**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

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

#define int long long

constexpr int N = 1e5 + 6;

void solve() {
    int n;
    std::cin >> n;
    std::vector<int> a(n), x(n);
    for (auto & v : a) std::cin >> v;
    for (auto & v : x) std::cin >> v;
    int ans = 0;
    for (int i = 0; i < n; ++i) {
        if (x[i] <= a[i] && !(x[i] >= 1 && (x[i] & (x[i] + 1)) == 0)) {
            ans ^= 1;
        }
    }
    std::cout << (ans ? "Alice" : "Bob") << "\n";
}

int32_t main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);

  int t;
  std::cin >> t;
  // t = 1;

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

  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the conditions for the bitwise AND operation are critical. The buggy code has a condition that checks if `x[i] <= a[i]` and if `x[i]` is a power of two or zero. This condition is incorrect because it doesn't properly handle the bitwise AND operation.

Let's consider a scenario where the condition `x[i] & d = d` is not correctly handled by the buggy code. Specifically, we can create a test case where `x[i]` is a number that can be reduced by a value `d` within the range `[1, a[i]]`, but the bitwise AND condition is not satisfied by the buggy code's logic.

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

```plaintext
1
1
3
5
```

Explanation:
- There is one test case.
- There is one pile with `a[0] = 3` and `x[0] = 5`.
- The correct logic should determine that Alice can win by taking 1 stone from the pile, leaving 4 stones. Since 4 is a power of 2, Bob cannot make a valid move, and Alice wins.
- The buggy code, however, will incorrectly determine that Bob wins because it does not properly handle the bitwise AND condition.

This test input should help identify the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
//#pragma optimize("O3")
//#pragma optimize("unroll-loops")
//#pragma traget("avx,avx2,tune=native")

#include <iostream>
#include <cmath>
#include <algorithm>
#include <stdio.h>
#include <cstdint>
#include <cstring>
#include <string>
#include <cstdlib>
#include <vector>
#include <bitset>
#include <map>
#include <queue>
#include <ctime>
#include <stack>
#include <set>
#include <list>
#include <random>
#include <deque>
#include <functional>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <complex>
#include <numeric>
#include <cassert>
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <thread>

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define el '\n'
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define point pair <ll, ll>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

using namespace std;
mt19937 rnd(time(0));

const ll INF = 2e18 + 10;
const ll inf = 2e9 + 10;
const ll N = 1e5 + 10;
const ll mod = 998244353;
const ll LOG = 18;

ll g(ll a, ll x) {
    ll b = 0;
    bool good = 0;
    for (ll i = 29; i >= 0; i--) {
        if (!(x >> i & 1) && (a >> i & 1)) {
            good = 1;
            continue;
        }

        if (!(x >> i & 1))
            continue;

        if ((a >> i & 1) || good)
            b |= 1 << i;
    }

    a = 0;
    ll id = 0;
    for (ll i = 0; i < 30; i++)
        if (x >> i & 1)
            a |= (1 << id) * (b >> i & 1), id++;

    for (ll i = 1; i < 30; i++)
        if ((1 << i) - 2 == a)
            return 0;
    for (ll i = 0; i < 30; i++)
        if ((1 << i) - 1 == a)
            return i;
    for (ll i = 0; i < 30; i++)
        if ((1 << i) == a)
            return i ^ 1;
    for (ll i = 2; i < 31; i++)
        if ((1 << i) > a)
            return i;
}

void solve() {
    ll n;
    cin >> n;

    vector <ll> a(n), x(n);
    for (ll i = 0; i < n; i++)
        cin >> a[i];
    for (ll i = 0; i < n; i++)
        cin >> x[i];

    ll nim = 0;
    for (ll i = 0; i < n; i++)
        nim ^= g(a[i], x[i]);

    cout << (nim ? "Alice" : "Bob") << el;
    return;
}
signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);

    int tests = 1;
    cin >> tests;

    while (tests--)
        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
2
2 3
7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<random>
#include<ctime>
using namespace std;
#define re int
#define ll long long
inline int read(){
    int x=0,ff=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')ff=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+(c^'0');c=getchar();}
    return x*ff;
}
int n,t,a[10005],c[10005],f[30][30][30];
inline int hbit(int x){
	for(re i=29;i>=0;i--)
		if((x>>i)&1)return i;
	return 23333;
}
inline int bit(int x){
	int y=0;
	while(x){y+=x&1;x>>=1;}
	return y;
}
int nimber(int x, int a) {
  int aprime = 0;
  bool goodbit = false;
  for (int bit=30; bit>=0; bit--) {
    if (x & (1 << bit)) {
      aprime *= 2;
      if (goodbit || (a & (1 << bit))) {
        aprime += 1;
      }
    } else if (a & (1 << bit)) {
      goodbit = true;
    }
  }

  // g(2^k - 2) = 0, for all k >= 1.
  for (int k=1; k<=30; k++) {
    if (aprime == (1 << k) - 2) {
      return 0;
    }
  }

  // g(2^k - 1) = k, for all k >= 1.
  for (int k=1; k<=30; k++) {
    if (aprime == (1 << k) - 1) {
      return k;
    }
  }

