**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
using namespace std;
template <typename T>
inline void inp(vector<T>& a) {
   for (auto& it : a) cin >> it;
}

int main() {
   cin.tie(0)->sync_with_stdio(0);
   int t = 1;
   cin >> t;
   while (t--) {
      int n, q;
      cin >> n >> q;
      vector<int> a(n);
      inp(a);
      vector<int> cnt(n + 1);
      map<int, int> mp;
      for (int i = 0; i < n; i++) {
         mp[a[i]] ^= 1;
         cnt[i + 1] = cnt[i] + (mp[a[i]] == 0 ? -a[i] : a[i]);
      }
      while (q--) {
         int l, r;
         cin >> l >> r;
         if (cnt[l - 1] == cnt[r])
            cout << "YES\n";
         else
            cout << "NO\n";
      }
   }
   return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define vi vector<int>
#define vii vector<vi>
#define ff first
#define ss second
#define int long long
#define read(a, n) for(int i = 0; i < n; i++) cin >> a[i]
#define pii pair<int, int>

using namespace std;

bool isprime(int m) {
    if (m < 2) return false;
    for (int i = 2; i * i <= m; i++) {
        if (m % i == 0) return false;
    }
    return true;
}

bool get_bit(int m, int i) {
    return m & (1ll << i);
}

int hesh(int m){
    return (m>>16)^m;
}

void solve() {
    int n,q;
    cin>>n>>q;
    vi a(n),pref(n+1);
    read(a,n);
    for(int i=0;i<n;i++){
        pref[i+1]=pref[i]^hesh(a[i]);
    }
    while(q--){
        int l,r;
        cin>>l>>r;
        if(pref[r]==pref[l-1]){
            cout<<"YES"<<'\n';
        }
        else{
            cout<<"NO"<<'\n';
        }
    }
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t=1;
    cin >> t;
    while (t--) {
        solve();
        cout<<'\n';
    }

    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 test input that exposes the bug in the provided buggy code, we need to focus on the logic of the code and identify where it might fail. The buggy code uses a prefix XOR approach to determine if the Sheriff can win or not. However, this approach is incorrect because it does not account for the optimal play strategy of both players.

### Analysis of the Bug:
The bug in the code lies in the use of the `hesh` function and the prefix XOR. The `hesh` function is used to hash the values, and the prefix XOR is used to check if the XOR of a subarray is zero. This approach is flawed because it does not consider the optimal play strategy of the players, which is crucial for determining the winner.

### Test Case to Expose the Bug:
We need a test case where the prefix XOR approach fails to correctly determine the winner. A good test case would be one where the subarray has an even number of equal elements, and the total score is such that the first player (Robin Hood) can always force a win or a tie.

### Example Test Case:
Let's create a test case with a small number of targets and queries to make it easy to verify.

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

### Explanation:
- **Test Case 1:**
  - **Number of targets (n):** 4
  - **Number of queries (q):** 2
  - **Target scores (a):** [1, 2, 2, 1]

- **Query 1: (1, 4)**
  - **Subarray:** [1, 2, 2, 1]
  - **Total score:** 1 + 2 + 2 + 1 = 6
  - **Optimal play:** Robin Hood can always force a tie by playing optimally. For example, if Robin Hood takes 1, Sheriff takes 2, Robin Hood takes 2, and Sheriff takes 1, both end up with a score of 3.
  - **Expected output:** NO (since the Sheriff can't guarantee a win)

- **Query 2: (2, 3)**
  - **Subarray:** [2, 2]
  - **Total score:** 2 + 2 = 4
  - **Optimal play:** Robin Hood can always force a tie by taking 2 first, and Sheriff takes the other 2.
  - **Expected output:** YES (since the Sheriff can at least tie)

### Expected Output:
```plaintext
NO
YES
```

This test case should expose the bug in the provided code because the prefix XOR approach will not correctly determine the outcome for these queries.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

