**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
t=int(input())
for z in range (t) :
    a=input()
    n=int(a[:a.find(' ')])
    k=int(a[a.find(' ')+1:])
    a=list(map(int, input().split(' ')))
    if k>=sum(a) :
        print(len(a))
    else :
        d=len(a)-1
        g=0
        for i in range (k//2) :
            try :
                a[g]=int(a[g])-1
                a[d]=int(a[d])-1
                if a[g]==0 :
                    g+=1
                if a[d]==0 :
                    d-=1
            except :
                break
        if k%2==1 :
            a[d]=int(a[d])-1
        print(a.count(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 expose the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has several issues, including:

1. **Incorrect handling of the attack sequence**: The code assumes that the Kraken always attacks the first and last ships in pairs, which is not correct. The Kraken alternates between the first and last ships, and the sequence can end with an attack on the first or last ship if the number of attacks is odd.
2. **Incorrect termination condition**: The code does not correctly handle the scenario where the Kraken's attacks exceed the total durability of the ships.
3. **Index out of bounds**: The code can raise an index out of bounds error if the Kraken's attacks cause the first or last ship to sink, and the loop continues without adjusting the indices.

To expose these issues, we can create a test input that:

- Has a small number of ships with low durability to quickly reduce the number of ships.
- Has a large number of attacks to ensure the Kraken's attacks exceed the total durability of the ships.
- Ensures the Kraken's attacks are odd to test the handling of the last attack.

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

```plaintext
1
4 7
1 2 1 2
```

Explanation:
- **Number of test cases**: 1
- **Number of ships**: 4
- **Number of attacks**: 7
- **Durability of ships**: [1, 2, 1, 2]

This input will cause the following sequence of attacks:
1. Attack the first ship (1 -> 0), remaining: [2, 1, 2]
2. Attack the last ship (2 -> 1), remaining: [2, 1, 1]
3. Attack the first ship (2 -> 1), remaining: [1, 1, 1]
4. Attack the last ship (1 -> 0), remaining: [1, 1]
5. Attack the first ship (1 -> 0), remaining: [1]
6. Attack the last ship (1 -> 0), remaining: []
7. Attack the last ship (no ships left, no action)

The expected output is 4, as all ships will sink. This input should expose the issues in the code's handling of the attack sequence and the termination condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
import bisect
import heapq
import sys
import math

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

############ ---- 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()))

def solve():
    n, k = invr()
    a = deque(inlt())
    turn = 0
    while a and k > 0:
        if len(a) == 1:
            a[0] -= k
            if a[0] <= 0:
                a.pop()
            break
        else:
            minimum = min(a[0], a[-1])
            if turn % 2 == 0:
                if a[0] <= a[-1] and a[0] * 2 - 1 > k:

                    k = 0
                elif a[-1] < a[0] and a[-1] * 2 > k:
                    k = 0
                elif a[0] <= a[-1] and a[0] * 2 - 1 == k:
                    k = 0
                    a[-1] -= a[0] - 1
                    a[0] = 0
                elif a[-1] <= a[0] and a[-1] * 2 == k:
                    k = 0
                    a[0] -= a[-1]
                    a[-1] = 0
                else:
                    if a[0] <= a[-1]:
                        turn = 0
                    else:
                        turn = 1
                    a[0] -= minimum
                    a[-1] -= minimum
                    k -= minimum * 2
            else:
                if a[-1] <= a[0] and a[-1] * 2 - 1 > k:
                    k = 0
                elif a[0] < a[-1] and a[0] * 2 > k:
                    k = 0
                elif a[-1] <= a[0] and a[-1] * 2 - 1 == k:
                    k = 0
                    a[0] -= a[-1] - 1
                    a[-1] = 0
                elif a[0] <= a[-1] and a[0] * 2 == k:
                    k = 0
                    a[-1] -= a[0]
                    a[0] = 0
                else:
                    if a[-1] <= a[0]:
                        turn = 1
                    else:
                        turn = 0
                    a[0] -= minimum
                    a[-1] -= minimum
                    k -= minimum * 2
            if a[0] <= 0:
                a.popleft()
            if a[-1] <= 0:
                a.pop()
    print(n - len(a))

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 expose the bug in the given code, we need to focus on edge cases where the logic of the code might fail. Here are some potential edge cases to consider:

1. **All ships have the same durability**: This can test the code's ability to handle uniform durability and the alternating attack pattern.
2. **High number of attacks**: This can test the code's ability to handle a large number of attacks efficiently.
3. **Mixed durability values**: This can test the code's ability to handle a mix of high and low durability values.
4. **Single ship**: This can test the code's ability to handle a single ship scenario.
5. **Edge case with very low durability**: This can test the code's ability to handle ships with very low durability values.

Given the constraints and the need for a small-scale test input, let's create a test input that combines some of these edge cases.

### Test Input
```plaintext
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
```

### Explanation
1. **Test Case 1**: `4 5 1 2 4 3`
   - This is a straightforward case where the Kraken attacks the ships in the order described in the problem statement.
   - Expected output: `2` (ships 1 and 4 sink).

2. **Test Case 2**: `4 6 1 2 4 3`
   - This is a similar case but with one more attack.
   - Expected output: `3` (ships 1, 4, and 2 sink).

3. **Test Case 3**: `5 20 2 7 1 8 2`
   - This case involves a large number of attacks and a mix of durability values.
   - Expected output: `5` (all ships sink).

4. **Test Case 4**: `2 2 3 2`
   - This case involves two ships with different durability values.
   - Expected output: `0` (no ships sink).

5. **Test Case 5**: `2 15 1 5`
   - This case involves a high number of attacks and low durability values.
   - Expected output: `2` (both ships sink).

6. **Test Case 6**: `2 7 5 2`
   - This case involves a mix of high and low durability values.
   - Expected output: `2` (both ships sink).

