**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boring Day
time_limit_ms: 2000
memory_limit_mb: 256
Description: On another boring day, Egor got bored and decided to do something. But since he has no friends, he came up with a game to play.

Egor has a deck of $n$ cards, the $i$-th card from the top has a number $a_i$ written on it. Egor wants to play a certain number of rounds until the cards run out. In each round, he takes a non-zero number of cards from the top of the deck and finishes the round. If the sum of the numbers on the cards collected during the round is between $l$ and $r$, inclusive, the round is won; otherwise, it is lost.

Egor knows by heart the order of the cards. Help Egor determine the maximum number of rounds he can win in such a game. Note that Egor is not required to win rounds consecutively.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of the test cases.

The first line of each test case contains three integers $n$, $l$, and $r$ ($1 \le n \le 10^{5}$, $1 \le l \le r \le 10^9$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the numbers on the cards from top to bottom.

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

Output: For each test case, output a single number — the maximum number of rounds Egor can win.

Examples:
input:
8
5 3 10
2 1 11 3 7
10 1 5
17 8 12 11 7 11 21 13 10 8
3 4 5
3 4 2
8 12 25
10 7 5 13 8 9 12 7
2 3 3
5 2
9 7 9
2 10 5 1 3 7 6 2 3
1 8 10
9
5 5 6
1 4 2 6 4
output:
3
0
1
4
0
3
1
2

Note: In the first test case, Egor can win $3$ rounds:

  * In the first round, take the top $2$ cards with values $2$ and $1$ and win, as their sum is $3$. After this, the deck will look like this: $[11, 3, 7]$.    * In the second round, take the top card and lose, as its value $11$ is greater than $r = 10$. After this, the deck will look like this: $[3, 7]$.    * In the third round, take the top card with value $3$ and win. After this, the deck will look like this: $[7]$.    * After this, in the fourth round, Egor only has to take the last card in the deck with value $7$ and win again.

In the second test case, Egor cannot win any rounds, no matter how hard he tries.

In the third test case, you can take one card in each round, then the first and third rounds will be losing, and the second round will be winning.

In the fourth test case, you can take two cards in each round and always win.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const long long mod=1e15;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--) {
		int n;
		long long l,r;
		cin>>n>>l>>r;
		long long  dp[n+1];
		int ans=0;
		long long a[n+1];
		a[0]=0;
		memset(dp,0,sizeof dp);
		for(int i=1; i<=n; i++) {
			cin>>a[i];
			a[i]+=a[i-1];
		}
		int k=0;
		for(int i=1; i<=n; i++) {
			dp[i]=dp[i-1];
			for(int j=i; j>=k; j--) {
				if(a[i]-a[j-1]>=l&&a[i]-a[j-1]<=r) {
					dp[i]=max(dp[i],dp[j-1]+1);
					break;
				}
			}
		}
		cout<<dp[n]<<'\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
10 10 10
1 1 1 1 1 1 1 1 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boring Day
time_limit_ms: 2000
memory_limit_mb: 256
Description: On another boring day, Egor got bored and decided to do something. But since he has no friends, he came up with a game to play.

Egor has a deck of $n$ cards, the $i$-th card from the top has a number $a_i$ written on it. Egor wants to play a certain number of rounds until the cards run out. In each round, he takes a non-zero number of cards from the top of the deck and finishes the round. If the sum of the numbers on the cards collected during the round is between $l$ and $r$, inclusive, the round is won; otherwise, it is lost.

Egor knows by heart the order of the cards. Help Egor determine the maximum number of rounds he can win in such a game. Note that Egor is not required to win rounds consecutively.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of the test cases.

The first line of each test case contains three integers $n$, $l$, and $r$ ($1 \le n \le 10^{5}$, $1 \le l \le r \le 10^9$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the numbers on the cards from top to bottom.

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

Output: For each test case, output a single number — the maximum number of rounds Egor can win.

Examples:
input:
8
5 3 10
2 1 11 3 7
10 1 5
17 8 12 11 7 11 21 13 10 8
3 4 5
3 4 2
8 12 25
10 7 5 13 8 9 12 7
2 3 3
5 2
9 7 9
2 10 5 1 3 7 6 2 3
1 8 10
9
5 5 6
1 4 2 6 4
output:
3
0
1
4
0
3
1
2

Note: In the first test case, Egor can win $3$ rounds:

  * In the first round, take the top $2$ cards with values $2$ and $1$ and win, as their sum is $3$. After this, the deck will look like this: $[11, 3, 7]$.    * In the second round, take the top card and lose, as its value $11$ is greater than $r = 10$. After this, the deck will look like this: $[3, 7]$.    * In the third round, take the top card with value $3$ and win. After this, the deck will look like this: $[7]$.    * After this, in the fourth round, Egor only has to take the last card in the deck with value $7$ and win again.

In the second test case, Egor cannot win any rounds, no matter how hard he tries.

In the third test case, you can take one card in each round, then the first and third rounds will be losing, and the second round will be winning.

In the fourth test case, you can take two cards in each round and always win.

Buggy Code:
#define ll long long
#define F pr.first
#define S pr.second
#define PB push_back
#define MP make_pair
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define SORT(v) sort(v.begin(),v.end())
#define REVERSE(v) reverse(v.begin(),v.end())
#define VEC(v) vector<int> v
#define VECs(v,n) vector<int> v(n)
#define VECv(v,n,val) vector<int> v(n,val)
#define IOTA(v,val) iota(v.begin(), v.end(), val)
#define MAP(mp) map<int,int> mp
#define MINI(v) *min_element(v.begin(),v.end())
#define MAXI(v) *max_element(v.begin(),v.end())

