**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, where each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,ans=0;
		cin>>n;
		int a[n];
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
		}
		for(int i=0;i<(n+1);i++)
		{
			for(int j=(i+1);j<n;j++)
			{
				if(((a[i]*a[j])%((i+1)*(j+1)))==0)
				{
					ans++;
				}

			}
		}
		cout<<ans<<endl;
	}
}

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

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

Think step by step.

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

1. **Identify the Bug**: The main issue in the provided code is the loop condition in the outer loop. The loop should run from `0` to `n-1`, but it currently runs from `0` to `n` (inclusive). This means the loop will go out of bounds, causing undefined behavior.

2. **Edge Cases**:
   - **Small Permutations**: Test with the smallest possible permutations to ensure the basic logic is correct.
   - **Large Permutations**: Test with a permutation close to the upper limit to check for performance issues and out-of-bounds errors.
   - **Special Permutations**: Test with permutations that have specific properties, such as all elements being in reverse order or a permutation that maximizes the number of valid pairs.

3. **Test Input**:
   - **Test Case 1**: A small permutation to ensure basic functionality.
   - **Test Case 2**: A permutation that causes the out-of-bounds error in the loop.
   - **Test Case 3**: A permutation that maximizes the number of valid pairs to check the correctness of the logic.

Here is a small-scale test input that includes these considerations:

```plaintext
3
1
1
5
1 2 3 4 5
10
10 9 8 7 6 5 4 3 2 1
```

**Explanation**:
- **Test Case 1**: A permutation of length 1. This should output `0` because there are no pairs.
- **Test Case 2**: A permutation of length 5 with elements in ascending order. This should output `3` (pairs (1, 2), (1, 4), and (2, 4)).
- **Test Case 3**: A permutation of length 10 with elements in descending order. This should output a larger number of valid pairs, testing the logic and the out-of-bounds error.

This test input is designed to be small enough to be easily verified but comprehensive enough to expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, where each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
import java.util.Scanner;
public class P1986G1 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int[] p = new int[n];
            java.util.Arrays.setAll(p, index -> sc.nextInt());
            System.out.println(count(n, p));
        }
        sc.close();
    }

    static int count(int n, int[] p) {
        int cnt = 0;
        int i = 0;
        int j = i + 1;
        while (i < n) {
            while (j < n) {
                if ((p[i] * p[j]) % ((i + 1) * (j + 1)) == 0) {
                    cnt++;
                }
                j++;
            }
            i++;
            j = i + 1;
        }
        return cnt;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code provided has a nested loop that iterates through the permutation, and it checks the condition `(p[i] * p[j]) % ((i + 1) * (j + 1)) == 0`. This condition can be tricky, especially for large values of `n` or specific permutations where the condition is more likely to fail.

### Steps to Generate the Test Input:
1. **Edge Case: Single Element Permutation**
   - A permutation of length 1 should always result in 0 pairs.
   - Example: `1\n1`

2. **Edge Case: Two Element Permutation**
   - A permutation of length 2 should result in 1 pair if the condition is met.
   - Example: `2\n1 2`

3. **Edge Case: Small Permutation with Specific Condition**
   - A small permutation where the condition is more likely to fail.
   - Example: `3\n2 3 1` (This permutation should result in 1 valid pair: (1, 2))

4. **Edge Case: Large Permutation with Many Valid Pairs**
   - A larger permutation where the condition is met for multiple pairs.
   - Example: `5\n2 4 1 3 5` (This permutation should result in 3 valid pairs: (1, 2), (1, 5), (2, 5))

5. **Edge Case: Permutation with No Valid Pairs**
   - A permutation where no pairs meet the condition.
   - Example: `4\n1 3 2 4` (This permutation should result in 0 valid pairs)

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

### Explanation:
- The first set of input data (1 element) should result in 0 pairs.
- The second set of input data (2 elements) should result in 1 pair.
- The third set of input data (3 elements) should result in 1 pair.
- The fourth set of input data (5 elements) should result in 3 pairs.
- The fifth set of input data (4 elements) should result in 0 pairs.

This test input is designed to cover a range of scenarios, including edge cases and larger permutations, to thoroughly test the provided code and expose any potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, where each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
import java.util.Scanner;
public class P1986G1 {
    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in)) {
            int len = sc.nextInt();

            for (int i = 0; i < len; i++) {
                int n = sc.nextInt();
                int[] p = new int[n];

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

                System.out.println(count(n, p));
            }
        }
    }

    static int count(int n, int[] p) {
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if ((p[i] * p[j]) % ((i + 1) * (j + 1)) == 0) {
                    cnt++;
                }
            }
        }
        return cnt;
    }
}

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

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

Think step by step.

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

1. **Small Permutations**: Test with small permutations to ensure the code handles basic cases correctly.
2. **Large Permutations**: Test with large permutations to ensure the code handles larger inputs efficiently.
3. **Specific Patterns**: Test with specific patterns that might cause issues, such as permutations where the elements are in a specific order that could lead to more or fewer valid pairs.

Given the constraints and the need for a small-scale test input, let's focus on a permutation that can expose the bug effectively.

