**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 ll long long
using namespace std;
const ll MOD = 1e9 + 7;  // Use ll for MOD
int main() {
  int t;
  cin >> t;
  while (t--) {
    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_sum = 0;
    for (ll i = 0; i < n; ++i) {
      ll valid_x = 0;
      for (ll d = 1; d <= a[i]; ++d) {
        if ((x[i] & d) == d) {
          valid_x ^= d;
        }
      }
      nim_sum ^= valid_x;
    }

    if (nim_sum == 0) {
      cout << "Alice" << endl;
    } else {
      cout << "Bob" << endl;
    }
  }

  return 0;
}

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

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

Think step by step.

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


**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(values):
    """Compute the minimum excludant of a list of non-negative integers."""
    m = 0
    while m in values:
        m += 1
    return m

def grundy(x, a):
    """Compute the Grundy number for a pile with x stones and max removal a."""
    if x == 0:
        return 0
    if (x, a) in memo:
        return memo[(x, a)]

    valid_moves = []
    for d in range(1, min(a, x) + 1):
        if x & d == d:
            valid_moves.append(grundy(x - d, a))

    g = mex(valid_moves)
    memo[(x, a)] = g
    return g

def solve_game(piles):
    global memo
    memo = {}
    nim_sum = 0
    for x, a in piles:
        nim_sum ^= grundy(x, a)
    return "Alice" if nim_sum > 0 else "Bob"

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

    index = 0
    T = int(data[index])
    index += 1
    results = []

    for _ in range(T):
        n = int(data[index])
        index += 1
        a_values = list(map(int, data[index:index+n]))
        index += n
        x_values = list(map(int, data[index:index+n]))
        index += n

        piles = list(zip(x_values, a_values))
        result = solve_game(piles)
        results.append(result)

    for result in results:
        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
1
1
1
1
2
2
2
2
3
1 2 3
3 2 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:
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:
```plaintext
1
2
1 6
10 7
```

This test input is designed to expose the bug in the provided code. The input consists of a single test case with two piles. The first pile has 10 stones and an associated value of 1, and the second pile has 7 stones and an associated value of 6.

In this scenario:
- For the first pile, the only possible move is to take 1 stone, but this would violate the condition \( x \, \& \, d = d \) because \( 10 \, \& \, 1 = 0 \).
- For the second pile, the possible moves are to take 1, 2, 4, or 6 stones. However, no matter which move Alice makes, Bob can always respond in a way that leaves Alice with no valid moves.

Thus, Bob should win this game if both players play optimally. This test case should help in identifying any issues in the code's logic for determining the winner.


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

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
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:
//Submission: 1732934885.1024504

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>

using namespace std;

unordered_map<long long, unordered_map<int, int>> memo;

int sg(int x, int a) {
    if (memo.count(x) && memo[x].count(a)) {
        return memo[x][a];
    }

    vector<bool> seen(200, false); // Assuming SG values are within this range (can adjust if needed)
    for (int d = 1; d <= a; ++d) {
        if ((x & d) == d) {
            seen[sg(x - d, a)] = true;
        }
    }

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

    memo[x][a] = mex;
    return mex;
}

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

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

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

        memo.clear(); // Clear memoization for each test case

        int nimSum = 0;
        for (int i = 0; i < n; ++i) {
            nimSum ^= sg(x[i], a[i]);
        }