#include<bits/stdc++.h>
using namespace std;
int main(){
   int N;
   cin>>N;
   while(N--){
      ll n,l,r;
      cin>>n>>l>>r;
      VEC(v);
      ll j = 0,k = j,sum = 0, count = 0;
      FOR(i,0,n){
         int x=0;
         cin>>x;
         if( (x >= l)&&(x <= r)){
            count++;
            x = -1;
         }
         else if(x > r){
            x = -1;
         }
         v.PB(x);
      }
      bool check =false;
      while(j < n){
         if((v[j] < l) && (v[j] > 0)){
            k = j;
            sum = 0;
            while((k >= 0)&&(v[k] != -1)){
               sum += v[k];
               k--;
               if((sum >= l)&&(sum <= r)){
                  count++;
                  check = true;
                  v[j] =-1;
                  break;
               }
               else if(sum > r){
                  // v[j] = -1;
                  // sum = 0;
                  break;
               }
            }
            j++;
         }
         else j++;
      }
      cout<<count<<endl;
   }
   return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 10 15
5 5 5 5 5
8 5 15
10 1 2 3 4 5 6 7
10 1 10
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

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

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;

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

    int strength = 0;
    int intelligence = 0;
    int checksPassed = 0;

    for (int i = 0; i < n; i++) {
        if (records[i] == 0) {
            if (m > 0) {
                strength++;
                m--;
            }
        } else if (records[i] > 0) {
            int requiredIntelligence = abs(records[i]);
            if (intelligence >= requiredIntelligence) {
                checksPassed++;
            } else if (m > 0) {
                int neededPoints = requiredIntelligence - intelligence;
                if (neededPoints <= m) {
                    intelligence += neededPoints;
                    m -= neededPoints;
                    checksPassed++;
                }
            }
        } else {
            int requiredStrength = abs(records[i]);
            if (strength >= requiredStrength) {
                checksPassed++;
            } else if (m > 0) {
                int neededPoints = requiredStrength - strength;
                if (neededPoints <= m) {
                    strength += neededPoints;
                    m -= neededPoints;
                    checksPassed++;
                }
            }
        }
    }

    cout << checksPassed << endl;
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
6 3
0 1 0 -1 0 1
```

**Explanation:**

- **n = 6** and **m = 3**: This means there are 6 records and 3 attribute points to distribute.
- **Records:**
  - **0**: Acquire one attribute point.
  - **1**: Intelligence check (requires Intelligence >= 1).
  - **0**: Acquire one attribute point.
  - **-1**: Strength check (requires Strength >= 1).
  - **0**: Acquire one attribute point.
  - **1**: Intelligence check (requires Intelligence >= 1).

**Expected Behavior:**
- The optimal strategy is to distribute the points as follows:
  1. **First 0**: Increase Intelligence by 1.
  2. **Second 0**: Increase Strength by 1.
  3. **Third 0**: Increase Intelligence by 1.

**Outcome:**
- **First check (1)**: Intelligence = 1, pass the check.
- **Second check (-1)**: Strength = 1, pass the check.
- **Third check (1)**: Intelligence = 2, pass the check.

**Total checks passed: 3**

**Why This Test Input is Challenging:**
- The test input is designed to expose the bug in the buggy code where the points are not optimally distributed. The buggy code might only increase Strength, leading to a suboptimal number of checks passed. This test input ensures that the code must correctly balance the distribution of points between Strength and Intelligence to pass the maximum number of checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

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

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> records(n);
    for (int i = 0; i < n; i++) {
        cin >> records[i];
    }

    int strength = 0, intelligence = 0, passedChecks = 0;
    vector<int> strengthChecks, intelligenceChecks;

    for (int record : records) {
        if (record == 0) {
            if (m > 0) {
                strength++;
                m--;
            }
        } else if (record > 0) {
            intelligenceChecks.push_back(record);
        } else {
            strengthChecks.push_back(-record);
        }
    }

    for (int req : intelligenceChecks) {
        if (intelligence >= req) {
            passedChecks++;
        }
    }

    for (int req : strengthChecks) {
        if (strength >= req) {
            passedChecks++;
        }
    }

    cout << passedChecks << endl;
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the logic that handles the allocation of attribute points. The code always increments the `strength` attribute when it encounters a `0` in the records, which is a significant flaw. We need to create a test case where this flaw results in suboptimal attribute allocation, leading to fewer passed checks.

Here's a step-by-step approach to generate a challenging test input:

1. **Initial Setup**: Define the number of records `n` and the number of attribute points `m`.
2. **Attribute Points**: Ensure there are exactly `m` zeros in the records.
3. **Checks**: Place checks that require both `strength` and `intelligence` to be maximized to pass them. This will expose the flaw in the buggy code where it always increases `strength` instead of dynamically choosing the best attribute to increase.

Let's create a test input:

- **n = 6, m = 3**: We need 3 attribute points and 3 checks.
- **Records**:
  - Place the attribute points (0) in positions where they can be allocated.
  - Place checks that require both `strength` and `intelligence` to be maximized.

Here's the test input:
```plaintext
6 3
0 1 0 -1 0 2
```

Explanation:
- **Records**: [0, 1, 0, -1, 0, 2]
  - The first record is an attribute point, so we can allocate it to either `strength` or `intelligence`.
  - The second record is an `intelligence` check requiring `intelligence >= 1`.
  - The third record is an attribute point.
  - The fourth record is a `strength` check requiring `strength >= 1`.
  - The fifth record is an attribute point.
  - The sixth record is an `intelligence` check requiring `intelligence >= 2`.

In this test case:
- If the code always increments `strength`, it will allocate all 3 points to `strength`.
- The final attributes will be `strength = 3` and `intelligence = 0`.
- The checks will be:
  - `intelligence >= 1`: Fail (0 < 1)
  - `strength >= 1`: Pass (3 >= 1)
  - `intelligence >= 2`: Fail (0 < 2)

Thus, the code will pass only 1 check, which is suboptimal. The optimal strategy would be to allocate 2 points to `intelligence` and 1 point to `strength`, passing all 3 checks.

This test input should effectively expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int N = 5010, M = 2e6 + 10;
const ll mod = 1e9 + 7;
int n, m;
int a[M], dp[N][N];

void solve()
{
    cin >> n >> m;
    int cnt = 0, ans = 0;
    for (int i = 1; i <= n; i ++ )
    {
        cin >> a[i];
        if (!a[i]) cnt ++ ;
        else if (a[i] > 0 && cnt >= a[i]) dp[cnt][a[i]] ++ ;
        else if (cnt >= -a[i]) dp[cnt][cnt + a[i]] ++ ;
    }
    for (int i = 1; i <= m; i ++ )
        for (int j = 0; j <= i; j ++ )
            if (!j) dp[i][j] += dp[i - 1][j];
            else dp[i][j] += max(dp[i - 1][j - 1], dp[i - 1][j]);
    for (int i = 0; i <= m; i ++ ) ans = max(ans, dp[m][i]);
    cout << ans << endl;
}

int main()
{
    int T = 1;
    // cin >> T;
    while (T -- ) solve();
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The code seems to have issues with how it handles the distribution of attribute points and the checks. Here are some key points to consider:

1. **Early Checks**: Checks that appear before any points are acquired.
2. **Large Checks**: Checks that require a high level of an attribute.
3. **Balanced Distribution**: Ensuring that both Strength and Intelligence checks are present and require careful allocation of points.
4. **Multiple Points in a Row**: Sequences where multiple points are acquired consecutively.

Given these considerations, let's construct a test input that includes these elements:

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

Explanation:
- **n = 10**: Total number of records.
- **m = 5**: Total number of points.
- **Records**:
  - `1`: Intelligence check (fail, as no points are acquired yet).
  - `-1`: Strength check (fail, as no points are acquired yet).
  - `0`: Acquire a point.
  - `2`: Intelligence check (fail, as only one point is acquired).
  - `0`: Acquire a point.
  - `-3`: Strength check (fail, as only two points are acquired).
  - `0`: Acquire a point.
  - `-4`: Strength check (fail, as only three points are acquired).
  - `0`: Acquire a point.
  - `5`: Intelligence check (pass, as four points are acquired and can be used for Intelligence).

This test input is designed to:
- Ensure early checks fail due to no points being acquired.
- Include checks that require a high level of an attribute, which are not immediately passable.
- Balance the distribution of Strength and Intelligence checks.
- Include a sequence where multiple points are acquired consecutively.

This should help in exposing any issues in the buggy code's logic for distributing points and handling checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

#Q1
# t=int(input())
# for _ in range(t):
#     s,t,sec=input(),input(),0
#     for i in range(min(len(s),len(t))):
#         if(s[i]!=t[i]):
#             break
#         else:
#             sec+=1
#     if sec==0:
#         print(len(s)+len(t)-sec)
#     else:
#         print(len(s)+len(t)-sec+1)

#Q2
# MOD = 10**9 + 7

# def solve(n,k):
#     ans=1
#     while k>0:
#         if k%2==1:
#             ans=(ans*n)%MOD
#         n=(n*n)%MOD
#         k//=2
#     return ans

# t = int(input())
# n=inlt()
# k=inlt()
# for i in range(t):
#     print(solve(2,k[i]))

def solve(n,m,records):
    str,int,pts,passed=0,0,m,0
    future_str=sum(1 for r in records if r<0)
    future_int=sum(1 for r in records if r>0)
    for r in records:
        if r==0:
            if future_str>0 and (future_int==0 or str<int):
                str+=1
                pts-=1
                future_int-=1
            else:
                int+=1
                pts-=1
                future_int-=1
        elif r>0:
            if int>=abs(r):
                passed+=1
            else:
                req=abs(r)-int
                if req<=pts:
                    int+=req
                    pts-=req
                    passed+=1
        elif r<0:
            if str>=abs(r):
                passed+=1
            else:
                req=abs(r)-str
                if req<=pts:
                    str+=req
                    pts-=req
                    passed+=1
    return passed

n,m=invr()
recs=inlt()
print(solve(n,m,recs))

**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
6 3
0 0 1 0 -2 1
```