This test input should help expose any issues in the logic of the given code, especially in handling edge cases and large numbers of attacks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
// Online Java Compiler
// Use this editor to write, compile and run your Java code online
import java.util.*;

public class HelloWorld {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){

         int n=sc.nextInt();
         long k=sc.nextLong();
         long a[]=new long[n];
         for(int i=0;i<n;i++){
             a[i]=sc.nextInt();
         }
         int f=0;
         int ans=0;
         int l=n-1;
         while(f<l){
             long min=Math.min(a[l],a[f]);
             if(min*2<=k){
                 k-=2*min;
                 a[l]-=min;
                 a[f]-=min;
                 if(a[l]==0){l--;ans++;}

                 if(a[f]==0){f++;ans++;}
             }
             else break;
         }
         if(a[f]>0&&a[f]<=k)ans++;

          System.out.println(ans);

        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.stream.IntStream;

public class Main {

    void solve() {
        int T = in.nextInt();
        while (T > 0) {
            T--;
            int n = in.nextInt(), k = in.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
            }
            int l = 0, r = n - 1, ans = 0, count = -1;
            if(T == 9918){
                out.println(Arrays.toString(arr));
                out.println("n === >"+n);
                out.println("k === >"+k);
            }
            while (l <= r && k > 0) {
                int min = Math.min(arr[l], arr[r]);
                count++;
                if (count % 2 == 0) {
                    if (l == r ) {
                        if(k<min){
                            break;
                        }
                        arr[l] -= min;
                        k -=  min;
                        if (arr[l] == 0) {
                            l++;
                        }
                        break;
                    }
                    if (k >= min * 2) {
                        arr[l] -= min;
                        arr[r] -= min;
                        k -= 2 * min;
                        if (arr[l] == 0) {
                            l++;
                        }
                        if (arr[r] == 0) {
                            r--;
                        }
                    } else {
                        arr[l] -= Math.ceil((double) k / 2);
                        arr[r] -= Math.floor((double) k / 2);
                        if (arr[l] == 0) {
                            l++;
                        }
                        if (arr[r] == 0) {
                            r--;
                        }
                        k = 0;
                    }
                } else {
                    if (l == r ) {
                        if(k<min){
                            break;
                        }
                        arr[l] -= min;
                        k -=  min;
                        if (arr[l] == 0) {
                            l++;
                        }
                        break;
                    }
                    if (k >= min * 2) {
                        arr[l] -= min;
                        arr[r] -= min;
                        k -= 2 * min;
                        if (arr[l] == 0) {
                            l++;
                        }
                        if (arr[r] == 0) {
                            r--;
                        }
                    } else {
                        arr[r] -= Math.ceil((double) k / 2);
                        arr[l] -= Math.floor((double) k / 2);
                        if (arr[l] == 0) {
                            l++;
                        }
                        if (arr[r] == 0) {
                            r--;
                        }
                        k = 0;
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                if (arr[i] == 0) {
                    ans++;
                }
            }
            out.println(ans);
        }
    }

    FastScanner in;

    PrintWriter out;

    void run() {
        in = new FastScanner();
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    class FastScanner {
        BufferedReader br;

        StringTokenizer st;

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

        public FastScanner(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

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

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

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

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

    public static void main(String[] args) {
        new Main().run();
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the key points to consider:

1. **Handling of Remaining Attacks**: The code should correctly handle the remaining attacks when the number of attacks is not sufficient to sink all ships.
2. **Symmetry in Attacks**: The code should handle the alternating pattern of attacks (first and last) correctly, especially when the number of attacks is odd or even.
3. **Edge Cases with Small Durability**: The code should handle cases where the durability of ships is very small, leading to quick sinking.
4. **Large Number of Attacks**: The code should handle a large number of attacks efficiently without running into performance issues.

Given these points, let's create a test input that can expose potential bugs:

1. **Small Number of Ships with Small Durability**:
   - This will test the code's ability to handle quick sinking and alternating attacks.
2. **Large Number of Attacks**:
   - This will test the code's efficiency and correctness in handling a large number of attacks.

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

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

### Explanation:
1. **First Test Case**:
   - **n = 4, k = 5, a = [1, 2, 4, 3]**
   - This is a typical case where the ships have small to medium durability and a moderate number of attacks. The expected output is `2` ships sunk.

2. **Second Test Case**:
   - **n = 5, k = 10, a = [2, 2, 2, 2, 2]**
   - This case has all ships with the same durability and a large number of attacks. The expected output is `5` ships sunk.

3. **Third Test Case**:
   - **n = 2, k = 100, a = [1, 1]**
   - This case has a very small number of ships with very small durability and a very large number of attacks. The expected output is `2` ships sunk.

This test input should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef     long long       ll;
void solve()
{
    ll n, k, i;
    cin >> n >> k;
    vector<ll> a(n), f(n + 1), b(n + 1);
    f[0] = 0;
    b[n] = 0;
    for(i = 0; i < n; i++) cin >> a[i];
    for(i = 1; i <= n; i++) f[i] = f[i-1] + a[i-1];
    for(i = n - 1; i >= 0; i--) b[i] = b[i+1] + a[i];
    ll x = k/2 + (k&1), y = k/2;
    ll findx = -1, bindx = -1;
    for(i = 0; i <= n; i++)
    {
        if(x >= f[i]) findx = i;
        else break;
    }
    for(i = n; i >= 0; i--)
    {
        if(y >= b[i]) bindx = i;
        else break;
    }
    // cout << findx << " " << n - bindx << endl;
    // cout << x << " " << y ;

    if(findx + 1 == (n - bindx))
    {
        if(f[findx + 1] - x + b[bindx] - y <= 0) {
            cout << n << '\n';
            return;
        }
        else cout << n - 1 << '\n';
    }

    ll ans = (findx + (n - bindx));
    if(ans > n) ans = n;
    cout << ans << endl;
}
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while(t--) solve();
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose a potential bug in the code. Here's why it is challenging:

1. **Initial Durability**: The ships have initial durabilities `[1, 2, 1]`.
2. **Number of Attacks**: The Kraken attacks 6 times.

### Step-by-Step Simulation:
1. **First Attack (First Ship)**: Durabilities become `[0, 2, 1]` (Ship 1 sinks).
2. **Second Attack (Last Ship)**: Durabilities become `[2, 0]` (Ship 3 sinks).
3. **Third Attack (First Ship)**: Durabilities become `[1, 0]` (Ship 2 sinks).
4. **Fourth Attack (Last Ship)**: Durabilities become `[1]`.
5. **Fifth Attack (First Ship)**: Durabilities become `[0]` (Ship 2 sinks again, but it was already sunk).
6. **Sixth Attack (First Ship)**: Durabilities remain `[0]` (no effect).

### Expected Output:
All 3 ships are sunk. The correct output should be `3`.

### Why This Test Case is Challenging:
- **Multiple Ships Sinking Early**: The first and last ships sink early, which can test how the code handles the removal of ships from the list.
- **Remaining Ships**: The middle ship is attacked multiple times, ensuring that the code correctly handles the alternating pattern of attacks.
- **Edge Case**: The total number of attacks is such that it should sink all ships, testing the boundary conditions of the algorithm.

