a>>b;
        int c[a];
        for(int i=0;i<a;i++){
            cin>>c[i];
        }
        for(int i=0;i<a;i++){
            if(c[i]>=b) {
                sum1+=c[i];
            }
            if(c[i]==0){
                sum1--;
                sum++;
            }
            if(sum1==0){
                br
            }
        }
        cout<<sum<<endl;
    }
    rt 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Let's analyze the code and identify potential issues:

1. **Initial Gold Distribution:**
   - Robin starts with 0 gold.
   - If a person has 0 gold, Robin gives 1 gold to that person if he has any.
   - If a person has gold greater than or equal to the threshold `k`, Robin takes all the gold.

2. **Bug Analysis:**
   - The code uses `sum1` to track the total gold Robin has.
   - The code uses `sum` to track the number of people Robin gives gold to.
   - The code has a condition `if(sum1 == 0) break;` which stops the loop if Robin runs out of gold.

3. **Potential Issues:**
   - The code might not handle the case where Robin gives gold to multiple people in a row without receiving any gold in between.
   - The code might not handle the case where Robin receives a large amount of gold from one person and then gives it away to multiple people.

4. **Challenging Test Input:**
   - We need a test case where Robin gives gold to multiple people in a row without receiving any gold in between.
   - We also need a test case where Robin receives a large amount of gold and then gives it away to multiple people.

Let's construct a test case that covers these scenarios:

```plaintext
2
5 1
0 0 0 100 0
3 2
0 0 0
```

**Explanation:**
1. **First Test Case:**
   - `n = 5`, `k = 1`
   - `a = [0, 0, 0, 100, 0]`
   - Robin starts with 0 gold.
   - Robin gives 1 gold to the first person (sum1 = -1, sum = 1).
   - Robin gives 1 gold to the second person (sum1 = -2, sum = 2).
   - Robin gives 1 gold to the third person (sum1 = -3, sum = 3).
   - Robin takes 100 gold from the fourth person (sum1 = 97).
   - Robin gives 1 gold to the fifth person (sum1 = 96, sum = 4).
   - Expected output: 4

2. **Second Test Case:**
   - `n = 3`, `k = 2`
   - `a = [0, 0, 0]`
   - Robin starts with 0 gold.
   - Robin gives 1 gold to the first person (sum1 = -1, sum = 1).
   - Robin gives 1 gold to the second person (sum1 = -2, sum = 2).
   - Robin gives 1 gold to the third person (sum1 = -3, sum = 3).
   - Expected output: 3

This test input should expose the bug in the provided code by ensuring that Robin gives gold to multiple people in a row without receiving any gold in between.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

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

int main() {

	int t;
	cin >> t;

	while(t--) {
		int n, k;
		cin >> n >> k;

		vector<int> arr(n);
		int gold = 0;
		for(int i = 0; i < n; i++) {
			cin >> arr[i];
			if(arr[i] >= k) {
				gold += arr[i];
				arr[i] = -1;
			}
		}

		int count = 0;
		bool rich = false;
		for(int i = 0; i < n; i++) {
			if(arr[i] == -1) {
				rich = true;
			}

			if(rich == true) {
				if(arr[i] == 0 && gold > 0) {
					gold--;
					count++;
				}
			}
		}

		cout << count << endl;
	}

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
p = int(input())
for i in range(p):
    k, m = map(int, input().split())
    A = []* k
    A = list(map(int, input().split()))
    y = 0
    l = 0
    for i in range(k):
       if A[i]>= 0:
        l = l + A[i]
        if l > 0:
          if A[i] == 0:
            y +=1
            l -= 1
    print(y)

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

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

Think step by step.