**Explanation:**
This test case is designed to expose potential bugs in the logic for handling the allocation of attribute points and the passing of checks. Here's the breakdown:

- **n = 6** and **m = 3**: There are 6 records and 3 attribute points to allocate.
- **Records**: [0, 0, 1, 0, -2, 1]

**Step-by-Step Analysis:**
1. **Initial State**: `str = 0`, `int = 0`, `pts = 3`, `passed = 0`
2. **Record 1 (0)**: Allocate a point. Since there are future checks for both Strength and Intelligence, we need to decide. Let's allocate to Intelligence.
   - `str = 0`, `int = 1`, `pts = 2`
3. **Record 2 (0)**: Allocate another point. Again, we have future checks for both attributes, so let's allocate to Intelligence.
   - `str = 0`, `int = 2`, `pts = 1`
4. **Record 3 (1)**: This is an Intelligence check requiring at least 1 Intelligence.
   - `int = 2 >= 1` (Pass)
   - `passed = 1`
5. **Record 4 (0)**: Allocate the last point. We have one future Strength check, so let's allocate to Strength.
   - `str = 1`, `int = 2`, `pts = 0`
6. **Record 5 (-2)**: This is a Strength check requiring at least 2 Strength.
   - `str = 1 < 2` (Fail)
7. **Record 6 (1)**: This is another Intelligence check requiring at least 1 Intelligence.
   - `int = 2 >= 1` (Pass)
   - `passed = 2`

**Expected Output:**
The maximum number of checks that can be passed is `2`.

This test case is small but comprehensive, ensuring that the algorithm correctly handles the allocation of points and the passing of checks. It also tests the scenario where the number of points is limited and the algorithm must make optimal decisions based on future checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
def max_checks():
    n, m = map(int, input().split())
    records = list(map(int, input().split()))

    strength, intelligence = 0, 0
    checks_passed = 0

    for record in records:
        if record == 0:
            # Prioritize the attribute needed for the next non-zero record.
            if records[min(n, records.index(record) + 2)] > 0:
                intelligence += 1
            else:
                strength += 1
        elif record > 0 and intelligence < record:
            continue
        elif record < 0 and strength < -record:
            continue
        else:
            checks_passed += 1

    return checks_passed

print(max_checks())