  // g(2^k) = k + (-1)^k, for all k >= 0.
  for (int k=1; k<=30; k++) {
    if (aprime == (1 << k)) {
      if (k % 2) return k - 1;
      else return k + 1;
    }
  }

  // g(2^k+1) = g(2^k+2) = ... = g(2^{k+1} - 3) = k + 1, for all k >= 2.
  for (int k=2; k<=30; k++) {
    if ((1 << k) < aprime && aprime <= (2 << k) - 3) {
      return k + 1;
    }
  }

  // should never get to this point
  return -1;
}
int main(){
	t=read();int x,y,u,v,p;
	memset(f,-1,sizeof(f));
	f[0][0][0]=0;
	for(re i=0;i<30;i++){
		for(re j=0;j<30;j++){
			for(re q=0;q<30;q++){
				if(i+j+q>=30)break;
				if(!j){
					f[i][j][q]=i+j+q;continue;
				}
				if(!i)continue;
				vector<int> qwq;
				for(re k=0;k<=i;k++){
					for(re l=0;l<=j;l++){
						if(k==i&&l)break;
						for(re r=0;r<=q;r++){
							if((k+l+r)==0)continue;
							if(k){
								// if(i==2&&j==1&&q==0){
								// 	cerr<<j-l+q-r<<" "<<k<<" "<<l<<" "<<r<<" 2333"<<endl;
								// }
								qwq.emplace_back(j-l+q-r+i-k);continue;
							}
							qwq.emplace_back(f[i-k][j-l][q-r]);
						}
					}
				}
				sort(qwq.begin(),qwq.end());
				unique(qwq.begin(),qwq.end());
				f[i][j][q]=0;
				for(re k:qwq){
					//if(k==-1)cerr<<i<<" "<<j<<" "<<q<<endl;
					if(k!=f[i][j][q])break;
					f[i][j][q]++;
				}
			}
		}
	}
	while(t--){
		n=read();y=0;
		for(re i=1;i<=n;i++)a[i]=read();
		for(re i=1;i<=n;i++)c[i]=read();
		for(re i=1;i<=n;i++){
			x=hbit(a[i]);c[i]&=(1<<(x+1))-1;
			int kk=c[i],tuu=nimber(c[i],a[i]);
			if(c[i]<=a[i]){
				c[i]=bit(c[i]);
			}
			else {u=v=p=0;p=bit(c[i]);
				for(;x>=0;x--){
					if(((a[i]&c[i])>>x)&1)u++;
					else if(((a[i]^c[i])>>x)&1)break;
				}
				for(;x>=0;x--){
					if((a[i]>>x)&1)break;
					if((c[i]>>x)&1)v++;
				}
				p-=u+v;
				c[i]=f[u][v][p];
			}
			if(tuu!=c[i]){
				cout<<a[i]<<" "<<kk<<endl;
			}
			y^=c[i];
		}
		if(y)puts("Alice");
		else puts("Bob");
	}
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
import sys

def main():
    sys.setrecursionlimit(2000)
    from sys import stdin
    input = sys.stdin.read().split()
    t = int(input[0])
    index =1
    result = []
    memo = {}
    def grundy(x, ai):
        if x ==0:
            return 0
        if ai >=x:
            return bin(x).count('1')
        if (x,ai) in memo:
            return memo[(x,ai)]
        moves = set()
        d = x
        while d >0:
            if d <= ai:
                moves.add(grundy(x - d, ai))
            d = (d -1) & x
        mex =0
        while mex in moves:
            mex +=1
        memo[(x,ai)] = mex
        return mex
    for _ in range(t):
        n = int(input[index])
        index +=1
        a = list(map(int, input[index:index+n]))
        index +=n
        x = list(map(int, input[index:index+n]))
        index +=n
        total_grundy =0
        for xi, ai in zip(x, a):
            total_grundy ^= grundy(xi, ai)
        if total_grundy !=0:
            result.append("Alice")
        else:
            result.append("Bob")
    print('\n'.join(result))

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
3
1
1
1
1
10
10
2
10
10
2
10
11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
def main():
    import sys
    from functools import lru_cache

    input = sys.stdin.read
    data = input().split()
    idx = 0

    @lru_cache(maxsize=None)
    def get_grundy(x, a):
        if x == 0:
            return 0
        possible_moves = set()
        for d in range(1, min(x + 1, a + 1)):
            if (x & d) == d:
                possible_moves.add(get_grundy(x - d, a))
        g = 0
        while g in possible_moves:
            g += 1
        return g

    T = int(data[idx])
    idx += 1
    for _ in range(T):
        n = int(data[idx])
        idx += 1
        a = list(map(int, data[idx:idx+n]))
        idx += n
        x = list(map(int, data[idx:idx+n]))
        idx += n
        grundy_xor = 0
        for xi, ai in zip(x, a):
            grundy_xor ^= get_grundy(xi, ai)
        if grundy_xor == 0:
            print("Bob")
        else:
            print("Alice")

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 code, we need to consider edge cases and scenarios where the algorithm might fail. The key points to focus on are:

1. **Edge Cases for Pile Values**: Values of `a_i` and `x_i` that are very close to the maximum limit or have specific bit patterns.
2. **Edge Cases for Number of Piles**: Small and large values of `n` to ensure the code handles both efficiently.
3. **Edge Cases for Test Cases**: Single test case and multiple test cases to ensure the code handles input parsing correctly.

