**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

using namespace std;
// using namespace __gnu_pbds;

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

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;

#define forn(i, n) for(int i = 0; i < n; i++)
#define ford(i, n) for(int i = n - 1; i >= 0; i--)
#define for1(i, n) for(int i = 1; i <= n; i++)
#define forr(i, l, r) for(int i = l; i <= r; i++)
#define forrd(i, l, r) for(int i = r; i >= l; i--)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define len(a) (int)(a).size()

#ifdef SARTHAK_LOCAL
#include "/Users/sarthakkumar/cpp/templates/CompCoding/debug.cpp"
#else
#define debug(...) 69
#endif

template<typename T> bool cmin(T& a, T b) { return b < a ? a = b, 1 : 0; }
template<typename T> bool cmax(T& a, T b) { return b > a ? a = b, 1 : 0; }

template<typename T>
T gcd(T a, T b) {
   while (b) { a %= b; swap(a, b); }
   return a;
}

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 rng(696969);

void Main() {
   int n;
   cin >> n;
   vi a(n);
   forn(i, n) cin >> a[i];

   vi type;
   for(int x: a) {
      type.push_back(x&1);
   }
   sort(rall(type));
   if(type.back() == 1) {
      cout << (len(type)%2 == 0 ? "Bob" : "Alice") << '\n';
   } else {
      cout << (len(type)%2 == 1 ? "Bob" : "Alice") << '\n';
   }

}

#define MULTI // comment for single test case

int main() {
	ios::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
	cout << setprecision(12) << fixed;
#ifdef SARTHAK_LOCAL
   clock_t start = clock();
#endif

	int tt = 1;
   #ifdef MULTI
      cin >> tt;
   #endif

	for(int test_case = 1; test_case <= tt; test_case++) {
		debug(test_case);

		Main();
      // cout << Main() << '\n';
      // cout << (Main() ? "YES" : "NO") << '\n';

      #ifdef SARTHAK_LOCAL
         cerr << "----------------------------------------------------------------------------------------------------------------------------------------" << '\n';
      #endif
	}

#ifdef SARTHAK_LOCAL
   clock_t end = clock();
   cerr << "Total Time: " << (double)(end - start) * 1e3 / CLOCKS_PER_SEC << "ms" << '\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 expose the bug in the given code, we need to focus on the logic used to determine the winner. The code appears to be using the parity of the number of piles and the type of the smallest pile to decide the winner. However, this logic is flawed because it doesn't consider the actual game mechanics and the constraints on the moves.

### Step-by-Step Analysis:
1. **Understanding the Bug**:
   - The code sorts the `type` array in descending order and checks the parity of the length of the array and the smallest element.
   - This approach is incorrect because it doesn't account for the game's rule that a player cannot take a number of stones `y` from a pile with `x` stones if `gcd(x, y) != 1`.

2. **Edge Cases**:
   - **Single Pile with Prime Number of Stones**: If there is only one pile with a prime number of stones, Alice can always win by taking all the stones.
   - **Multiple Piles with Small Numbers**: If there are multiple piles with small numbers, the GCD condition can significantly limit the moves, making the game more complex.

3. **Test Input**:
   - We need a test case where the number of piles and the number of stones in each pile expose the bug in the logic.

### Test Input:
Let's create a test case with multiple piles where the number of stones in each pile is small and the GCD condition plays a crucial role.

```plaintext
1
4
4 6 8 10
```

### Explanation:
- **Piles**: 4 piles with 4, 6, 8, and 10 stones.
- **GCD Condition**:
  - For the pile with 4 stones, Alice can only take 1 or 3 stones (since gcd(4, 2) = 2 and gcd(4, 4) = 4).
  - For the pile with 6 stones, Alice can only take 1, 5 stones (since gcd(6, 2) = 2, gcd(6, 3) = 3, gcd(6, 6) = 6).
  - For the pile with 8 stones, Alice can only take 1, 3, 5, 7 stones (since gcd(8, 2) = 2, gcd(8, 4) = 4, gcd(8, 8) = 8).
  - For the pile with 10 stones, Alice can only take 1, 3, 7, 9 stones (since gcd(10, 2) = 2, gcd(10, 5) = 5, gcd(10, 10) = 10).

This test case is designed to make the game complex and expose the bug in the code's logic. The code's current approach of sorting and checking parity will not correctly determine the winner in this scenario.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import math

def is_prime(n):
    if n < 2:
        return False
    for i in range(2, int(math.sqrt(n)) + 1):
        if n % i == 0:
            return False
    return True

def solve():
    n = int(input())
    a = list(map(int, input().split()))
    primes = 0
    for x in a:
        if is_prime(x):
            primes += 1
    if n == 1 or n % 2 == 1 and primes != n:
        print("Bob")
    else:
        print("Alice")

t = int(input())
for _ in range(t):
    solve()

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases or scenarios where the code might fail. The code has a logical flaw in its win condition determination, particularly in how it handles the number of prime piles and the parity of the number of piles.

### Step-by-Step Analysis:
1. **Prime Piles and Win Condition**:
   - The code checks if the number of prime piles is not equal to the total number of piles and if the total number of piles is odd. If both conditions are true, it concludes that Bob wins.
   - This logic is incorrect because it doesn't consider the actual game rules and the optimal play strategy.

2. **Optimal Play Strategy**:
   - In the game, a player can only take a number of stones such that the GCD with the current number of stones in the pile is 1.
   - This means that if a pile has a prime number of stones, the player can only take 1 stone from it, making it a special case.

3. **Edge Cases**:
   - **Single Pile**: If there is only one pile, the player who starts (Alice) will always win if the pile has more than 1 stone, because they can take all the stones and win immediately.
   - **Multiple Piles with Special Conditions**: If there are multiple piles, the game can become more complex. We need to consider the parity of the number of piles and the nature of the stones in each pile.

### Constructing the Test Input:
To create a test input that exposes the bug, we can use a scenario where the number of piles is odd, and the number of prime piles is not equal to the total number of piles. This should ideally make the code output "Bob" incorrectly.