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

public class CodeForcesSolution {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;
    /*static int mod = (int) 1e9+7;*/
    public static void main(String[] args) throws IOException {
        int t = readInt();
        while (t-- > 0) {
            int n = readInt(), q = readInt();
            int[] prefXor = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                int v = readInt();
                prefXor[i] = prefXor[i - 1] ^ v;
            }
            for (int i = 0; i < q; i++) {
                int l = readInt(), r = readInt();
                if (l != r && (prefXor[r] ^ prefXor[l - 1]) == 0) {
                    System.out.println("YES");
                } else {
                    System.out.println("NO");
                }
            }
        }
    }
    static int bs(List<Integer> list, int l, int r, boolean lw) {
        int lo = 0, hi = list.size()-1, res = -1;
        while (lo<=hi) {
            int mid = (lo+hi)/2;
            if (list.get(mid) < l) {
                lo = mid + 1;
            } else if (list.get(mid) > r) {
                hi = mid - 1;
            } else {
                res = mid;
                if (lw) {
                    hi = mid - 1;
                } else {
                    lo = mid + 1;
                }
            }
        }
        return res;
    }
    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    static long readLong() throws IOException {
        return Long.parseLong(next());
    }
    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter() throws IOException {
        return next().charAt(0);
    }
    static String readLine() throws IOException {
        return br.readLine().trim();
    }
    static int readALotInt() throws IOException{
        int x = 0, c;
        while((c = br.read()) != ' ' && c != '\n')
            x = x * 10 + (c - '0');
        return x;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
import java.util.*;

public class Main{

    public static void main(String args[]){

        Scanner s = new Scanner(System.in);

        int t = s.nextInt();

        while(t-- > 0){

            int n = s.nextInt();
            int k = s.nextInt();

            int a[] = new int[n];

            for(int i = 0 ; i < n ; i++)
                a[i] = s.nextInt();

            int pre[] = new int[n];

            pre[0] = a[0];

            for(int i = 1; i < n ; i++){
                pre[i] = pre[i - 1] ^ a[i];
            }

            while(k-- > 0){

                int l = s.nextInt();
                int r = s.nextInt();

                if((r - l + 1) % 2 != 0)
                    System.out.println("NO");

                else{
                    if(l > 1){

                        if((pre[l - 2] ^ pre[r - 1]) == 0)
                            System.out.println("YES");
                        else
                            System.out.println("NO");
                    }
                    else{
                        if(pre[r - 1] == 0)
                            System.out.println("YES");
                        else
                            System.out.println("NO");
                    }
                }

            }

        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
const int N = 1e6 + 9, mod = 1e9 + 7;
void solve() {
    int n, q;
    cin >> n >> q;
    vector<int> v(n + 1);
    for(int i = 1; i <= n; i++) {
        cin >> v[i];
    }
    while(q--) {
        int l, r;
        cin >> l >> r;
        map<int, int> mp;
        for(int i = l; i <= r; i++) {
            mp[v[i]]++;
        }
        bool flag = true;
        for(auto [x, y] : mp) {
            if(y & 1) {
                flag = false;
                break;
            }
        }
        if(flag) {
            cout << "YES\n";
        }
        else {
            cout << "NO\n";
        }
    }
}
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t = 1, cs = 1;
    cin >> t;
    while (t--) {
        //cout << "Case " << cs++ << ":\n";
        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
2
5 3
1 2 3 4 5
1 5
2 4
1 3
4 1
2 1 1 2
1 2
1 3
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
//#pragma GCC optimize(2)

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int>PII;

#define fi first
#define se second
#define ls u<<1
#define rs u<<1|1
#define all(ss) ss.begin(),ss.end()
#define pb push_back

int const mod=1e9+7;
int const B=507;
//int const mod=998244353;
//int const base=131,mod=2e9+11;
int const N=2e5+7,M=2e6+7;
int const INF=0x3f3f3f3f;
LL const INFF=0x3f3f3f3f3f3f3f3f;

/*
-fexec-charset=GBK
-finput-charset=UTF-8

*/
int n,m;
int x,y,z;
int a[N];
struct Node{  //500*2e5=1e8   2e5/500*2e5=8e7
	int l,r,id;
	bool operator<(const Node &p)const{
		return l/B==p.l/B?r<p.r:l<p.l;
	}
}Q[N];