Let's construct a test input that includes these edge cases:

1. **Pile with `x_i` and `a_i` close to the maximum limit**:
   - This will test the ability of the code to handle large numbers and bit operations.
2. **Pile with specific bit patterns**:
   - This will test the correctness of the bitwise AND operation.
3. **Single pile and multiple piles**:
   - This will test the general logic of the game.
4. **Multiple test cases**:
   - This will test the input parsing and the overall structure of the code.

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

```plaintext
3
1
2147483647
2147483647
2
1
1
2
2
3
3
1
2
4
15
15
```

Explanation:
1. **First Test Case**:
   - 1 pile with `a_1 = 2147483647` and `x_1 = 2147483647`.
   - This is a large number close to the maximum limit, testing the handling of large numbers and bitwise operations.
2. **Second Test Case**:
   - 2 piles with `a_1 = 1`, `x_1 = 1` and `a_2 = 2`, `x_2 = 2`.
   - Simple test case with small values, ensuring the basic logic works.
3. **Third Test Case**:
   - 3 piles with `a_1 = 3`, `x_1 = 1`, `a_2 = 2`, `x_2 = 4`, and `a_3 = 15`, `x_3 = 15`.
   - Mixed values to test a variety of scenarios, including specific bit patterns.

This test input should help expose any bugs in the given code, especially in handling large numbers and specific bit patterns.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
def mex(s):
    s = set(s)
    i = 0
    while i in s:
        i += 1
    return i

def possible_d(x, a):
    mask = x
    while mask:
        if mask <= a:
            yield mask
        mask = (mask - 1) & x
    if x <= a:
        yield x

def get_grundy(x, a, grundy):
    if (x, a) in grundy:
        return grundy[(x, a)]
    if x == 0:
        grundy[(x, a)] = 0
    else:
        moves = [get_grundy(x - d, a, grundy) for d in possible_d(x, a)]
        grundy[(x, a)] = mex(moves)
    return grundy[(x, a)]

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    grundy = {}
    for _ in range(t):
        n = int(data[idx])
        idx += 1
        a = list(map(int, data[idx:idx+n]))
        idx += n
        x = list(map(int, data[idx:idx+n]))
        idx += n
        xor_sum = 0
        for ai, xi in zip(a, x):
            g = get_grundy(xi, ai, grundy)
            xor_sum ^= g
        if xor_sum:
            print("Alice")
        else:
            print("Bob")

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Understand the Bug**: The provided code uses a Grundy number (Nimber) approach to determine the winner. The `possible_d` function generates possible moves, and the `get_grundy` function calculates the Grundy number for a given state. The bug might be in the `possible_d` function, which generates the possible moves.

2. **Edge Cases**: Consider edge cases where the number of stones in a pile is very small or very large, and where the associated value `a_i` is very small or very large. Also, consider cases where the bitwise AND condition is particularly restrictive.

3. **Small-Scale Test Input**: Create a small-scale test input that is complete and valid, but still challenging for the code to handle correctly.

Given these considerations, a challenging test input could be:

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

### Explanation:
1. **First Test Case**:
   - Piles: 1, 1
   - Values: 1, 1
   - This is a simple case where each pile has only one stone and the associated value is also 1. The only move possible is to take the single stone from each pile. Alice can take the stone from the first pile, and Bob can take the stone from the second pile, resulting in Bob's win.

2. **Second Test Case**:
   - Piles: 2, 2, 2
   - Values: 3, 3, 3
   - Each pile has 3 stones, and the associated value is 3. The possible moves are 1, 2, and 3. Alice can take 1, 2, or 3 stones from any pile. The game is more complex, and the Grundy number approach needs to handle this correctly. If the Grundy numbers are calculated incorrectly, this case can expose the bug.

3. **Third Test Case**:
   - Piles: 1, 2, 3, 4
   - Values: 1, 2, 3, 4
   - This case has a mix of small and medium-sized piles. The associated values are the same as the number of stones, which simplifies the possible moves but still requires correct Grundy number calculation. If the code fails to handle the bitwise AND condition correctly, this case can expose the bug.

This test input is small-scale but covers a range of scenarios that can challenge the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

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

public class Main {
    // Memoization map for Grundy numbers
    static HashMap<Long, Integer> memo = new HashMap<>();