This test case will help identify if the code correctly handles the removal of ships and the alternating attack pattern.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

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

using namespace std;

#define ll  long long
#define db  long double
#define sz(x) (int)x.size()

void _3zzat() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

using namespace __gnu_pbds;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template<typename T>
istream &operator>>(istream &input, vector<T> &data) {
    for (T &x: data)
        input >> x;
    return input;
}

template<typename T>
ostream &operator<<(ostream &output, const vector<T> &data) {
    for (const T &x: data)
        output << x << " ";
    return output;
}

int dx8[] = {1, -1, 0, 0, 1, -1, -1, 1};
int dy8[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
string dir = "RLDU";
const ll N = 1e6 + 2, M = 900 + 2, MOD = 1e9 + 7, inf = 3e18, inf2 = 1e9;

void code() {
    int n;
    ll k;
    cin >> n >> k;
    deque<ll> v(n);
    for (auto &i: v) cin >> i;
    ll cnt = 0;
    while (sz(v) > 1) {
        ll l = v.front(), r = v.back();
        if (k < min(l, r) * 2)break;
        v.pop_front();
        v.pop_back();
        if (l <= r) {
            k -= l * 2;
            if (r != l) {
                v.push_back(r - l);
                cnt++;
                continue;
            }
            cnt += 2;
        } else {
            k -= r * 2;
            if (r != l) {
                v.push_front(l - r);
                cnt++;
                continue;
            }
            cnt += 2;
        }
    }
    if ((v.front() <= k && sz(v) == 1) or (k >= min(v.front(), v.back()) * 2 - 1))cnt++;
    cout << cnt << '\n';
}

signed main() {
    _3zzat();

    int tes = 1;
    cin >> tes;
    for (int i = 1; i <= tes; ++i) {
        code();
    }

}

**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 edge cases and scenarios that can cause the algorithm to fail. Here are some key points to consider:

1. **Small Number of Ships**: Test with a small number of ships to ensure the algorithm handles basic cases correctly.
2. **Large Number of Attacks**: Test with a large number of attacks to see how the algorithm handles large values.
3. **Uneven Durability**: Test with ships having very different durabilities to see how the algorithm handles uneven distributions.
4. **Single Ship**: Test with a single ship to see how the algorithm handles edge cases.
5. **All Ships Sunk**: Test with a scenario where all ships are sunk to see if the algorithm correctly counts the number of sunk ships.

Let's create a test input that combines these scenarios:

```plaintext
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
```

This input covers:
1. A case where not all ships are sunk.
2. A case where more ships are sunk.
3. A case with a large number of attacks.
4. A case with a single ship.
5. A case where all ships are sunk.
6. A case with a small number of attacks.

Each of these scenarios will help identify potential 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: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
veces = int(input())

for _ in range(veces):

    cant, atac = map(int, input().split())
    lista = (list(map(int, input().split())))

    nu = atac // 2
    np = atac // 2

    if atac % 2 != 0:
        np += 1

    p = 0
    u = 0

    if sum(lista) > atac:
        for i in range(cant):
            u += lista[i]

            if nu - u <= 0:
                if nu - u < 0:
                    lista[i] = abs(nu - u)
                else:
                    i += 1
                break

        lista = lista[i:][::-1]

        for i in range(len(lista)):
            p += lista[i]

            if np - p <= 0:
                if np - p < 0:
                    lista[i] = abs(np - p)
                else:
                    i += 1
                break

        print(cant - len(lista[i:]))

    else:
        print(cant)

**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 buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. Here are the key points to consider:

1. **Uneven Distribution of Durability**: The code might handle attacks differently if the durability of ships is unevenly distributed.
2. **Odd and Even Attacks**: The code splits the attacks into two parts, so we need to test with an odd and even number of attacks to see if it handles both cases correctly.
3. **All Ships Sunk**: Test cases where all ships are sunk to see if the code correctly counts the number of sunk ships.
4. **No Ships Sunk**: Test cases where no ships are sunk to ensure the code handles this scenario correctly.
5. **Single Ship**: Test cases with a single ship to see if the code handles this edge case correctly.

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

```plaintext
4
4 7
1 2 4 3
3 6
1 2 1
2 1
1 1
1 10
1
```

### Explanation:
1. **Test Case 1**: `4 7 1 2 4 3`
   - This is a modified version of the example provided, with an odd number of attacks. The expected output is `3` because the Kraken will sink the first, last, and first ships again.
2. **Test Case 2**: `3 6 1 2 1`
   - This has an even number of attacks and ships with low durability. The expected output is `3` because all ships will be sunk.
3. **Test Case 3**: `2 1 1 1`
   - This has a single attack on two ships with low durability. The expected output is `1` because only the first ship will be sunk.
4. **Test Case 4**: `1 10 1`
   - This has a single ship and a large number of attacks. The expected output is `1` because the single ship will be sunk.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
t =int(input())

while t:
    t -= 1
    n,k = [int(a) for a in input().split()]
    l = [int(a) for a in input().split()]
    td = int(n*(n+1)/2)
    shipsink = 0

    # if k < td:
    #     while True:
    #         if k - l[0] -l[n-1] < 0:
    #             break
    #         else:
    #             k -= (l[0]+l[n-1])
    #             l.pop()
    #             l.pop(0)
    #             shipsink += 2
    #             n -= 2
    if k%2 == 0:
        right = int(k/2)
        left = right
    else:
        right = int((k+1)/2)
        left = k-right

    r=0

    if k < td:
        while right-l[r] >= 0:
            right -= l[r]
            l[r] = 0
            r += 1
            if r == n:
                break

        le=-1
        while left-l[le] >= 0:
            left -= l[le]
            l[le] = 0
            le -= 1
            if le < -n:
                break
        # print(l)
        for a in range(n):
            if l[a] == 0:
                shipsink +=1

    else:
        shipsink = n

    print(shipsink)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

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

	public static void main(String[] args) {
		// TODO Auto-generated method stub
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0)
        {
        	int n=sc.nextInt();
        	long k=sc.nextLong();
        	long a[]=new long[n];
        	long sum1=0;
        	for(int i=0;i<n;i++)
        	{
        		a[i]=sc.nextLong();
        		sum1+=a[i];
        	}
        	if(sum1<=k)
        	{
        	  System.out.println(n);
        	  continue;
        	}
        	int left=0;int right=n-1;
        	int ans=0;

        	while(left<=right && k>0)
        	{
        		long minus=(long)Math.min(a[left], a[right]);
        		if(2*minus<=k)
        		{
        			k=k-2*minus;
        			if(minus<=a[left])
        			{
        				if(minus<a[left])
        				{
        					a[left]-=minus;
        				}
        				else
        				{
        					a[left]-=minus;
        				left++;
        				ans++;
        				}
        			}
        			if(minus<=a[right])
        			{
        				if(minus<a[right])
        				{
        					a[right]-=minus;
        				}
        				else
        				{
        			   right--;
        			   ans++;
        				}
        			}

        		}
        		else
        		{
        			if(k>=a[0])
        			{
        				ans++;
        				k-=a[0];
        				left=n-1;
        				right=0;
        			}
        			else
        			{
        				left=n-1;
        				right=0;
        			}

        		}
        		//System.out.println("hello");

        	}
        	System.out.println(ans);

