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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
N = 2*10**5+1
sieve = [-1]*N
primes = []
for i in range(2,N):
    if sieve[i] == -1:
        for j in range(i,N,i):
            sieve[j] = i
        primes.append(i)

p = [0]*N
for i in primes:
    if sieve[i-2] == i-2:
        p[i] = p[i-2]+1
        p[i] %= 2
l = [0]*N
for y in range(len(primes)):
    for z in range(y,len(primes)):
        i = primes[y]
        j = primes[z]
        if i+j<=2*10**5:
            l[i+j] |= (1<<max(p[i],p[j]))
for i in primes:
    l[i] = (1<<1-p[i])

# 1 bit is odd, 2 bit is even

for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    if _ == 6027:
        print(a)
    elif _ > 5:
        continue
    b = [l[i] for i in a]
    if n % 2 == 0:
        if 3 in b or 1 in b:
            print('Alice')
        else:
            print('Bob')
    else:
        if b == [3]*n:
            print('Bob')
        elif 3 in b:
            print('Alice')
        elif 1 in b and (0 in b or 2 in b):
            print('Alice')
        else:
            print('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
6
4 6 8 10 12 14
```

### Explanation:
This test case is designed to challenge the buggy code in the following ways:

1. **Even Number of Piles**: The number of piles is even (6), which triggers the first branch of the logic in the buggy code.
2. **Non-Prime Sums**: The numbers 4, 6, 8, 10, 12, and 14 are all even and can be split into two prime numbers (e.g., 4 can be split into 2 and 2, 6 can be split into 3 and 3). However, the sums of these numbers (e.g., 4+6=10, 8+10=18) are not prime, which might cause issues with the precomputed `l` array.
3. **Mixed Values in `b`**: The values in the `b` array (which are derived from the `l` array) should be a mix of 1, 2, and possibly 0, which can trigger the conditions in the logic.

This test case should expose any issues in the logic where the code incorrectly determines the winner based on the values in the `b` array.


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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f3f3f3f3fll
#define debug(x) cerr<<#x<<"="<<x<<endl
using namespace std;
using ll=long long;
using ld=long double;
using pli=pair<ll,int>;
using pi=pair<int,int>;
template<typename A>
using vc=vector<A>;
inline int read()
{
    int s=0,w=1;char ch;
    while((ch=getchar())>'9'||ch<'0') if(ch=='-') w=-1;
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    return s*w;
}
inline ll lread()
{
    ll s=0,w=1;char ch;
    while((ch=getchar())>'9'||ch<'0') if(ch=='-') w=-1;
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    return s*w;
}
const int L=200000;
bool is[200001];
int pri[200001];
bool w0[200001];
bool w1[200001];
int a[200001];
int n;
inline void init()
{
    is[1]=1;
    for(int i=2;i<=L;i++)
    {
        if(!is[i]) pri[++pri[0]]=i;
        for(int j=1;j<=pri[0]&&i*pri[j]<=L;j++)
        {
            is[i*pri[j]]=1;
            if(i%pri[j]==0) break;
        }
    }
    //可以分解为两个质数l0的,是w0
    for(int i=3;i<=L;i+=2) if(!is[i-2]&&!w0[i-2]) w0[i]=1;
    bitset<100001>jl0,ow0;//奇数的l0,偶数的w0
    for(int i=1;i<=L;i+=2) if(!w0[i]&&!is[i]) jl0[(i+1)/2]=1;
    for(int i=1;i<=L;i+=2) if(!w0[i]&&!is[i]) ow0|=jl0<<(i/2);
    for(int i=2;i<=L;i+=2) if((int)ow0[i>>1]) w0[i]=1;

    //本身是w0,且可以被分解为两个w0的,是w1
    bitset<100001>jw0,ow1;
    for(int i=1;i<=L;i+=2) if(w0[i]) jw0[(i+1)/2]=1;
    for(int i=1;i<=L;i+=2) if(w0[i]) ow1|=jw0<<(i/2);
    for(int i=2;i<=L;i+=2) if((int)ow1[i>>1]&&w0[i]) w1[i]=1;

    // for(int i=1;i<=20;i++) printf("%d : %d %d\n",i,w0[i],w1[i]);
}
inline void solve()
{
    int cw0=0,cw1=0;n=read();
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
        if(w0[a[i]]) cw0++;
        if(w1[a[i]]) cw1++;
    }
    if(!cw0) printf("Bob\n");
    else if(n%2==0||cw0!=n) printf("Alice\n");
    else if(!cw1||cw1==n) printf("Bob\n");
    else printf("Alice\n");
}
int main()
{
    init();int T=read();
    while(T--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
6
4 6 8 10 12 14
```


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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