    public static void main(String[] args) throws IOException {
        // Use BufferedReader for fast input
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // Read all input at once
        StringBuilder sb = new StringBuilder();
        String line;
        // Read the first line: number of test cases
        line = br.readLine();
        int t = Integer.parseInt(line.trim());
        for (int test = 0; test < t; test++) {
            // Read n
            line = br.readLine();
            while (line != null && line.trim().isEmpty()) {
                line = br.readLine(); // Skip empty lines
            }
            if (line == null) break;
            int n = Integer.parseInt(line.trim());
            // Read a_i
            long[] a = new long[n];
            int count = 0;
            while (count < n) {
                line = br.readLine();
                String[] parts = line.trim().split("\\s+");
                for (String part : parts) {
                    if (part.isEmpty()) continue;
                    a[count++] = Long.parseLong(part);
                    if (count == n) break;
                }
            }
            // Read x_i
            long[] x = new long[n];
            count = 0;
            while (count < n) {
                line = br.readLine();
                String[] parts = line.trim().split("\\s+");
                for (String part : parts) {
                    if (part.isEmpty()) continue;
                    x[count++] = Long.parseLong(part);
                    if (count == n) break;
                }
            }
            // Compute XOR of Grundy numbers
            int totalXOR = 0;
            for (int i = 0; i < n; i++) {
                int g = computeGrundy(x[i], a[i]);
                totalXOR ^= g;
            }
            if (totalXOR != 0) {
                sb.append("Alice\n");
            } else {
                sb.append("Bob\n");
            }
        }
        // Print all results at once
        System.out.print(sb.toString());
    }

    // Function to compute Grundy number with memoization
    static int computeGrundy(long x, long a) {
        if (x == 0) return 0;
        if (a == 0) return 0;
        long key = (x << 32) | a;
        if (memo.containsKey(key)) {
            return memo.get(key);
        }
        // Set to store reachable Grundy numbers
        boolean[] reachable = new boolean[32]; // Assume mex won't exceed 31
        long subset = x;
        while (subset != 0) {
            if (subset <= a) {
                long next_x = x - subset;
                int g = computeGrundy(next_x, a);
                if (g < reachable.length) {
                    reachable[g] = true;
                }
            }
            subset = (subset - 1) & x;
        }
        // Find mex
        int mex = 0;
        while (mex < reachable.length && reachable[mex]) {
            mex++;
        }
        // Memoize and return
        memo.put(key, mex);
        return mex;
    }
}

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

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

Think step by step.

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

1. **Edge Cases for Grundy Numbers**:
   - Cases where the Grundy number calculation might be incorrect.
   - Cases where the bitwise operations might not behave as expected.

2. **Large Values for `a` and `x`**:
   - Values close to the upper limit of `2^30` to test the limits of the algorithm.

3. **Special Cases for Bitwise Operations**:
   - Values where the bitwise AND operation might not produce the expected results.

4. **Multiple Piles**:
   - Different combinations of `a` and `x` values to test the XOR combination.

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

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

### Explanation:
1. **Test Case 1**:
   - Single pile with `a = 1` and `x = 1`.
   - This is a simple case where both `a` and `x` are 1, and the game should be trivial.

2. **Test Case 2**:
   - Two piles with `a = 1` and `x = 2` for the first pile, and `a = 2` and `x = 2` for the second pile.
   - This tests the case where the number of stones in the first pile is not a valid move for the first player.

3. **Test Case 3**:
   - Three piles with `a = 1, 2, 3` and `x = 7, 6, 5`.
   - This tests a combination of different `a` and `x` values to see if the Grundy number calculation is correct.

4. **Test Case 4**:
   - Three piles with `a = 2, 3, 4` and `x = 7, 6, 5`.
   - This tests a different combination of `a` and `x` values to ensure the XOR combination is correct.

5. **Test Case 5**:
   - Three piles with `a = 3, 4, 5` and `x = 7, 6, 5`.
   - This is another combination to test the robustness of the algorithm.

These test cases cover a range of scenarios and should help expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;
#define int long long
signed main() {
    int t;
    cin >> t;

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

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

        int nim_sum = 0;

        // Calculate the valid moves for each pile
        for (int i = 0; i < n; ++i) {
            int max_remove = x[i] > a[i] ? a[i] : x[i]; // the maximum removable stones
            // Calculate possible moves
            for (int d = 1; d <= max_remove; ++d) {
                if ((x[i] & d) == d) { // Valid move condition
                    nim_sum ^= d; // Use d for nim-sum calculation
                }
            }
        }

        // Determine the winner based on nim_sum
        if (nim_sum != 0) {
            cout << "Bob" << endl; // Bob wins
        } else {
            cout << "Alice" << endl; // Alice wins
        }
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
#include<bits/stdc++.h>
#include<iostream>
//#include<cstdio>
using namespace std;
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
#define int long long
typedef long long ll;
#define F first
#define S second
#define pb emplace_back
int a[10005],b[10005];

signed main()
{
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int now=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        for(int i=0;i<n;i++)
        {
            cin>>b[i];
        }
        for(int i=0;i<n;i++)
        {
            if(b[i]>a[i])
            {
                for(int j=30;j>=0;j--)
                {
                    //cout<<(1<<j)<<' ';
                    if((b[i]>=(1<<j))&&(a[i]<(1<<j)))
                    {
                        b[i]-=(1<<j);
                    }
                }
                if(b[i]>0)
                {
                    if(b[i]<=a[i])
                    {
                        now^=__builtin_popcount(b[i]);
                        //cout<<__builtin_popcount(b[i])<<' ';
                    }
                    else
                    {
                        for(int j=30;;j--)
                        {
                            if((a[i]&(1<<j))>0)
                            {
                                // j >= 1
                                if((a[i]&(1<<(j-1)))>0)
                                {
                                    if(((b[i]&a[i])==a[i]))
                                    {
                                        bool ye=0;
                                        for(int k=0;k<30;k++)
                                        {
                                            if((a[i]&(1<<k))==0&&a[i]+(1<<k)==b[i])
                                            {
                                                ye=1;
                                                break;
                                            }
                                        }
                                        if(!ye) now^=__builtin_popcount(b[i]);
                                    }
                                    else now^=__builtin_popcount(b[i]);