**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
5 3
0 1 0 -2 0
```

### Explanation:
- **Input Description:**
  - `n = 5` (number of records)
  - `m = 3` (number of attribute points)
  - `records = [0, 1, 0, -2, 0]`

- **Test Case Explanation:**
  - The first record is `0`, so we can spend a point on either Strength or Intelligence.
  - The second record is `1`, an Intelligence check. To pass this, we need at least 1 point in Intelligence.
  - The third record is `0`, another point to spend.
  - The fourth record is `-2`, a Strength check. To pass this, we need at least 2 points in Strength.
  - The fifth record is `0`, the final point to spend.

- **Expected Behavior:**
  - The correct strategy is to spend the first point on Intelligence to pass the second record (`1`).
  - The second point should be spent on Strength to pass the fourth record (`-2`).
  - The third point can be spent on either attribute, but it won't affect the number of checks passed.

- **Expected Output:**
  - The maximum number of checks passed should be `2`.

- **Why This Test Case is Challenging:**
  - The buggy code incorrectly assumes that the next non-zero record after a `0` is the one to prioritize. This can lead to incorrect attribute point allocation.
  - In this test case, the buggy code might incorrectly spend the first point on Strength, leading to failure of the Intelligence check and only passing one check instead of two.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
#from math import ceil,log,lcm,gcd,isfinite
#from collections import deque,Counter,defaultdict
#from bisect import bisect
#from heapq import heappush,heappop
#from sortedcontainers import SortedList,SortedSet,SortedDict
#from random import randint

from sys import stdin,stdout
input=lambda:stdin.readline().rstrip()
print=lambda *x,sep=' ',end='\n':stdout.write(sep.join(map(str,x))+end)

def out(l):
    print(' '.join(map(str,l)))

def yes():
    print('Yes')

def no():
    print('No')

def alice():
    print('Alice')

def bob():
    print('Bob')

def solve():
    n,m=map(int,input().split())
    r=list(map(int,input().split()))
    zero=0
    dp=[0]*(m+1)
    for j in r:
        if j>0:
            for i in range(j,zero+1):
                dp[i]+=1
        elif j<0:
            for i in range(m+j,-1,-1):
                dp[i]+=1
        else:
            zero+=1
    print(max(dp))

for _ in range(1):
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios where the code might fail. The main issues in the code seem to be related to how it handles the points and checks, especially in the dynamic programming (DP) part.

Here are some key points to consider:
1. **Order of Checks and Points**: The code should handle the order of checks and points correctly. If a check comes before the points are available, it should be handled appropriately.
2. **Negative and Positive Checks**: The code should correctly handle both positive (Intelligence) and negative (Strength) checks.
3. **Maximizing the Number of Passed Checks**: The code should be able to maximize the number of checks passed by distributing points optimally.

Given these considerations, let's create a test input that challenges the code:

```plaintext
8 3
1 0 -1 0 2 0 -2 1
```

### Explanation:
- **n = 8**: There are 8 records.
- **m = 3**: There are 3 points to distribute.
- **r = [1, 0, -1, 0, 2, 0, -2, 1]**:
  - **1**: An Intelligence check requiring 1 point.
  - **0**: Acquire a point.
  - **-1**: A Strength check requiring 1 point.
  - **0**: Acquire a point.
  - **2**: An Intelligence check requiring 2 points.
  - **0**: Acquire a point.
  - **-2**: A Strength check requiring 2 points.
  - **1**: An Intelligence check requiring 1 point.

### Expected Output:
The optimal strategy is:
1. Spend the first point on Intelligence.
2. Spend the second point on Strength.
3. Spend the third point on Intelligence.

This will pass the following checks:
- The first Intelligence check (1).
- The first Strength check (-1).
- The second Intelligence check (2).

Thus, the expected output is:
```plaintext
3
```

This test input is small-scale, complete, and valid, and it should expose any issues in the buggy code related to handling the order of checks and points, and maximizing the number of passed checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

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

public class d {
    static FastReader in = new FastReader();
    static final Random random = new Random();
    static long mod = 1000000007;
    static HashMap<String, Integer> map = new HashMap<>();

    public static void main(String[] args) throws java.lang.Exception {
        StringBuilder res = new StringBuilder();
        int n = in.nextInt();
        int m = in.nextInt();

        int[][][] dp = new int[n + 1][m + 1][m + 1];
        int ans = 0;

        for (int i = 1; i <= n; i++) {
            int r = in.nextInt();
            for (int s = 0; s <= m; s++) {
                for (int t = 0; t <= m; t++) {
                    if (s + t > m)
                        continue;

                    dp[i][s][t] = dp[i - 1][s][t];

                    if (r == 0) {
                        if (s > 0)
                            dp[i][s][t] = Math.max(dp[i][s][t], dp[i - 1][s - 1][t]);
                        if (t > 0)
                            dp[i][s][t] = Math.max(dp[i][s][t], dp[i - 1][s][t - 1]);
                    } else if (r > 0) {
                        if (t >= r)
                            dp[i][s][t] = Math.max(dp[i][s][t], dp[i - 1][s][t] + 1);
                    } else {
                        if (s >= -r)
                            dp[i][s][t] = Math.max(dp[i][s][t], dp[i - 1][s][t] + 1);
                    }

                    if (s + t <= m) {
                        ans = Math.max(ans, dp[n][s][t]);
                    }
                }
            }
        }

        res.append(ans);
        print(res);
    }

    static long pow(long base, long num) {
        long res = 1;

        while (num > 0) {
            if (num % 2 == 1) {
                res = (res * base) % mod;
            }

            base = (base * base) % mod;
            num /= 2;
        }

        return res;
    }

    static int max(int a, int b) {
        if (a < b)
            return b;
        return a;
    }

    static int min(int a, int b) {
        if (a < b)
            return a;
        return b;
    }

    static void ruffleSort(int[] a) {
        int n = a.length;
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n), temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }

    static <E> void print(E res) {
        System.out.println(res);
    }

    static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    static int lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    static int abs(int a) {
        if (a < 0)
            return -1 * a;
        return a;
    }

    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 {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] readintarray(int n) {
            int res[] = new int[n];
            for (int i = 0; i < n; i++)
                res[i] = nextInt();
            return res;
        }

        long[] readlongarray(int n) {
            long res[] = new long[n];
            for (int i = 0; i < n; i++)
                res[i] = nextLong();
            return res;
        }
    }
}

**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
6 3
0 2 0 -1 0 -2
```

### Explanation:
1. **Input Description:**
   - `n = 6` (number of records)
   - `m = 3` (number of attribute points)

2. **Sequence of Records:**
   - `0`: Acquire an attribute point.
   - `2`: Intelligence check (requires Intelligence >= 2).
   - `0`: Acquire an attribute point.
   - `-1`: Strength check (requires Strength >= 1).
   - `0`: Acquire an attribute point.
   - `-2`: Strength check (requires Strength >= 2).