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

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

bool cansplit(int x,stack<int>&st)
{
    for(int i = x -2 ; i > 1 ;i--)
    {
        if(is_prime(i))
        {
            if(is_prime(x-i))
            {
                if(i>3)
                st.push(i);
                if(x-1>3)
                st.push(x-i);
                return true;
            }
        }
    }
    return false;
}
void solve() {

        int n;
        cin >> n;
        vector<int> a(n);
        int k;
        for (int &x : a)
        {
            cin >> x;
            if(x>3)k++;
        }

        if(k == 0)
        {
            cout<<"Bob";
            return;
        }

        int prime_count = 0;
        stack<int>st;
        for (int x : a) {
            if(k==0)break;
            if(x>3)
            {
                if(cansplit(x,st))
                {
                    prime_count++;
                }
                k--;
            }
        }

        while(!st.empty())
        {
            int x = st.top();
            st.pop();
            if(cansplit(x,st))
            {
                prime_count++;
            }

        }

        if (prime_count % 2 == 1) {
            cout << "Alice";
        } else {
            cout << "Bob";
        }

}

int main() {
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while(t--)
    {
        solve();
        if(t>0)cout<<endl;
    }
    return 0;
}

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

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

Think step by step.

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


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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
N = 2*10**5+1
sieve = [-1]*N
primes = []
for i in range(2,N):
    if sieve[i] == -1:
        for j in range(i,N,i):
            sieve[j] = i
        primes.append(i)

p = [0]*N
for i in primes:
    if sieve[i-2] == i-2:
        p[i] = p[i-2]+1
        p[i] %= 2
l = [0]*N
l1 = [0]*N
for y in range(len(primes)):
    for z in range(y,len(primes)):
        i = primes[y]
        j = primes[z]
        if i+j<=2*10**5:
            l[i+j] |= (1<<max(p[i],p[j]))
            if p[i] == p[j] == 1:
                l1[i+j] = 1
        else:
            break
for i in primes:
    l[i] = (1<<1-p[i])

# 1 bit is odd, 2 bit is even

for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    b = [l[i] for i in a]
    x = sum(l1[i] for i in a)
    if n % 2 == 0:
        if 3 in b or 1 in b:
            print('Alice')
        else:
            print('Bob')
    else:
        if b == [3]*n:
            if x == n:
                print('Bob')
            elif x > 0:
                print('Alice')
            else:
                print('Bob')
        elif (1 in b or 3 in b) and (0 in b or 2 in b) or 0 < x < n:
            print('Alice')
        else:
            print('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
4
2
2 2
3
3 5 7
4
4 6 8 10
5
8 8 8 8 8
```


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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
N = 2*10**5+1
sieve = [-1]*N
primes = []
for i in range(2,N):
    if sieve[i] == -1:
        for j in range(i,N,i):
            sieve[j] = i
        primes.append(i)

p = [0]*N
for i in primes:
    if sieve[i-2] == i-2:
        p[i] = p[i-2]+1
        p[i] %= 2
l = [0]*N
l1 = [0]*N
for y in range(len(primes)):
    for z in range(y,len(primes)):
        i = primes[y]
        j = primes[z]
        if i+j<=2*10**5:
            l[i+j] |= (1<<max(p[i],p[j]))
            if p[i] == p[j] == 1:
                l1[i+j] = 1
for i in primes:
    l[i] = (1<<1-p[i])

# 1 bit is odd, 2 bit is even

for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    b = [l[i] for i in a]
    if n % 2 == 0:
        if 3 in b or 1 in b:
            print('Alice')
        else:
            print('Bob')
    else:
        if b == [3]*n:
            x = [l1[i] for i in a]
            if x == [1]*n:
                print('Bob')
            elif 1 in x:
                print('Alice')
            else:
                print('Bob')
        elif (1 in b or 3 in b) and (0 in b or 2 in b):
            print('Alice')
        else:
            print('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
5
2 4 6 8 10
```


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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

        }
        System.out.println(sb);

    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code has issues with determining the winning conditions, especially when dealing with prime numbers and their combinations.

2. **Edge Cases**:
   - Piles with small numbers that cannot be split into two primes.
   - Piles with numbers that can be split into two primes.
   - Piles with numbers that are prime themselves.
   - Piles with numbers that are sums of two primes.

3. **Test Case Construction**:
   - Create a test case with a small number of piles to ensure it fits within the output limit.
   - Include piles with numbers that are difficult to split into two primes.
   - Include piles with numbers that are sums of two primes.

Given these considerations, let's construct a test case:

```plaintext
3
2
2 1
4
4 6 8 10
6
8 11 13 17 19 23
```

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

2. **Test Case 2**:
   - Piles: 4, 6, 8, 10
   - Alice can choose `k = 2`, remove two piles (e.g., 8 and 10), and split the remaining piles (4 into 2 and 2, 6 into 3 and 3). Bob will have no valid moves, so Alice wins.

3. **Test Case 3**:
   - Piles: 8, 11, 13, 17, 19, 23
   - All numbers are either prime or can be split into two primes. Alice can make optimal moves to ensure she wins. For example, she can choose `k = 3`, remove three piles (e.g., 11, 13, 17), and split the remaining piles (8 into 3 and 5, 19 into 2 and 17, 23 into 2 and 21). Bob will have no valid moves, so Alice wins.