### Test Input

We will use a permutation where the elements are in a specific order that can create a large number of valid pairs, but also check for edge cases.

```plaintext
3
1
1
5
2 4 1 3 5
10
1 2 3 4 5 6 7 8 9 10
```

### Explanation

1. **First Set**:
   - Length: 1
   - Permutation: [1]
   - Expected Output: 0 (No valid pairs)

2. **Second Set**:
   - Length: 5
   - Permutation: [2, 4, 1, 3, 5]
   - Expected Output: 3 (Valid pairs: (1, 2), (1, 5), (2, 5))

3. **Third Set**:
   - Length: 10
   - Permutation: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   - Expected Output: 10 (Valid pairs: (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3))

This test input includes a small permutation, a medium permutation, and a larger permutation, each with a different pattern to ensure the code handles various cases correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, where each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
from collections import defaultdict

def gcd(a, b):
  if b == 0:
    return a
  return gcd(b, a % b)

def solve():
  n = int(input())
  a = list(map(int, input().split()))
  f = defaultdict(int)
  d = [[] for i in range(n + 1)]
  for i in range(1, n + 1):
    for j in range(i, n + 1, i):
      d[j].append(i)
  ans = 0
  for i in range(n):
    g = gcd(a[i], i + 1)
    x = a[i] // g
    y = (i + 1) // g
    for j in d[x]:
      ans += f[(j, y)]
    for j in d[x]:
      f[(y, j)] += 1
  print(ans)

tc = int(input())
while tc > 0:
  solve()
  tc -= 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 generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Test with the smallest and largest possible values for `n` and `t`.
2. **Special Permutations**: Test with permutations that might cause issues in the gcd calculations or the divisor lists.
3. **Large Test Cases**: Test with a large number of test cases to ensure the algorithm handles it efficiently.

Given the constraints and the nature of the problem, let's focus on a permutation that might cause issues in the gcd calculations and the divisor lists. Specifically, we can use a permutation where the gcd values are not straightforward, which might expose issues in the code's handling of gcd and divisors.

Here's a small-scale test input that should challenge the buggy code:

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

### Explanation:
1. **Test Case 1**: `n = 1`, `p = [1]`
   - This is a simple edge case where there are no index pairs.
2. **Test Case 2**: `n = 2`, `p = [2, 1]`
   - This is a simple permutation with one valid index pair: (1, 2).
3. **Test Case 3**: `n = 5`, `p = [5, 4, 3, 2, 1]`
   - This permutation is the reverse of the natural order, which might cause issues in the gcd calculations and the divisor lists.

This test input should help expose any bugs in the provided buggy code by testing edge cases and a more complex permutation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, where each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
#author : Sushmanth

from sys import stdin , stdout
input = stdin.readline

inp = lambda : list(map(int , input().split()))

def gcd(a , b):
    if(b == 0):return a
    return gcd(b , a % b)

sz = (10 ** 5) + 1
div = [[] for i in range(sz)]
for i in range(1 , sz):
    for j in range(i , sz , i):
        div[j].append(i)

def answer():

    num , den = [] , []
    c = 0
    for i in range(n):
        g = gcd(a[i] , i + 1)
        num.append(a[i] // g)
        den.append((i + 1) // g)
        if(num[i] % den[i] == 0):
            c += 1

    den_to_num = [dict() for i in range(n + 1)]
    for i in range(n):
        for d in div[num[i]]:
            key = str(d)
            den_to_num[den[i]][key] = den_to_num[den[i]].get(key , 0) + 1

    ans = 0
    for i in range(n):
        for d in div[num[i]]:
            ans += den_to_num[d].get(str(den[i]) , 0)

    ans = (ans - c) // 2
    return ans

for T in range(int(input())):

    n = int(input())
    a = inp()

    print(answer())

**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 consider edge cases and scenarios that can maximize the failure probability. The code has several potential issues, such as incorrect handling of the gcd calculation, incorrect counting of valid pairs, and potential issues with the data structures used.

Here are the steps to generate a challenging test input:

1. **Edge Case with Small Permutation**:
   - Test with a small permutation where the gcd calculation might fail.
   - Example: `n = 3` with permutation `2 3 1`.

2. **Edge Case with Large Permutation**:
   - Test with a larger permutation that might expose issues in the gcd and counting logic.
   - Example: `n = 10` with permutation `8 9 7 12 1 10 6 3 2 4`.

3. **Edge Case with All Elements Equal to Their Indices**:
   - Test with a permutation where each element is equal to its index.
   - Example: `n = 5` with permutation `1 2 3 4 5`.

4. **Edge Case with All Elements in Reverse Order**:
   - Test with a permutation where each element is in reverse order.
   - Example: `n = 5` with permutation `5 4 3 2 1`.

5. **Edge Case with Mixed Values**:
   - Test with a permutation that has a mix of values to ensure the gcd and counting logic works correctly.
   - Example: `n = 7` with permutation `1 3 2 5 4 7 6`.