3. **Optimal Strategy:**
   - Spend the first point on Intelligence.
   - Spend the second point on Strength.
   - Spend the third point on Strength.

4. **Check Results:**
   - For the Intelligence check at index 2 (`r_2 = 2`), you will fail because you only have 1 point in Intelligence.
   - For the Strength check at index 4 (`r_4 = -1`), you will pass because you have 1 point in Strength.
   - For the Strength check at index 6 (`r_6 = -2`), you will pass because you have 2 points in Strength.

5. **Expected Output:**
   - The maximum number of checks you can pass is 2.

This test input is designed to expose the bug in the given code by creating a scenario where the order of attribute points and checks is crucial. The code should correctly handle the scenario where spending points in a specific order maximizes the number of passed checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Solve {
    static int n, m;
    static List<List<Integer>> strengthPositions = new ArrayList<>();
    static List<List<Integer>> intelligencePositions = new ArrayList<>();

    static long calculateMaxScore() {
        // DP table to store results
        long[][] dp = new long[m + 1][m + 1];

        // Initialize the DP table
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= m; j++) {
                dp[i][j] = Long.MIN_VALUE; // Use Long.MIN_VALUE to represent uncomputed states
            }
        }

        // Base case
        dp[m][0] = 0; // No items to consider

        // Bottom-up approach to fill the DP table
        for (int index = m - 1; index >= 0; index--) {
            for (int position = 0; position <= m; position++) {
                if (position <= index) {
                    long sameScenario = dp[index + 1][position];
                    long differentScenario = dp[index + 1][position + 1];

                    dp[index][position] = Math.max(
                        sameScenario + strengthPositions.get(index).get(position) - strengthPositions.get(index + 1).get(position)
                                + intelligencePositions.get(index).get(index + 1 - position),
                        differentScenario + strengthPositions.get(index).get(position)
                                + intelligencePositions.get(index).get(index + 1 - position)
                                - intelligencePositions.get(index).get(index + 1 - position)
                    );
                }
            }
        }

        // The answer is the maximum value computed starting from the initial scenarios
        return Math.max(dp[0][0], dp[0][1]);
    }

    static void initializeStrengthAndIntelligenceCounts(int[] values) {
        int[] strengthCount = new int[m + 1];
        int[] intelligenceCount = new int[m + 1];

        for (int value : values) {
            if (value > 0) strengthCount[value]++;
            else if (value < 0) intelligenceCount[-value]++;
        }

        for (int value : values) {
            if (value > 0) {
                strengthCount[value]--;
            } else if (value < 0) {
                intelligenceCount[-value]--;
            } else {
                List<Integer> currentStrength = new ArrayList<>();
                List<Integer> currentIntelligence = new ArrayList<>();
                for (int k = 0; k <= m; k++) {
                    currentStrength.add(strengthCount[k]);
                    currentIntelligence.add(intelligenceCount[k]);
                }
                strengthPositions.add(currentStrength);
                intelligencePositions.add(currentIntelligence);
            }
        }

        List<Integer> finalStrength = new ArrayList<>();
        List<Integer> finalIntelligence = new ArrayList<>();
        for (int k = 0; k <= m; k++) {
            finalStrength.add(strengthCount[k]);
            finalIntelligence.add(intelligenceCount[k]);
        }
        strengthPositions.add(finalStrength);
        intelligencePositions.add(finalIntelligence);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        n = scanner.nextInt();
        m = scanner.nextInt();

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

        initializeStrengthAndIntelligenceCounts(values);

        long result = calculateMaxScore();
        System.out.println(result);

        scanner.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases:**
   - **Zero Points (m = 0):** This should result in no checks being passed.
   - **Single Attribute Point (m = 1):** This should test the scenario where the point is used optimally.
   - **Multiple Attribute Points (m > 1):** This should test the scenario where points are distributed between Strength and Intelligence.

2. **Scenarios:**
   - **Early Checks:** Checks that come before any points are acquired.
   - **Late Checks:** Checks that come after all points are acquired.
   - **Mixed Checks:** A mix of Strength and Intelligence checks.

3. **Input Constraints:**
   - The sequence must contain exactly `m` elements equal to `0`.
   - The number of records `n` must be between `m + 1` and `2 * 10^6`.

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

```plaintext
6 2
1 -1 0 0 2 -2
```

### Explanation:
- **n = 6, m = 2:** We have 6 records and 2 attribute points.
- **Records:**
  - `1`: An Intelligence check requiring Intelligence >= 1.
  - `-1`: A Strength check requiring Strength >= 1.
  - `0`: Acquire an attribute point.
  - `0`: Acquire another attribute point.
  - `2`: An Intelligence check requiring Intelligence >= 2.
  - `-2`: A Strength check requiring Strength >= 2.

### Expected Output:
- **Output: 2**
- **Explanation:** The optimal strategy is to spend the first point on Intelligence and the second point on Strength. This way, you can pass the first Intelligence check and the first Strength check, but you will fail the second Intelligence check and the second Strength check.