This test input should effectively expose the bug in the buggy code by challenging its ability to handle different types of piles and their combinations.


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

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

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

The player who is unable to make a move loses.

Determine who will win if both players play optimally.

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import java.util.*;
public class PrimeNumbersGame {
    public static ArrayList<Integer> getPrimes(int n) {
        boolean[] isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        for (int p = 2; p * p <= n; p++) {
            if (isPrime[p]) {
                for (int i = p * p; i <= n; i += p) {
                    isPrime[i] = false;
                }
            }
        }
        ArrayList<Integer> primeNumbers = new ArrayList<>();
        for (int i = 2; i <= n; i++) {
            if (isPrime[i]) {
                primeNumbers.add(i);
            }
        }
        return primeNumbers;
    }
    public static boolean canSplit(int stones, Set<Integer> primeSet) {
        for (Integer prime : primeSet) {
            if (primeSet.contains(stones - prime) && prime <= stones - prime) {
                return true;
            }
        }
        return false;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int total = sc.nextInt();
        List<int[]> allPiles = new ArrayList<>();
        int maxPileSize = 0;

        for (int si = 0; si < total; si++) {
            int aa = sc.nextInt();
            int[] piles = new int[aa];
            for (int i = 0; i < aa; i++) {
                piles[i] = sc.nextInt();
                if (piles[i] > maxPileSize) {
                    maxPileSize = piles[i];
                }
            }
            allPiles.add(piles);
        }
        ArrayList<Integer> primes = getPrimes(maxPileSize);
        Set<Integer> primeSet = new HashSet<>(primes);

        for (int[] piles : allPiles) {
            boolean isAliceTurn = true;
            boolean canMove = true;
            int pileCount = piles.length;

            while (canMove) {
                canMove = false;
                for (int i = 0; i < pileCount; i++) {
                    if (piles[i] > 1 && canSplit(piles[i], primeSet)) {
                        canMove = true;
                        int stones = piles[i];
                        piles[i] = 0;
                        for (Integer prime : primeSet) {
                            if (primeSet.contains(stones - prime) && prime <= stones - prime) {
                                piles = Arrays.copyOf(piles, piles.length + 2);
                                piles[pileCount] = prime;
                                piles[pileCount + 1] = stones - prime;
                                pileCount += 2;
                                break;
                            }
                        }
                        break;
                    }
                }
                isAliceTurn = !isAliceTurn;
            }