        if (nimSum == 0) {
            cout << "Bob" << endl;
        } else {
            cout << "Alice" << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
1
1
1
2
1
2
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: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
#include<stdio.h>
#include<stdlib.h>
#define MAXN 100001
int cmp(const void *a,const void *b){
	return (*(int*)a-*(int*)b);
}
int main(){
	int N,n,i,j,str[MAXN],arr[MAXN],sum,flag;
	scanf("%d",&N);
	while(N--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%d",&str[i]);
		}
		for(int i=1;i<=n;i++){
			scanf("%d",&arr[i]);
		}
		qsort(str+1,n,sizeof(str[1]),cmp);
		qsort(arr+1,n,sizeof(arr[1]),cmp);
		sum=0,flag=0;
		int j=n,num=0,a,b;
		for(int i=1;i<=n;i++){
			if(str[j]>arr[i]){
				sum+=(str[j]-arr[i]);
				j--;
			}else{
				flag=1;
				a=i;
				break;
			}
		}
		if(flag==1){
			sum+=str[j];
			num=sum-str[j]-(str[j+1]-arr[a-1])+str[j+1];
		}else{
			num=sum-(str[1]-arr[n])+str[1];
		}
		printf("%d\n",sum>num?sum:num);
	}
	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
10 10 10
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
#include <iostream>
using namespace std;

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

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

        int count = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] < b[i]) {
                count++;
            }
        }

        if (count == n) {

            cout << a[n - 1] << endl;
        } else if (n == 1) {
            cout << a[0] << endl;
        } else if (n == 2) {
            cout << a[0] + a[1] - b[1] << endl;
        } else {
            int sum1 = 0, sum2 = 0;

            for (int i = 0; i < n - 1; i++) {
                if (a[i] > b[i + 1]) {
                    sum1 += a[i];
                }
            }

            for (int i = 1; i < n; i++) {
                if (b[i] < a[i - 1]) {
                    sum2 += b[i];
                }
            }

            cout << sum1 +a[n-1]- sum2 << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
1
100
1
2
1 1
2 1
3
1 100 1
1 1 100
4
100 1 100 1
1 100 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
#include<iostream>
#include<string>
using namespace std ;

void solve(){
  int n ;
  cin >> n ;
  int *a = new int[n] , *b = new int[n] ;
  for(int i = 0 ; i < n ; i++) cin >> a[i] ;
  for(int i = 0 ; i < n ; i++) cin >> b[i] ;
  int ans = a[n-1] ;
  for(int i = 0 ; i < n-1 ; i++){
    int x = (a[i]-b[i+1]);
    int y = 0 ;
    bool flag = false ;
    if((i+2) < n){
       y = a[i+1] - b[i+2] ;
       flag = true ;
    }
    if(x < 0 && y < 0) continue ;
    if(y > x){
      i++ ;
      ans += y ;
    }else ans += x;
  }
  cout << ans << endl ;
}

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

    int t ;
    cin >> t ;
    while(t--){
      solve();
    }
	return 0 ;
}

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

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

Think step by step.