/*
转化：在一段区间内，每一个数出现的次数都为偶数
*/
void solve(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)	scanf("%d",a+i);
    for(int i=0;i<m;i++){
		scanf("%d%d",&x,&y);
		Q[i]={x,y,i};
	}

	int tot=0;
	map<int,int>mp;
	auto add=[&](int x){
		 if(++mp[x]%2)	tot++;  //奇数++
		 else tot--;   //偶数--
	};
	auto del=[&](int x){
		 if(--mp[x]%2) tot++;   //奇数++
		 else tot--;	//偶数--
	};

	vector<bool>vis(m);
	for(int l=1,r=0,i=0;i<m;i++){
		while(l>Q[i].l)	add(a[--l]);
		while(r<Q[i].r)	add(a[++r]);
		while(l<Q[i].l)	del(a[l++]);
		while(r>Q[i].r) del(a[r--]);
		vis[Q[i].id]=tot==0;
	}
	for(int i=0;i<m;i++){
		puts(vis[i]?"YES":"NO");
	}
}

void init(){

}

int main()
{
    //std::ios::sync_with_stdio(false);   cin.tie(0); cout.tie(0);
    //init();
    int T=1;
    //cin>>T;
    scanf("%d",&T);

   	for(int i=1;i<=T;i++){
	   	solve();
	}

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios where the code might fail. The code is designed to determine if the Sheriff can avoid losing the game by checking if the total score can be divided equally between the two players. The bug in the code lies in how it handles the counts of scores and their parity.

### Steps to Generate the Test Input:
1. **Edge Cases:**
   - **Single Target:** When there is only one target, the player who starts (Robin Hood) will always win.
   - **Two Targets:** When there are two targets, the player who starts can always choose the higher score target and win.
   - **Three Targets:** When there are three targets, the player who starts can always choose the middle target and force the other player to choose the lower scoring targets.

2. **Parity and Sum:**
   - The code checks if the total sum of scores in a given range can be divided equally. If the total sum is odd, it is impossible for the scores to be divided equally, and the player who starts (Robin Hood) will always win.
   - If the total sum is even, the code should check if the scores can be distributed such that both players end up with the same score. This is where the code might fail if it does not correctly handle the distribution of scores.

3. **Query Ranges:**
   - Queries should include ranges that are both even and odd in length to test the parity check.
   - Include ranges where the total sum is both even and odd to test the correctness of the parity check.

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

### Explanation:
- **Test Case 1:**
  - **Number of targets (n):** 5
  - **Number of queries (q):** 3
  - **Scores (a):** 2, 1, 2, 1, 1

- **Query 1:**
  - **Range:** 1 to 2
  - **Scores in range:** 2, 1
  - **Total sum:** 3 (odd)
  - **Expected output:** NO (Robin Hood wins)

- **Query 2:**
  - **Range:** 1 to 3
  - **Scores in range:** 2, 1, 2
  - **Total sum:** 5 (odd)
  - **Expected output:** NO (Robin Hood wins)

- **Query 3:**
  - **Range:** 4 to 5
  - **Scores in range:** 1, 1
  - **Total sum:** 2 (even)
  - **Expected output:** YES (Sheriff can avoid losing)

This test input is designed to challenge the code by including both even and odd sums and ranges, which should expose any issues in the parity check and score distribution logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
import sys

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

    idx = 0
    t = int(data[idx])
    idx += 1
    results = []

    for _ in range(t):
        n, q = int(data[idx]), int(data[idx + 1])
        idx += 2

        targets = list(map(int, data[idx:idx + n]))
        idx += n

        # Process each query
        for __ in range(q):
            l, r = int(data[idx]) - 1, int(data[idx + 1]) - 1
            idx += 2

            # Extract the relevant subarray and sort it
            subarray = targets[l:r + 1]
            subarray.sort(reverse=True)  # Sort in descending order

            robin_score = 0
            sheriff_score = 0

            # Alternate turns: Robin starts first
            for i in range(len(subarray)):
                if i % 2 == 0:  # Robin's turn (even index)
                    robin_score += subarray[i]
                else:  # Sheriff's turn (odd index)
                    sheriff_score += subarray[i]

            # Determine if the Sheriff does not lose
            if sheriff_score >= robin_score:
                results.append("YES")
            else:
                results.append("NO")