                                }
                                else
                                {
                                    int b_min=0,a_max=-1;
                                    for(int k=0;;k++)
                                    {
                                        if((b[i]&(1<<k))>0)
                                        {
                                            b_min=k;
                                            break;
                                        }
                                    }
                                    for(int k=j-1;k>=0;k--)
                                    {
                                        if((a[i]&(1<<k))>0)
                                        {
                                            a_max=k;
                                            break;
                                        }
                                    }
                                    if(a_max>=b_min) now^=__builtin_popcount(b[i]);
                                    else
                                    {
                                        int x=__builtin_popcount(b[i]);
                                        if(x%2==0) x-=2;
                                        now^=x;
                                    }
                                }
                                break;
                            }
                        }

                    }

                }
            }
            else
            {
                now^=__builtin_popcount(b[i]);
                //cout<<__builtin_popcount(b[i])<<' ';
            }
        }
        if(now==0) cout<<"Bob"<<'\n';
        else cout<<"Alice"<<'\n';

    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the conditions for removing stones might be tricky. Specifically, we need to consider cases where the bitwise AND condition is not straightforward, and the values of \(a_i\) and \(x_i\) are such that the algorithm might fail to correctly determine the winner.

Here are the steps to generate such a test input:

1. **Small Test Case**: Since we need a small-scale test input, we will use a single test case with a small number of piles.
2. **Edge Cases**: We will choose values for \(a_i\) and \(x_i\) such that the bitwise AND condition is not trivial and might expose the bug.
3. **Optimal Play**: Ensure that the values are such that the optimal play is not immediately obvious, requiring the algorithm to correctly handle the conditions.

Let's consider the following test case:

- **Number of Piles**: 3
- **Values of \(a_i\)**: 15, 7, 1
- **Values of \(x_i\)**: 14, 7, 1

Explanation:
- For the first pile (14 stones, 15 limit):
  - Alice can take 14 stones (14 & 14 = 14), leaving 0 stones.
  - Bob cannot make a move, so Alice wins.
- For the second pile (7 stones, 7 limit):
  - Alice can take 7 stones (7 & 7 = 7), leaving 0 stones.
  - Bob cannot make a move, so Alice wins.
- For the third pile (1 stone, 1 limit):
  - Alice can take 1 stone (1 & 1 = 1), leaving 0 stones.
  - Bob cannot make a move, so Alice wins.

This test case should expose the bug if the algorithm incorrectly handles the bitwise AND condition or the optimal play.

Here is the test input:

```plaintext
1
3
15 7 1
14 7 1
```

This input will ensure that the algorithm correctly determines that Alice wins the game.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
using namespace std;
#include <bits/stdc++.h>
struct VectorHash {
    std::size_t operator()(const std::vector<int>& v) const {
        std::hash<int> hasher;
        std::size_t seed = 0;
        for (int i : v) {
            // Combine the hash of the current element with the accumulated hash
            seed ^= hasher(i) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
        }
        return seed;
    }
};
vector<unordered_map<vector<int>, int, VectorHash>>dp(2);

int recsolve(vector<int>&x, vector<int>&a, int turn){
    int res = (turn+1)%2;

    if(dp[turn].count(x)){
        return dp[turn][x];
    }

    for(int i=0; i<x.size(); i++){
        for(int d=1; d<=min(x[i], a[i]); d++){
            if((x[i]&d)==d){
                x[i]-=d;
                if(turn){
                    res = max(res, recsolve(x, a, 0));
                }
                else{
                    res = min(res, recsolve(x, a, 1));
                }
                x[i]+=d;
            }
        }
    }

    return dp[turn][x] = res;
}

int grundy(int x, int a, map<int, map<int, int>> &mp){
    if(x==0){
        return 0;
    }
    if(mp.count(x)&& mp[x].count(a)){
        return mp[x][a];
    }
    if(x<=a && mp.count(x)&& mp[x].count(x)){
        return mp[x][x];
    }

    bitset<10001> childGrundy;

    for(int d=1; d<=min(x, a); d++){
        if((x&d)==d){
            childGrundy|=(1<<grundy(x-d, a, mp));
        }
    }

    int sol = 10001;
    for(int k=0; k<10001; k++){
        if(!childGrundy.test(k)){
            sol = k;
            break;
        }
    }

    if(x<=a){
        mp[x][x] = sol;
    }

    return mp[x][a] = sol;
}

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

    map<int, map<int, int>> mp;
    // cout<<grundy(1, 1, mp)<<endl;
    int res = grundy(x[0], a[0], mp);
    for(int i=1; i<n; i++){
        res^=grundy(x[i], a[i], mp);
    }

    // int res = recsolve(x, a, 0);

    if(!res){
        cout<<"Bob"<<endl;
    }
    else{
        cout<<"Alice"<<endl;
    }
}

int main(){
    int t;
    cin>>t;

    for(int i=0; i<t; i++){
        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
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
import sys
sys.setrecursionlimit(1 << 25)