### Example Test Input:
```plaintext
1
3
4 6 8
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Piles**: 3 (odd)
- **Pile Sizes**: 4, 6, 8 (none of these are prime)

In this case:
- The code will count the number of prime piles (0) and see that it is not equal to the total number of piles (3).
- Since the number of piles is odd, the code will incorrectly conclude that Bob wins.

However, the correct output should be "Alice" because Alice can always make a move that leaves Bob in a losing position. For example, Alice can take 1 stone from the pile of 4, leaving 3 stones (a prime number). Bob then cannot take any stones from this pile without violating the GCD condition.

This test input should 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: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import math
from typing import List

def gcd(a: int, b: int) -> int:
    """Compute the Greatest Common Divisor of a and b."""
    while b != 0:
        a, b = b, a % b
    return a

def primeFactors(n: int) -> List[int]:
    """Return list of prime factors of n, excluding 'n' itself."""
    factors = []
    # Check for the already checked factor
    i = 2
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return factors

def nimWinner(a: List[int]) -> str:
    max_factor = -1
    # Find the maximum unique factor of all elements
    for i in range(len(a)):
        f = set()
        e = primeFactors(a[i])
        for j in e:
            f.add(j)
        if len(f) > max_factor:
            max_factor = len(f)

    # If the number of unique factors is odd, Bob wins; otherwise, Alice wins
    return "Bob" if max_factor % 2 == 1 else "Alice"

# Main function to handle inputs and outputs
def main():
    t = int(input().strip())
    for _ in range(t):
        n = int(input().strip())
        a = list(map(int, input().split()))
        print(nimWinner(a))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the logic that determines the winner based on the number of unique prime factors. The buggy code incorrectly assumes that the player who has an odd number of unique prime factors will lose, which is not always correct.

Let's break down the problem and identify a test case that can expose the bug:

1. **Understanding the Bug**:
   - The code incorrectly uses the number of unique prime factors to determine the winner.
   - The correct approach should consider the Nim game strategy, where the XOR of the pile sizes determines the winner.

2. **Creating a Test Case**:
   - We need a test case where the number of unique prime factors does not correctly predict the winner.
   - A good test case would be one where the Nim sum (XOR of pile sizes) determines the winner, but the number of unique prime factors does not.

3. **Example Test Case**:
   - Consider a test case with piles that have sizes that are powers of prime numbers, but the Nim sum (XOR) results in a non-zero value, indicating that Alice should win.
   - For example, piles of sizes 3, 5, and 7. The Nim sum is 3 XOR 5 XOR 7 = 1, which means Alice wins.

4. **Test Input**:
   - We will use a single test case with these pile sizes.

Here is the test input:

```plaintext
1
3
3 5 7
```

This test input should expose the bug in the provided code, as the correct output should be "Alice" based on the Nim game strategy, but the buggy code will output "Bob" based on the incorrect logic of counting unique prime factors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    input()
    print("Alice" if n % 2 == 0 else "Bob")

**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
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import java.util.*;

public class Codechef {
 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int t = sc.nextInt();
  while (t-- > 0) {
   int n = sc.nextInt(), a[] = new int[n];
   HashSet<Integer> hs = new HashSet<>();
   for (int i = 0; i < n; i++) {
    a[i] = sc.nextInt();
    if (hs.contains(a[i])) {
     hs.remove(a[i]);
    } else {
     hs.add(a[i]);
    }
   }
   long sum = 0;
   for (int x : hs) {
    sum += x;
   }

   if (sum % 2 != 0) {
    if (!hs.contains(3))
     System.out.println("Alice");
    else
     System.out.println("Bob");
   } else {
    System.out.println("Bob");
   }
  }
 }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//stack=new Stack<Integer>();
		//set=new HashSet<List<Integer>>();
		//f(0, new int[3]);System.out.println(set.size());
		//p1=false;
		ss=new int[(int) (1e7+1)];Arrays.fill(ss, 1);
		size=new int[(int) (1e7+1)];size[1]=1;
		List<Integer>list=new ArrayList<Integer>();
		for (int i = 2; i < ss.length; i++) {
			if (ss[i]==1) {
				int j=2;
				while (i*j<ss.length) {
					ss[i*j]=0;j++;
				}
			}
			size[i]=size[i-1]+ss[i];
		}

		  int t=sc.nextint(); while (t-->0) {

		  sovle2(); }

		  out.close();

	}
	static long[] ram,v;
	static int []size,ss;

	static void sovle2() {
		int n=sc.nextint();
		int []arr=new int[n];
		long ans=0;
		for (int i = 0; i < arr.length; i++) {
			arr[i]=sc.nextint();
			if (arr[i]%2!=0) {
				ans+=size[arr[i]];
				//System.out.println(size[arr[i]]);
			}

		}
		//System.out.println(ans);
		if (ans%2==0) {
			out.println("Bob");
		}else {
			out.println("Alice");
		}
	}

	static boolean p1;
	static int[][] va,l,r;
	static int n,m;
	static void sovle() {
		int l=sc.nextint();
		int r=sc.nextint();
		int L=sc.nextint();
		int R=sc.nextint();
		if (R<l||r<L) {
			out.println(1);
		}else if((l>=L&&r<=R)||(l<=L&&r>=R)){

			if (l==L&&r==R) {
				//System.out.println("aaa");
				out.println(r-l);
			}else if(l!=L&&R!=r){
				out.println(Math.min(r-l, R-L)+2);
			}else {
				out.println(Math.min(r-l, R-L)+1);
			}

		}else {

			out.println(Math.min(r-L, R-l)+2);
		}
	}

	static long  pow(long x,long a) {
		if (a==0) {
			return 1;
		}else {
			long c=pow(x, a/2);
			if (a%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*x%mod;
			}
		}
	}
	static long mod = (long) (1e9+7);

	static void sovle1() {

	}

	static long []c;
	static int lowbit(int x) {
		return x&-x;
	}
	static void add(int x,long v) {
		while (x<c.length) {
			//System.out.println(x);
			c[x]^=v;
			x+=lowbit(x);
		}
	}
	static long sum(int x) {
		long ans=0;
		//System.out.println(x+"dasdsafadf");
		while (x>0) {
			//System.out.println(x);
			ans^=c[x];
			x-=lowbit(x);
		}
		return ans;
	}
	static Map<Integer, List<Integer>>map;
	static void mapinit(int chu,int n) {
		map=new HashMap<Integer, List<Integer>>();
		for (int i = chu; i < n+chu; i++) {
			map.put(i, new ArrayList<Integer>());
		}
	}
}

class kattio extends PrintWriter {

	BufferedReader r;
	StringTokenizer st;

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

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

	public kattio(String in, String out) throws IOException {
		super(out);
		r = new BufferedReader(new FileReader(in));
	}

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

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

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

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

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios where the code might fail. The key points to consider are:

1. **Piles with a single stone**: These should be considered as positions where the player cannot make a move.
2. **Piles with composite numbers**: The code might incorrectly handle these piles if it doesn't correctly account for the GCD condition.
3. **Piles with prime numbers**: The code should handle these correctly, but we should still test them to ensure the logic is sound.
4. **Piles with numbers that have multiple divisors**: This can test the GCD condition more rigorously.

Let's construct a test input that includes these scenarios:

```plaintext
3
3
1 2 3
4
4 6 8 9
5
1 4 9 16 25
```

### Explanation:

1. **First Test Case**:
   - Piles: 1, 2, 3
   - The first pile has 1 stone, so the player cannot make a move from it.
   - The second pile has 2 stones, and the only valid move is to take 1 stone.
   - The third pile has 3 stones, and the only valid moves are to take 1 or 2 stones.
   - Alice will lose because the first pile is a losing position.

2. **Second Test Case**:
   - Piles: 4, 6, 8, 9
   - The first pile has 4 stones, and the valid moves are to take 1 stone.
   - The second pile has 6 stones, and the valid moves are to take 1, 5 stones.
   - The third pile has 8 stones, and the valid moves are to take 1, 7 stones.
   - The fourth pile has 9 stones, and the valid moves are to take 1, 2, 4, 8 stones.
   - Alice can make a move from the first pile, but the GCD condition might not be handled correctly, leading to a potential bug.

3. **Third Test Case**:
   - Piles: 1, 4, 9, 16, 25
   - The first pile has 1 stone, so the player cannot make a move from it.
   - The second pile has 4 stones, and the valid moves are to take 1 stone.
   - The third pile has 9 stones, and the valid moves are to take 1, 8 stones.
   - The fourth pile has 16 stones, and the valid moves are to take 1, 15 stones.
   - The fifth pile has 25 stones, and the valid moves are to take 1, 24 stones.
   - Alice will lose because the first pile is a losing position, and the other piles have valid moves that might not be handled correctly.

This test input should help expose any bugs in the provided code, especially those related to the GCD condition and the handling of piles with a single stone.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

using namespace std;

#define fastio std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define nl "\n"
typedef long long ll;
#define all(a) (a).begin(), (a).end()
#define show(C) for (auto J: C) cout << J << " ";cout << endl;
const ll MAX = 1e7+3;

vector<ll>grundy(MAX, -1);
vector<ll>prime(MAX,-1);

void sieve()
{
    grundy[0] = 0;
    grundy[1] = 1;
    prime[1] = 1;
    for(ll i=2; i<MAX; i+=2){
        grundy[i] = 0;
        prime[i] = 2;
    }

    ll cnt = 2;
    for(ll i=3; i<MAX; i+=2)
    {
        if(prime[i]==-1)
        {
            prime[i] = i;
            grundy[i] = cnt++;
            for(ll j=i*i; j<MAX; j+=i)
            {
                prime[j] = i;
                if(j%2)
                grundy[j] = grundy[prime[j]];
            }
        }
    }
}

int main()
{
    fastio
    ll t; cin>>t;
    while(t--)
    {
        int n; cin>>n;
        vector<int>v(n);
        for(int i=0; i<n; i++) cin>>v[i];
        ll ans = 0;
        for(auto x : v) ans ^= grundy[x];
        if(ans){
            cout<<"Bob"<<nl;
        }else cout<<"Alice"<<nl;
    }

    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
1
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
#ifndef VLSMB_VS
#include <bits/stdc++.h>
#else
#include <iostream>
#include <iomanip>
#include <set>
#include <algorithm>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <cstdio>
#include <cmath>
#endif

#define int long long
#define pii pair<int,int>
#define endl '\n'
#define veci vector<int>
#define vecpi vector<pii >
#define mpi map<int,int>
#define mpc map<char,int>
#define mppi map<pii,int>
#define inf 0x7fffffffffffffff
#define ninf (long long)(0x8000000000000000)
#define eps 1e-8
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define seti set<int>
#define mulseti multiset<int>
#define setpi set<pii >
#define priqli priority_queue<int,veci,less<int> >
#define priqgi priority_queue<int,veci,greater<int> >
#define priqgpi priority_queue<pii,vecpi,greater<pii > >
#define ull unsigned long long
using namespace std;
//const int N = 1000000;
void solve();
void init();
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
int extgcd(int a,int b,int&x,int&y)
{
    if(!b)
    {
        x=1,y=0;
        return a;
    }
    int d=extgcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
int remod(int a,int m)
{
    int x,y;
    extgcd(a,m,x,y);
    return (x%m+m)%m;
}
int mul(int a,int b,int m)
{
    int ans=0;
    while(b)
    {
        if(b&1)ans=(ans+a)%m;
        a=(a+a)%m;
        b>>=1;
    }
    return ans;
}
int extcrt(veci&a,veci&m)
{
    int x=0,y=0;
    int m1=m[0],a1=a[0];
    int ans=0;
    for(int i=1;i<min(a.size(),m.size());i++)
    {
        int a2=a[i],m2=m[i];
        int a=m1,b=m2,c=(a2-a1%m2+m2)%m2;
        int d=extgcd(a,b,x,y);
        if(c%d!=0)return -1;
        x=mul(x,c/d,b/d);
        ans=a1+x*m1;
        m1=m2/d*m1;
        ans=(ans%m1+m1)%m1;
        a1=ans;
    }
    return ans;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    init();
    cin>>t;
    while(t--)solve();
    cout.flush();
    return 0;
}
const int N=2e7;
bool notpri[N];
void solve()
{
    int n;
    cin>>n;
    veci a(n);
    int ans=0;
    for(auto&x:a)cin>>x;
    for(const auto&x:a)ans^=x;
    cout<<(notpri[ans]?"Bob":"Alice")<<endl;
}
void init()
{
    notpri[1]=true;
    for(int i=2;i<N;i++)
    {
        if(!notpri[i])
        {
            for(int j=2*i;j<N;j+=i)notpri[j]=true;
        }
    }
}

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

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

Think step by step.

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

1. **GCD Condition**: The condition that the number of stones taken must have a GCD of 1 with the current number of stones in the pile.
2. **XOR Calculation**: The code uses the XOR of the pile sizes to determine the winner, which might not always be correct due to the GCD condition.

Given these points, let's construct a test case where the XOR of the pile sizes might lead to an incorrect conclusion due to the GCD condition.

### Step-by-Step Reasoning:

1. **Small Number of Piles**: Start with a small number of piles to keep the input manageable.
2. **Pile Sizes with Specific GCD Properties**: Choose pile sizes such that the GCD condition is critical for determining the winner.

### Example Test Case:

- **Number of Piles**: 2
- **Pile Sizes**: 3 and 4

### Explanation:

- **Pile 1**: 3 stones
  - Possible moves: 1, 2 (both have GCD 1 with 3)
- **Pile 2**: 4 stones
  - Possible moves: 1, 3 (both have GCD 1 with 4)

In this scenario, Alice can take 1 stone from either pile, and the game will proceed. However, the XOR of 3 and 4 is 7, which is not a prime number. The code might incorrectly determine the winner based on this XOR value.

### Test Input:

```plaintext
1
2
3 4
```

This test case is small and valid, and it should expose the bug in the faulty solution by highlighting the importance of the GCD condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

def is_prime(num):
    if num == 1:
        return False

    for i in range(2, int(num**0.5)+1):
        if num%i == 0:
            return False

    return True

t = int(input())

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

    xor = 0

    for i in a:
        xor ^= i

    print("Bob" if xor == 1 or xor%2 == 0 else "Alice")

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the code's logic fails. The buggy code uses the XOR of all pile sizes to determine the winner, but this approach is incorrect because it does not account for the specific game rules involving the greatest common divisor (GCD).

### Key Points to Consider:
1. **GCD Condition**: The player cannot take a number of stones `y` from a pile with `x` stones if `gcd(x, y) != 1`.
2. **Optimal Play**: Both players play optimally, so the strategy should consider the GCD condition.
3. **XOR Calculation**: The buggy code uses XOR to determine the winner, which is not appropriate for this game.

### Edge Case Analysis:
- **Single Pile with Prime Number**: If there is only one pile with a prime number of stones, Alice can always win by taking all the stones.
- **Multiple Piles with Small Numbers**: If there are multiple piles with small numbers, the GCD condition can make it difficult to win.
- **Piles with Composite Numbers**: Composite numbers can have multiple divisors, making it challenging to determine the winner using XOR.

### Challenging Test Input:
Let's consider a test case with multiple piles where the GCD condition plays a significant role:

```plaintext
1
5
4 6 8 10 12
```

### Explanation:
- **Pile Sizes**: 4, 6, 8, 10, 12
- **GCD Condition**: For each pile, the player cannot take a number of stones that is not coprime with the current number of stones in the pile.
- **Optimal Play**: Alice will need to carefully choose her moves to ensure she can always make a valid move, considering the GCD condition.

### Why This Test Case is Challenging:
- The pile sizes are all even and composite, which means there are multiple divisors to consider.
- The GCD condition will significantly limit the possible moves, making it difficult for the buggy code's XOR logic to correctly determine the winner.

This test input should expose the bug in the buggy code and demonstrate that the XOR approach is incorrect for this game.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import sys
input = sys.stdin.buffer.readline

from collections import Counter

def gcd(x, y):
    """greatest common divisor of x and y"""
    while y:
        x, y = y, x % y
    return x

def memodict(f):
    """memoization decorator for a function taking a single argument"""
    class memodict(dict):
        def __missing__(self, key):
            ret = self[key] = f(key)
            return ret

    return memodict().__getitem__

def pollard_rho(n):
    """returns a random factor of n"""
    if n & 1 == 0:
        return 2
    if n % 3 == 0:
        return 3

    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            prev = p
            p = (p * p) % n
            if p == 1:
                return gcd(prev - 1, n)
            if p == n - 1:
                break
        else:
            for i in range(2, n):
                x, y = i, (i * i + 1) % n
                f = gcd(abs(x - y), n)
                while f == 1:
                    x, y = (x * x + 1) % n, (y * y + 1) % n
                    y = (y * y + 1) % n
                    f = gcd(abs(x - y), n)
                if f != n:
                    return f
    return n

@memodict
def prime_factors(n):
    """returns a Counter of the prime factorization of n"""
    if n <= 1:
        return Counter()
    f = pollard_rho(n)
    return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)

def distinct_factors(n):
    """returns a list of all distinct factors of n"""
    factors = [1]
    for p, exp in prime_factors(n).items():
        factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
    return factors

def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [], []
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n // i)
    if small[-1] == large[-1]:
        large.pop()
    large.reverse()
    small.extend(large)
    return small

def process(A):
    A = [x for x in A if x > 0]
    d = {}
    for x in A:
        pf = prime_factors(x)
        if len(pf)==0:
            p = 1
        else:
            p = min(pf)
        if p not in d:
            d[p] = 0
        d[p]+=1
    L = [p for p in d if d[p] % 2==1 and p != 2]
    if len(L)==0:
        print("Bob")
    elif len(L)==1:
        print("Alice")
    elif len(L) % 2==0:
        print("Alice")
    else:
        print("Bob")
t = int(input())
for i in range(t):
    n = int(input())
    A = [int(x) for x in input().split()]
    process(A)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The key points to consider are:

1. **Small Piles with Special Properties**: Piles with small numbers that have specific properties, such as being prime or having a small set of prime factors.
2. **Piles with Large Numbers**: Piles with large numbers to test the factorization and GCD calculations.
3. **Piles with Specific GCD Constraints**: Piles where the GCD constraints might be particularly restrictive, such as piles with even numbers or numbers with specific prime factors.

Given these considerations, a small-scale test input that could expose the bug might look like this:

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

### Explanation:
1. **Test Case 1:**
   - Piles: 1, 2, 3
   - Alice can only take from piles with 1 or 3 stones. If Alice takes 1 stone from the pile with 3 stones, Bob can take the remaining 2 stones. If Alice takes 1 stone from the pile with 2 stones, Bob can take the remaining 1 stone. This should result in Bob winning.

2. **Test Case 2:**
   - Piles: 2, 4, 6, 8
   - All piles are even numbers. Alice can only take 1 stone from each pile, and Bob can then take the remaining stones. This should result in Alice winning because she can force Bob into a position where he cannot make a valid move.

3. **Test Case 3:**
   - Piles: 1, 3, 5, 7, 9
   - All piles are odd numbers. Alice can take 1 stone from any pile, and Bob can then take the remaining stones. This should result in Bob winning because Alice cannot force Bob into a position where he cannot make a valid move.

This test input is designed to cover a variety of edge cases and should help expose any bugs in the algorithm, particularly in the factorization and GCD calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

public class Main {
          static HashSet<Integer> primeSet;
          static int[] firstPrimeArr;
          static int maxNum = (int) (1e7);

          static void findAllprimes(int maxNum) {
                    boolean[] arr = new boolean[maxNum + 1];
                    for (long i = 2; i <= maxNum; i++) {
                              if (arr[(int)(i)])
                                        continue;
                              for (long j = i * i; j <= maxNum; j += i) {
                                        arr[(int)(j)] = true;
                              }
                    }
                    for (int i = 2; i <= maxNum; i++) {
                              if (!arr[i]) {
                                        primeSet.add(i);
                              }
                    }
          }

          public static void fillFirstPrimeArr(int maxNum) {
                    firstPrimeArr = new int[maxNum + 1];
                    for (int i = 1; i <= maxNum; i++)
                              firstPrimeArr[i] = -1;
                    for (int i = 2; i <= maxNum; i++) {
                              if (firstPrimeArr[i] != -1)
                                        continue;
                              for (int j = i; j <= maxNum; j += i) {
                                        if (firstPrimeArr[j] == -1)
                                                  firstPrimeArr[j] = i;
                              }
                    }
          }

          public static String fun(int n, int[] arr) {
                    int xor = 0;
                    for(int x: arr) xor ^= x;
                    if(xor == 0 || xor == 2) return "Bob";
                    if(xor == 1) return "Bob";
                    if(primeSet.contains(xor)) {
                              return "Alice";
                    }
                    int myFirstPrimeDivisor = firstPrimeArr[xor];
                    if(myFirstPrimeDivisor == 2) return "Bob";
                    return "Alice";
          }

          public static void main(String[] args) throws IOException {
                    primeSet = new HashSet<>();
                    fillFirstPrimeArr(maxNum);
                    findAllprimes(maxNum);
                    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                    BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
                    StringTokenizer st = new StringTokenizer(br.readLine());
                    int t = Integer.parseInt(st.nextToken());

                    while (t-- > 0) {
                              st = new StringTokenizer(br.readLine());
                              int n = Integer.parseInt(st.nextToken());
                              st = new StringTokenizer(br.readLine());
                              int[] arr = new int[n];
                              for (int i = 0; i < n; i++) {
                                        arr[i] = Integer.parseInt(st.nextToken());
                              }
                              bw.write(fun(n, arr) + "\n");
                    }

                    br.close();
                    bw.flush();
                    bw.close();
          }
}

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

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

Think step by step.

Response:
```plaintext
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
```

This test input is a small-scale, complete, and valid set of test cases that can help expose the bug in the buggy code. The input includes:

1. A test case where Alice should lose (`3 3 2 9`).
2. A test case where Alice should win (`3 3 6 1`).
3. A test case where Alice should lose (`1 2 3 4 5`).

The first and third test cases are expected to return "Bob" based on the problem's constraints and the provided example. The second test case is expected to return "Alice". This set of test cases will help identify issues in the logic of the buggy code, particularly in the function `fun` where the XOR and prime checks are performed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import java.util.*;

public class Codechef {
 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int t = sc.nextInt();
  while (t-- > 0) {
   int n = sc.nextInt(), a[] = new int[n];
   int r[] = new int[10000001];
   for (int i = 0; i < n; i++) {
    a[i] = sc.nextInt();
    r[a[i]] = (r[a[i]] + 1) % 2;
   }
   long sum = 0;
   for (int i = 1; i < 10000001; i++) {
    if (r[i] == 1) {
     sum += i;
    }
   }
   if (sum % 2 != 0) {
    if (r[3] == 0)
     System.out.println("Alice");
    else
     System.out.println("Bob");
   } else {
    System.out.println("Bob");
   }
  }
 }
}

**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 buggy code, we need to focus on the logic that determines the winner. The buggy code has a few key issues:

1. It incorrectly uses a fixed array `r` of size 10,000,001 to track the number of piles with each number of stones.
2. It uses the sum of indices where `r[i] == 1` to determine the winner, which is not a valid strategy for this game.
3. It has a special condition for `r[3]` that doesn't make sense in the context of the game rules.

To create a test input that exposes these issues, we can use a test case where the piles have values that should lead to a specific winner, but the buggy code will incorrectly determine the result.

Let's consider a test case where the piles are such that the correct winner should be Bob, but the buggy code incorrectly outputs Alice due to its flawed logic.

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

```plaintext
1
3
3 3 6
```

Explanation:
- The test case has 3 piles with 3, 3, and 6 stones.
- According to the game rules, Alice cannot make a valid move from the pile with 6 stones because the only valid move would be to take 1 stone, but the GCD of 6 and 1 is 1, which is not allowed.
- Bob will then win because Alice cannot make a valid move.

However, the buggy code will incorrectly output "Alice" because it uses the sum of indices where `r[i] == 1` and the special condition for `r[3]` to determine the winner, which is not a valid strategy for this game.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

const int N = 1e7 + 10, M = 1e7;
int t, n, primes[N], sg[N];

int main() {
    sg[0] = 0, sg[1] = 1, sg[2] = 0;
    int cnt = 0, tot = 1;
    for (int i = 3; i <= M; i += 2 ) {
        if (!sg[i]) {
            sg[i] = ++ tot;
            primes[cnt ++ ] = i;
            for (int j = 0; j < cnt && primes[j] <= M / i; j ++ ) {
                sg[primes[j] * i] = sg[primes[j]];
                if (i % primes[j] == 0) break;
            }
        }
    }
    cin >> t;
    while (t -- ) {
        cin >> n;
        int res = 0;
        for (int i = 0; i < n; i ++ ) {
            int x;
            cin >> x;
            res ^= sg[x];
        }
        if (res) puts("Alice");
        else puts("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 generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases or scenarios that might cause the code to fail. The buggy code uses a precomputed Sprague-Grundy (SG) function, which might have issues with certain edge cases or specific sequences of numbers.

### Key Points to Consider:
1. **Prime Numbers and Composites**: The SG function is precomputed in a way that might not handle composite numbers correctly, especially those with multiple prime factors.
2. **Small Values**: Small values like 1, 2, 3, and 4 are critical as they have specific properties in the game.
3. **Large Values**: Large values that are not prime and have multiple prime factors can also be challenging.
4. **Edge Cases**: Values that are powers of primes, or have specific properties that might not be handled correctly by the precomputed SG function.

### Test Input:
We will create a test case that includes a mix of small and large values, including some that are powers of primes and some that are composite with multiple prime factors.