            if (isAliceTurn) {
                System.out.println("Alice");
            } else {
                System.out.println("Bob");
            }
        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
#include<cstdio>
struct node{
	int n;
	node *nxt;
}*mp[1000005],use[1000005],*nw;
void add(int u,int v)
{
	node *p=nw++;
	p->n=v;
	p->nxt=mp[u];
	mp[u]=p;
}
int type[1000005];
int qu1[1000005],qu2[1000005],qf1,qf2,ql1,ql2;
bool dfs(int now)
{
	node *p=mp[now];
	while(p)
	{
		if(type[p->n]==0)
		{
			type[p->n]=-type[now];
			if(dfs(p->n)) return 1;
		}
		else
		{
			if(type[p->n]!=-type[now]) return 1;
		}
		p=p->nxt;
	}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m;
		scanf("%d%d",&n,&m);
		qf1=qf2=ql1=ql2=0;
		for(int i=1;i<=n;i++)
			mp[i]=0,type[i]=0;
		nw=use;
		for(int i=1;i<=n;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			add(u,v);
			add(v,u);
		}
		type[1]=1;
		if(dfs(1))
		{
			printf("Alice\n");
			fflush(stdout);
			for(int i=1;i<=n;i++)
				printf("1 2\n"),fflush(stdout),scanf("%*d %*d");
		}
		else
		{
			printf("Bob\n");
			fflush(stdout);
			for(int i=1;i<=n;i++)
				if(type[i]==1) qu1[ql1++]=i;
				else qu2[ql2++]=i;
			for(int i=1;i<=n;i++)
			{
				int x,y;
				scanf("%d%d",&x,&y);
				if(x>y)
				{
					int z=x;
					x=y;
					y=z;
				}
				if(x==1)
				{
					if(qf1<ql1)
					{
						printf("%d %d\n",qu1[qf1++],1);
						fflush(stdout);
					}
					else
					{
						printf("%d %d\n",qu2[qf2++],y);
						fflush(stdout);
					}
				}
				else
				{
					if(qf2<ql2)
					{
						printf("%d %d\n",qu2[qf2++],3);
						fflush(stdout);
					}
					else
					{
						printf("%d %d\n",qu1[qf1++],x);
						fflush(stdout);
					}
				}
			}
		}
	}
	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
2
4 4
1 2
2 3
3 4
4 1
5 5
1 2
2 3
3 4
4 5
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
#include <iostream>
#include <set>
#include <map>
#define ll long long
using namespace std;
void solve() {
	ll n, m, u, v;
	cin >> n >> m;
	map<ll, ll> graph;
	set<ll> part_1, part_2;
	ll cnt = 1;
	bool is_double = true;
	for (ll i = 0; i < m; i++) {
		cin >> u >> v;
		if (graph.find(v) == graph.end()) {
			if (graph.find(u) == graph.end()) {
				graph[u] = cnt;
				graph[v] = -cnt;
				cnt++;
				part_1.insert(u);
				part_2.insert(v);
			}
			else {
				graph[v] = -graph[u];
				if (graph[v] < 0) {
					part_2.insert(v);
				}
				else {
					part_1.insert(v);
				}
			}
		}
		else {
			if (graph.find(u) == graph.end()) {
				graph[u] = -graph[v];
				if (graph[u] < 0) {
					part_2.insert(u);
				}
				else {
					part_1.insert(u);
				}
			}
			else {
				if ((graph[u] > 0) == (graph[v] > 0)) {
					is_double = false;
					for (int j = i + 1; j < m; j++) {
						cin >> u >> v;
					}
					break;
				}
			}
		}
	}
	if (!is_double) {
		cout << "Alice\n"; cout.flush();
		for (int i = 0; i < n; i++) {
			cout << "1 2\n";
			cout.flush();
			cin >> u >> v;
		}
	}
	else {
		cout << "Bob\n"; cout.flush();
		for (ll i = 0; i < n; i++) {
			cin >> u >> v; // now u and v are colours
			int x = min(u, v);
			if (x == 1) {
				if (part_1.size()) {
					cout << *part_1.begin() << " 1" << '\n';
					part_1.erase(part_1.begin());
				}
				else {
					cout << *part_2.begin() << ' ' << max(u, v) << '\n';
					cout.flush();
					part_2.erase(part_2.begin());
					for (ll j = i + 1; j < n; j++) {
						cin >> u >> v;
						cout << *part_2.begin() << ' ' << max(u, v) << '\n';
						cout.flush();
						part_2.erase(part_2.begin());
					}
					break;
				}
			}
			else {
				if (part_2.size()) {
					cout << *part_2.begin() << " 2" << '\n';
					part_2.erase(part_2.begin());
				}
				else {
					cout << *part_1.begin() << ' ' << (u == 2 ? v : u) << '\n';
					cout.flush();
					part_1.erase(part_1.begin());
					for (ll j = i + 1; j < n; j++) {
						cin >> u >> v;
						cout << *part_1.begin() << ' ' << (u == 2 ? v : u) << '\n';
						cout.flush();
						part_1.erase(part_1.begin());
					}
					break;
				}
			}
			cout.flush();
		}
	}
}
int main() {
	ll t = 1; cin >> t;
	while (t--) {
		solve();
	}
}

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

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

Think step by step.

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

4 4
1 2
2 3
3 4
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
#include <bits/stdc++.h>
#define lb(x) (x&-x)
#define L(i,j,k) for(int i=(j);i<=(k);++i)
#define R(i,j,k) for(int i=(j);i>=(k);--i)
#define swap(a,b) (a^=b^=a^=b)

using namespace std;
using i64 = long long;

typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
void chmin(int &x, int c) {
    x = min(x, c);
}
void chmax(int &x, int c) {
    x = max(x, c);
}

const int maxn = 1e6 + 10, mod = 998244353;
int h[maxn],ne[maxn],e[maxn],idx,co[maxn],n,m;
void add(int a,int b){idx++,e[idx]=b,ne[idx]=h[a],h[a]=idx;}
bool dfs(int u,int c){
    co[u]=c;
    for(int i=h[u];i;i=ne[i]){
        int j=e[i];
        if(!co[j]){
            if(!dfs(j,3-c))return 0;
        }else if(co[j]==c)return 0;
    }
    return 1;
}
int A[maxn],B[maxn];

void solve() {
	int n,m;cin>>n>>m;
	L(i,1,m){int a,b;cin>>a>>b,add(a,b),add(b,a);}
	L(i,1,n){if(!co[i]){if(!dfs(i,1)){
		cout<<"Alice"<<endl;
		L(i,1,n){int a,b;cout<<"1 2"<<endl,cin>>a>>b;}
		L(i,1,m*2)e[i]=0,ne[i]=0,h[i]=0,co[i]=0;idx=0;
		return;
	}}}
	L(i,1,m*2+10)e[i]=0,ne[i]=0,h[i]=0,co[i]=0;idx=0;
	cout<<"Bob"<<endl;
	int cnt1=0,cnt2=0;
	L(i,1,n)if(co[i]==1)A[++cnt1]=i;else B[++cnt2]=i;
	L(i,1,n){
		int a,b;cin>>a>>b;
		if(a>b)swap(a,b);
		if(a==1){
			//还有才放
			if(cnt1)cout<<A[cnt1]<<' '<<1<<endl,cnt1--;
			else cout<<B[cnt2]<<' '<<b<<endl,cnt2--;
		}else{
			if(cnt2)cout<<B[cnt2]<<' '<<2<<endl,cnt2--;
			else cout<<A[cnt1]<<' '<<3<<endl,cnt1--;
		}
	}
	L(i,1,m*2+10)e[i]=0,ne[i]=0,h[i]=0,co[i]=0;idx=0;
}

signed main() {
    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 code, we need to consider edge cases and scenarios where the algorithm might fail. The code seems to have issues with how it handles the coloring process and the conditions under which Alice or Bob wins. Specifically, the code might not handle certain graph structures or sequences of moves correctly.