//        	long b[]=new long[n];
//        	int temp=0;
////        	for(int i=0;i<n;i++)
////        		System.out.print(a[i]+" ");
////        	System.out.println();
//        	if(n%2==0)
//        	{
//        		 int mid=(n+1)/2 -1;
//        	for(int i=0;i<=mid;i++)
//        	{
//        		b[temp]=a[i];
//        		temp+=2;
//        	}
//        	int temp1=1;
//        	for(int i=n-1;i>mid;i--)
//        	{
//        		b[temp1]=a[i];
//        		temp1+=2;
//        	}
//        	}
//        	else
//        	{
//        		int mid=n/2;
//        		int tp=0;
//        		for(int i=0;i<=mid;i++)
//        		{
//        			b[tp]=a[i];
//        			tp+=2;
//        		}
//        		tp=1;
//        		for(int i=mid+1;i<n;i++)
//        		{
//        			//System.out.println("-->"+a[i]);
//        			b[tp]=a[i];
//        			tp+=2;
//        		}
//        	}
//        	long ans=0;
//        	long sum=0;
//        	for(int i=0;i<n;i++)
//        	{
//        	   	sum+=b[i];
//        	   	if(sum<=k)
//        	   	{
//        	   		ans++;
//        	   	}
//        	}
//        	System.out.println(ans);
////        	for(int i=0;i<n;i++)
////        		System.out.print(b[i]+" ");
////        	System.out.println();
//

        }
	}

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class C_Inhabitant_of_the_Deep_Sea{
  public static void main(String args[]){
    Scanner sc = new Scanner(System.in);
    int tt = sc.nextInt();
    for(int ii = 0;ii<tt;ii++){
      solution(sc);
    }
  }

  private static void solution(Scanner sc){
    int n = ip(sc);
    long k = ip(sc);
   int arr[] = ipa(n,sc);
   int l = 0;
   int r = n-1;
   int ans = 0;
   while(k>0 && l<=r){
    if(l == r){
      if(k>=arr[l]){
        ans++;
      p(ans);
      return;
      }
      else{
        p(ans);
        return;
      }
    }
    if(arr[l] == arr[r]){
      k -= 2*arr[l];
      if(k>=0){
        ans+=2;
        l++;
        r--;
      }
      else{
        k += 2*arr[l];
        long b = (k/2) + (k%2);
        long c = k/2;

        arr[l] -= b;
        arr[r] -= c;
        if(arr[l] <= 0) {
          ans++;
          k-= b;
          l++;
      }
      else if(arr[r]<=0){
        ans++;
        k-=c;
        r--;
      }
      }
    }
    else{
      int x = min(arr[l],arr[r]);
      k -= 2*x;
      if(k>=0){
        ans++;
        if(arr[l] == x){
          arr[r] -= x;
          l++;
        }
        else{
          arr[l] -= x;
          r--;
        }
      }
      else{
        long b = (k/2) + (k%2);
        long c = k/2;

        arr[l] -= b;
        arr[r] -= c;
        if(arr[l] <= 0) {
          ans++;
          k-= b;
          l++;
      }
      else if(arr[r]<=0){
        ans++;
        k-=c;
        r--;
      }
    }
   }
  }
   p(ans);
  }

  private static void ps(String s){System.out.println(s);}
  private static void p(int n){ System.out.println(n);}
  private static void pl(long n){ System.out.println(n);}
  private static long ipl(Scanner sc){ long a = sc.nextLong(); return a;}
  private static int ip(Scanner sc){ int a = sc.nextInt(); return a;}

  private static long[] ipal(int n,Scanner sc){
    long arr[] = new long[n];
    for(int i = 0;i<n;i++) arr[i] = sc.nextLong();
    return arr;
  }

  private static int[] ipa(int n,Scanner sc){
    int arr[] = new int[n];
    for(int i = 0;i<n;i++) arr[i] = sc.nextInt();
    return arr;
  }

  private static String ips(Scanner sc){
    String s = sc.next();
    return s;
  }

  private static boolean prime(long n){
    if(n<2) return false;
    if(n==2 || n==3) return true;
    if(n%2 == 0 || n%3 == 0) return false;
    long sqrtN = (long)Math.sqrt(n)+1;
    for(long i = 6L;i<=sqrtN;i+=6){
      if(n%(i-1) == 0 || n%(i+1) == 0) return false;
    }
    return true;
  }

  private static void sort(int arr[]){
    ArrayList<Integer> list = new ArrayList<>();
    for(int i = 0;i<arr.length;i++){
      list.add(arr[i]);
    }
    Collections.sort(list);
    for(int i = 0;i<arr.length;i++){
      arr[i] = list.get(i);
    }
  }

  private static void sortl(long arr[]){
    ArrayList<Long> list = new ArrayList<>();
    for(int i = 0;i<arr.length;i++){
      list.add(arr[i]);
    }
    Collections.sort(list);
    for(int i = 0;i<arr.length;i++){
      arr[i] = list.get(i);
    }
  }

  private static boolean issorted(int arr[]){
    int n = arr.length;
    for(int i = 0;i<n-1;i++){
      if(arr[i]>arr[i+1]) return false;
    }
    return true;
  }

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

  private static long gcd(long a,long b){
	      if(a > b) a = (a+b)-(b=a);
	      if(a == 0L) return b;
	      return gcd(b%a,a);
	    }

	private static int countdigit(long l) {
	        if (l >= 1000000000000000000L) return 19;
	        if (l >= 100000000000000000L) return 18;
	        if (l >= 10000000000000000L) return 17;
	        if (l >= 1000000000000000L) return 16;
	        if (l >= 100000000000000L) return 15;
	        if (l >= 10000000000000L) return 14;
	        if (l >= 1000000000000L) return 13;
	        if (l >= 100000000000L) return 12;
	        if (l >= 10000000000L) return 11;
	        if (l >= 1000000000L) return 10;
	        if (l >= 100000000L) return 9;
	        if (l >= 10000000L) return 8;
	        if (l >= 1000000L) return 7;
	        if (l >= 100000L) return 6;
	        if (l >= 10000L) return 5;
	        if (l >= 1000L) return 4;
	        if (l >= 100L) return 3;
	        if (l >= 10L) return 2;
	        return 1;
	  }

    private static int kadane(int[] arr) {
      int n = arr.length;
      int maxSum = Integer.MIN_VALUE;
      int currSum = 0;
      for (int i = 0; i <= n - 1; i++) {
          currSum += arr[i];
          if (currSum > maxSum) {
          maxSum = currSum;
          }
          if (currSum < 0) {
          currSum = 0;
          }
      }
      return maxSum;
  }

    private static long sum(int arr[]){
      long sum = 0;
      int n = arr.length;
      for(int i = 0;i<n;i++){
          sum += arr[i];
      }
      return sum;
    }

private static int c(int arr[], int k){
  int count = 0;
  for(int i = 0;i<arr.length;i++){
    if(arr[i] == k) count++;
  }
  return count;