```plaintext
1
5
4 9 15 16 25
```

### Explanation:
- **4**: A composite number that is a power of a prime (2^2).
- **9**: A composite number that is a power of a prime (3^2).
- **15**: A composite number with multiple prime factors (3 * 5).
- **16**: A composite number that is a power of a prime (2^4).
- **25**: A composite number that is a power of a prime (5^2).

This test case includes a variety of numbers that might expose issues in the precomputed SG function, especially how it handles composite numbers with multiple prime factors and powers of primes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

const int N = 1E3 + 5;
int sg[N];

void init() {
	for (int i = 1 ; i < N ; ++i) {
		set<int> P;
		for (int j = 1 ; j <= i ; ++j) {
			if (__gcd(i , j) == 1) {
				P.insert(sg[i - j]);
			}
		}
		while (P.count(sg[i])) sg[i] += 1;
	}
}

void solve() {

	int n;
	cin >> n;

	int ans = 0;
	for (int i = 1 ; i <= n ; ++i) {
		int x;
		cin >> x;
		ans ^= sg[x];
	}

	if (ans) {
		cout << "Alice\n";
	} else {
		cout << "Bob\n";
	}

}

int main() {

	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	init();

	int t = 1;
	cin >> t;

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

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The buggy code initializes the Sprague-Grundy (SG) function for a limited range (up to 1000), but the problem constraints allow for values up to \(10^7\). This means that the code will not correctly handle inputs where the number of stones in any pile is greater than 1000.

Let's construct a test case that includes a pile with a number of stones greater than 1000. This will ensure that the code's SG function, which is only initialized up to 1000, will not have the correct values for these larger numbers.

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