Combining these edge cases, we can generate the following test input:

```plaintext
5
3
2 3 1
10
8 9 7 12 1 10 6 3 2 4
5
1 2 3 4 5
5
5 4 3 2 1
7
1 3 2 5 4 7 6
```

This test input covers a variety of scenarios that can help expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

bool is_prime(int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0) return false;
    }
    return true;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> piles(n);
        int sum = 0;
        for (int i = 0; i < n; ++i) {
            cin >> piles[i];
            if (!is_prime(piles[i])) {
                sum ^= piles[i];
            }
        }

        if (sum == 0) {
            cout << "Alice" << endl;
        } else {
            cout << "Bob" << 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
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

using namespace std;
using ll = long long;

#ifdef _DEBUG
#include "utils/debug.h"
#else
#define debug(...) 0
#endif

// Returns all the primes <= N
// Runtime: O(NloglogN)
vector<int> sieve(int N) {
	vector<int> prime(N+1, 1);
	vector<int> primes;
	prime[0] = prime[1] = 0;
	for (int i = 2; i <= N; ++i) {
		if (!prime[i]) continue;
		primes.push_back(i);
		for (ll j = 1LL * i * i; j <= N; j += i) {
			prime[j] = 0;
		}
	}
	return primes;
}

int main() {
	cin.tie(0)->sync_with_stdio(0);
	const int MN = 2e5+10;
	vector<int> prime(MN), val(MN), can1(MN), can02(MN), can11(MN);
	auto primes = sieve(MN);
	for (int p : primes) {
		prime[p] = 1;
	}
	val[1] = 0;
	for (int i = 3; i < MN; i += 2) {
		if (prime[i - 2]) val[i] = val[i - 2] + 1;
	}
	for (int p1 : primes) {
		for (int p2 : primes) {
			if (p1 + p2 >= MN) continue;
			if (val[p1] == 1 && val[p2] == 1) can11[p1+p2] = 1;
			if (val[p1] == 1 || val[p2] == 1) can1[p1+p2] = 1;
			else can02[p1+p2] = 1;
		}
	}

	int T; cin >> T;
	while (T--) [&]() {
		int N; cin >> N;
		int cnt0 = 0;
		int cnt1 = 0;
		int cnt11 = 0;
		for (int i = 0; i < N; ++i) {
			int x; cin >> x;
			if (x % 2 == 1) {
				if (val[x] % 2 == 1) cnt1++;
				else cnt0++;
			}
			else { // even
				if (x == 2) cnt0++;
				else {
					assert(can1[x] || can02[x]);
					if (!can02[x]) { // x must be split into a 1
						cnt0++;
					}
					else {
						if (can11[x]) cnt11++;
						cnt1++;
					}
				}
			}
		}
		if (N % 2 == 0) {
			if (cnt1) cout << "Alice" << '\n';
			else cout << "Bob" << '\n';
		}
		else {
			if (cnt0 == N) cout << "Bob" << '\n';
			else if (cnt1 != N) cout << "Alice" << '\n';
			else {
				// cnt1 == N
				if (cnt11 != N) cout << "Alice" << '\n';
				else cout << "Bob" << '\n';
			}
		}
	}();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

// #pragma GCC optimize("O3")

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Size(a) (int)a.size()
#define ll long long
#define ld long double

// #define int long long

using namespace std;

const int N = 2e5 + 5;

bool prime[N];
vector<int> primes;
bool dp[N], dp2[N];

void solve() {
    int n;
    cin >> n;
    int cnt = 0;
    int cnt2 = 0;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        cnt += dp[a[i]];
        cnt2 += dp2[a[i]];
    }
    if (cnt > 0 && (n % 2 == 0 || cnt < n || cnt2 > 0))
        cout << "Alice\n";
    else
        cout << "Bob\n";
}

void precalc() {
    fill(prime + 2, prime + N, true);
    for (int i = 2; i < N; i++) {
        if (prime[i]) {
            primes.push_back(i);
            for (int j = 2 * i; j < N; j += i)
                prime[j] = false;
        }
    }
    cerr << "primes.size(): " << primes.size() << "\n";
    for (int i = 0; i < primes.size(); i++) {
        int p = primes[i];
        for (int j = 0; j <= i; j++) {
            int q = primes[j];
            if (p + q >= N)
                break;
            dp[p + q] |= !dp[p] && !dp[q];
            dp2[p + q] |= dp[p] && dp[q];
        }
    }
    // cerr << "dp:\n";
    // for (int i = 2; i < N; i++)
    //     cerr << "i, dp[i], dp2[i]: " << i << " " << dp[i] << " " << dp2[i] << "\n";
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    #ifdef dimss
        freopen("test.txt", "r", stdin);
    #endif

    precalc();

    int T = 1;
    cin >> T;
    while (T--) {
        solve();
        #ifdef dimss
            cerr << "--------\n";
        #endif
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are some key points to consider:

1. **Prime Number Handling**: The code relies on precomputed prime numbers and their sums. We need to ensure that the input includes numbers that can and cannot be split into two prime numbers.

2. **Optimal Play**: Both players play optimally, so the input should test the logic of the game where both players make the best possible moves.

3. **Pile Counts and Values**: The number of piles and the values of stones in each pile should be chosen to challenge the logic of the game, especially when the number of piles is even or odd.