}

private static int maxarr(int arr[]){
  int n = arr.length;
  int max = Integer.MIN_VALUE;
  for(int i = 0;i<n;i++){
    max = max(max,arr[i]);
  }
  return max;
}

private static int minarr(int arr[]){
  int n = arr.length;
  int min = Integer.MAX_VALUE;
  for(int i = 0;i<n;i++){
    min = min(min,arr[i]);
  }
  return min;
}
}

**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 might cause the algorithm to fail. Here are some potential edge cases to consider:

1. **Single Ship Case**: When there is only one ship.
2. **Equal Durability for All Ships**: When all ships have the same durability.
3. **High Durability with Many Attacks**: When the ships have high durability and the number of attacks is very large.
4. **Low Durability with Few Attacks**: When the ships have low durability and the number of attacks is small.
5. **Mixed Durability and Attacks**: When the ships have mixed durability values and the number of attacks is moderate.

Given the constraints and the need for a small-scale test input, let's focus on a scenario that is likely to cause issues in the buggy code. We will use a case where the ships have mixed durability and the number of attacks is moderate.

### Test Input
```plaintext
4
3 7
1 2 3
4 10
1 1 1 1
5 15
1 2 3 4 5
2 10
1 1
```

### Explanation
1. **3 7 1 2 3**: This case has three ships with mixed durability. The Kraken attacks 7 times. The expected output is 3, as all ships will sink.
2. **4 10 1 1 1 1**: This case has four ships with low durability. The Kraken attacks 10 times. The expected output is 4, as all ships will sink.
3. **5 15 1 2 3 4 5**: This case has five ships with mixed durability. The Kraken attacks 15 times. The expected output is 5, as all ships will sink.
4. **2 10 1 1**: This case has two ships with low durability. The Kraken attacks 10 times. The expected output is 2, as both ships will sink.

These test cases are designed to cover a range of scenarios and should help expose 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: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
int32_t main(){
    int t;
    cin>>t;
    while(t--){
       int n,k;
       cin>>n>>k;
       vector<int> v(n);
       deque<int> g;
       for(int i=0;i<n;i++){
           cin>>v[i];
           g.push_back(v[i]);
       }
       int ct=0;bool flag=true;
       while(!g.empty() && k>0){
           if(flag==true){
               k--;
               int h=g.front();
               g.pop_front();
               h--;
               if(h==0)ct++;
               else g.push_front(h);
               flag=false;

           }
           else{
               k--;
               int h=g.back();
               g.pop_back();
               h--;
               if(h==0)ct++;
               else g.push_back(h);
               flag=true;
           }
       }
       cout<<ct<<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:
To expose the bug in the provided code, we need to focus on edge cases where the behavior of the Kraken's attacks and the durability of the ships might lead to unexpected results. Here are some key points to consider:

1. **Edge Case 1: All Ships with Durability 1**:
   - If all ships have a durability of 1, the Kraken will sink a ship with each attack. This can help us verify if the code correctly handles multiple ships being sunk in quick succession.

2. **Edge Case 2: Large Number of Attacks**:
   - If the number of attacks is very large, we need to ensure that the code can handle the case where the Kraken attacks more times than there are ships, and it correctly stops when all ships are sunk.

3. **Edge Case 3: Ships with Varying Durability**:
   - If the ships have varying durability, we need to ensure that the code correctly alternates between the first and last ships and handles the sinking of ships correctly.

4. **Edge Case 4: Single Ship**:
   - If there is only one ship, the Kraken will only attack that ship. This can help us verify if the code handles a single ship correctly.

5. **Edge Case 5: No Ships Sunk**:
   - If the number of attacks is less than the total durability of all ships, no ships should be sunk. This can help us verify if the code correctly handles this scenario.

Based on these considerations, let's create a small-scale test input that can expose the bug in the code:

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

Explanation of the test cases:
1. **Test Case 1**:
   - 3 ships with durabilities [1, 1, 1] and 6 attacks.
   - Expected output: 3 (all ships are sunk).

2. **Test Case 2**:
   - 5 ships with durabilities [1, 2, 3, 4, 5] and 10 attacks.
   - Expected output: 5 (all ships are sunk).

3. **Test Case 3**:
   - 2 ships with durabilities [10, 10] and 1 attack.
   - Expected output: 0 (no ships are sunk).

4. **Test Case 4**:
   - 2 ships with durabilities [1, 1] and 1 attack.
   - Expected output: 1 (one ship is sunk).

These test cases cover a range of scenarios that can help identify issues in the code's logic and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

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

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

    int t;
    cin >> t;

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

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

        deque<int> d;
        for (int i = 0; i < n; i++) {
            d.pb(i);
        }

        int sunk = 0;
        for (int i = 0; i < k; i++) {
            int idx = (i % 2 == 0) ? d.front() : d.back();
            a[idx]--;
            if (a[idx] == 0) {
                sunk++;
                if (i % 2 == 0) {
                    d.pop_front();
                    if (d.empty()) break;
                } else {
                    d.pop_back();
                    if (d.empty()) break;
                }
            }
        }

        cout << sunk << 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
3 6
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
num_test = [int(x) for x in input().split(' ')]

for _ in range(num_test[0]):
  n,k = [int(x) for x in input().split(' ')]
  a = [int(x) for x in input().split(' ')]

  idx = 0
  done = False
  while k > 0 and not done:
    a[idx] -=1
    if a[idx] <= 0:
        del a[idx]
        if len(a) == 0:
          print(n)
          done = True
    if idx == 0:
      idx = -1
    else:
      idx = 0
    k-=1

  if not done:
    print(int(n-len(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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: The code might fail when the number of attacks `k` is very large compared to the number of ships `n` and their durability.
2. **Boundary Conditions**: The code might fail when the number of ships is very small, or when the durabilities are very low or very high.
3. **Sequence of Attacks**: The code might fail when the sequence of attacks causes the ships to sink in a specific order that the code does not handle correctly.

Let's construct a test case that maximizes the failure probability:

- **Small Number of Ships**: This will make the sequence of attacks more predictable and easier to debug.
- **High Durability for Some Ships**: This will ensure that the Kraken has to attack multiple times before a ship sinks, which can expose issues in the attack sequence.
- **Low Durability for Some Ships**: This will ensure that ships sink quickly, which can expose issues in the removal of ships from the list.

Here is a small-scale test input that should be challenging for the buggy code:

```plaintext
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
```

This test input includes:
1. A case where the number of attacks is just enough to sink some ships.
2. A case where the number of attacks is more than enough to sink all ships.
3. A case with a large number of attacks and a mix of high and low durability ships.
4. A case with a small number of ships and attacks.
5. A case with a small number of ships and a large number of attacks.
6. A case with a small number of ships and a mix of high and low durability.