    # Output all results at once
    sys.stdout.write("\n".join(results) + "\n")

# Read input and solve the problem
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
5 5
1 2 3 4 5
1 1
1 2
1 3
1 4
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
def solve():
    t = int(input())  # Number of test cases

    for _ in range(t):
        # Read inputs
        n, q = map(int, input().split())  # n: number of targets, q: number of queries
        a = list(map(int, input().split()))  # Points for hitting each target

        # Process each query
        for _ in range(q):
            l, r = map(int, input().split())  # The range [l, r]

            # Extract the relevant subarray and sort it in descending order
            subarray = sorted(a[l-1:r], reverse=True)

            # Calculate Robin Hood's and Sheriff's scores
            robin_score = 0
            sheriff_score = 0
            for i in range(len(subarray)):
                if i % 2 == 0:  # Robin's turn (even indices)
                    robin_score += subarray[i]
                else:  # Sheriff's turn (odd indices)
                    sheriff_score += subarray[i]

            # Compare scores and output result
            if sheriff_score >= robin_score:
                print("YES")
            else:
                print("NO")

# Run the function to process inputs
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
5 3
1 2 3 4 5
1 3
1 5
3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
import java.util.*;

public class RobinHoodArchery {
    public static void main(String[] args) {
        int i = 0;
        int j = 0;
        List<List<String>> ll = new ArrayList<>();
        List<String> lis = new ArrayList<>();
        Scanner sc = new Scanner(System.in);
        while (true) {
            if (i == 0) {
                i = Integer.valueOf(sc.nextLine());
            } else {
                if (j == 0) {
                    String[] ss = sc.nextLine().split(" ");
                    j = Integer.valueOf(ss[1]) + 1;
                } else {
                    lis.add(sc.nextLine());
                    j--;
                    if (j == 0) {
                        ll.add(lis);
                        if (--i == 0) {
                            break;
                        }
                        lis = new ArrayList<>();
                    }
                }
            }
        }
        int p = ll.size();
        for (int k = 0; k < p; k++) {
            List<String> li = ll.get(k);
            String[] ss0 = li.get(0).split(" ");
            int[] points = new int[ss0.length];
            for (int q = 0; q < ss0.length; q++) {
                points[q] = Integer.valueOf(ss0[q]);
            }
            int p1 = li.size();
            int[][] arr = new int[p1 - 1][2];
            for (int q = 1; q < p1; q++) {
                String[] ss = li.get(q).split(" ");
                for (int q1 = 0; q1 < 2; q1++) {
                    arr[q - 1][q1] = Integer.valueOf(ss[q1]);
                }
            }
            archery(points, arr);
        }
    }

    static void archery(int[] nums, int[][] arr) {
        int n = arr.length;
        loop: for (int i = 0; i < n; i++) {
            if ((arr[i][1] - arr[i][0]) % 2 == 0) {
                System.out.println("No");
            } else {
                Map<Integer, Integer> mm = divideConquer(nums, arr[i][0] - 1, arr[i][1] - 1);
                for (Map.Entry<Integer, Integer> en : mm.entrySet()) {
                    if (en.getValue() % 2 != 0) {
                        System.out.println("No");
                        continue loop;
                    }
                }
                System.out.println("Yes");
            }
        }
    }

    static Map<Integer, Integer> divideConquer(int[] nums, int i, int j) {
        if (i == j) {
            Map<Integer, Integer> mm = new HashMap<>();
            mm.put(nums[i], 1);
            return mm;
        }
        int mid = i + (j - i) / 2;
        Map<Integer, Integer> left = divideConquer(nums, i, mid);
        Map<Integer, Integer> right = divideConquer(nums, mid + 1, j);
        Map<Integer, Integer> mm = new HashMap<>();
        mm.putAll(left);
        for (Map.Entry<Integer, Integer> en : right.entrySet()) {
            if (mm.containsKey(en.getKey())) {
                mm.put(en.getKey(), en.getValue() + mm.get(en.getKey()));
            } else {
                mm.put(en.getKey(), en.getValue());
            }
        }
        return mm;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
output:
NO
NO
YES
NO
NO
YES

Note:

Buggy Code:
import java.util.*;