Response:
```plaintext
3
3
10 20 30
1 2 3
4
10 1 1 10
1 10 10 1
5
1 2 3 4 5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
# cook your dish here
import sys
import heapq
import math
from itertools import permutations
from typing import Dict, Tuple, List

# Node of segmentTree
class TreeNode:
    def __init__(self, val:int ,arr : List[int], left = None, right = None):
        self.val = val
        self.left = left
        self.right = right
        self.range = arr

# full segmentTree
class segmentTree:
    def __init__(self):
        return

    def build(self , arr : List[int] , l ,r):
        n = len(arr)
        if r-l == 0:
            return TreeNode(arr[l] , [l,r])
        if -1 < l < r < n:
            mid = (l+r)//2
            left = self.build(arr , l , mid)
            right = self.build(arr,mid+1,r)
            total = left.val+right.val
            root = TreeNode(total , [l,r] , left , right)
            return root

    def update(self, root ,index: int, diff:int ):
        if root == None:
            return
        if root.range[0] == root.range[1] == index:
            root.val-=diff
            return
        if root.range[0] <= index and index <= root.range[1]:
            root.val-=diff
            self.update(root.left ,index, diff )
            self.update(root.right ,index, diff )

    def range(self, root ,index1: int, index2:int ):
        if root == None:
            return 0
        if index1 < index2 < root.range[0] < root.range[1]:
            return 0
        if index1 <= root.range[0] <= root.range[1] <= index2:
            return root.val
        l = self.range(root.left , index1 ,index2)
        r = self.range(root.right , index1 , index2)
        return l+r

    def display(self, root):
        if root != None:
            print(root.val , root.range)
            self.display(root.left)
            self.display(root.right)

class Solution:
    def minimum_swap(self,nums):
        def myfun(x):
            return x[0]
        arr = []
        for i in range(len(nums)):
            arr.append([nums[i] , i])
        arr.sort(key = myfun)
        ans= 0
        for i in range(len(nums)):
            while arr[i][1] != i:
                temp = arr[i]
                t = arr[i][1]
                arr[i] = arr[t]
                arr[t] = temp
                ans+=1
        return ans

    def permutation(self , arr):
        permutation = [''.join(p) for p in permutations(arr)]
        return permutation

    def substring(self , arr):
        res = set()
        res = [arr[i: j] for i in range(len(arr))
            for j in range(i + 1, len(arr) + 1)]
        return res

    def lexicographically(self, arr):
        strings = [sub for sub in arr if isinstance(sub, str)]
        res = heapq.nsmallest(1, strings)[0]
        return res

    def printPowerSet(self ,set,set_size):
        res = []
        a =""
        s = []
        for i in range(len(set)):
            for j in range(i+1 , len(set)):
                s.append(set[i])
                s.append(set[j])
                aa =self.lexicographically(s)
                a+= aa
                res.append(a)
                if aa != set[j]:
                    a+=set[j]
                if aa != set[i]:
                    a+=set[i]
        return res

    def CalculatePower(self,N,X):
      P=1
      for i in range(1, X+1):
        P=P*N
      return P

    def BFS_in_grid(self, grid: List[List[int]] , n:int , m:int , cost: int, match , dic , s,e) ->int:
        queue = [(cost,s,e)]
        vist = set()
        vist.add((s,e))
        while queue:
            c , ind1 , ind2 = heapq.heappop(queue)
            for i,j in dic:
                x , y = ind1+i , ind2+j
                if -1 < x < n and -1 < y < m and match == grid[x][y] and (x,y) not in vist:
                    heapq.heappush(queue , (c+1 , x ,y))
                    vist.add((x,y))
            if ind1 == n-1 and ind2 == m-1:
                return c
        return -1

    def call(self):
        n = int(input())-1
        a = list(map(int,input().split()))
        b = list(map(int,input().split()))
        a.sort()
        b.sort()
        maxa = summ1 = summ2 = 0
        for i in range(n+1):
            summ1+=a[n-i]
            # print(summ1 , summ2)
            maxa = max(maxa , (summ1-summ2))
            summ2+=b[i]

        return [maxa]

obj = Solution()
t = int(input())
for i in range(t):
    arr = obj.call()
    for i in arr:
        print(i ,end="")
    print()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
# https://codeforces.com/contest/2051/problem/A
t = int(input())

for _ in range(t):
    n = int(input())
    m = list(map(int,input().split()))
    s = list(map(int,input().split()))
    s.append(0)
    m.sort(reverse = True)
    s.sort()
    result = 0
    for i in range(n):
        if m[i] > s[i]:
            result += m[i]-s[i]
    print(result)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify its issues. The main problem in the buggy code is that it sorts the lists `m` and `s`, which does not respect the original order of the days. This can lead to incorrect results because the training schedule of Monocarp and Stereocarp is dependent on the original order of the days.

### Steps to Generate the Test Input:
1. **Create a small-scale test case** where the original order of the days is crucial.
2. **Ensure that sorting the lists will produce a different and incorrect result** compared to the correct solution.
3. **Use values that highlight the issue with the sorting**.

### Example Test Input:
Let's create a test case with 3 days where the values of `m` and `s` are such that sorting will give a different result.