This should help in exposing 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: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
def krakenAttack(n, k, arr):
    attackIndex = 0
    count = 0

    for attack in range(k):

        if attack%2 == 0:
            attackIndex = 0
        else:
            attackIndex = -1
        #check if empty array
        if not arr:
            break

        if len(arr) == 1:
            arr[0] -= 1
            if arr[0] == 0:
                count += 1
                arr.pop(0)
                break
        else:
            arr[attackIndex] -= 1

            if arr[attackIndex] == 0:
                arr.pop(attackIndex)
                count += 1

    return count

def main():
    tc = int(input())
    results = []

    for _ in range(tc):
        n, k = map(int, input().split())
        arr = list(map(int, input().split()))

        results.append(krakenAttack(n, k, arr))

    for result in results:
        print(result)

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, we need to consider edge cases and scenarios where the code might fail. The buggy code has a few issues:

1. **Handling of `attackIndex`**: The code alternates between the first and last ship, but it doesn't correctly handle the case where the first or last ship is removed, which can lead to incorrect indices.
2. **Array length changes**: The array length changes as ships are sunk, but the code doesn't correctly adjust the `attackIndex` based on the new array length.
3. **Breaking out of the loop**: The code breaks out of the loop if the array becomes empty, but it doesn't handle the case where the Kraken still has attacks left.

Given these issues, a challenging test input would be one where the Kraken attacks a large number of times, and the ships' durability is such that the array length changes frequently. This will test how the code handles dynamic changes in the array length and the correct alternation of attacks.

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

```plaintext
1
4 7
1 2 1 2
```

Explanation:
- **Number of test cases**: 1
- **Number of ships (n)**: 4
- **Number of attacks (k)**: 7
- **Durability of ships (a)**: [1, 2, 1, 2]

This input is designed to:
- Cause the array to change length frequently as ships are sunk.
- Test the alternation of attacks between the first and last ship.
- Ensure that the Kraken has more attacks than the number of ships, which will test how the code handles the remaining attacks after some ships are sunk.

Let's break down the expected behavior:
1. First attack: Ship 1 (durability 1) -> Sinks, array becomes [2, 1, 2]
2. Second attack: Ship 3 (durability 2) -> Durability becomes 1, array becomes [2, 1, 1]
3. Third attack: Ship 1 (durability 2) -> Durability becomes 1, array becomes [1, 1, 1]
4. Fourth attack: Ship 3 (durability 1) -> Sinks, array becomes [1, 1]
5. Fifth attack: Ship 1 (durability 1) -> Sinks, array becomes [1]
6. Sixth attack: Ship 1 (durability 1) -> Sinks, array becomes []
7. Seventh attack: No ships left, Kraken stops.

Expected output: 4 (since all 4 ships sink).

This test input should expose the bugs in the code, especially the handling of dynamic array length and correct alternation of attacks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
import java.util.*;

public class Code{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int uiop = sc.nextInt();
        for(int ghop=0;ghop<uiop;ghop++){
           int n = sc.nextInt();
           int k = sc.nextInt();
           int[] arr = new int[n];
           for(int i=0;i<n;i++){
             arr[i] = sc.nextInt();
           }
           int  i = 0, j = n-1;
           long curr = 0;
           long x = k%2 == 0 ? k/2 : k/2+1;
           while(i<n && x>0){
            if(arr[i]<=x){
              x -= arr[i];
            }else{
              arr[i] -= x;
              x = 0;
              break;
            }
            i++;
           }
           x = k/2;
           while(j>=i && x>0){
            if(arr[j]<=x){
              x -= arr[j];
            }else{
              arr[j] -= x;
              x = 0;
              break;
            }
            j--;
           }
           System.out.println((i+(n-j-1)));

        }
    }
}

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

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

Think step by step.

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

1. **Small Number of Ships with High Durability**: This can help in checking the logic of the while loops and the handling of the remaining attacks.
2. **Large Number of Attacks**: This can help in checking the logic of the modulo operation and the distribution of attacks.
3. **Edge Case with Odd and Even Number of Attacks**: This can help in checking the logic of the alternating attacks.
4. **Zero Durability Ships**: This can help in checking the logic of handling ships that sink immediately.

Given these considerations, let's create a small-scale test input that can expose potential bugs in the code:

```plaintext
3
3 7
1 2 3
4 10
1 2 2 1
2 5
1 1
```

### Explanation:
1. **Test Case 1:**
   - **Ships:** 3
   - **Attacks:** 7
   - **Durabilities:** [1, 2, 3]
   - **Expected Output:** 3
   - **Reason:** The Kraken will attack the ships in the following order: 1, 3, 1, 3, 1, 3, 1. All ships will sink.

2. **Test Case 2:**
   - **Ships:** 4
   - **Attacks:** 10
   - **Durabilities:** [1, 2, 2, 1]
   - **Expected Output:** 4
   - **Reason:** The Kraken will attack the ships in the following order: 1, 4, 1, 4, 1, 4, 1, 3, 1, 3. All ships will sink.

3. **Test Case 3:**
   - **Ships:** 2
   - **Attacks:** 5
   - **Durabilities:** [1, 1]
   - **Expected Output:** 2
   - **Reason:** The Kraken will attack the ships in the following order: 1, 2, 1, 2, 1. Both ships will sink.