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

        int t = sc.nextInt();
        StringBuilder result = new StringBuilder();

        for (int testCase = 0; testCase < t; testCase++) {
            int n = sc.nextInt();
            int q = sc.nextInt();

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

            for (int query = 0; query < q; query++) {

                int l = sc.nextInt() - 1;
                int r = sc.nextInt() - 1;

                List<Integer> scores = new ArrayList<>();
                for (int i = l; i <= r; i++) {
                    scores.add(a[i]);
                }

                Collections.sort(scores, Collections.reverseOrder());

                int robinScore = 0;
                int sheriffScore = 0;

                for (int i = 0; i < scores.size(); i++) {

                    if (i % 2 == 0) {
                        robinScore += scores.get(i);
                    } else {
                        sheriffScore += scores.get(i);
                    }
                }

                if (sheriffScore >= robinScore) {
                    result.append("YES\n");
                } else {
                    result.append("NO\n");
                }
            }
        }

        System.out.print(result);
        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
4 4
3 1 2 4
1 2
1 3
1 4
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
void alien()
{
    ll n, k;
    cin >> n >> k;
    vector<ll> arr(n);
    ll a = 0, b = 0;
    for (size_t i = 0; i < n; i++)
    {
        /* code */
        cin >> arr[i];
    }
    sort(arr.begin(),arr.end(), greater<ll>());
    for (size_t i = 0; i < n; i++)
    {
        /* code */
        if (i % 2 == 0)
        {
            a += arr[i];
        }
        else
        {
            b += arr[i];
        }
    }

    cout<<max(0LL, a - b - k);
}
int main()
{
    ll t;
    cin >> t;
    while (t--)
    {
        alien();
        cout << "\n";
    }
    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
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define nline endl
using namespace std;

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

    int t = 1 ;
    cin >> t ;

    while(t--){

        int n, k;
        cin >> n >> k;
        int arr[n + 1];
        for (int i = 1; i <= n;i++){
            cin >> arr[i];
        }
        sort(arr+1, arr + n+1);

        int a = 0;
        int b = 0;
        bool flag = true;

        for (int i = n; i >= 1;i--){
            if(flag){
                a += arr[i];
            }else{
                b += arr[i];
            }

            if(flag){
                flag = false;
            }else{
                flag = true;
            }
        }
        if(a-b-k>=0){
        cout << a - b-k<< endl;
    } else{
        cout << 0 << endl;
    }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
/*
Written by Subhajit Bera.
Country:-INDIA
*/
#include <bits/stdc++.h>
typedef long long ll;
#include <iostream>
using namespace std;

int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int t;
cin>>t;
while(t>0)
{
  ll n,k;
  cin>>n>>k;
  vector<ll> arr(n);
  for(ll i=0;i<n;i++){
    cin>>arr[i];
  }
  sort(arr.rbegin(),arr.rend());
  for(ll i=1;i<n;i++){
    if(i%2LL){
        arr[i]=arr[i]+min(k,arr[i-1]-arr[i]);
        k-=min(k,arr[i-1]-arr[i]);
        if(k==0) break;
    }
  }
  ll sum=0;
  for(ll i=0;i<n;i++){
    if(i%2LL){
        sum-=arr[i];
    }else{
        sum+=arr[i];
    }
  }
  cout<<sum<<"\n";
t--;
}
}

**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
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
import sys
from heapq import nlargest, heappop as pop, heappush as push, heapify as hfy
from collections import Counter as ctr, defaultdict, deque
from itertools import permutations, combinations, product
from bisect import bisect_left as bl, bisect_right as br
from functools import lru_cache

# Commonly used utility functions
int1 = lambda x: int(x) - 1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def ni(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
def MI(): return map(int, sys.stdin.readline().split())
def FI(): return float(sys.stdin.readline())
def LLI_gen(rows_number):
    for _ in range(rows_number):
        yield LI()
def yn(cond): print("YES" if cond else "NO")
def ynr(cond): return "YES" if cond else "NO"
def dd(): return defaultdict(int)

# Direction vectors for grid problems
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij2 = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]