```plaintext
1
5
1001 1002 1003 1004 1005
```

Explanation:
- We have one test case.
- The number of piles is 5.
- Each pile contains a number of stones greater than 1000 (specifically, 1001, 1002, 1003, 1004, and 1005).

This input will force the code to use the SG values for numbers that are not initialized correctly, leading to incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import math
from functools import lru_cache
import sys

@lru_cache(None)
def gcd(a, b):
    return math.gcd(a, b)

@lru_cache(None)
def valid_moves(x):
    moves = []
    for y in range(1, x + 1):
        if gcd(x, y) == 1:
            moves.append(y)
    return moves

@lru_cache(None)
def calculate_grundy(x):
    if x == 0:
        return 0

    grundy_set = set()
    for move in valid_moves(x):
        grundy_set.add(calculate_grundy(x - move))

    grundy = 0
    while grundy in grundy_set:
        grundy += 1

    return grundy

def determine_winner(piles):
    xor_sum = 0
    for pile in piles:
        xor_sum ^= calculate_grundy(pile)

    return "Alice" if xor_sum != 0 else "Bob"

for tc in range(int(input())):
    n = int(input())
    stones = [int(x) for x in sys.stdin.readline().split()]
    print(determine_winner(stones))

**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
2
1
1
3
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

def compute_dp(N):
    dp = [0] * N

    for i in range(1, N):
        moves = set()
        for j in range(1, i + 1):
            if gcd(i, j) == 1:
                moves.add(dp[i - j])

        mex = 0
        while mex in moves:
            mex += 1
        dp[i] = mex

    return dp

def find_winner(dp, piles):
    xor_sum = 0
    for pile in piles:
        xor_sum ^= dp[pile]
    return "Alice" if xor_sum != 0 else "Bob"

t = int(input())
test_cases = []

max_value = 0
for _ in range(t):
    n = int(input())
    piles = list(map(int, input().split()))
    max_value = max(max_value, max(piles))
    test_cases.append(piles)

dp = compute_dp(max_value + 1)

for piles in test_cases:
    print(find_winner(dp, 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
3
1
1
2
2 3
3
6 10 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

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

public class E {
    String filename = null;
    InputReader sc;

    int MAX = (int)1e7 + 2;
    int[] nb = new int[MAX + 1];
    void precompute() {
        nb[0] = 0;
        nb[1] = 1;

        int mex = 2;
        for (int i = 3; i < MAX; i += 2) {
            if (nb[i] == 0) {
                nb[i] = mex;
                mex++;
            }

            for (long j = (long)i * i; j < MAX; j += i) {
                int jj = (int)j;
                if (nb[jj] == 0) {
                    nb[jj] = nb[i];
                }
            }
        }
//        for (int i = 0; i < 10; i++) System.out.println(i + " -> " + nb[i]);
    }

    void solve() {
        int n = sc.nextInt();
        int[] a = sc.nextArray(n);

        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum ^= nb[a[i]];
        }

        if (sum == 0) {
            System.out.println("Bob");
        } else {
            System.out.println("Alice");
        }
    }

    public void run() throws FileNotFoundException {
        if (filename == null) {
            sc = new InputReader(System.in);
        } else {
            sc = new InputReader(new FileInputStream(filename));
        }

        int nTests = sc.nextInt();
        precompute();
        for (int test = 0; test < nTests; test++) {
            solve();
        }
    }

    public static void main(String[] args) {
        E sol = new E();
        try {
            sol.run();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

        public float nextFloat() {
            return Float.parseFloat(next());
        }

        public double nextDouble() {
            return Float.parseFloat(next());
        }

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

        public int[] nextArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the precomputed values in the `nb` array might be incorrect or not handle the game rules correctly.

### Analysis:
1. **Precomputation Bug**: The precomputation logic in the `precompute` method seems to be incorrect. It only initializes values for odd indices and then uses a sieve-like approach to fill in the values for multiples of these indices. This approach might not correctly handle all values, especially for even numbers and numbers with more complex prime factorizations.

2. **Game Rules**: The game rules state that a player cannot take a number of stones `y` from a pile with `x` stones if `gcd(x, y) != 1`. This means that for a pile with an even number of stones, the player cannot take an even number of stones, and for a pile with a prime number of stones, the player can only take 1 stone.

### Challenging Test Input:
To expose the bug, we can use a test case where the precomputed values are likely to be incorrect. A good candidate is a test case with a small number of piles, each containing a number of stones that are either even or have a specific structure that the precomputation logic might not handle correctly.

### Test Input:
```plaintext
1
3
4 6 8
```

### Explanation:
- **Pile 1 (4 stones)**: The player can only take 1 stone (since `gcd(4, 2) != 1` and `gcd(4, 4) != 1`).
- **Pile 2 (6 stones)**: The player can only take 1 stone (since `gcd(6, 2) != 1`, `gcd(6, 3) != 1`, and `gcd(6, 6) != 1`).
- **Pile 3 (8 stones)**: The player can only take 1 stone (since `gcd(8, 2) != 1`, `gcd(8, 4) != 1`, and `gcd(8, 8) != 1`).

This test case is designed to highlight the issue with the precomputation logic, which might not correctly handle even numbers or numbers with specific prime factorizations. If the precomputed values are incorrect, the XOR sum used in the `solve` method will also be incorrect, leading to the wrong output.

### Expected Output:
- The correct output for this test case should be `Bob` because each pile can only be reduced by 1 stone, and since Alice goes first, Bob will always have a valid move until the last pile, where Alice will be forced to take the last stone and lose. However, the buggy code might incorrectly determine the winner due to the flawed precomputation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Not a Nim Problem
time_limit_ms: 2000
memory_limit_mb: 512
Description: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.

On their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:

  * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$.

The player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.

Determine who will win.

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

Each test case consists of two lines:

  * the first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output Alice if Alice wins, or Bob if Bob wins.

Examples:
input:
3
3
3 2 9
4
3 3 6 1
5
1 2 3 4 5
output:
Bob
Alice
Bob

Note:

Buggy Code:
import java.util.*;

// sprague grundy theorem
// if xor of grundy values of all elements is not zero then alice (first chance) wins.
// grundy value of 1 is 1.
// grundy value of a number is MEX (Minimum Excludant or smallest non-negative integer not present) of greedy values of those elements where we could reach from current number (in this case, those numbers with which gcd==1 && obviously smaller number than current number because we cannot remove number of piles more than its remaining existing quantity).
// grundy value of 1->1 , 2->0 (mex=2 but gcd(2,mex)!=1), 3->2, 5->3,......so on.
// grundy values of even numbers is 0.

public class Codechef {
 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int c = 2;
  int pre[] = new int[10000001];
  pre[1] = 1;