This test input should help in exposing issues with the logic of handling attacks and the sinking of ships 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: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
// 我与z的感情，是从我的告白开始。
// 我第一次记住他，是因为他姣好的容貌，在大学新生里脱颖而出。
// 但可别因此以为我是见色起意。
// 人是视觉动物，对于美好的事物都会下意识的注意。
// 比起他我会更在意一片落叶，因为我可以收藏这没有第二片相同的尸体。
// 得益于周围，我也加上了他的微信。
// 但那时我对他并无想法。
// 我喜欢观察我感兴趣的人，所以默默观察过他。
// 发现他很有让人快乐的天赋。
// 每一次见到他时，身边的人不是与他说笑就是打闹。
// 不论是谁，呆在他身边一会儿后都能与他开开玩笑。
// 我羡慕这样的能力。
// 因为我拧巴又内耗，时常将关系相处的一塌糊涂。
// 某天我刚好坐在他的不远处，听见他身边的人问起他的前女友。
// “你为啥和你前女友分啊?”
// “嗯 ，不喜欢了。”
// “哈哈天，她和她闺蜜没把你骂死啊?”
// “或许?但我们当时是吃了顿饭好聚好散的，我不喜欢了就直接告诉她，也没有吊着她。”
// 我心想这个人还蛮诚实的，不喜欢了就大大方方表达，没有情感纠纷与藕断丝连，也没有冷暴力与优柔寡断。
// 如果我是他前女友，虽然会不可避免的伤心，但的确会选择好聚好散的。
// 或许是我看向他的目光太贪婪，他开始不自在的左顾右盼。
// 明亮的笑容闯入我的视野，他嘴角若隐若现的梨涡，像爱情的唇珠。我想戳一下。
// 我发现他上课时有一条固定的路径。
// 于是我户外阅读时就会特地选择在他必经之路附近的椅子，从而每次都能见他从我的眼前走过。
// 久而久之，我还发现了他其他的特点。
// 他与我一样，走路不会踩落叶，会特地绕开。
// 见到好看的小花时，会拿手机饶有兴致拍照。
// 时不时的会抬头，不知道是在看天空还是树枝。
// 他与我不同，又相同。
// 他是每时每刻悬挂的太阳，我是忽明忽暗站立的路灯。
// 我被他吸引，开始在我的文字里叙述他的一举一动。
// 我想过写书信，也想过发微信搭讪。
// 可我始终觉得不够礼貌，我不够了解他，也还没有让他了解我。
// 但是转念一想，我不主动，他怎么会有机会了解我呢?
// 在窗外的多个夜色里，我曾翻阅无数次他的朋友圈。
// 他喜欢在清吧里微醺的感觉，喜欢听周杰伦的歌，喜欢记录夕阳晚霞，还喜欢各种各样鲜花。
// 于是白天思虑再三，我决定按响爱情的门铃。请求给我一个可以彼此了解的机会。
// 可别误会，这不是告白。
// 我选择了一种我认为最妥当的方式。
// 在清晨的露珠摇摇欲坠时，我去往花店，买了一束向日葵。
// 向日葵多好呀，向阳而生。
// 不过我不会亲手送给他，因为我没这勇气。
// 所以我涂涂改改，写了一封小小的卡片：“z，你好。接下来的话可能会有些冒失，先请求你的原谅。
// 我是x，心仪你有一段时间了。思虑再三我想要请求你给我一个了解你的机会，也希望你可以试着来了解我。
// 如果你愿意，这束向日葵请收下，是送给你的。
// 如果你不愿意，也没关系，你可以将这束花退回女生宿舍楼下，因为我也喜欢鲜花，所以我可以送给我自己。
// 希望不会让你有负担，祝你天天开心。”
// 然后我把鲜花交给了外卖员，让他将我的心意送去。
// 期间我有忐忑后悔，在宿舍里坐立不安。白天的光直晃眼睛，我阴暗的蜷缩在角落。
// 但没几分钟我又会挣扎着下楼，查看鲜花有没有被退回。
// 我知道，我这样看起来有些神经质。
// 可直到我下晚课时，鲜花都杳无音信。
// 这时我又有些窃喜，庆幸是一个好兆头。
// 第二天是周末，我中午11点睡醒。睡醒第一件事就是风风火火的下楼查看。
// 结果令我目光一颤。鲜花安静的躺在楼下。
// 虽然早已做好被拒绝的准备，但还是不可避免的有些失落。爱情拒绝了我的打扰。
// 我拿起鲜花，忽然有一张小卡片落入了我的视线。
// 这不是我的那张蓝色卡片，而是崭新的粉色。
// 我有些手足无措的拾起。卡片干干净净，字迹工整舒适：
// “鲜花我只要一朵就够了，剩下的你应该送给勇敢真诚的自己。
// 哦对了，我愿意。”

// 注意类名必须为 Main, 不要有任何 package xxx 信息
 // package Dduo;
import java.util.*;
import java.util.List;
import java.math.*;
import java.time.*;
import java.io.*;
import java.util.regex.*;

import java.awt.*;
import java.awt.event.InputEvent;

// Eclipse IDE 2020 08
// OpenJDK 1.8
// 点击下方网页可访问个人主页
// http://gczdy.cn

/**
 *
 * @Author Dduo
 * @Date 2024-11-30
 */
public class Main {

	// 普通流
    static Scanner sc = new Scanner(System.in);

    // 数据流快读
//    static Read sc=new Read();

    static long MOD = (long) (1e9 + 7);
    static final int[] dx = {1, -1, 0, 0};
    static final int[] dy = {0, 0, 1, -1};

    // 一切新鲜 有点冒险 请告诉我怎么走到终点
    // 没有人了解 没有人像我和陌生人的爱恋
    // 我想我会开始想念你 可是我刚刚才遇见了你
    // 我想我已慢慢喜欢你 因为我拥有爱情的勇气

    /**
     *
     * @param args
     * @return
     * @throws Exception
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {

        int t = 1;
        t = sc.nextInt();

//        预处理函数
//        preconditioning();

        while (t-- > 0) {
            solve();
        }

   	 	sc.close();

        return;
    }

//  偷懒... 纪念由于某场div2写了print结果wa了四次.
    static <T> void soutln(T t) {
    	System.out.println(t);
    }

    static <T> void sout(T t) {
    	System.out.print(t);
    }

    // 要养成开静态变量的习惯
    static int cnt=0;

    // 凡事都有偶然的凑巧结局却又如宿命的必然
    static void solve() throws Exception {

    	int n=sc.nextInt();
    	int k=sc.nextInt();

    	// 后面
    	long k2 = k / 2;
    	// 前面
        long k1 = k - k2;

        long h1 = 0;
        long h2 = 0;

        long ans = 0;

        long[] a = new long[(int) n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = sc.nextLong();
            if (k1 >= a[i]) {
                k1 -= a[i];
                h1 = i;
            } else {
                a[i] -= k1;
                k1 = 0;
            }
        }

        ans += h1;

        for (int i = n; i > h1; i--) {
            if (a[i] <= k2) {
                k2 -= a[(int) i];
                h2++;
            } else {
                break;
            }
        }

        ans += h2;

        soutln(ans);
    }

  // 小时候的我
  // 做着美的梦
  // 梦想中 那片天空
  // 好像挂满笑容
  // 长大后的我
  // 坚定着美梦
  // 因为我 已慢慢懂
  // 努力就能成功
  // 我一个人的力量
  // 能去闯 就算要飞越海洋
  // 也不能 阻挡我对梦的渴望
  // 我一个人的信仰
  // 像太阳 散发着炽热的光
  // 能照亮 所有内心 灰暗的地方
  // 天边的彩虹
  // 倒映我的梦
  // 路边的 彩色霓虹
  // 流淌着感动
  // 风吹的温柔
  // 带走了沉重
  // 我要和别人不同
  // 做自己的英雄
  // 我一个人的力量
  // 能去闯 就算要飞越海洋
  // 也不能 阻挡我对梦的渴望
  // 我一个人的信仰
  // 像太阳 散发着炽热的光
  // 能照亮 所有内心 灰暗的地方
  // 我一个人的力量
  // 能去闯 就算要飞越海洋
  // 也不能 阻挡我对梦的渴望
  // 我一个人的信仰
  // 像太阳 散发着炽热的光
  // 能照亮 所有内心 灰暗的地方
  // 小时候的我
  // 做着美的梦
  // 梦想中 那片天空
  // 好像挂满笑容
  // 长大后的我
  // 坚定着美梦
  // 因为我 已慢慢懂
  // 努力就能成功
  // 就能成功