# Large constants for edge cases
inf = float('inf')
md = 10**9 + 7

# Additional utilities
def modinv(x, m=md): return pow(x, m - 2, m)  # Modular Inverse
def factorial_mod(n, m=md): return 1 if n == 0 else n * factorial_mod(n-1, m) % m
def nCr_mod(n, r, m=md):
    if r > n: return 0
    num = den = 1
    for i in range(r):
        num = num * (n - i) % m
        den = den * (i + 1) % m
    return num * modinv(den, m) % m

def solve():
    # n = ni()
    # s1=SI()
    n,k=LI()
    a1 = LI()
    ans = 0
    # res=[]
    if n%2==1:
        a1.append(0)
        n+=1
    a1.sort(reverse=True)
    for i in range(0,n-1,2):
        ans+=a1[i]-a1[i+1]

    return max(0,ans-k)

if __name__ == "__main__":
    testnumber = ni()
    results = []
    for _ in range(testnumber):
        results.append(solve())

    #sys.stdout.write("\n".join( results) + "\n")
    sys.stdout.write("\n".join(map(str, results)) + "\n")

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the problem and the code's logic:

### Problem Analysis:
- Alice and Bob are playing a game where they take turns picking items from a list of items with costs.
- Alice wants to maximize the score \( A - B \), while Bob wants to minimize it.
- Bob can increase the costs of some items by a total of at most \( k \).

### Code Analysis:
- The code sorts the items in descending order and calculates the score by summing the differences between the costs of items picked by Alice and Bob.
- It then subtracts \( k \) from the calculated score to account for Bob's ability to increase the costs.

### Potential Bugs:
- The code does not correctly handle the scenario where Bob can increase the costs of items to minimize the score.
- The code assumes that the optimal strategy for Bob is to always subtract \( k \) from the score, which is not necessarily true.

### Challenging Test Input:
- We need a test case where Bob can strategically increase the costs to significantly change the score.

### Test Input:
```plaintext
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
```

### Explanation:
1. **Test Case 1:**
   - \( n = 2 \), \( k = 5 \)
   - Items: \( [1, 10] \)
   - Bob can increase the cost of the first item by 5, making the costs \( [6, 10] \).
   - Alice will take 10, Bob will take 6.
   - Score: \( 10 - 6 = 4 \)

2. **Test Case 2:**
   - \( n = 3 \), \( k = 0 \)
   - Items: \( [10, 15, 12] \)
   - Bob cannot change the costs.
   - Alice will take 15, Bob will take 12, Alice will take 10.
   - Score: \( (15 + 10) - 12 = 13 \)

3. **Test Case 3:**
   - \( n = 4 \), \( k = 6 \)
   - Items: \( [3, 1, 2, 4] \)
   - Bob can increase the costs to make them equal, e.g., \( [4, 4, 4, 4] \).
   - Score: \( (4 + 4) - (4 + 4) = 0 \)

4. **Test Case 4:**
   - \( n = 2 \), \( k = 4 \)
   - Items: \( [6, 9] \)
   - Bob can increase the cost of the first item by 3, making the costs \( [9, 9] \).
   - Score: \( 9 - 9 = 0 \)

This test input should help expose the bug in the provided code by demonstrating scenarios where Bob's strategic increases in costs can significantly affect the final score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
for _ in range(int(input())):

    n,k=[int(x) for x in input().split()]

    lst=list(map(int,input().split()))
    lst.sort()

    i=len(lst)-1
    while(i>0):
        if lst[i]== lst[i-1]:
            i-=2
        else:
            diff = lst[i]-lst[i-1]
            if diff<=k:
                i-=2
            else:
                lst[i-1]+=k
                break
    s1=0
    s2=0
    count=0
    for j in range(i,-1,-1):
        if count == 0:
            s1+=lst[j]
            count=1
        else:
            s2+=lst[j]
            count=0
    print(s1-s2)

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

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

Think step by step.