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    idx = 0
    T = int(data[idx])
    idx += 1
    res = []
    for _ in range(T):
        N = int(data[idx])
        idx += 1
        A = list(map(int, data[idx:idx+N]))
        idx += N
        X = list(map(int, data[idx:idx+N]))
        idx += N
        memo = {}

        def get_grundy(x, a):
            if x == 0:
                return 0
            key = (x, a)
            if key in memo:
                return memo[key]
            moves = set()
            # Find all possible moves d
            for d in range(1, min(a, x)+1):
                if x & d == d:
                    moves.add(get_grundy(x - d, a))
            # Find the mex (minimum excludant)
            g = 0
            while g in moves:
                g += 1
            memo[key] = g
            return g

        total_grundy = 0
        for a, x in zip(A, X):
            total_grundy ^= get_grundy(x, a)

        if total_grundy == 0:
            res.append("Bob")
        else:
            res.append("Alice")

    print("\n".join(res))

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:
```plaintext
1
1
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
def compute_grundy(x, a):
    """
    Compute the Grundy number for a single pile with current stones x and maximum removal a.
    Allowed moves are d where x & d = d and 1 <= d <= a.
    """
    # Initialize a set to hold Grundy numbers of reachable positions
    reachable = set()

    # Enumerate all subsets of x (represented as bits)
    subset = x
    while subset:
        # Ensure d <= a
        if subset <= a:
            # Compute the next position
            next_x = x - subset
            # Recursively compute the Grundy number for the next position
            reachable.add(compute_grundy(next_x, a))
        # Move to the next subset
        subset = (subset - 1) & x

    # Compute the mex (minimum excludant) of reachable Grundy numbers
    mex = 0
    while mex in reachable:
        mex += 1
    return mex

def determine_winner(test_cases):
    results = []
    for case in test_cases:
        n, a, x = case
        total_xor = 0
        for ai, xi in zip(a, x):
            g = compute_grundy(xi, ai)
            total_xor ^= g
        if total_xor != 0:
            results.append("Alice")
        else:
            results.append("Bob")
    return results

# Read input
import sys
from functools import lru_cache

def main():
    sys.setrecursionlimit(1 << 25)
    input = sys.stdin.read
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    test_cases = []
    for _ in range(t):
        n = int(data[idx])
        idx += 1
        a = list(map(int, data[idx:idx + n]))
        idx += n
        x = list(map(int, data[idx:idx + n]))
        idx += n
        test_cases.append((n, a, x))

    # Optimize compute_grundy with memoization
    @lru_cache(maxsize=None)
    def compute_grundy_memo(x, a):
        reachable = set()
        subset = x
        while subset:
            if subset <= a:
                next_x = x - subset
                reachable.add(compute_grundy_memo(next_x, a))
            subset = (subset - 1) & x
        mex = 0
        while mex in reachable:
            mex += 1
        return mex

    results = []
    for case in test_cases:
        n, a, x = case
        total_xor = 0
        for ai, xi in zip(a, x):
            g = compute_grundy_memo(xi, ai)
            total_xor ^= g
        if total_xor != 0:
            results.append("Alice")
        else:
            results.append("Bob")
    for res in results:
        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Maximum removal constraint (`a_i`)**: The code should handle cases where `a_i` is very small or very large.
2. **Bitwise AND condition**: The code should handle cases where the bitwise AND condition is particularly tricky.
3. **Grundy number computation**: The code should handle cases where the Grundy number calculation is complex or has edge cases.

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

### Test Case:
- **Number of test cases (`t`)**: 1
- **Number of piles (`n`)**: 3
- **Values of `a_i`**: [1, 31, 2]
- **Values of `x_i`**: [1, 31, 2]

### Explanation:
- **Pile 1**: `a_1 = 1`, `x_1 = 1`
  - Alice can only remove 1 stone, and the pile will be empty. Bob cannot make a move, so Alice wins.
- **Pile 2**: `a_2 = 31`, `x_2 = 31`
  - Alice can remove any number of stones from 1 to 31. If Alice removes 31 stones, the pile will be empty, and Bob cannot make a move, so Alice wins.
- **Pile 3**: `a_3 = 2`, `x_3 = 2`
  - Alice can remove either 1 or 2 stones. If Alice removes 2 stones, the pile will be empty, and Bob cannot make a move, so Alice wins.

### Test Input:
```plaintext
1
3
1 31 2
1 31 2
```

This test case is designed to expose the following issues:
- The code should correctly handle the case where `a_i` is very small (1).
- The code should correctly handle the case where `a_i` is very large (31).
- The code should correctly handle the case where the bitwise AND condition is straightforward (1, 31, 2).

This input should result in Alice winning, and if the code fails to correctly compute the Grundy numbers or handle the bitwise AND condition, it will fail to produce the correct result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class BitGame {

    private static Map<String, Integer> memo = new HashMap<>();

    public static int goku(int naruto, int luffy) {
        if (naruto == 0) return 0;
        String key = naruto + "," + luffy;
        if (memo.containsKey(key)) return memo.get(key);