This test input is designed to challenge the buggy code by testing a mix of early and late checks, and ensuring that the points are used optimally.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
#include <iostream>
#include <cmath>
#include <vector>
#include <set>
#include <array>
#include <unordered_set>

using namespace std;
int checks[200000];
int map[5000][5000];
int mennyi;

int main() {
    set<array<short,2>> utolsopoziciok;
    array<short,2> a{0,0};
    utolsopoziciok.emplace(a);
    int pontok;
    int hany = 0;
    cin >> mennyi >> pontok;
    for (int i = 0; i < mennyi; ++i) {
        cin >> checks[i];
    }
    map[0][0] = 0;
    for (int i = 0; i < mennyi; ++i) {
        if(checks[i] == 0)
        {

            auto pp = utolsopoziciok;
            utolsopoziciok.clear();
            for (int k = 0; k <= hany; ++k) {
                map[k][hany-k+1] = map[k][hany-k];
                map[k+1][hany-k] = map[k][hany-k];
            }
            hany++;
        }
        else if(checks[i] < 0)
        {
            for (int k = 0; k < hany; ++k) {
                if(hany-k >= abs(checks[i]))
                {
                    map[k][hany-k]++;
                }
            }
        }
        else if(checks[i] > 0)
        {
            for (int k = 0; k < hany; ++k) {
                if(hany-k >= abs(checks[i]))
                {
                    map[hany-k][k]++;
                }
            }
        }
    }
    int legnagyobb = 0;
    for (int k = 0; k <= hany; ++k) {
        int szam = map[hany - k][k];
        if (szam > legnagyobb) {
            legnagyobb = szam;
        }
    }
    cout << legnagyobb << endl;
//        for(int a = 0; a < 20; a++)
//        {
//            for(int b = 0; b < 40; b++)
//            {
//                cout << map[a][b] << " ";
//            }
//            cout << 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
7 3
0 1 0 -1 0 1 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define maxn 5050
using namespace std;
int dp[maxn];
int main(){
    for(int i=0;i<maxn;++i)dp[i]=-(1<<30);
    dp[0]=0;
    int n,m,p=0;
    cin>>n>>m;
    for(int i=1;i<=n;++i){
        int a;scanf("%d",&a);
        if(a==0){
            ++p;
            dp[p]=dp[p-1];
            for(int j=p-1;j>=1;--j)dp[j]=max(dp[j],dp[j-1]);
        }
        else if(a>0){
            for(int j=a;j<=p;++j)++dp[j];
        }
        else{
            a=-a;
            for(int j=0;p-j>=a;++j)++dp[j];
        }
    }
    int mx=0;
    for(int i=0;i<=p;++i)mx=max(mx,dp[i]);
    cout<<mx<<endl;
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5 3
0 0 1 0 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
from collections import defaultdict

n,m = map(int, input().split())
a = list(map(int, input().split()))
stren = [0]*m
tot = 0
for c in a:
    if c == 0:
        for i in range(m-1,0,-1): stren[i] = max(stren[i], stren[i-1])
        tot += 1
    elif c > 0:
        for i in range(c,m): stren[i] += 1
    else:
        c = -c
        for i in range(tot-c+1): stren[i] += 1

print(max(stren))

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

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

Think step by step.

Response:
```plaintext
7 3
1 -1 0 0 0 2 -2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
n, m = map(int, input().split())
r = list(map(int, input().split()))

# Initialize DP table
m = int(m)
dp = [[-1] * (m + 1) for _ in range(m + 1)]
dp[0][0] = 0

for i in range(n):
    if r[i] == 0:
        # Copy dp to dp_next
        dp_next = [list(row) for row in dp]
        for k in range(m):
            for s in range(k + 1):
                if dp[k][s] >= 0:
                    # Increase Strength
                    dp_next[k + 1][s + 1] = max(dp_next[k + 1][s + 1], dp[k][s])
                    # Increase Intelligence
                    dp_next[k + 1][s] = max(dp_next[k + 1][s], dp[k][s])
        dp = dp_next
    else:
        for k in range(m + 1):
            for s in range(k + 1):
                if dp[k][s] >= 0:
                    t = k - s
                    if r[i] > 0 and t >= r[i]:
                        dp[k][s] += 1
                    elif r[i] < 0 and s >= -r[i]:
                        dp[k][s] += 1

# Find the maximum number of checks passed
max_checks = 0
for k in range(m + 1):
    for s in range(k + 1):
        if dp[k][s] > max_checks:
            max_checks = dp[k][s]

print(max_checks)

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

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

Think step by step.

Response:
```plaintext
7 3
0 0 2 0 -2 -1 1
```

**Explanation:**

- **Test Case Design:**
  - **Number of Records (n) and Attribute Points (m):** 7 records and 3 attribute points.
  - **Sequence (r):** `0 0 2 0 -2 -1 1`

- **Scenario:**
  - The first two records are attribute points, which means you can spend these points to increase either Strength or Intelligence.
  - The third record is an Intelligence check requiring a level of 2.
  - The fourth record is another attribute point.
  - The fifth record is a Strength check requiring a level of 2.
  - The sixth record is a Strength check requiring a level of 1.
  - The seventh record is an Intelligence check requiring a level of 1.

- **Expected Behavior:**
  - If the algorithm correctly handles the allocation of attribute points, it should be able to pass the following checks:
    - The Strength check at record 5 (if you allocate 2 points to Strength).
    - The Strength check at record 6 (if you allocate 2 points to Strength).
    - The Intelligence check at record 7 (if you allocate 1 point to Intelligence).

- **Challenging Aspects:**
  - This test case is designed to challenge the algorithm's ability to handle the allocation of points efficiently, especially when checks are mixed and require specific levels of both attributes.
  - The order of checks and points is such that a greedy approach (e.g., always increasing Strength or Intelligence) might fail to pass the maximum number of checks.
  - The test case includes a mix of positive and negative checks, which can expose issues in the dynamic programming logic, especially in how it handles the transition between different states.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                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 {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    static class DisjointSet {
        List<Integer> rank = new ArrayList<>();
        List<Integer> parent = new ArrayList<>();
        List<Integer> size = new ArrayList<>();
        public DisjointSet(int n) {
            for (int i = 0; i <= n; i++) {
                rank.add(0);
                parent.add(i);
                size.add(1);
            }
        }

        public int findUPar(int node) {
            if (node == parent.get(node)) {
                return node;
            }
            int ulp = findUPar(parent.get(node));
            parent.set(node, ulp);
            return parent.get(node);
        }

        public void unionBySize(int u, int v) {
            int ulp_u = findUPar(u);
            int ulp_v = findUPar(v);
            if (ulp_u == ulp_v) return;
            if (size.get(ulp_u) < size.get(ulp_v)) {
                parent.set(ulp_u, ulp_v);
                size.set(ulp_v, size.get(ulp_v) + size.get(ulp_u));
            } else {
                parent.set(ulp_v, ulp_u);
                size.set(ulp_u, size.get(ulp_u) + size.get(ulp_v));
            }
        }
    }
    static long power(long a, long b, long m) {
        a %= m;
        long res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = (res * a) % m;
            a = (a * a) % m;
            b >>= 1;
        }
        return res;
    }
    static long gcd(long a, long b)  {
        return b == 0? a: gcd(b, a % b);
    }
    static int MOD = 1000000007;
    static long modinv(long a) {
        return a <= 1 ? a : MOD - (MOD / a) * modinv(MOD % a) % MOD;
    }

    static int ans;
    public static int helper(int i, int a[], int n, int str, int intel, HashMap<String, Integer> map) {
        if(i == n) {
            return 0;
        }
        String s = i+" "+Integer.toString(str)+" "+Integer.toString(intel);
        if(map.containsKey(s)) return map.get(s);

        int ans = 0;
        if(a[i] == 0) {
            int a1 = helper(i+1, a, n, str+1, intel, map);
            int a2 = helper(i+1, a, n, str, intel+1, map);
            ans =  Math.max(a1, a2);
        } else if(a[i] < 0) {
            if(str >= Math.abs(a[i])) ans =  1 + helper(i+1, a, n, str, intel, map);
            else ans = helper(i+1, a, n, str, intel, map);
        } else {
            if(intel >= a[i]) ans = 1 + helper(i+1, a, n, str, intel, map);
            else ans = helper(i+1, a, n, str, intel, map);
        }
        map.put(s, ans);
        return ans;
    }
    public static void main(String args[]) throws IOException {
        FastReader sc = new FastReader();
        // int T = sc.nextInt();
        // while (T-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();

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

            HashMap<String, Integer> map = new HashMap<>();
            int x = helper(0, a, n, 0, 0,  map);

            System.out.println(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
7 3
0 1 0 -1 0 2 -2
```