4. **Edge Cases**: Consider scenarios with the minimum and maximum allowed values for the number of piles and stones in each pile.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
```

This input includes:
- A case with 2 piles, one of which is 1 (which cannot be split into two primes).
- A case with 3 piles, all of which are prime numbers.
- A case with 4 piles, all of which can be split into two primes.
- A case with 5 piles, all of which are the same and can be split into two primes.

These test cases should expose any issues in the logic of the game, especially in how the code handles the splitting of piles and the determination of the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

Buggy Code:
def determine_winner(test_cases):
    results = []
    for n, piles in test_cases:

        if n == 2 and piles == [2, 1]:
            results.append("Bob")
        elif n == 3 and piles == [3, 5, 7]:
            results.append("Alice")
        elif n == 4 and piles == [4, 6, 8, 10]:
            results.append("Alice")
        elif n == 5 and piles == [8, 8, 8, 8, 8]:
            results.append("Bob")
        else:
            results.append("Bob" if n % 2 == 0 else "Alice")
    return results

t = int(input())
test_cases = []
for _ in range(t):
    n = int(input())
    piles = list(map(int, input().split()))
    test_cases.append((n, piles))

results = determine_winner(test_cases)

for result in results:
    print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

Buggy Code:
def sieve_of_eratosthenes(n):
    is_prime = [True] * (n + 1)
    is_prime[0] = is_prime[1] = False
    for i in range(2, int(n**0.5) + 1):
        if is_prime[i]:
            for j in range(i * i, n + 1, i):
                is_prime[j] = False
    return {x for x in range(2, n + 1) if is_prime[x]}

def precompute(max_stones):
    primes = sieve_of_eratosthenes(max_stones)
    winning_positions = [False] * (max_stones + 1)
    for i in range(2, max_stones + 1):
        if not winning_positions[i - 1] or any(i - p in primes for p in primes if p < i):
            winning_positions[i] = True
    return winning_positions

def game_winner(piles, winning_positions):
    return "Bob" if sum(winning_positions[p] for p in piles) % 2 else "Alice"

t = int(input())
piles_list = []
max_stones = 0
for _ in range(t):
    input()
    piles = list(map(int, input().split()))
    piles_list.append(piles)
    max_stones = max(max_stones, *piles)

winning_positions = precompute(max_stones)
for piles in piles_list:
    print(game_winner(piles, winning_positions))

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
3
2 3 5
4
4 6 8 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

Buggy Code:
def sieve_of_eratosthenes(n):
    is_prime = [True] * (n + 1)
    is_prime[0] = is_prime[1] = False
    for i in range(2, int(n**0.5) + 1):
        if is_prime[i]:
            for j in range(i * i, n + 1, i):
                is_prime[j] = False
    return {x for x in range(2, n + 1) if is_prime[x]}

def game_winner(piles):
    max_stones = max(piles)
    primes = sieve_of_eratosthenes(max_stones)
    winning_positions = [False] * (max_stones + 1)

    for i in range(2, max_stones + 1):
        if not winning_positions[i - 1] or any(i - p in primes for p in primes if p < i):
            winning_positions[i] = True

    return "Bob" if sum(winning_positions[p] for p in piles) % 2 else "Alice"

# Input processing
for _ in range(int(input())):
    input()  # Skip reading 'n' as it's not needed
    piles = list(map(int, input().split()))
    print(game_winner(piles))

**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
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static boolean[] isPrimes;
    private static int sup = 200_001;
    private static List<Integer> primes;
    private static boolean[] isWins;
    private static boolean[] twoWins;

    private static void init() {
        isPrimes = new boolean[sup];
        primes = new ArrayList<>();
        isPrimes[2] = true;
        isWins = new boolean[sup];
        twoWins = new boolean[sup];
        for (int i = 2; i < sup; ++i) {
            if (!isPrimes[i]) primes.add(i);
            for (int p : primes) {
                if (p * i >= sup) break;
                isPrimes[p * i] = true;
                if ((i % p) == 0) break;
            }
        }
        isWins[1] = isWins[2] = false;
        isWins[4] = true;

        for (int i = 3; i < sup; i += 2) isWins[i] = !isPrimes[i - 2] && !isWins[i - 2];
        int prePrime = 2;
        for (int i = 6; i < sup; i += 2) {
            for (int j = prePrime; j >= 0; --j) {
                int prime = primes.get(j);
                if (!isPrimes[prime] && !isPrimes[i - prime] && !isWins[prime] && !isWins[prime]) {
                    isWins[i] = true;
                    break;
                }
            }
            if (isWins[i]) {
                for (int j = prePrime; j >= 0; --j) {
                    int prime = primes.get(j);
                    if (!isPrimes[prime] && !isPrimes[i - prime] && isWins[prime] && isWins[prime]) {
                        twoWins[i] = true;
                        break;
                    }
                }
            }

            if (prePrime < primes.size() - 1 && primes.get(prePrime + 1) <= i) ++prePrime;
        }
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        init();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt();
            int c1 = 0, c2 = 0;
            for (int i = 0; i < n; ++i) {
                int v = nextInt();
                if (isWins[v]) ++c1;
                if (twoWins[v]) ++c2;
            }
            if (c1 == 0) {
                sb.append("Bob").append("\n");
            } else if (c1 < n) {
                sb.append("Alice").append("\n");
            } else if (c2 < n) {
                sb.append("Alice").append("\n");
            } else {
                sb.append("Bob").append("\n");
            }

        }
        System.out.println(sb);

    }
}

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

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

Think step by step.

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

1. **Prime Numbers and Splitting**: The game involves splitting piles into two piles with a prime number of stones. The buggy code has a flawed logic for determining whether a pile can be split into two prime numbers.

2. **Winning and Losing States**: The code uses precomputed arrays `isWins` and `twoWins` to determine the winning and losing states. However, the logic for updating these arrays might be incorrect.

3. **Edge Cases**:
   - **Small Piles**: Piles with very small numbers (e.g., 1, 2, 3) that cannot be split into two primes.
   - **Large Piles**: Piles with large numbers that can be split into two primes in multiple ways.
   - **Mixed Piles**: A combination of small and large piles to test the logic for different scenarios.

4. **Test Case Structure**:
   - **Multiple Test Cases**: To ensure the code is tested thoroughly, we will include multiple test cases.
   - **Different Pile Configurations**: Each test case will have different configurations of piles to cover various scenarios.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
5
2
1 2
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
6
1 2 3 4 5 6
```

### Explanation:
1. **Test Case 1**:
   - Piles: 1, 2
   - Both piles cannot be split into two primes, so Alice cannot make a move. Bob wins.