```plaintext
1
3
10 1 1
1 10 1
```

### Explanation:
- **Day 1**: Monocarp solves 10 problems, Stereocarp solves 1 problem.
- **Day 2**: Monocarp solves 1 problem, Stereocarp solves 10 problems.
- **Day 3**: Monocarp solves 1 problem, Stereocarp solves 1 problem.

### Correct Solution:
- If Monocarp trains on Day 1 and Day 3, he will solve 10 + 1 = 11 problems.
- Stereocarp will train on Day 2 and Day 3, solving 10 + 1 = 11 problems.
- The difference is 11 - 11 = 0.

### Buggy Code Output:
- The buggy code sorts `m` and `s`:
  - `m` becomes `[10, 1, 1]`
  - `s` becomes `[1, 1, 10]`
- The result calculation will be:
  - Day 1: 10 - 1 = 9
  - Day 2: 1 - 1 = 0
  - Day 3: 1 - 10 = -9
- The total result is 9 + 0 - 9 = 0, which is incorrect.

### Challenging Test Input:
```plaintext
1
3
10 1 1
1 10 1
```

This test input will expose the bug in the buggy code because the sorting will produce a different and incorrect result compared to the correct solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
t = input('')
for i in range(int(t)):
    n = int(input(''))
    a_i = input('')
    b_i= input('')
    a_i = a_i.split(' ')
    b_i = b_i.split(' ')
    max_val = 0
    for i in range(n-1):
        if a_i[i] > b_i[i+1]:
            max_val+=int(a_i[i])-int(b_i[i+1])
    max_val+=int(a_i[n-1])
    print(max_val)

**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 scenarios where the code's logic fails. The main issue in the code is that it only considers the difference between the number of problems Monocarp solves on a day and the number of problems Stereocarp solves on the next day. It does not properly account for the optimal strategy to maximize the difference.

Here are the steps to generate a challenging test input:

1. **Small-scale Test Case**: Since the input should be small-scale, we will use a small value of `n`.
2. **Edge Case Analysis**:
   - **Case 1**: Monocarp should train on the last day to maximize the difference.
   - **Case 2**: Monocarp should train on a day where the difference between `a_i` and `b_{i+1}` is maximized.
   - **Case 3**: Monocarp should train on multiple days to maximize the difference, but the code might miss some of these days.

Let's create a test case where the optimal strategy involves training on the last day and a few other days to maximize the difference.

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

**Explanation**:
1. **First Test Case**:
   - `n = 4`
   - `a = [10, 1, 1, 10]`
   - `b = [1, 10, 1, 1]`
   - Monocarp should train on days 1 and 4 to maximize the difference:
     - Day 1: Monocarp solves 10 problems, Stereocarp solves 10 problems on day 2.
     - Day 4: Monocarp solves 10 problems, Stereocarp does not train on day 5.
   - Total difference: `(10 - 10) + (10 - 0) = 10`

2. **Second Test Case**:
   - `n = 3`
   - `a = [1, 2, 3]`
   - `b = [3, 2, 1]`
   - Monocarp should train on the last day:
   - Day 3: Monocarp solves 3 problems, Stereocarp does not train.
   - Total difference: `3 - 0 = 3`

3. **Third Test Case**:
   - `n = 5`
   - `a = [5, 1, 5, 1, 5]`
   - `b = [1, 5, 1, 5, 1]`
   - Monocarp should train on days 1, 3, and 5:
     - Day 1: Monocarp solves 5 problems, Stereocarp solves 5 problems on day 2.
     - Day 3: Monocarp solves 5 problems, Stereocarp solves 5 problems on day 4.
     - Day 5: Monocarp solves 5 problems, Stereocarp does not train.
   - Total difference: `(5 - 5) + (5 - 5) + (5 - 0) = 5`

This test input should expose the bug in the provided code by highlighting scenarios where the code's logic fails to maximize the difference.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