Let's break down the problem and think about potential edge cases:

1. **Small Graphs with Specific Structures:**
   - A graph where Bob can always win if he plays optimally.
   - A graph where Alice can always win if she plays optimally.

2. **Graphs with Cycles:**
   - A graph with a cycle where Alice can force a win by choosing specific colors.

3. **Graphs with High Connectivity:**
   - A complete graph or a highly connected graph where Alice can easily force a win.

4. **Graphs with Specific Edge Cases:**
   - A graph with a single edge or a linear structure where Bob can always win if he plays optimally.

Let's consider a small graph where Bob can always win if he plays optimally. A linear graph (a path) with 3 vertices is a good candidate. In this graph, Bob can always avoid creating an edge between two vertices of the same color.

### Test Input
```plaintext
1
3 2
1 2
2 3
```

### Explanation
- **Test Case:**
  - **Number of Test Cases:** 1
  - **Number of Vertices (n):** 3
  - **Number of Edges (m):** 2
  - **Edges:**
    - 1 2
    - 2 3

### Expected Behavior
- **Alice's Strategy:**
  - Alice can choose any two colors in each round.
- **Bob's Strategy:**
  - Bob can always avoid creating an edge between two vertices of the same color by choosing the color that does not create a conflict.

### Expected Output
- **Output:**
  - Bob
  - 1 2
  - 2 3

This test case is designed to expose the bug in the code by ensuring that Bob can always win if he plays optimally, which the provided code might not handle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
from collections import deque
import sys

def thing(n, edges):
	adj_list = [[] for i in range(n + 1)]
	colors = [-1] * (n + 1)

	for (u, v) in edges:
		adj_list[u].append(v)
		adj_list[v].append(u)

	bob_can_win = True
	color = 1

	queue = deque([(1, 1)]) #(node, color)

	while queue:
		node, color = queue.popleft()

		if colors[node] == -1:
			for i in adj_list[node]:
				queue.append((i, 1 - color))
			colors[node] = color

		else:
			if colors[node] != color:
				bob_can_win = False

	if bob_can_win:
		print("Bob")
		sys.stdout.flush()

		queue = deque([(1, 1)]) #(node, color)
		colors = [-1] * (n + 1)

		while queue:
			node, color = queue.popleft()

			if colors[node] == -1:
				alice_nums = [int(i) for i in input().split(" ")]
				#input()
				new_color = 0
				if alice_nums[0] == color:
					new_color = alice_nums[1]
				else:
					new_color = alice_nums[0]

				for i in adj_list[node]:
					queue.append((i, new_color))

				colors[node] = color
				print(node, new_color)
				sys.stdout.flush()

	else:
		print("Alice")
		sys.stdout.flush()

		for i in range(n):
			print(1, 3)
			sys.stdout.flush()
			input()
			#input()

#[int(i) for i in input().split(" ")]

testcases = range(int(input()))

for case in testcases:
	n, m = [int(i) for i in input().split(" ")]
	edges = []

	for x in range(m):
		edges.append([int(i) for i in input().split(" ")])

	thing(n, edges)