2. **Test Case 2**:
   - Piles: 3, 5, 7
   - Alice can choose \( k = 1 \), remove the pile of 7, and split the pile of 5 into 2 and 3. Bob cannot make a move, so Alice wins.

3. **Test Case 3**:
   - Piles: 4, 6, 8, 10
   - Alice can choose \( k = 2 \), remove two piles (e.g., 8 and 10), and split the remaining piles (e.g., 4 into 2 and 2, and 6 into 3 and 3). Bob cannot make a move, so Alice wins.

4. **Test Case 4**:
   - Piles: 8, 8, 8, 8, 8
   - It can be shown that if both players play optimally, Bob will win.

5. **Test Case 5**:
   - Piles: 1, 2, 3, 4, 5, 6
   - Mixed piles with different sizes. Alice can make a move by removing and splitting appropriately, but the exact optimal strategy will depend on the code's logic.

This test input should help expose any bugs in the provided code, especially in the logic for determining winning and losing states.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static boolean[] isPrimes;
    private static int sup = 200_001;
    private static List<Integer> primes;
    private static boolean[] isWins;
    private static boolean[] twoWins;

    private static void init() {
        isPrimes = new boolean[sup];
        primes = new ArrayList<>();
        isWins = new boolean[sup];
        twoWins = new boolean[sup];
        isPrimes[1] = true;
        for (int i = 2; i < sup; ++i) {
            if (!isPrimes[i]) primes.add(i);
            for (int p : primes) {
                if (p * i >= sup) break;
                isPrimes[p * i] = true;
                if ((i % p) == 0) break;
            }
        }
        isWins[4] = true;

        for (int i = 3; i < sup; i += 2) isWins[i] = !isPrimes[i - 2] && !isWins[i - 2];
        List<Integer> isPrimeLosing = new ArrayList<>();
        List<Integer> isPrimeWinning = new ArrayList<>();
        for (int i = 3; i < sup; i += 2) {
            if (!isPrimes[i]) {
                if (isWins[i]) {
                    isPrimeWinning.add(i);
                } else {
                    isPrimeLosing.add(i);
                }
            }
        }
        for (int i = 0; i < isPrimeLosing.size(); ++i) {
            for (int j = i; j < isPrimeLosing.size() && isPrimeLosing.get(j) + isPrimeLosing.get(i) < sup; ++j) isWins[isPrimeLosing.get(j) + isPrimeLosing.get(i)] = true;
        }
        for (int i = 0; i < isPrimeWinning.size(); i++) {
            for (int j = 0; j < isPrimeWinning.size() && isPrimeWinning.get(j) + isPrimeWinning.get(i) < sup; j++) isWins[isPrimeWinning.get(j) + isPrimeWinning.get(i)] = true;
        }
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        init();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt();
            int c1 = 0, c2 = 0;
            for (int i = 0; i < n; ++i) {
                int v = nextInt();
                if (isWins[v]) ++c1;
                if (twoWins[v]) ++c2;
            }
            if (c1 == 0) {
                sb.append("Bob").append("\n");
            } else if (c1 < n || (n % 2) == 0) {
                sb.append("Alice").append("\n");
            } else if (c2 < n) {
                sb.append("Alice").append("\n");
            } else {
                sb.append("Bob").append("\n");
            }

        }
        System.out.println(sb);

    }
}

**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 buggy code, we need to focus on edge cases and scenarios where the faulty logic in the code might fail. Here are some considerations:

1. **Prime Numbers**: The code deals with splitting piles into prime numbers. We should include piles with numbers that can and cannot be split into two primes.
2. **Number of Piles**: The code has specific logic for even and odd numbers of piles. We should test both even and odd numbers of piles.
3. **Winning and Losing States**: The code uses precomputed winning and losing states. We should include cases where these states might be incorrectly handled.

Let's construct a small-scale test input that covers these scenarios:

```plaintext
5
2
1 2
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
6
3 5 7 11 13 17
```

### Explanation:
1. **Test Case 1**:
   - Piles: 1, 2
   - Both 1 and 2 cannot be split into two primes. Alice cannot make a move, so Bob wins.
2. **Test Case 2**:
   - Piles: 3, 5, 7
   - Alice can remove the pile of 7 and split the pile of 5 into 2 and 3. Bob cannot make a move, so Alice wins.
3. **Test Case 3**:
   - Piles: 4, 6, 8, 10
   - Alice can remove two piles (8 and 10) and split the remaining piles (4 into 2 and 2, 6 into 3 and 3). Bob cannot make a move, so Alice wins.
4. **Test Case 4**:
   - Piles: 8, 8, 8, 8, 8
   - It can be shown that if both players play optimally, Bob will win.
5. **Test Case 5**:
   - Piles: 3, 5, 7, 11, 13, 17
   - All these numbers are prime and cannot be split. Alice can remove some piles and split others, but the exact sequence of moves needs to be considered carefully. This test case is designed to challenge the precomputed winning and losing states.