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

public class Main {
    public static void maximize(int ar[],int br[],int n){
        int ans1=0,ans2=0;
        if(n==1){
            System.out.println(ar[0]);
            return ;
        }
        int flag=0;
        for(int i=0;i<n;i++){
            //System.out.println(i);
            if(i==0 && i+1<n){
                if(ar[i]+ar[i+1]-br[i+1]>0){
                    ans1+=ar[i];
                    ans2+=br[i+1];
                    flag=1;
                }
                // System.out.println(ans1);
                // System.out.println(ans2);
            }
            else if(i>0 && i+1<n){
                // System.out.println("i->"+i);
                // System.out.println("ar[i]->"+ar[i]);
                // System.out.println("br[i]->"+br[i+1]);

                if(flag==1 && ar[i]+ar[i+1]-br[i+1]-br[i]>0 &&ar[i]>=br[i]){
                    //System.out.println(i);
                    ans1+=ar[i];
                    ans2+=br[i+1];
                    flag=1;
                }
                else if(flag==0 && ar[i]+ar[i+1]-br[i+1]>0){
                    ans1+=ar[i];
                    ans2+=br[i+1];
                    // System.out.println(ans1);
                    // System.out.println(ans2);
                    flag=1;
                }

                else{
                    flag=0;
                }

            }
            if(i==n-1){
                    ans1+=ar[i];
                    //System.out.println(ans1);
            }

        }
        // System.out.println(ans1);
        // System.out.println(ans2);
        System.out.println(ans1-ans2);
        return;
    }
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t= Integer.parseInt(br.readLine());
        while(t-->0){
            int n=Integer.parseInt(br.readLine());
            int []ar= new int[n];
            int []cr= new int[n];
            String[] input = br.readLine().split(" ");
            for(int i=0;i<n;i++){
                ar[i]=Integer.parseInt(input[i]);
                //bw.write(ar[i]+" ");

            }
            // bw.newLine();
            String[] input1 = br.readLine().split(" ");
            for(int i=0;i<n;i++){
                cr[i]=Integer.parseInt(input1[i]);
                //bw.write(cr[i]+" ");

            }
            // bw.newLine();
            maximize(ar,cr,n);
            //bw.newLine();
        }
        bw.flush();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
2
1 100
100 1
1
100
100
3
1 2 3
3 2 1
6
100 2 3 4 5 6
1 2 3 4 5 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
import java.util.Scanner;

public class Olympiad {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        if (t >= 1 && t < 1000) {
            while (t-- > 0) {
                int n = sc.nextInt();
                if (n >= 1 && n <= 100) {
                    int[] a = new int[n];
                    int[] b = new int[n];
                    for (int i = 0; i < n; i++) {
                        a[i] = sc.nextInt();
                    }
                    for (int i = 0; i < n; i++) {
                        b[i] = sc.nextInt();
                    }
                    int maxDifference = 0;
                    for (int i = 0; i < n - 1; i++) {
                        if (a[i] > b[i + 1]) {
                            maxDifference += (a[i] - b[i + 1]);
                        }
                    }
                    maxDifference += a[n - 1];
                    System.out.println(maxDifference);
                }

            }
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3
10 1 10
1 10 1
```