### Explanation:
This test input is designed to challenge the algorithm by presenting a sequence where the optimal distribution of attribute points is not immediately obvious. Here's the breakdown:

- **Input:** `7 3` (7 records and 3 attribute points)
- **Sequence:** `0 1 0 -1 0 2 -2`

### Step-by-Step Analysis:
1. **Initial State:**
   - Strength: 0
   - Intelligence: 0
   - Points available: 3

2. **Record 1: `0`**
   - You can choose to increase either Strength or Intelligence.
   - Let's increase Strength: Strength = 1, Intelligence = 0

3. **Record 2: `1`**
   - This is an Intelligence check ( Intelligence >= 1 to pass).
   - Current Intelligence = 0, so you fail this check.

4. **Record 3: `0`**
   - You can choose to increase either Strength or Intelligence.
   - Let's increase Intelligence: Strength = 1, Intelligence = 1

5. **Record 4: `-1`**
   - This is a Strength check (Strength >= 1 to pass).
   - Current Strength = 1, so you pass this check.

6. **Record 5: `0`**
   - You can choose to increase either Strength or Intelligence.
   - Let's increase Intelligence: Strength = 1, Intelligence = 2

7. **Record 6: `2`**
   - This is an Intelligence check (Intelligence >= 2 to pass).
   - Current Intelligence = 2, so you pass this check.

8. **Record 7: `-2`**
   - This is a Strength check (Strength >= 2 to pass).
   - Current Strength = 1, so you fail this check.

### Expected Output:
- **Output:** `2`
- **Reason:** You can pass the Strength check at record 4 and the Intelligence check at record 6, but you fail the Intelligence check at record 2 and the Strength check at record 7.

This test input is designed to ensure that the algorithm correctly handles the distribution of attribute points and the order of checks. It also tests the edge case where the optimal strategy is not to always increase the same attribute.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

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

public class Main
{
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static  BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    public static StringTokenizer tokenizer;
    public static void read() throws IOException {
        tokenizer = new StringTokenizer(br.readLine());
    }
    public static int Int() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) read();
        return Integer.parseInt(tokenizer.nextToken());
    }
    public static long Long() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) read();
        return Long.parseLong(tokenizer.nextToken());
    }
    public static String Str() throws IOException {
        return br.readLine();
    }
    //-----------------------------------------------

    public static long quickPow(long b, long x, long mod) {
        if (x == 0) {
            return 1;
        }
        if (x == 1) {
            return b % mod;
        }
        long half = quickPow(b, x / 2, mod) % mod;
        long result = (half * half) % mod;
        result = (result * quickPow(b, x % 2, mod)) % mod;
        return result;
    }

    public static HashMap<Integer, HashMap<Integer, Integer>> memo;
    public static int f(int[] points, ArrayList<Integer> zeroIndexes, int zI, int I) {
        if (memo.containsKey(zI) && memo.get(zI).containsKey(I)) {
            return memo.get(zI).get(I);
        }
        int res;
        if (zI >= zeroIndexes.size()) {
            res = 0;
        } else {
            int resIPlus = 0;
            int resSPlus = 0;
            for (int i = zeroIndexes.get(zI) + 1; i < (zI + 1 >= zeroIndexes.size() ? points.length : zeroIndexes.get(zI + 1)); i++) {

                if (points[i] > 0) {
                    resIPlus += ((I + 1) >= points[i] ? 1 : 0);
                    resSPlus += (I >= points[i] ? 1 : 0);
                }
                int S = (zI + 1) - (I + 1);
                int SPlus = (zI + 1) - I;
                if (points[i] < 0) {
                    resIPlus += (S >= Math.abs(points[i]) ? 1 : 0);
                    resSPlus += (SPlus >= Math.abs(points[i]) ? 1 : 0);
                }
            }
            resIPlus += f(points, zeroIndexes, zI + 1, I + 1);
            resSPlus += f(points, zeroIndexes, zI + 1, I);
            res = Math.max(resIPlus, resSPlus);
        }
        if (!memo.containsKey(zI)) {
            memo.put(zI, new HashMap<>());
        }
        memo.get(zI).put(I, res);
        return res;
    }

    public static void main (String[] args) throws java.lang.Exception {
        int T = 1;
        for (int t = 0; t < T; t++) {
            memo = new HashMap<>();
            int N = Int(), K = Int();
            int[] n = new int[N];
            ArrayList<Integer> zeroIndexes = new ArrayList<>();
            for (int i = 0; i < N; i++) {
                n[i] = Int();
                if (n[i] == 0) {
                    zeroIndexes.add(i);
                }
            }

            System.out.println(f(n, zeroIndexes, 0, 0));
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
7 3
0 0 1 0 -2 1 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Attribute Checks
time_limit_ms: 2500
memory_limit_mb: 512
Description: Imagine a game where you play as a character that has two attributes: "Strength" and "Intelligence", that are at zero level initially.