Response:
```plaintext
3
2 0
1 1
4 1
1 2 3 4
3 2
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

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

    # Sort in descending order
    a.sort(reverse=True)

    for i in range(n):
        if k == 0:
            break
        if i == n - 1:
            break

        diff = a[i] - a[i + 1]
        r = min(diff, k)
        a[i + 1] += r
        k -= r
        i += 1  # Increment i for the next iteration

    odd_index_sum = 0
    even_index_sum = 0

    for i in range(n):
        if i % 2 == 0:
            even_index_sum += a[i]
        else:
            odd_index_sum += a[i]

    print(abs(odd_index_sum - even_index_sum))

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

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
2 0
5 5
4 10
1 2 3 4
5 15
1 2 3 4 5
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `2 0` and `5 5`
   - **Description:** This case tests the scenario where Bob cannot modify the costs (`k = 0`). Alice and Bob will take items in turns, and the score will be `5 - 5 = 0`.

2. **Test Case 2:**
   - **Input:** `4 10` and `1 2 3 4`
   - **Description:** This case tests a scenario where Bob has enough budget (`k = 10`) to significantly increase the costs. Bob can increase the costs to make the game end with a score of `0`. For example, Bob can increase `a_1` by `1`, `a_2` by `2`, `a_3` by `3`, and `a_4` by `4`, making the costs `[2, 4, 6, 8]`. Alice will take `8` and `4`, and Bob will take `6` and `2`, resulting in a score of `0`.

3. **Test Case 3:**
   - **Input:** `5 15` and `1 2 3 4 5`
   - **Description:** This case tests a scenario with a larger number of items and a larger budget (`k = 15`). Bob can increase the costs to make the game end with a score of `0`. For example, Bob can increase `a_1` by `2`, `a_2` by `3`, `a_3` by `4`, `a_4` by `5`, and `a_5` by `1`, making the costs `[3, 5, 7, 9, 6]`. Alice will take `9` and `7`, and Bob will take `6` and `5`, and the last item will be `3`, resulting in a score of `0`.

These test cases cover different edge cases and scenarios to ensure the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    while(t-->0)
    {
       int n=sc.nextInt(),k=sc.nextInt();
       int nums[]=new int[n];
       for(int i=0;i<n;i++)
       {
           nums[i]=sc.nextInt();
       }
       Arrays.sort(nums);
       int a=n-1,b=n-2;
       int alice=0,bob=0;
       while(b>=0)
       {

           if(nums[a]==nums[b])
           {
              a-=2;
               b=a-1;
           }else if(nums[b]+k>=nums[a])
           {
               k-=(nums[a]-nums[b]);
               a-=2;
               b=a-1;
           }else
           {
               alice+=nums[a];
               bob+=(nums[b]+k);
               a-=2;
               b=a-1;
           }
           if(a<0||b<0)break;
       }
       if(a>=0)
       {
           alice+=nums[a];
       }
       if(b>=0)
       {
           bob+=nums[b];
       }
       System.out.println(alice-bob);
    }
	}
}

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

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

Think step by step.

Response:
```plaintext
3
2 0
1 1
4 10
1 2 3 4
5 15
10 10 10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class SplittingItems {

	public static void main(String[] args) {
		int t = ii();
		while(t-- > 0) solve();
	}

	public static void solve() {
		int n = ii(), k = ii();
		int [] A = iir(n);
		Arrays.sort(A);
		for(int i = n - 2; i >= 0; i -= 2) {
			A[i] += Math.min(k, A[i + 1] - A[i]);
			k -= Math.min(k, A[i + 1] - A[i]);
		}
		long x = 0, alice = 0, bob = 0;
		for(int i = n - 1; i >= 0; --i) {
			if(x % 2 == 0) alice += A[i];
			if(x++ % 2 != 0) bob += A[i];
		}
		System.out.println(alice - bob);
	}

	static FastReader sc = new FastReader();
	final static int mod = 1_000_000_007;
	final static int inf = Integer.MAX_VALUE;
	final static long INF = Long.MAX_VALUE;

	static class FastReader {
    	BufferedReader br;
    	StringTokenizer st;
    	public FastReader() {
    		br = new BufferedReader(new InputStreamReader(System.in));
        }
    	String next() {
    		while (st == null || !st.hasMoreElements()) {
    			try {
    				st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
    		}
            return st.nextToken();
        }
        int nextInt() {return Integer.parseInt(next());}
        long nextLong() {return Long.parseLong(next());}
        double nextDouble() {return Double.parseDouble(next());}
        String nextLine(){
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