Response:
```plaintext
1
5 1
0 0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
t = int(input())
for _ in range(t):
    n,k = list(map(int,input().split(" ")))
    nums = list(map(int,input().split(" ")))
    zero = 0
    gold = 0
    for j in range(n):
        if(nums[j]>0):
            if(nums[j]>=k):
                gold+=nums[j]
        else:
            zero+=1
    print(min(gold,zero))

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

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

Think step by step.

Response:
```plaintext
1
5 1
0 1 0 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
// package Codeforces974;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class ProblemA {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int test=Integer.parseInt(br.readLine());
        StringBuilder print=new StringBuilder();
        while(test--!=0){
            StringTokenizer st=new StringTokenizer(br.readLine());
            int n=Integer.parseInt(st.nextToken());
            int k=Integer.parseInt(st.nextToken());
            int a[]=new int[n];
            st=new StringTokenizer(br.readLine());
            int gold=0,ans=0;
            for(int i=0;i<n;i++){
                a[i]=Integer.parseInt(st.nextToken());
                if(a[i]>0){
                    gold+=a[i];
                }
                else{
                    if(gold>0){
                        ans++;
                        gold--;
                    }
                }
            }
            print.append(ans).append("\n");
        }
        System.out.println(print);
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The bug in the code is that it incorrectly handles the condition where `a[i] >= k` (Robin takes all the gold) and `a[i] == 0` (Robin gives 1 gold if he has any). The code currently only checks if `a[i] > 0` to add gold to Robin's stash, but it doesn't correctly handle the threshold `k`.

### Step-by-Step Analysis:
1. **Threshold Handling**: The code should check if `a[i] >= k` to take the gold, not just if `a[i] > 0`.
2. **Edge Case with Zero Gold**: If `a[i] == 0`, Robin should give 1 gold if he has any, but the code only checks if `gold > 0` and doesn't consider the threshold `k`.

### Test Input to Expose the Bug:
- **Test Case 1**: A person with gold exactly equal to the threshold `k`.
- **Test Case 2**: A person with gold less than the threshold `k` but greater than zero.
- **Test Case 3**: A person with zero gold when Robin has just enough gold to give one.

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

### Explanation:
1. **Test Case 1**:
   - `n = 1`, `k = 1`
   - `a = [1]`
   - Robin should take the gold (1 gold) and not give any gold.
   - Expected output: `0`
   - Actual output (buggy code): `0` (correct, but we need to check other cases)

2. **Test Case 2**:
   - `n = 1`, `k = 2`
   - `a = [1]`
   - Robin should not take the gold (1 < 2) and not give any gold.
   - Expected output: `0`
   - Actual output (buggy code): `0` (correct, but we need to check other cases)

3. **Test Case 3**:
   - `n = 1`, `k = 1`
   - `a = [0]`
   - Robin should give 1 gold if he has any.
   - Expected output: `1`
   - Actual output (buggy code): `1` (correct, but we need to check other cases)

### More Challenging Test Input:
To further expose the bug, let's add a more complex test case:

```plaintext
1
3 2
2 0 1
```

### Explanation:
- **Test Case**:
  - `n = 3`, `k = 2`
  - `a = [2, 0, 1]`
  - Robin should take 2 gold from the first person.
  - Robin should give 1 gold to the second person.
  - Robin should not give any gold to the third person because he only has 1 gold left and the third person has 1 gold, which is less than the threshold `k`.
  - Expected output: `1`
  - Actual output (buggy code): `2` (incorrect, as the buggy code incorrectly handles the threshold)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
import java.io.*;
import java.util.*;
public class Solution1  {

    public static long solve(int n,long[] arr, long th){

        long g=0;
        long z=0;
        for(int i=0;i<n;i++){
            if(arr[i]==0){
                z++;
            }else if(arr[i]>=th){
                g+=arr[i];
            }
        }
        return Math.min(z,g);
    }
    public static void main(String[] args) throws IOException{