This test input should help expose any bugs in the faulty code by covering a variety of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

// ================================================================================================== FFT
const long long mod = 998244353;
const long long base = 3;

long long modpow(long long a, long long b, long long m) {
	long long p = 1, q = a;
	for (int i = 0; i < 30; i++) {
		if ((b >> i) & 1) { p *= q; p %= m; }
		q *= q; q %= m;
	}
	return p;
}

long long Division(long long a, long long b, long long m) {
	return (a * modpow(b, m - 2, m)) % m;
}

vector<long long> ntt(vector<long long> dat, int typ) {
	long long root = (typ == 1 ? base : Division(1, base, mod));
	int size_ = dat.size();

	// Step 1. Reorder
	for (int i = 0, j = 1; j < size_ - 1; j++) {
		for (int k = size_ / 2; k > (i ^= k); k /= 2);
		if (i > j) swap(dat[i], dat[j]);
	}

	// Step 2. Calculate
	for (int b = 2; b <= size_; b *= 2) {
		vector<long long> pows(b, 1);
		pows[1] = modpow(root, (mod - 1) / b, mod);
		for (int i = 2; i < b; i++) pows[i] = (pows[1] * pows[i - 1]) % mod;

		// Main Part
		for (int stt = 0; stt < size_; stt += b) {
			for (int i = 0; i < b / 2; i++) {
				long long a1 = dat[stt + i] + pows[i + 0 * b / 2] * dat[stt + i + b / 2];
				long long a2 = dat[stt + i] + pows[i + 1 * b / 2] * dat[stt + i + b / 2];
				a1 %= mod;
				a2 %= mod;
				dat[stt + i + 0 * b / 2] = a1;
				dat[stt + i + 1 * b / 2] = a2;
			}
		}
	}

	// Step 3. Finalize
	if (typ == 2) {
		long long mult = Division(1, size_, mod);
		for (int i = 0; i < size_; i++) {
			dat[i] = (dat[i] * mult) % mod;
		}
	}
	return dat;
}

vector<long long> convolve(vector<long long> v1, vector<long long> v2) {
	int size_ = 1;
	while (size_ < v1.size() + v2.size()) size_ *= 2;
	while (v1.size() < size_) v1.push_back(0);
	while (v2.size() < size_) v2.push_back(0);
	vector<long long> r1 = ntt(v1, 1);
	vector<long long> r2 = ntt(v2, 1);
	vector<long long> r3(size_, 0);
	for (int i = 0; i < size_; i++) r3[i] = (r1[i] * r2[i]) % mod;
	vector<long long> res = ntt(r3, 2);
	return res;
}

// ================================================================================================== Functions
bool prime[1 << 20];
int ranks[1 << 20];
bool Num1[1 << 20];
bool Num2[1 << 20];

void Initialize() {
	for (int i = 1; i <= 1000000; i++) prime[i] = true;
	prime[1] = false;
	for (int i = 2; i <= 1000000; i++) {
		for (int j = 2 * i; j <= 1000000; j += i) prime[j] = false;
	}
	for (int i = 1; i <= 1000000; i++) {
		if (i <= 3 || prime[i - 2] == false) ranks[i] = 0;
		else if (i <= 5 || prime[i - 4] == false) ranks[i] = 1;
		else ranks[i] = 2;
	}

	// Get Arrays
	vector<long long> Array0(200001, 0);
	vector<long long> Array1(200001, 0);
	vector<long long> Array2(200001, 0);
	for (int i = 3; i <= 200000; i += 2) {
		if (prime[i] == false) continue;
		if (ranks[i] == 0) Array0[i] = 1;
		if (ranks[i] == 1) Array1[i] = 1;
		if (ranks[i] == 2) Array2[i] = 1;
	}
	vector<long long> res1 = convolve(Array0, Array0);
	vector<long long> res2 = convolve(Array0, Array2);
	vector<long long> res3 = convolve(Array2, Array2);
	vector<long long> res4 = convolve(Array1, Array1);
	for (int i = 2; i <= 200000; i += 2) {
		if (res1[i] >= 1 || res2[i] >= 1 || res3[i] >= 1) Num1[i] = true;
		if (res4[i] >= 1) Num2[i] = true;
	}
}

string Solve(int N, vector<int> A) {
	// Step 1. Init
	int zeros = 0;
	int ones = 0;
	int twos = 0;
	int evens = N;
	for (int i = 0; i < N; i++) {
		if (A[i] % 2 == 0 && A[i] >= 6) continue;
		evens -= 1;
		if (A[i] % 2 == 1) {
			if (ranks[A[i]] == 0) zeros += 1;
			if (ranks[A[i]] == 1) ones += 1;
			if (ranks[A[i]] == 2) twos += 1;
		}
		if (A[i] == 2) zeros += 1;
		if (A[i] == 4) ones += 1;
	}
	// cout << zeros << " " << ones << " " << twos << " " << evens << " " << endl;
	/*if (evens == 0) {
		if (twos >= 1) {
			if (ones == 0) return "Bob";
			return "Alice";
		}
		if (ones == 0) return "Bob";
		if (zeros == 0 && ones % 2 == 1) return "Bob";
		return "Alice";
	}*/

	// Step 2. First Check
	int num_step1 = ones + evens;
	int impossibles = 0;
	for (int i = 0; i < N; i++) {
		if (A[i] % 2 == 0 && Num1[A[i]] == false) impossibles += 1;
	}
	for (int k = 1; k <= N / 2; k++) {
		if (num_step1 <= 2 * k && impossibles <= k && twos <= N - k && ones + twos + evens >= k) return "Alice";
	}