**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
4 4
1 2
2 3
3 4
4 1
5 5
1 2
2 3
3 4
4 5
5 1
```

### Explanation:

1. **Test Case 1:**
   - **Graph:** A simple cycle graph with 4 vertices (1-2, 2-3, 3-4, 4-1).
   - **Challenge:** This graph is a simple cycle, and Bob can easily win by coloring the vertices in an alternating pattern (e.g., 1, 2, 1, 2). The buggy code might fail to recognize this and might incorrectly claim that Bob can win in all cases.

2. **Test Case 2:**
   - **Graph:** A simple cycle graph with 5 vertices (1-2, 2-3, 3-4, 4-5, 5-1).
   - **Challenge:** This graph is also a simple cycle, but with an odd number of vertices. In this case, Bob cannot win by coloring the vertices in an alternating pattern. The buggy code might fail to recognize this and might incorrectly claim that Bob can win.

### Why These Test Cases Are Challenging:
- **Cycle Graphs:** Both test cases involve cycle graphs, which are critical in determining the winner. In a cycle graph with an even number of vertices, Bob can always win by alternating colors. In a cycle graph with an odd number of vertices, Alice can always win because Bob cannot avoid creating a monochromatic edge.
- **Small Scale:** The test cases are small enough to be manageable within the constraints but are still challenging enough to expose potential bugs in the code.
- **Edge Cases:** The test cases cover both even and odd cycle graphs, which are edge cases that can easily trip up a buggy algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
#!/usr/bin/env python
from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

import sys

def is_bipartite(graph):
    n = len(graph)
    color = [-1] * n

    for start in range(n):
        if color[start] == -1:
            color[start] = 0
            stack = [start]

            while stack:
                parent = stack.pop()

                for child in graph[parent]:
                    if color[child] == -1:
                        color[child] = 1 - color[parent]
                        stack.append(child)
                    elif color[parent] == color[child]:
                        return False, color

    return True, color

def play_as_alice(n):
    for _ in range(n):
        print("1 2")
        sys.stdout.flush()
        input()  # Read Bob's move, but we don't need to use it

def play_as_bob(n, graph):
    import sys

    colors = [0] * n  # 0: uncolored, 1, 2, 3: actual colors
    bipartite_class = [-1] * n  # -1: unassigned, 0 or 1: class

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

        # Find an uncolored vertex
        v = next(i for i in range(n) if colors[i] == 0)

        if bipartite_class[v] == -1:
            # If this vertex hasn't been assigned a class yet
            neighbor_classes = set(bipartite_class[u] for u in graph[v] if bipartite_class[u] != -1)
            if not neighbor_classes:
                bipartite_class[v] = 0  # Assign arbitrarily if no colored neighbors
            else:
                bipartite_class[v] = 1 - next(iter(neighbor_classes))  # Assign opposite of neighbor

        # Choose color based on bipartite class and neighbor colors
        neighbor_colors = set(colors[u] for u in graph[v] if colors[u] != 0)
        if a not in neighbor_colors and (bipartite_class[v] == 0 or b in neighbor_colors):
            c = a
        else:
            c = b

        colors[v] = c

        # Propagate bipartite class to unassigned neighbors
        for u in graph[v]:
            if bipartite_class[u] == -1:
                bipartite_class[u] = 1 - bipartite_class[v]

        print(f"{v+1} {c}")
        sys.stdout.flush()

def solve_case():
    n, m = map(int, input().split())
    graph = [[] for _ in range(n)]
    for _ in range(m):
        u, v = map(int, input().split())
        u -= 1  # Convert to 0-indexed
        v -= 1  # Convert to 0-indexed
        graph[u].append(v)
        graph[v].append(u)

    is_bip, coloring = is_bipartite(graph)

    if is_bip:
        print("Bob")
        sys.stdout.flush()
        play_as_bob(n, graph)
    else:
        print("Alice")
        sys.stdout.flush()
        play_as_alice(n)

def main():
    t = int(input())
    for _ in range(t):
        solve_case()

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

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 given buggy code, we need to focus on edge cases and scenarios that can make the algorithm fail. The main issue in the code is how it handles the bipartite graph checking and the strategy for Bob to avoid losing.

### Step-by-Step Analysis:

1. **Bipartite Graph Check**: The code checks if the graph is bipartite. If it is, Bob is chosen to play. If not, Alice is chosen to play. The bug in the code might be in how it handles the bipartite check or the strategy for Bob.

2. **Bob's Strategy**: Bob's strategy is to avoid creating a monochromatic edge. The code assigns colors based on the bipartite class and the colors of the neighbors. The bug might be in how it assigns colors or propagates the bipartite class.

3. **Edge Cases**:
   - **Complete Graph**: A complete graph (where every vertex is connected to every other vertex) is not bipartite. This can force Alice to play and potentially expose issues in her strategy.
   - **Cycle Graph**: A cycle graph with an odd number of vertices is not bipartite. This can also force Alice to play and expose issues.
   - **Tree Graph**: A tree graph is bipartite. This can force Bob to play and expose issues in his strategy.

### Challenging Test Input:

A complete graph with 4 vertices (K4) is a good candidate because it is not bipartite and can force Alice to play. This will test how Alice's strategy handles a fully connected graph.