**Explanation:**

- **Test Case:**
  - Number of days: 3
  - Problems solved by Monocarp: [10, 1, 10]
  - Problems solved by Stereocarp: [1, 10, 1]

- **Reasoning:**
  - The buggy code only considers the difference between `a[i]` and `b[i+1]` for `i < n-1`, and adds `a[n-1]` directly to the total difference.
  - This approach fails to correctly account for the optimal strategy when the last day has a high number of problems that Monocarp can solve.
  - In this test case, if Monocarp trains on days 1 and 3, he will solve 10 + 10 = 20 problems. Stereocarp will train on days 2 and 3, solving 10 + 1 = 11 problems.
  - The correct difference should be 20 - 11 = 9, but the buggy code will incorrectly output 11 (because it only considers the first two days and adds the last day's problems without considering the impact on Stereocarp).

This test case should expose the bug in the given code by highlighting the incorrect handling of the last day's problems.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

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

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

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

        int a[100], b[100];

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

        int max_diff = INT_MIN;

        // Check every subset of days Monocarp can train
        for (int mask = 1; mask < (1 << n); mask++) {
            int sum1 = 0, sum2 = 0;

            for (int i = 0; i < n; i++) {
                if (mask & (1 << i)) { // If day i is selected
                    sum1 += a[i];
                    if (i + 1 < n) sum2 += b[i + 1];
                }
            }

            max_diff = max(max_diff, sum1 - sum2);
        }

        cout << max_diff << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4
10 10 10 10
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: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n), b(n);
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        for (int i = 0; i < n; ++i) {
            cin >> b[i];
        }

        int max_diff = -1e9;
        for (int mask = 0; mask < (1 << n); ++mask) {
            int monocarp_sum = 0;
            int stereocarp_sum = 0;
            for (int i = 0; i < n; ++i) {
                if ((mask >> i) & 1) {
                    monocarp_sum += a[i];
                    if (i < n - 1 && ((mask >> i) & 1)) { // Corrected condition
                        stereocarp_sum += b[i + 1];
                    }
                }
            }
            max_diff = max(max_diff, monocarp_sum - stereocarp_sum);
        }
        cout << max_diff << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
2
1 100
100 1
1
100
1
3
100 1 1
1 100 1
6
100 1 1 1 1 100
1 100 1 100 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    max_diff = -float('inf')

    for i in range(2**n):
        mono_trains = [False] * n
        for j in range(n):
            if (i >> j) & 1:
                mono_trains[j] = True

        m = 0
        for k in range(n):
            if mono_trains[k]:
                m += a[k]

        s = 0
        stereo_trains = [False] * n
        for k in range(n - 1):
            if mono_trains[k]:
                stereo_trains[k + 1] = True

        for k in range(n):
            if stereo_trains[k]:
                s += b[k]

        max_diff = max(max_diff, m - s)

    print(max_diff)

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

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

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

Think step by step.

Response:
```plaintext
1
3
10 10 10
1 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
def max_difference(t, test_cases):
    results = []
    for case in test_cases:
        n, a, b = case
        max_diff = 0

        for mask in range(1 << n):
            m, s = 0, 0

            stereocarp_days = set()

            for i in range(n):
                if mask & (1 << i):
                    m += a[i]
                    if i + 1 < n:
                        stereocarp_days.add(i + 1)

            for day in stereocarp_days:
                s += b[day]

            max_diff = max(max_diff, m - s)

        results.append(max_diff)
    return results
t = int(input())
test_cases = []
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    test_cases.append((n, a, b))

output = max_difference(t, test_cases)
print(*output, sep="\n")

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

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

Think step by step.

Response:
```plaintext
4
2
100 1
1 100
1
100
100
3
1 100 1
1 1 100
6
100 1 100 1 100 1
100 100 1 100 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                b[i] = sc.nextInt();
            }

            int maxDifference = Integer.MIN_VALUE;

            for (int mask = 0; mask < (1 << n); mask++) {
                int m = 0, s = 0;
                for (int i = 0; i < n; i++) {
                    if ((mask & (1 << i)) != 0) {
                        m += a[i];
                        if (i + 1 < n) {
                            s += b[i + 1];
                        }
                    }
                }
                maxDifference = Math.max(maxDifference, m - s);
            }

            System.out.println(maxDifference);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
10 1 1 1 1
1 10 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
import java.util.Scanner;

public class problem1a {
      public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    int ans=0;
  for(int j=0;j<t;j++){
      int  n=sc.nextInt();
    int[] a =new int[n];
    int [] b= new int [n];

    for(int i=0;i<n;i++){
        a[i]=sc.nextInt();
    }
    for(int i=0;i<n;i++){
        b[i]=sc.nextInt();
    }
     ans=helper(0, 0, 0, a, b);
     System.out.println(ans);
}
;
    }
    public static int helper(int i,int m,int s,int[] a,int [] b){

    int n=a.length;
        if(i==n-1){
            m+=a[n-1];
            return m-s;
        }
        int take=helper(i+1,m+a[i],s+b[i+1],a,b);
        int nottake=helper(i+1,m,s,a,b);
        int ans=Math.max(take,nottake);
        return ans;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
const int N = 100 + 10;
int n;
int a[N], b[N];