During the game, you'll acquire $m$ attribute points that allow you to increase your attribute levels — one point will increase one of the attributes by one level. But sometimes, you'll encounter a so-called "Attribute Checks": if your corresponding attribute is high enough, you'll pass it; otherwise, you'll fail it.

Spending some time, you finally prepared a list which contains records of all points you got and all checks you've met. And now you're wondering: what is the maximum number of attribute checks you can pass in a single run if you'd spend points wisely?

Note that you can't change the order of records.

Input: The first line contains two integers $n$ and $m$ ($1 \le m \le 5000$; $m < n \le 2 \cdot 10^6$) — the number of records in the list and the total number of points you'll get during the game.

The second line contains $n$ integers $r_1, r_2, \dots, r_n$ ($-m \le r_i \le m$), where $r_i$ encodes the $i$-th record:

  * If $r_i = 0$, then the $i$-th record is an acquiring one attribute point. You can spend to level up either Strength or Intelligence;    * If $r_i > 0$, then it's an Intelligence check: if your Intelligence level is greater than or equal to $|r_i|$, you pass.    * If $r_i < 0$, then it's a Strength check: if your Strength level is greater than or equal to $|r_i|$, you pass.

Additional constraint on the input: the sequence $r_1, r_2, \dots, r_n$ contains exactly $m$ elements equal to $0$.

Output: Print one integer — the maximum number of checks you can pass.

Examples:
input:
10 5
0 1 0 2 0 -3 0 -4 0 -5
output:
3

input:
3 1
1 -1 0
output:
0

input:
9 3
0 0 1 0 2 -3 -2 -2 1
output:
4

Note: In the first test, it's optimal to spend each point in Strength, so you'll fail $2$ Intelligence checks but pass $3$ Strength checks.

In the second test, you'll fail both checks, since the first point you get comes after the checks.

In the third test, one of the optimal strategies is:

  1. spend the first point on Intelligence;    2. spend the second point on Strength;    3. spend the third point on Strength;

As a result, you'll pass $2$ Intelligence checks $r_3$ and $r_9$ and $2$ Strength checks $r_7$ and $r_8$.

Buggy Code:
#include<bits/stdc++.h>
//#include<windows.h>
// #define double long double
//#define int long long
#define signed bool __end;signed
#define PII pair<int,int>
using namespace std;
bool __begin;
namespace _IO{inline int read(){char ch=getchar();int f=1,res=0;while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){res*=10;res+=ch-'0';ch=getchar();}return res*f;}inline void write(int x){if(x<0){putchar('-');write(-x);}else if(x<10){putchar(x+'0');}else{write(x/10);putchar((x%10)+'0');}return;}}
namespace _Min{inline int Min(int x,int y){return x<=y?x:y;}inline int Max(int x,int y){return x>=y?x:y;}}

using namespace _IO;
using namespace _Min;

int n,m,a[2000005];
int bt[5005];
int f[5005];
int lowbit(int x){
	return (x&(-x));
}
void Add(int x,int v){
	x++;
	while(x<=m+1){
		bt[x]+=v;
		x+=lowbit(x);
	}
	return;
}
int query(int x){
	int res=0;
	x++;
	while(x>0){
		res+=bt[x];
		x-=lowbit(x);
	}
	return res;
}
void add(int l,int r,int v){
	if(l>r)return;
	Add(l,v),Add(r+1,-v);
	return;
}

signed main(){
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    n=read(),m=read();
    int sum=0;
    for(int i=1;i<=m;i++){
    	add(i,i,-0x3f3f3f3f);
	}
	add(0,0,0);
    for(int i=1;i<=n;i++){
    	a[i]=read();
    	if(a[i]==0){
    		sum++;
    		f[0]=query(0);
    		for(int j=sum;j>=1;j--){
    			f[j]=query(j);
    			add(j,j,-f[j]);
				f[j]=Max(f[j],f[j-1]);
				add(j,j,f[j]);
			}
		}else if(a[i]>0){
			add(a[i],sum,1);
		}else{
			add(0,sum+a[i],1);
		}
	}
	int ans=0;
	for(int i=0;i<=m;i++){
		ans=Max(ans,query(i));
	}
	write(ans);
	// cerr<<(abs((&__end)-(&__begin))>>20)<<"MB\n"<<clock()<<"ms";
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to create a test input that challenges the logic of the algorithm, particularly focusing on edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Understand the Bug**: The buggy code seems to have issues with handling the updates and queries in the Fenwick tree (Binary Indexed Tree) and the dynamic programming part. It might not correctly handle the updates and queries for attribute checks, especially when the attribute points are not used optimally.

2. **Create a Small-Scale Test Input**: We need a small-scale test input that is complete and valid, and that can expose the bug. Let's consider a scenario where the points are distributed in a way that the optimal strategy is not straightforward.

3. **Test Case**: We will create a test case where the attribute points are acquired in a way that the optimal strategy is to distribute the points between Strength and Intelligence in a balanced manner, but the code might fail to handle this correctly.

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

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

**Explanation**:
- **Input**:
  - `n = 10` and `m = 5`, meaning there are 10 records and 5 attribute points.
  - The sequence of records is `0, 1, 0, -1, 0, 2, 0, -2, 0, -3`.
- **Records**:
  - `0` indicates an attribute point.
  - `1` and `2` are Intelligence checks.
  - `-1`, `-2`, and `-3` are Strength checks.
- **Optimal Strategy**:
  - Distribute the 5 attribute points between Strength and Intelligence to maximize the number of checks passed.
  - One optimal strategy is to spend 3 points on Strength and 2 points on Intelligence, which would pass the checks for `1`, `2`, and `-1`, `-2`.

This test input should help expose any issues in the buggy code's handling of attribute points and checks.