```plaintext
1
4 6
1 2
1 3
1 4
2 3
2 4
3 4
```

### Explanation:

- **Test Case 1**:
  - **Number of Vertices (n)**: 4
  - **Number of Edges (m)**: 6
  - **Edges**:
    - 1-2
    - 1-3
    - 1-4
    - 2-3
    - 2-4
    - 3-4

This graph is a complete graph with 4 vertices, which is not bipartite. Therefore, Alice should play. The goal is to see if Alice's strategy can correctly handle this fully connected graph and force a win by creating a monochromatic edge.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

Buggy Code:
from collections import deque
from sys import stdout
t = int(input())
for _ in range(t):
    n, m = list(map(int, input().split()))
    neigh = [[] for u in range(n+1)]
    for i in range(m):
        u, v = list(map(int, input().split()))
        neigh[u].append(v)
        neigh[v].append(u)
    q = deque([1])
    seen = [0 for u in range(n+1)]
    color = [None for i in range(n+1)]
    color[1] = 1
    size = 1
    is_bipartite = 1
    while size > 0:
        u = q.popleft()
        size -= 1
        for v in neigh[u]:
            if seen[v] == 0:
                q.append(v)
                size += 1
                seen[v] = 1
                color[v] = -color[u]
            if color[v] == color[u]:
                is_bipartite = -1
    if is_bipartite == -1:
        print('Alice')
        for i in range(n):
            print(1, 2)
            print()
            stdout.flush()
            j, c = list(map(int, input().split()))
    else:
        print('Bob')
        first = deque([u for u in range(n+1) if color[u] == 1])
        second = deque([u for u in range(n+1) if color[u] == -1])
        N_1 = len(first)
        N_2 = len(second)
        for i in range(n):
            a, b = list(map(int, input().split()))
            if ((a == 1) or (b == 1)) and (N_1 > 0):
                u = first.popleft()
                N_1 -= 1
                print(u, 1)
                print()
                stdout.flush()
            elif N_1 > 0:
                u = second.popleft()
                N_2 -= 1
                print(u, 2)
                print()
                stdout.flush()
            else:
                u = second.popleft()
                N_2 -= 1
                print(u, max(a, b))
                print()
                stdout.flush()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

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

public class ColorGame {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(in.readLine());
        for (int z = 0; z < t; z++) {
            StringTokenizer st = new StringTokenizer(in.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            ArrayList<Integer>[] edges = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                edges[i] = new ArrayList<Integer>();
            }
            for (int i = 0; i < m; i++) {
                st = new StringTokenizer(in.readLine());
                int a = Integer.parseInt(st.nextToken()) - 1;
                int b = Integer.parseInt(st.nextToken()) - 1;
                edges[a].add(b);
                edges[b].add(a);
            }

            int[] nodes = new int[n];
            boolean[] visited = new boolean[n];
            boolean alice = false; // Alice can force a win
            Stack<Integer> next = new Stack<Integer>();
            Stack<Integer> color = new Stack<Integer>();
            next.push(0);
            color.push(0);

            while (!next.isEmpty()) {
                int curr = next.pop();
                int col = color.pop();
                nodes[curr] = col;
                visited[curr] = true;
                for (int i = 0; i < edges[curr].size(); i++) {
                    int pos = edges[curr].get(i);
                    if (!visited[pos]) {
                        next.add(pos);
                        color.add((col + 1) % 2);
                    }
                    else {
                        if (nodes[pos] == col) {
                            // An already visited neighbor has the same color
                            // Odd cycle
                            alice = true;
                        }
                    }
                }
            }

            if (alice) {
                System.out.println("Alice");
                for (int i = 0; i < n; i++){
                    System.out.println("1 2");
                    in.readLine();
                }
            }
            else {
                int firstEven = -1;
                int firstOdd = -1;
                for (int i = 0; i < n; i++) {
                    if (nodes[i] % 2 == 0 && firstEven < 0){
                        firstEven = i;
                    }
                    else if (firstOdd < 0){
                        firstOdd = i;
                    }
                    if (firstEven >= 0 && firstOdd >= 0) break;
                }

                System.out.println("Bob");
                for (int i = 0; i < n; i++) {
                    st = new StringTokenizer(in.readLine());
                    int choiceA = Integer.parseInt(st.nextToken());
                    int choiceB = Integer.parseInt(st.nextToken());
                    int swap = choiceA;
                    choiceA = Math.min(choiceA, choiceB);
                    choiceB = Math.max(swap, choiceB);

                    if (choiceA == 1) {
                        if (firstOdd >= 0) {
                            System.out.println((firstOdd + 1) + " 1");
                            int temp = firstOdd;
                            firstOdd = -1;
                            // Find the next odd value
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 1) {
                                    firstOdd = j;
                                    break;
                                }
                            }
                        }
                        else {
                            // No odds left
                            System.out.println((firstEven + 1) + " " + choiceB);
                            int temp = firstEven;
                            firstEven = -1;
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 0) {
                                    firstEven = j;
                                    break;
                                }
                            }
                        }
                    }
                    else if (choiceA == 2) {
                        if (firstEven >= 0) {
                            System.out.println((firstEven + 1) + " 2");
                            int temp = firstEven;
                            firstEven = -1;
                            // Find the next even value
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 0) {
                                    firstEven = j;
                                    break;
                                }
                            }
                        }
                        else {
                            // No evens left
                            System.out.println((firstOdd + 1) + " " + choiceB);
                            int temp = firstOdd;
                            firstOdd = -1;
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 1) {
                                    firstOdd = j;
                                    break;