	// Step 3. Second Check
	if (N % 2 == 1) {
		int num_step2 = N - ones;
		int impossible2 = 0;
		for (int i = 0; i < N; i++) {
			if (A[i] % 2 == 0 && Num2[A[i]] == false) impossible2 += 1;
		}
		for (int k = 1; k <= N / 2; k++) {
			if (num_step2 <= 2 * k && impossible2 + zeros + twos <= k && ones + twos + evens >= k) return "Alice";
		}
	}
	return "Bob";
}

// ================================================================================================== Main

int main() {
	int T; cin >> T;
	Initialize();
	for (int t = 1; t <= T; t++) {
		int N; cin >> N;
		vector<int> A(N, 0);
		for (int i = 0; i < N; i++) cin >> A[i];
		cout << Solve(N, A) << 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
4
6 6 6 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

Buggy Code:
// LUOGU_RID: 169378759
// what is matter? never mind.
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,sse4,popcnt,abm,mmx,avx,avx2")
#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
//#define int long long
#define ull unsigned long long
#define SZ(x) ((int)((x).size()))
#define ALL(x) (x).begin(),(x).end()
using namespace std;
inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}

#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi;

#define maxn 200006
#define inf 0x3f3f3f3f

bool vis[maxn],is[maxn];
bitset<maxn> odd,win,even;

#define bint bitset<maxn>
bint mul(bint a,bint b,int n){
	bint c;
	For(i,0,n) if(b[i]) c|=(a<<i);
	return c;
}

void sieve(int n){
	For(i,2,n) if(!vis[i]){
		is[i]=1;
		for(int j=i;j<=n;j+=i) vis[j]=1;
	}
	for(int i=5;i<=n;i+=2) {
		odd[i]=odd[i-2]^is[i-2];
		win[i]=odd[i];
	}
	bint los;
	for(int i=3;i<=n;i+=2) los[i]=(!odd[i]);
	even=mul(los,los,n);
	win|=even;
	win[4]=1;
	win[2]=0;
	even=mul(odd,odd,n);
}

int n,a[maxn];
bool work()
{
	n=read();
	int cwin=0;
	For(i,1,n){
		a[i]=read();
	//	cout<<"win "<<a[i]<<" "<<win[a[i]]<<"\n";
		cwin+=win[a[i]];
	}
	if(!cwin) return 0;
	if(n%2==0) return 1;
	if(n%2==1 && cwin!=n) return 1;
	cwin=0;
	For(i,1,n) cwin+=even[a[i]];
	if(cwin>=1 && cwin<n) return 1;
	return 0;
}

signed main()
{
	sieve(2e5);
	int T=read();
	while(T--)puts(work()?"Alice":"Bob");
	return 0;
}
/*

*/

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might not be handled correctly. The code seems to have issues with how it handles the splitting of piles into prime numbers and the overall game strategy. Here are some key points to consider:

1. **Prime Number Splitting**: The code needs to ensure that a pile can be split into two prime numbers. This is a crucial step, and the code might not handle it correctly for certain edge cases.
2. **Winning Strategy**: The code uses a bitset to determine the winning and losing positions, but it might not handle the edge cases where the number of piles and the stones in each pile are critical.

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

- **Test Case 1**: A single test case with a small number of piles, where each pile has a number of stones that cannot be split into two prime numbers.
- **Test Case 2**: A single test case with a small number of piles, where each pile has a number of stones that can be split into two prime numbers.
- **Test Case 3**: A single test case with an odd number of piles, where some piles can be split and some cannot.

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

```plaintext
3
2
2 1
3
3 5 7
5
4 6 8 10 12
```

Explanation:
- **Test Case 1**: 2 piles with 2 and 1 stones. Neither 1 nor 2 can be split into two prime numbers, so Alice cannot make a move, and Bob wins.
- **Test Case 2**: 3 piles with 3, 5, and 7 stones. Alice can choose \( k = 1 \), remove the pile of 7 stones, and split the pile of 5 stones into 2 and 3. Bob cannot make a move, so Alice wins.
- **Test Case 3**: 5 piles with 4, 6, 8, 10, and 12 stones. Alice can choose \( k = 2 \), remove two piles (e.g., 8 and 10), and split the remaining piles into prime numbers. Bob cannot make a move, so Alice wins.

This test input should help expose any issues in the code's handling of prime number splitting and the overall game strategy.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

Buggy Code:
def is_prime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True

def split_pile(pile):
    for i in range(2, pile):
        if is_prime(i) and is_prime(pile - i):
            return (i, pile - i)
    return None

def optimal(pile, memo):
    pile_tuple = tuple(sorted(pile))
    if pile_tuple in memo:
        return memo[pile_tuple]

    for k in range(1, len(pile) // 2 + 1):
        for i in range(len(pile)):
            new_piles = pile[:i] + pile[i+1:]
            split = split_pile(pile[i])
            if split:
                new_piles.extend(split)
                if not optimal(new_piles, memo):
                    memo[pile_tuple] = True
                    return True
    memo[pile_tuple] = False
    return False

def determine_winner(pile):
    memo = {}
    return "Alice" if optimal(pile, memo) else "Bob"

n = int(input())
for _ in range(n):
    t = int(input())
    pile = list(map(int, input().split()))
    print(determine_winner(pile))

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

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

Think step by step.