        Reader sc=new Reader();
        PrintWriter out = new PrintWriter(System.out);

        int t=sc.nextInt();
        while(t-->0)
        {
            int n =sc.nextInt();
            long k = sc.nextLong();
            long arr[] = new long[n];
            for(int i=0;i<n;i++){
                arr[i] = sc.nextLong();
            }
            out.println(solve(n,arr,k));
        }
        out.flush();
    }

    public static long power(long x,long power)
    {
        if(power==0) return 1;

        long ans1 = power(x,power/2);
        ans1=ans1*ans1;

        if(power%2==0) return ans1;
        return x*ans1;
    }

    //    (a*a_inverse) = 1(mod m)
//    finding a_inverse
    public static long modInverse(long a,long m)
    {
//        works only when m is prime
        long g=gcd(a,m);
        if(g==1) return power(a,m-2,m);
        else return -1;
    }

    //    (x^power)mod(m)
    public static long power(long x,long power,long m)
    {
        if(power==0) return 1;

        long p = power(x,power/2,m)%m;
        p = (p*p)%m;

        if(power%2==0) return p;
        return (x*p)%m;
    }
    public static long gcd(long small,long large)
    {
        if(small==0) return large;
        return gcd(large % small,small);
    }

    public static boolean isprime(long no)
    {
        if(no<=1) return false;
        if(no<=3) return true;
        if(no%2==0 || no%3==0) return false;
        // 6k+1 6k+5 can be prime
        for(long i=5;i*i<=no;i+=6)
        {
            if(no%(i)==0 || no%(i+2)==0)
                return false;
        }
        return true;
    }

    //    prime no smaller than or equal to n
    public static boolean[] prime_nos_till_no(int no)
    {
        boolean prime[]=new boolean[no+1];
//        prime[i]== true means prime
//        prime[i]== false means not a prime
//        initialize prime array as true

        Arrays.fill(prime,true);
        prime[0]=false;
        for(int i=2;i*i<=no;i++)
        {
            if(prime[i]==true)
            {
                for(int j=i*i;j<=no;j+=i)
                {
                    prime[j]=false;
                }
            }
        }
        return prime;

    }
    public static void shufflearray(long arr[])
    {

        int n=arr.length;
        Random rand=new Random();
        for(int i=0;i<n;i++)
        {
            long temp=arr[i];
            int randomPos=i+rand.nextInt(n-i);
            arr[i]=arr[randomPos];
            arr[randomPos]=temp;
        }
    }
    //       Arrays.sort(arr, new Comparator<pair>() {
//        //@Override
//        public int compare(pair o1, pair o2) {
//            long l1=o1.a-o2.a;
//            if(l1<0L) return -1;
//            if(l1==0) return 0;
//            return 1;
//        }
//    });
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String nextLine() throws IOException {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() throws IOException{
            int c = read();
            while (isSpaceChar(c)) c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }

    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

### Analysis of the Buggy Code
1. **Gold Calculation and Zero Count**:
   - The code correctly counts the number of people with zero gold (`z`).
   - The code also correctly calculates the total gold taken by Robin (`g`).

2. **Return Value**:
   - The code returns `Math.min(z, g)`, which is the minimum of the number of people with zero gold and the total gold taken by Robin.

### Potential Bugs
1. **Gold Distribution**:
   - The code does not correctly handle the scenario where Robin has to distribute gold to people with zero gold. It only considers the minimum of `z` and `g`, which is incorrect if Robin has more gold than the number of people with zero gold.

### Challenging Test Input
To expose the bug, we need a test case where the number of people with zero gold is greater than the total gold taken by Robin. This will highlight the issue with the incorrect return value.

### Test Case
Let's create a test case with:
- `n = 5` (number of people)
- `k = 5` (threshold for taking gold)
- `a = [0, 0, 0, 10, 10]` (gold of each person)

In this test case:
- Robin will take `10 + 10 = 20` gold from the 4th and 5th people.
- There are 3 people with zero gold.
- The correct output should be 3 because Robin can give 1 gold to each of the 3 people with zero gold.

### Output
```plaintext
1
5 5
0 0 0 10 10
```

This test case will expose the bug in the buggy code, as the correct output is 3, but the buggy code will return 2 (which is the minimum of 3 and 20).