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coloring Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Consider an undirected connected graph consisting of $$$n$$$ vertices and $$$m$$$ edges. Each vertex can be colored with one of three colors: $$$1$$$, $$$2$$$, or $$$3$$$. Initially, all vertices are uncolored.

Alice and Bob are playing a game consisting of $$$n$$$ rounds. In each round, the following two-step process happens:

1. Alice chooses two different colors.
2. Bob chooses an uncolored vertex and colors it with one of the two colors chosen by Alice.

Alice wins if there exists an edge connecting two vertices of the same color. Otherwise, Bob wins.

You are given the graph. Your task is to decide which player you wish to play as and win the game.

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 10^4$$$, $$$n - 1 \le m \le \min(\frac{n \cdot (n - 1)}{2}, 10^4)$$$) — the number of vertices and the number of edges in the graph, respectively.

Each of the next $$$m$$$ lines of each test case contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$) — the edges of the graph. It is guaranteed that the graph is connected and there are no multiple edges or self-loops.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
3 3
1 2
2 3
3 1

3 1

2 2

1 1
4 4
1 2
2 3
3 4
4 1

2 3

1 2

2 1

3 1
output:
Alice
3 1

1 2

2 1

Bob

1 2

2 1

4 1

3 3

Note: Note that the sample test cases are example games and do not necessarily represent the optimal strategy for both players.

In the first test case, you choose to play as Alice.

1. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$1$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$2$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$1$$$.

Alice wins because the edge $$$(3, 1)$$$ connects two vertices of the same color.

In the second test case, you choose to play as Bob.

1. Alice chooses two colors: $$$2$$$ and $$$3$$$. Bob chooses vertex $$$1$$$ and colors it with color $$$2$$$.
2. Alice chooses two colors: $$$1$$$ and $$$2$$$. Bob chooses vertex $$$2$$$ and colors it with color $$$1$$$.
3. Alice chooses two colors: $$$2$$$ and $$$1$$$. Bob chooses vertex $$$4$$$ and colors it with color $$$1$$$.
4. Alice chooses two colors: $$$3$$$ and $$$1$$$. Bob chooses vertex $$$3$$$ and colors it with color $$$3$$$.

Bob wins because there are no edges with vertices of the same color.

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

public class ColorGame {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(in.readLine());
        for (int z = 0; z < t; z++) {
            StringTokenizer st = new StringTokenizer(in.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            ArrayList<Integer>[] edges = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                edges[i] = new ArrayList<Integer>();
            }
            for (int i = 0; i < m; i++) {
                st = new StringTokenizer(in.readLine());
                int a = Integer.parseInt(st.nextToken()) - 1;
                int b = Integer.parseInt(st.nextToken()) - 1;
                edges[a].add(b);
                edges[b].add(a);
            }

            int[] nodes = new int[n];
            boolean[] visited = new boolean[n];
            boolean alice = false; // Alice can force a win
            Stack<Integer> next = new Stack<Integer>();
            Stack<Integer> color = new Stack<Integer>();
            next.push(0);
            color.push(0);

            while (!next.isEmpty()) {
                int curr = next.pop();
                int col = color.pop();
                nodes[curr] = col;
                visited[curr] = true;
                for (int i = 0; i < edges[curr].size(); i++) {
                    int pos = edges[curr].get(i);
                    if (!visited[pos]) {
                        next.add(pos);
                        color.add((col + 1) % 2);
                    }
                    else {
                        if (nodes[pos] == col) {
                            // An already visited neighbor has the same color
                            // Odd cycle
                            alice = true;
                        }
                    }
                }
            }

            if (alice) {
                System.out.println("Alice");
                for (int i = 0; i < n; i++){
                    System.out.println("1 2");
                    in.readLine();
                }
            }
            else {
                int firstEven = -1;
                int firstOdd = -1;
                for (int i = 0; i < n; i++) {
                    if (nodes[i] % 2 == 0){
                        firstEven = i;
                    }
                    else{
                        firstOdd = i;
                    }
                    if (firstEven >= 0 && firstOdd >= 0) break;
                }

                System.out.println("Bob");
                for (int i = 0; i < n; i++) {
                    st = new StringTokenizer(in.readLine());
                    int choiceA = Integer.parseInt(st.nextToken());
                    int choiceB = Integer.parseInt(st.nextToken());
                    int swap = choiceA;
                    choiceA = Math.min(choiceA, choiceB);
                    choiceB = Math.max(swap, choiceB);

                    if (choiceA == 1) {
                        if (firstOdd >= 0) {
                            System.out.println((firstOdd + 1) + " 1");
                            int temp = firstOdd;
                            firstOdd = -1;
                            // Find the next odd value
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 1) {
                                    firstOdd = j;
                                    break;
                                }
                            }
                        }
                        else {
                            // No odds left
                            System.out.println((firstEven + 1) + " " + choiceB);
                            int temp = firstEven;
                            firstEven = -1;
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 0) {
                                    firstEven = j;
                                    break;
                                }
                            }
                        }
                    }
                    else if (choiceA == 2) {
                        if (firstEven >= 0) {
                            System.out.println((firstEven + 1) + " 2");
                            int temp = firstEven;
                            firstEven = -1;
                            // Find the next even value
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 0) {
                                    firstEven = j;
                                    break;
                                }
                            }
                        }
                        else {
                            // No evens left
                            System.out.println((firstOdd + 1) + " " + choiceB);
                            int temp = firstOdd;
                            firstOdd = -1;
                            for (int j = temp + 1; j < n; j++) {
                                if (nodes[j] % 2 == 1) {
                                    firstOdd = j;
                                    break;
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

**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
3 3
1 2
2 3
3 1
4 4
1 2
2 3
3 4
4 1
```