int ans;
void serch(int x, int sa, int sb)
{
    ans = std::max(ans, sa - sb);
    if (x > n)
        return;
    if (x < n)
        serch(x + 1, sa + a[x], sb + b[x + 1]);
    else if(x==n)
        serch(x + 1, sa + a[x], sb);
    serch(x + 1, sa, sb);
}

void solve()
{
    std::cin >> n;
    for (int i = 1; i <= n; i++)
    {
        std::cin >> a[i];
    }
    for (int i = 1; i <= n; i++)
    {
        std::cin >> b[i];
    }

    ans = 0;
    serch(1, 0, 0);
    std::cout << ans << "\n";
}

signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int t;
    std::cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

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

1. The recursive function `serch` does not correctly handle the base case when `x` exceeds `n`.
2. The recursive function `serch` does not correctly handle the scenario where Monocarp trains on the last day, as it incorrectly adds `b[n+1]` to `sb`.

Given these issues, a good test input should:

1. Include a case where `n` is 1 to test the base case.
2. Include a case where `n` is greater than 1, and the optimal solution involves Monocarp training on the last day.

Here is a small-scale test input that meets these criteria:

```plaintext
2
1
5
8
3
1 1 1
2 2 2
```

Explanation:
1. The first test case (`n = 1`) ensures that the base case is tested. Here, Monocarp trains on the only day, and Stereocarp does not train at all. The expected output is `5`.
2. The second test case (`n = 3`) includes a scenario where the optimal solution involves Monocarp training on the last day. Here, Monocarp should train on the last day to maximize the difference, and Stereocarp will not train on the last day. The expected output is `1`.

This input should help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

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

int maximize_difference(int n, const vector<int>& a, const vector<int>& b) {
    int max_diff = INT_MIN;

    // Generate all subsets using bitmasking
    for (int mask = 0; mask < (1 << n); ++mask) {
        int m = 0, s = 0;

        for (int i = 0; i < n; ++i) {
            if (mask & (1 << i)) { // Monocarp trains on day i
                m += a[i];
                if (i + 1 < n) { // Stereocarp trains on day i + 1
                    s += b[i + 1];
                }
            }
        }

        max_diff = max(max_diff, m - s);
    }

    return max_diff;
}

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

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

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

        cout << maximize_difference(n, a, b) << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the current logic might fail. The main issue in the buggy code is that it does not correctly handle the condition where Monocarp and Stereocarp's training days overlap in a way that maximizes the difference in the number of problems solved.

### Key Points to Consider:
1. **Overlapping Training Days**: The code might not correctly account for the scenario where Monocarp trains on a day that causes Stereocarp to train on a day with a high problem count, potentially reducing the difference.
2. **Edge Cases**: Consider cases with small values of `n` and specific patterns in `a` and `b` that can expose the logic flaw.

### Challenging Test Input:
Let's create a test case where the difference in problem counts is maximized, but the code might not handle it correctly due to the overlapping training days.

```plaintext
1
4
10 20 30 40
1 2 3 4
```

### Explanation:
- **Day 1**: Monocarp solves 10 problems. If Monocarp trains on this day, Stereocarp will train on Day 2.
- **Day 2**: Stereocarp solves 2 problems. If Monocarp trains on Day 1, Stereocarp will train on Day 2.
- **Day 3**: Monocarp solves 30 problems. If Monocarp trains on this day, Stereocarp will train on Day 4.
- **Day 4**: Stereocarp solves 4 problems. If Monocarp trains on Day 3, Stereocarp will train on Day 4.

### Expected Output:
- If Monocarp trains on Day 1 and Day 3:
  - Monocarp solves: 10 (Day 1) + 30 (Day 3) = 40 problems.
  - Stereocarp solves: 2 (Day 2) + 4 (Day 4) = 6 problems.
  - Difference: 40 - 6 = 34.

This test case is designed to expose the bug where the code might not correctly account for the optimal days for Monocarp to train, leading to a suboptimal difference in the number of problems solved.