Response:
```plaintext
3
2
2 3
4
4 6 8 10
5
8 8 8 8 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

Buggy Code:
def determine_winner(test_cases):
    results = []
    for n, piles in test_cases:
        valid_piles_count = sum(1 for a in piles if a >= 4)
        if valid_piles_count == 0:
            results.append("Bob")
        elif valid_piles_count % 2 == 0:
            results.append("Alice")
        else:
            results.append("Bob")
    return results

t = int(input())
test_cases = []
for _ in range(t):
    n = int(input())
    piles = list(map(int, input().split()))
    test_cases.append((n, piles))

results = determine_winner(test_cases)
for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
3
2
2 3
3
5 7 11
4
4 6 9 14
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prime Split Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game with $n$ piles of stones, where the $i$-th pile has $a_i$ stones. Players take turns making moves, with Alice going first.

On each move, the player does the following three-step process:

  1. Choose an integer $k$ ($1 \leq k \leq \frac n 2$). Note that the value of $k$ can be different for different moves.    2. Remove $k$ piles of stones.    3. Choose another $k$ piles of stones and split each pile into two piles. The number of stones in each new pile must be a prime number.

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of piles of stones.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the number of stones in the piles.

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

Output: For each test case, output "Alice" (without quotes) if Alice wins and "Bob" (without quotes) otherwise.

You can output each letter in any case (upper or lower). For example, the strings "alIcE", "Alice", and "alice" will all be considered identical.

Examples:
input:
4
2
2 1
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
output:
Bob
Alice
Alice
Bob

Note: In the first test case, there are $2$ piles of stones with $2$ and $1$ stones respectively. Since neither $1$ nor $2$ can be split into two prime numbers, Alice cannot make a move, so Bob wins.

In the second test case, there are $3$ piles of stones with $3$, $5$, and $7$ stones respectively. Alice can choose $k = 1$, remove the pile of $7$ stones, and then split the pile of $5$ stones into two piles of prime numbers of stones, $2$ and $3$. Then, the piles consist of $3$ piles of stones with $3$, $2$, and $3$ stones respectively, leaving Bob with no valid moves, so Alice wins.

In the third test case, there are $4$ piles of stones with $4$, $6$, $8$, and $10$ stones respectively. Alice can choose $k = 2$, removing two piles of $8$ and $10$ stones. She splits the pile of $4$ stones into two piles of prime numbers of stones, $2$ and $2$, and the pile of $6$ stones into two piles of $3$ and $3$ stones. Then, Bob has no valid moves, so Alice wins.

In the fourth test case, there are $5$ piles of stones, each containing $8$ stones. It can be shown that if both players play optimally, Bob will win.

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

public class Codechef {

    static boolean[] isPrime = new boolean[200001];

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

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder result = new StringBuilder();

        int t = Integer.parseInt(br.readLine().trim());
        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine().trim());
            String[] stonesInput = br.readLine().trim().split(" ");
            int[] stones = new int[n];

            for (int i = 0; i < n; i++) {
                stones[i] = Integer.parseInt(stonesInput[i]);
            }

            result.append(findWinner(stones)).append("\n");
        }
        System.out.print(result);
    }

    private static String findWinner(int[] stones) {
        int xorSum = 0;

        for (int stone : stones) {
            xorSum ^= calculateGrundy(stone);
        }

        return xorSum != 0 ? "Alice" : "Bob";
    }

    private static int calculateGrundy(int stones) {
        if (stones <= 1) return 0; // Base case: can't split further

        Set<Integer> reachableGrundyNumbers = new HashSet<>();

        for (int p = 2; p <= stones / 2; p++) {
            int q = stones - p;
            if (isPrime[p] && isPrime[q]) {
                reachableGrundyNumbers.add(calculateGrundy(p) ^ calculateGrundy(q));
            }
        }

        int grundy = 0;
        while (reachableGrundyNumbers.contains(grundy)) {
            grundy++;
        }
        return grundy;
    }

    private static void precomputePrimes() {
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;

        for (int i = 2; i * i <= 200000; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= 200000; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The primary issues in the code are related to the prime number splitting and the Grundy number calculation. Let's consider the following scenarios:

1. **Piles with numbers that cannot be split into two primes**: The code should handle cases where a pile cannot be split into two prime numbers.
2. **Piles with numbers that can be split in multiple ways**: The code should correctly handle cases where a pile can be split into different pairs of prime numbers.
3. **Special cases with small numbers**: Numbers like 1, 2, 3, and 4 should be handled correctly, as they have specific constraints.

Given these considerations, let's construct a test input that includes these scenarios:

```plaintext
3
2
1 2
4
3 5 7 11
6
4 6 8 10 12 14
```

### Explanation:
1. **Test Case 1:**
   - Piles: 1 and 2
   - Neither 1 nor 2 can be split into two prime numbers.
   - Alice cannot make a move, so Bob wins.

2. **Test Case 2:**
   - Piles: 3, 5, 7, and 11
   - All these numbers are prime and can be split into two prime numbers (e.g., 5 can be split into 2 and 3).
   - Alice can make a move, and the game continues. The expected result is Alice winning if she plays optimally.

3. **Test Case 3:**
   - Piles: 4, 6, 8, 10, 12, and 14
   - These numbers can be split into different pairs of prime numbers (e.g., 4 can be split into 2 and 2).
   - The game should be more complex, and the Grundy number calculation should handle these cases correctly.
   - The expected result is Alice winning if she plays optimally.

This test input should help expose any bugs in the code related to prime number splitting and Grundy number calculation.