        boolean[] ichigo = new boolean[naruto + 1];
        for (int d = 1; d <= luffy; d++) {
            if (naruto >= d && (naruto & d) == d) {
                ichigo[goku(naruto - d, luffy)] = true;
            }
        }

        int mex = 0;
        while (ichigo[mex]) {
            mex++;
        }

        memo.put(key, mex);
        return mex;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        StringBuilder result = new StringBuilder();

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int[] akira = new int[n];
            int[] sakura = new int[n];

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

            int zoro = 0;
            for (int j = 0; j < n; j++) {
                zoro ^= goku(sakura[j], akira[j]);
            }

            result.append(zoro == 0 ? "Bob\n" : "Alice\n");
        }
        System.out.print(result.toString());
        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:
To expose the bug in the provided code, we need to focus on edge cases where the bitwise AND condition and the constraints on the number of stones and the allowed moves can lead to unexpected behavior. The bug in the code might be related to how it handles the memoization and the game states.

Let's consider the following test case:

1. A test case where the number of stones in a pile is a power of 2, and the allowed moves are also a power of 2.
2. A test case where the number of stones and the allowed moves are such that no valid move can be made.
3. A test case where the number of stones and the allowed moves are such that the game state is complex, but still solvable.

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

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

Explanation of the test cases:
1. **Test Case 1:**
   - 1 pile with 1 stone and 1 allowed move.
   - Alice can take the only stone, and Bob loses.
   - Expected output: "Alice"

2. **Test Case 2:**
   - 1 pile with 2 stones and 1 allowed move.
   - Alice cannot take 2 stones because 2 > 1.
   - Bob loses because no valid move can be made.
   - Expected output: "Bob"

3. **Test Case 3:**
   - 2 piles with 1 and 2 stones, and 1 and 2 allowed moves.
   - Alice can take 1 stone from the first pile, leaving 0 stones.
   - Bob cannot make a move from the first pile, and the second pile has 2 stones and 2 allowed moves.
   - Alice can take 2 stones from the second pile, and Bob loses.
   - Expected output: "Alice"

This test input should help in identifying the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
def is_valid_move(stones, max_take):
    """
    Возвращает список допустимых ходов для текущей кучи
    stones: текущее количество камней
    max_take: максимальное количество камней, которое можно взять (a[i])
    """
    valid_moves = []
    for d in range(1, min(stones + 1, max_take + 1)):
        # Проверяем оба условия:
        # 1. 1 ≤ d ≤ a[i]  (уже учтено в range)
        # 2. x & d == d, где x - текущее количество камней
        if (stones & d) == d:
            valid_moves.append(d)
    return valid_moves

def grundy_number(stones, max_take, memo=None):
    """
    Вычисляет число Гранди для позиции
    """
    if memo is None:
        memo = {}

    if stones == 0:
        return 0

    if stones in memo:
        return memo[stones]

    # Получаем все возможные ходы
    moves = is_valid_move(stones, max_take)

    if not moves:
        return 0

    # Находим все возможные следующие значения Гранди
    next_grundy = set()
    for move in moves:
        next_grundy.add(grundy_number(stones - move, max_take, memo))

    # Находим минимальное исключенное число (mex)
    grundy = 0
    while grundy in next_grundy:
        grundy += 1

    memo[stones] = grundy
    return grundy

def solve_game():
    n = int(input())
    a = list(map(int, input().split()))
    x = list(map(int, input().split()))

    # Вычисляем XOR всех чисел Гранди
    xor_sum = 0
    for i in range(n):
        xor_sum ^= grundy_number(x[i], a[i])

    # Если XOR sum != 0, первый игрок (Алиса) выигрывает
    # Если XOR sum == 0, второй игрок (Боб) выигрывает
    return "Alice" if xor_sum != 0 else "Bob"

# Чтение количества тестов и решение каждого из них
t = int(input())
for _ in range(t):
    print(solve_game())

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

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

Think step by step.

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

1. **Identify Potential Bugs**:
   - The code uses a memoization technique to store previously computed Grundy numbers, which is generally efficient.
   - The main potential issue lies in the `is_valid_move` function, where the conditions for valid moves are checked. Specifically, the condition `if (stones & d) == d` might be the source of the problem.

2. **Edge Cases**:
   - **Single Pile with Small Values**: Test with a single pile where the number of stones and the maximum take are small.
   - **Single Pile with Large Values**: Test with a single pile where the number of stones and the maximum take are large.
   - **Multiple Piles with Mixed Values**: Test with multiple piles where the values of stones and maximum take are mixed to see if the XOR sum calculation is correct.

3. **Challenging Scenarios**:
   - **No Valid Moves**: A scenario where no valid moves can be made from a pile.
   - **Symmetric Piles**: Piles with symmetric values that should result in a draw (XOR sum of 0).

Based on these considerations, here is a small-scale test input that is complete and valid:

```plaintext
4
1
1
1
1
2
2
1
3
3
3
1
5
15
15
```

Explanation of the test cases:
1. **Single Pile with 1 stone and 1 maximum take**:
   - Alice cannot make a move, so Bob wins.
2. **Single Pile with 2 stones and 2 maximum take**:
   - Alice can take 1 or 2 stones, but Bob can always win by taking the remaining stones.
3. **Single Pile with 3 stones and 3 maximum take**:
   - Alice can take 1 or 3 stones, but Bob can always win by taking the remaining stones.
4. **Three Piles with 1, 5, and 15 stones and 15 maximum take**:
   - This is a more complex scenario where the XOR sum needs to be calculated correctly.

This test input should help in exposing any bugs in the provided code, especially in the `is_valid_move` function and the Grundy number computation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bit Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of this problem. The only difference is that you need to output the winner of the game in this version, and the number of stones in each pile are fixed. You must solve both versions to be able to hack.

Alice and Bob are playing a familiar game where they take turns removing stones from $n$ piles. Initially, there are $x_i$ stones in the $i$-th pile, and it has an associated value $a_i$. A player can take $d$ stones away from the $i$-th pile if and only if both of the following conditions are met:

  * $1 \le d \le a_i$, and    * $x \, \& \, d = d$, where $x$ is the current number of stones in the $i$-th pile and $\&$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The player who cannot make a move loses, and Alice goes first.

You're given the $a_i$ and $x_i$ values for each pile, please determine who will win the game if both players play optimally.

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

The first line of each test case contains $n$ ($1 \le n \le 10^4$) — the number of piles.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i < 2^{30}$).

The third line of each test case contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \le x_i < 2^{30}$).

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

Output: Print a single line with the winner's name. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
2
1 6
10 7
3
10 8 15
25 4 14
4
8 32 65 64
7 45 126 94
3
20 40 1
23 55 1
5
12345 9876 86419 8641 1
6789 54321 7532 97532 1
2
20 64
44 61
3
57 109 55
69 90 85
output:
Bob
Bob
Bob
Bob
Bob
Alice
Alice

Note: In the first test case, neither player can take any stones from the first pile since there is no value of $d$ satisfying the conditions. For the second pile, to begin with, Alice can remove between $1$ and $6$ stones. No matter which move Alice performs, Bob can remove the rest of the stones on his turn. After Bob's move, there are no more moves that Alice can perform, so Bob wins.

In the second test case, here is one example of how the game might go. Alice moves first, and she decides to remove from the first pile. She cannot take $17$ stones, because $17 > 10$, which fails the first condition. She cannot take $10$ stones, because $25 \, \& \, 10 = 8$ which fails the second condition. One option is to take $9$ stones; now the pile has $16$ stones left. On Bob's turn he decides to take stones from the second pile; the only option here is to take all $4$. Now, no more stones can be taken from either of the first two piles, so Alice must take some stones from the last pile. She decides to take $12$ stones, and Bob then follows by taking the last $2$ stones on that pile. Since Alice now has no legal moves left, Bob wins. It can be shown that no matter which strategy Alice follows, Bob will always be able to win if he plays optimally.

Buggy Code:
//#pragma optimize("O3")
//#pragma optimize("unroll-loops")
//#pragma traget("avx,avx2,tune=native")

#include <iostream>
#include <cmath>
#include <algorithm>
#include <stdio.h>
#include <cstdint>
#include <cstring>
#include <string>
#include <cstdlib>
#include <vector>
#include <bitset>
#include <map>
#include <queue>
#include <ctime>
#include <stack>
#include <set>
#include <list>
#include <random>
#include <deque>
#include <functional>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <complex>
#include <numeric>
#include <cassert>
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <thread>

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define el '\n'
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define point pair <ll, ll>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

using namespace std;
mt19937 rnd(time(0));

const ll INF = 2e18 + 10;
const ll inf = 2e9 + 10;
const ll N = 1e5 + 10;
const ll mod = 998244353;
const ll LOG = 18;

ll g(ll a, ll x) {
    ll b = 0;
    bool good = 0;
    for (ll i = 29; i >= 0; i--) {
        if (!(x >> i & 1) && (a >> i & 1)) {
            good = 1;
            continue;
        }

        if (!(x >> i & 1))
            continue;

        if ((a >> i & 1) || good)
            b |= 1 << i;
    }

    a = 0;
    ll id = 0;
    for (ll i = 0; i < 30; i++)
        if (x >> i & 1)
            a |= (1 << id) * (b >> i & 1), id++;

    for (ll i = 0; i < 30; i++)
        if ((1 << i) == a)
            return i ^ 1;
    for (ll i = 0; i < 30; i++)
        if ((1 << i) - 1 == a)
            return i;
    for (ll i = 1; i < 30; i++)
        if ((1 << i) - 2 == a)
            return 0;
    for (ll i = 0; i < 31; i++)
        if ((1 << i) > a)
            return i;
}

void solve() {
    ll n;
    cin >> n;

    vector <ll> a(n), x(n);
    for (ll i = 0; i < n; i++)
        cin >> a[i];
    for (ll i = 0; i < n; i++)
        cin >> x[i];

    ll nim = 0;
    for (ll i = 0; i < n; i++)
        nim ^= g(a[i], x[i]);

    cout << (nim ? "Alice" : "Bob") << el;
    return;
}
signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);

    int tests = 1;
    cin >> tests;

    while (tests--)
        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
2
1 6
10 7
```