	public static int ii() {
		return sc.nextInt();
	}

	public static long il() {
		return sc.nextLong();
	}

	public static String is() {
		return sc.next();
	}

	public static int [] iir(int n) {
		int [] A = new int[n];
		for(int i=0;i<n;++i)
			A[i] = ii();
		return A;
	}

	public static long [] ilr(int n) {
		long [] A = new long[n];
		for(int i=0;i<n;++i)
			A[i] = il();
		return A;
	}

	public static void sort(int [] A) {
		Arrays.sort(A);
	}

	public static void sort(List<Integer> A) {
		Collections.sort(A);
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The code sorts the array and then attempts to increase the costs of the items in a way that minimizes the score. However, there are a few potential issues:

1. **Incorrect Calculation of Increases**: The code tries to increase the cost of items in a specific pattern, but it might not be optimal.
2. **Handling of Even and Odd Length Arrays**: The code assumes a specific pattern for increasing costs, which might not work well for arrays of even length.
3. **Edge Cases with Small or Large Values of k**: The code might fail when k is very small or very large.

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

- **Small k and Even Length Array**: This can test the handling of small increases and the pattern of increasing costs.
- **Large k and Odd Length Array**: This can test the handling of large increases and the pattern of increasing costs.

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

```plaintext
3
4 1
1 2 3 4
5 10
1 2 3 4 5
3 0
1 2 3
```

### Explanation:
1. **Test Case 1**:
   - **Input**: 4 1
   - **Array**: [1, 2, 3, 4]
   - **k**: 1
   - **Expected Output**: 2
   - **Explanation**: Bob can increase the cost of the third item (3) by 1, making it [1, 2, 4, 4]. Alice will take 4 and 4, Bob will take 2 and 1. The score will be (4 + 4) - (2 + 1) = 5 - 3 = 2.

2. **Test Case 2**:
   - **Input**: 5 10
   - **Array**: [1, 2, 3, 4, 5]
   - **k**: 10
   - **Expected Output**: 0
   - **Explanation**: Bob can increase the cost of the fifth item (5) by 5, the fourth item (4) by 4, the third item (3) by 1, making it [1, 2, 4, 8, 10]. Alice will take 10 and 2, Bob will take 8, 4, and 1. The score will be (10 + 2) - (8 + 4 + 1) = 12 - 13 = -1, but since Bob can only increase, the best he can do is 0.

3. **Test Case 3**:
   - **Input**: 3 0
   - **Array**: [1, 2, 3]
   - **k**: 0
   - **Expected Output**: 1
   - **Explanation**: Bob cannot change the costs. Alice will take 3 and 1, Bob will take 2. The score will be (3 + 1) - 2 = 2 - 1 = 1.

This test input should help expose the bug in the provided code by testing different scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int N=2e5+30;
long long a[N];
long long S[N];
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--){
        int n,k;
        cin>>n>>k;
        for (int i=1; i<=n; i++)
            cin>>a[i];
        sort(a+1,a+n+1,greater<long long>());
        for (int i=1; i<=n; i++)
        if(i%2==0){
            if(k==0) break;
            if(k>=a[i-1]-a[i])
                {
                    a[i]=a[i-1];
                    k-=(a[i-1]-a[i]);
                }
                else
                    {
                    a[i]+=k;
                    k=0;
                    break;
                }
        }

        long long ans=0;
        for (int i=1; i<=n; i++)
            if(i%2) ans+=a[i];
                else ans-=a[i];
        cout<<ans<<'\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
3
2 0
1 2
4 10
1 2 3 4
5 5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

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

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void tcase() {
	int n, k;
	cin >> n >> k;
	priority_queue<int>pq;
	for (int i = 0; i < n; ++i) {
		int x; cin >> x;
		pq.emplace(x);
	}
	ll bob = 0 , alice = 0;
	while (!pq.empty()) {
		bob += pq.top();
		pq.pop();
		if (pq.empty())
			break;
		alice += pq.top();
		pq.pop();
	}
	cout << max(abs(alice - bob) - k , 0LL) << '\n';
}
int32_t main() {

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

	int32_t t = 1;
	cin >> t;

	while (t-- > 0)
		tcase();
}

**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
1 2 3 4
```