  for (int i = 3; i < 10000001; i += 2) {// pre computing grundy values of all elements
   if (pre[i] == 0) {
    pre[i] = c;
    for (int j = i * i; j < 10000001 && j > 0; j += i) {
     if (pre[j] == 0) {
      pre[j] = c;
     }
    }
    c++;
   }
  }
  int t = sc.nextInt();
  while (t-- > 0) {
   int n = sc.nextInt(), a[] = new int[n], b[] = new int[n];
   int xor = 0;
   for (int i = 0; i < n; i++) {
    a[i] = sc.nextInt();
    b[i] = pre[a[i]];
    xor ^= b[i];
   }
   if (xor != 0) {
    System.out.println("Alice");
   } else {
    System.out.println("Bob");
   }
  }
 }
}

**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
1
10000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

Buggy Code:
#include <bits/stdc++.h>
#define pb push_back
#define F first
#define S second
#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
typedef long long ll;
typedef long double ld;
using namespace std;

const int MAX_N = 5e5 + 10;
const int LOG = 25;

vector<int> G[MAX_N];
int par[LOG][MAX_N];
int dep[MAX_N];
int seen[MAX_N];

int getDist(int u, int v) {
    int ret = 0;

    if (dep[u] > dep[v]) swap(u, v);

    int diff = dep[v] - dep[u];
    for (int i = 0; i < LOG; i++) {
        if (diff & (1 << i)) {
            v = par[i][v];
            ret += (1 << i);
        }
    }

    if (u == v) return ret;

    for (int i = LOG - 1; i >= 0; i--) {
        if (par[i][v] == par[i][u]) continue;
        v = par[i][v];
        u = par[i][u];

        ret += (1 << (i + 1));
    }

    ret += 2;
    return ret;
}

int bfs2(int i, vector<int> st) {
    queue<pair<int, int>> bat;
    for (auto i : st) bat.push({i, 0});
    while (!bat.empty()) {
        auto v = bat.front();
        bat.pop();

        if (G[v.F].size() < 3) {
            return v.S;
        }

        if (seen[v.F] == i) continue;
        seen[v.F] = i;

        for (auto i : G[v.F]) {
            if (seen[v.F] != i)
                bat.push({i, v.S + 1});
        }
    }

    return -1000;
}

void solve(int iter, int iter_count)
{
    int n;
    cin >> n;

    for (int i = 0; i < n + 10; i++) {
        G[i].clear();
        dep[i] = 0;
        seen[i] = 0;
        for (int j = 0; j < LOG; j++) par[j][i] = 0;
    }

    vector<int> current_mid;
    current_mid.pb(1);
    int current_d = 1;
    bool fail = false;

    cout << 1 << ' ';
    for (int i = 2; i <= n; i++) {
        int p;
        cin >> p;

        G[p].pb(i);
        G[i].pb(p);

        dep[i] = dep[p] + 1;

        if (G[p].size() > 3)
            fail = true;

        if (fail) {
            cout << -1 << ' ';
            continue;
        }

        par[0][i] = p;
        for (int j = 1; j < LOG; j++)
            par[j][i] = par[j - 1][par[j - 1][i]];

        vector<int> finals;
        for (auto j : current_mid) {
            int dist = getDist(i, j) + 1;
            if (dist <= current_d) finals.pb(j);
        }

        if (finals.size() == 0) {
            current_d++;
            finals.pb(current_mid[0]);
            for (auto j : G[current_mid[0]]) {
                int dist = getDist(i, j) + 1;
                if (dist == current_d - 1) finals.pb(j);
            }
        }
        current_mid = finals;

        cout << current_d + bfs2(i, current_mid) << ' ';
    }

    cout << '\n';
}

int main()
{
    fast;
    int iter_count = 1;
    cin >> iter_count;
    for (int iter = 0; iter < iter_count; iter++)
    {
        solve(iter, iter_count);
    }
    return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `3 1 1`
   - **Output:** `1 2 2`
   - **Description:** This is a simple case where the tree is a full binary tree of depth 2 after adding the third vertex. This test case ensures that the basic functionality is working.

2. **Test Case 2:**
   - **Input:** `6 1 2 3 4 5`
   - **Output:** `1 2 2 3 3 4`
   - **Description:** This test case checks the incremental construction of the tree and ensures that the binary depth is correctly calculated at each step. The tree is a full binary tree of depth 4 after adding the sixth vertex.

3. **Test Case 3:**
   - **Input:** `10 1 1 2 1 4 2 4 5 8 8`
   - **Output:** `1 2 2 3 3 4 4 4 5 5`
   - **Description:** This test case is designed to expose issues with the code's handling of more complex tree structures. The tree is a full binary tree of depth 5 after adding the tenth vertex. The last vertex (10) is added to a leaf node (8), which should not cause the binary depth to change.

These test cases are small but cover a range of scenarios, including simple and complex tree structures, to help identify potential bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

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

using namespace std;

const int N = 510000;
const int LOGN = 20;

int jmp[N][LOGN], dep[N], d[N], p[N], sub[N];
vector<pair<int, int>> cent[N];
multiset<int> dists[N];
bool bad, used[N];
vector<int> g[N];

void dfs(int v, int p = -1) {
	dep[v] = (p == -1 ? 0 : dep[p] + 1);
	jmp[v][0] = (p == -1 ? v : p);
	for (int k = 1; k < LOGN; k++)
		jmp[v][k] = jmp[jmp[v][k - 1]][k - 1];
	for (int u : g[v]) if (u != p)
		dfs(u, v);
}

int lca(int u, int v) {
	if (dep[u] < dep[v])
		swap(u, v);
	for (int k = LOGN - 1; k >= 0; k--)
		if (dep[u] - (1 << k) >= dep[v])
			u = jmp[u][k];
	if (u == v) return u;
	for (int k = LOGN - 1; k >= 0; k--)
		if (jmp[u][k] != jmp[v][k]) {
			u = jmp[u][k];
			v = jmp[v][k];
		}
	return jmp[u][0];
}

int goup(int v, int k) {
	for (int t = LOGN - 1; t >= 0; t--)
		if (k >= (1 << t)) {
			k -= (1 << t);
			v = jmp[v][t];
		}
	return v;
}

int vert(int u, int v, int k) {
	int l = lca(u, v);
	int d1 = dep[u] - dep[l];
	int d2 = dep[v] - dep[l];
	if (k <= d1) return goup(u, k);
	k -= d1;
	k = d2 - k;
	return goup(v, k);
}

int dist(int u, int v) {
	int l = lca(u, v);
	return dep[u] + dep[v] - 2 * dep[l];
}

void add_set(int v) {
	for (auto [u, w] : cent[v])
		dists[u].insert(w);
}

void rem_set(int v) {
	for (auto [u, w] : cent[v])
		dists[u].extract(w);
}

int query(int v) {
	int ans = (int)(1e9);
	for (auto [u, w] : cent[v]) {
		if (dists[u].empty())
			continue;
		ans = min(ans, *dists[u].begin() + w);
	}
	return ans;
}

void add(int v) {
	d[v]++;
	if (d[v] == 1 && v != 1) add_set(v);
	if (d[v] == 3) rem_set(v);
	if (d[v] > 3) bad = true;
}

void dfs2(int v, int p = -1) {
	sub[v] = 1;
	for (int u : g[v]) if (p != u && !used[u]) {
		dfs(u, v);
		sub[v] += sub[u];
	}
}

int centroid(int v, int t, int p = -1) {
	for (int u : g[v]) if (p != u && !used[u] && sub[u] > t / 2)
		return centroid(u, t, v);
	return v;
}

void push(int v, int x, int p = -1, int d = 0) {
	cent[v].push_back({x, d});
	for (int u : g[v]) if (p != u && !used[u])
		push(u, x, v, d + 1);
}

void build(int v) {
	dfs2(v);
	int x = centroid(v, sub[v]);
	push(x, x);
	used[x] = true;
	for (int u : g[x]) if (!used[u])
		build(u);
}

void test_case() {
	int n; cin >> n;
	for (int i = 1; i <= n; i++) {
		cent[i].clear();
		dists[i].clear();
		g[i].clear();
	}
	for (int i = 2; i <= n; i++) {
		cin >> p[i];
		g[p[i]].push_back(i);
		g[i].push_back(p[i]);
	}
	for (int i = 1; i <= n; i++)
		d[i] = 0;
	dfs(1);
	for (int i = 1; i <= n; i++)
		used[i] = false;
	build(1);
	bad = false;
	int l = 1, r = 1;
	add_set(1);
	for (int i = 1; i <= n; i++) {
		int cur = dist(l, r);
		if (i > 1) {
			if (dist(l, i) > cur) r = i;
			else if (dist(i, r) > cur) l = i;
			cur = dist(l, r);
			add(i);
			add(p[i]);
		}
		if (bad) {
			cout << -1 << ' ';
			continue;
		}
		int a = vert(l, r, cur / 2);
		int b = vert(r, l, cur / 2);
		cout << min(query(a), query(b)) + (cur + 1) / 2 + 1 << ' ';
	}
	cout << '\n';
}

int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int t; cin >> t;
	while (t--) test_case();
}

**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
10
1 1 2 2 4 4 5 5 7 6
```