	// by 可爱抱抱呀
	// 取模次数过多耗时会大幅增加，可以设一个取模的阈值，并用减，最后答案在取总模
	// 取模的数字，别少写(int)1e9+7，inf别写错类型(long)8e18
	// 初始化数据的时候，最小值不要傻乎乎的默认为0，因为可能有负数
	// 记住看数字范围，需要开long吗，需要用BigInteger吗，需要手动处理字符串吗，复杂度数量级控制在1e7或者以下了吗
	// 开数组的数据范围最高不能超过1e7，数据范围再大就要用哈希表离散化了
	// 基本数据类型不能自定义sort排序，二维数组就可以了，顺序排序的时候是小减大，注意返回值应该是int
	// BFS的时候记得在循环里要往队列里push，拓扑排序要确认无环
	// “>>”运算的时候，int和long类型得到的数字可能是不同的
	class Read{
	    BufferedReader bf;
	    StringTokenizer st;
	    BufferedWriter bw;
	    public Read(){
	        bf=new BufferedReader(new InputStreamReader(System.in));
	        st=new StringTokenizer("");
	        bw=new BufferedWriter(new OutputStreamWriter(System.out));
	        //什么时候才能持续稳定ak力扣、AcWing呢？
	        //什么时候才能ak cf的div2，（div1）呢？才能打div2不计rating呢？
	        //什么时候才能ak  abc  arc 不计rating呢？
	        //什么时候才能ak 牛客练习赛不计rating呢?
	        //什么时候才能ak 洛谷的div2呢？才能打div2不计rating呢？
	    }

	    //输入部分：
	    public String nextLine() throws IOException{
	        return bf.readLine();
	    }
	    public String next() throws IOException{
	        while(!st.hasMoreTokens()){
	            st=new StringTokenizer(bf.readLine());
	        }
	        return st.nextToken();
	    }
	    public char nextChar() throws IOException{
	        //确定下一个token只有一个字符的时候再用
	        return next().charAt(0);
	    }
	    public int nextInt() throws IOException{
	        return Integer.parseInt(next());
	    }
	    public long nextLong() throws IOException{
	        return Long.parseLong(next());
	    }
	    public double nextDouble() throws IOException{
	        return Double.parseDouble(next());
	    }
	    public float nextFloat() throws IOException{
	        return Float.parseFloat(next());
	    }
	    public byte nextByte() throws IOException{
	        return Byte.parseByte(next());
	    }
	    public short nextShort() throws IOException{
	        return Short.parseShort(next());
	    }
	    public BigInteger nextBigInteger() throws IOException{
	        return new BigInteger(next());
	    }
	    public BigDecimal nextDecimal() throws IOException{
	        return new BigDecimal(next());
	    }

	    //以下为输出部分：：

	    public void println(int a) throws IOException{
	        print(a);
	        println();
	        return;
	    }
	    public void print(int a) throws IOException{
	        bw.write(String.valueOf(a));
	        return;
	    }
	    public void println(String a) throws IOException{
	        print(a);
	        println();
	        return;
	    }
	    public void print(String a) throws IOException{
	        bw.write(a);
	        return;
	    }
	    public void println(long a) throws IOException{
	        print(a);
	        println();
	        return;
	    }
	    public void print(long a) throws IOException{
	        bw.write(String.valueOf(a));
	        return;
	    }
	    public void println(double a) throws IOException{
	        print(a);
	        println();
	        return;
	    }
	    public void print(double a) throws IOException{
	        bw.write(String.valueOf(a));
	        return;
	    }
	    public void print(BigInteger a) throws IOException{
	        bw.write(a.toString());
	        return;
	    }
	    public void print(char a) throws IOException{
	        bw.write(String.valueOf(a));
	        return;
	    }
	    public void println(char a) throws IOException{
	        print(a);
	        println();
	        return;
	    }
	    public void println() throws IOException{
	        bw.newLine();
	        return;
	    }
	    //其他调试命令：
	    public void flush() throws IOException{
	        //交互题分组调试，或者提前退出的情况下可以先运行此语句再推出
	        bw.flush();
	        return;
	    }
	    public boolean hasNext() throws IOException{
	        //本地普通IDE难以使用这个方法调试，需要按照数据组flush，刷新语句:
	        //sc.flush()
	        //调试完可删去
	        return bf.ready();
	    }
	}
	//Trie插入新节点后，下降的命令放在if外面
}

// 如果你讨厌我 那我就永远消失
// 被爱的人不用道歉
// 下次再见面 不要冷冰冰 也不要说反话!

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

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

Think step by step.

Response:
To generate a challenging test input for the given problem, we need to focus on edge cases and scenarios that can expose potential bugs in the buggy code. Here are some key points to consider:

1. **Edge Cases:**
   - **Small Number of Ships and Attacks:** Test with small values of `n` and `k` to ensure the basic logic works.
   - **Large Number of Ships and Attacks:** Test with large values of `n` and `k` to ensure the algorithm handles large inputs efficiently.
   - **Equal Durability:** Test with ships having equal durability to see how the Kraken's attacks are distributed.
   - **Mixed Durability:** Test with ships having varying durabilities to check the attack order and sinking conditions.

2. **Specific Scenarios:**
   - **All Ships Sunk:** Ensure all ships can be sunk by the Kraken.
   - **No Ships Sunk:** Ensure the Kraken can attack without sinking any ships.
   - **Partial Ships Sunk:** Ensure the Kraken can sink some ships but not all.

Given these points, a small-scale test input that can be challenging is:

- **Test Case 1:** Small number of ships and attacks.
- **Test Case 2:** Large number of ships and attacks.
- **Test Case 3:** Equal durability for all ships.
- **Test Case 4:** Mixed durability with some ships having very high durability.

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

```plaintext
4
3 4
1 1 1
4 10
2 1 3 4
5 20
2 2 2 2 2
6 15
1 2 3 4 5 6
```

Explanation:
1. **Test Case 1:** 3 ships, 4 attacks. All ships will be sunk.
2. **Test Case 2:** 4 ships, 10 attacks. The ships will be sunk in a specific order, and some will remain.
3. **Test Case 3:** 5 ships, 20 attacks. All ships have equal durability, and all will be sunk.
4. **Test Case 4:** 6 ships, 15 attacks. Mixed durability, some ships will be sunk, and some will remain.

This test input should help expose any bugs in the buggy code.