**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;
template<typename T>
using vec = vector<T>;
using vi = vec<int>;
using ii = pair<int, int>;
using vii = vec<ii>;
using i64 = long long;
using vi64 = vec<i64>;
using iii = tuple<int, int, int>;
using str = string;
template<typename Key, typename Value>
using hash_map = unordered_map<Key, Value>;
template<typename T>
using hash_set = unordered_set<T>;
template<typename T>
using min_heap = priority_queue<T, vec<T>, greater<T>>;
template<typename T>
using max_heap = priority_queue<T>;

// Code below

template<typename T>
class RangeOperation{
    using vT = vector<T>;
    using opT = function<T(T,T)>;

    private:
        int n;
        vT A;
        opT op, inv;

    public:
    RangeOperation(int _n, T initialVal, opT _op, opT _inv){
        n = _n;
        A = vT(n+1, initialVal);
        op = _op;
        inv = _inv;
    }

    void range_update(int l, int r, T val){
        if(r < l) return;
        A[l] = op(A[l], val);
        A[r+1] = inv(A[r+1], val);
    }

    vT get_vector(){
        vT ans(n);
        ans[0] = A[0];
        for(int i = 1; i < n; i++)
            ans[i] = op(ans[i-1], A[i]);
        return ans;
    }

    void clear(T initialVal){
        A = vT(n+1, initialVal);
    }
};

void solve(){
    int n, m;
    cin >> n >> m;
    vi a(n); for(auto& a_i : a) cin >> a_i;

    vi dp(m+1, 0);
    int P = 0;

    RangeOperation<int> intel_check(m+1, 0, [](int a, int b){return a+b;}, [](int a, int b){return a-b;});
    RangeOperation<int> force_check(m+1, 0, [](int a, int b){return a+b;}, [](int a, int b){return a-b;});

    for(auto x : a){
        if(x > 0)
            intel_check.range_update(x, min(P, m), 1);
        else if(x < 0)
            force_check.range_update(0, min(P + x, m), 1);
        else if(x == 0 && P < 2*m){
            vi new_dp = dp;
            vi intel_passes = intel_check.get_vector();
            vi force_passes = force_check.get_vector();
            for(int intel = 0; intel <= P; intel++)
                new_dp[intel] += intel_passes[intel] + force_passes[intel];
            dp = new_dp;
            P++;
            intel_check.clear(0);
            force_check.clear(0);
        }
    }

    vi intel_passes = intel_check.get_vector();
    vi force_passes = force_check.get_vector();
    for(int intel = 0; intel <= P; intel++)
        dp[intel] += intel_passes[intel] + force_passes[intel];

    int ans = dp[0];
    for(int candidate : dp)
        ans = max(ans, candidate);
    cout << ans << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int testCases = 1;
    // cin >> testCases;

    while(testCases--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the code to fail. Here are some potential issues to target:

1. **Early Checks Before Points**: The code might not handle checks that appear before any attribute points are acquired.
2. **Late Points**: The code might not handle points that come very late in the sequence.
3. **Mixed Checks**: The code might not handle a sequence with both Intelligence and Strength checks mixed in.
4. **High-Level Checks**: The code might not handle checks that require a high level of attributes.

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

- Early checks before any points are acquired.
- Late points that come after many checks.
- Mixed checks (both Intelligence and Strength checks).
- High-level checks that require a high level of attributes.

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

```plaintext
15 5
1 -1 2 -2 0 0 0 0 0 3 -3 4 -4 5 -5
```

### Explanation:
- **1, -1, 2, -2**: These are early checks that come before any points are acquired.
- **3, -3, 4, -4, 5, -5**: These are high-level checks that require a high level of attributes.
- **0 0 0 0 0**: These are the points that can be used to level up the attributes.

This input should challenge the code to handle early checks, mixed checks, and high-level checks correctly.


**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 sys
from collections import Counter, deque, defaultdict
from itertools import accumulate
import functools
from functools import lru_cache
li = lambda: sys.stdin.readline().rstrip()
lli = lambda: sys.stdin.readline().rstrip().split()
Q = lambda: list(map(int, lli()))
from math import inf
from bisect import bisect_left
from random import randint
rd = randint(10 ** 9, 2 * 10 ** 9)
n, m = Q()
a = Q() + [0]
pi = [0] * (n+1)  # 前i个属性点
for i,x in enumerate(a):
    if i > 0:
        pi[i] = pi[i-1]
    if x == 0:
        pi[i] += 1

dp = [0] * (m + 2)
p, q = [], []
for i,x in enumerate(a):
    # n_d = defaultdict(int)
    # for j in range(p[i], -1, -1):
        # if x == 0:
            # dp[j] = max(dp[j], dp[j-1])
        # elif x > 0:
            # dp[j] = max(dp[j], dp[j] + (j >= x))
        # else:
            # dp[j] = max(dp[j], dp[j] + (p[i] - j >= -x))
    if x == 0:
        n_d = [0] * (m+2)
        p.sort(); q.sort()
        # 第一次碰到x=0时 无法加点
        for j in range(pi[i] - 1, -1, -1):
            v = bisect_left(p, j+1) + bisect_left(q, pi[i]-1-j+1)  # [1, 3, 4]
            # if v: print(j, dp[j], v, dp[j] + v, p, q)
            # n_d[j+1] = max(n_d[j+1], dp[j+1], dp[j] + v)  # 点智力
            # n_d[j] = max(n_d[j], dp[j] + v)  # 点力量
            n_d[j] = max(n_d[j], dp[j] + v)
        for j in range(pi[i] - 1, -1, -1):
            n_d[j+1] = max(n_d[j+1], n_d[j])
        dp = n_d
        # print(dp)
        p, q = [], []
    elif x > 0:
        p.append(x)
    else:
        q.append(-x)
    # print(dp)
print(max(dp))

**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 -2 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:
# https://codeforces.com/contest/2025

import sys

input = lambda: sys.stdin.readline().rstrip()  # faster!

n, m = map(int, input().split())  # 1 <= m <= 5000, m < n <= 2 * 10**6
r = list(map(int, input().split()))

p = [0] * n
p[0] = int(r[0] == 0)
for i in range(1, n):
    p[i] = p[i - 1] + int(r[i] == 0)

score = [0] * (m + 1)
add = [0] * (m + 2)

for i in range(n):
    k = p[i]
    if r[i] == 0:
        for j in range(m + 1):
            score[j] += add[j]
            add[j + 1] += add[j]
            add[j] = 0
    elif r[i] > 0:
        if r[i] <= k:
            add[r[i]] += 1
            add[k + 1] -= 1
    elif r[i] < 0:
        if k + r[i] >= 0:
            add[0] += 1
            add[k + r[i] + 1] -= 1

for j in range(m + 1):
    score[j] += add[j]
    add[j + 1] += add[j]
    add[j] = 0

ans = max(score)
print(ans)

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

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

Think step by step.

Response:
```plaintext
6 3
0 1 0 -1 0 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:
import java.util.*;
import java.io.*;

public class AttributeCheck {
    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;
        }
    }
    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    public static void main(String[] args) {
        try {
            FastReader in=new FastReader();
            FastWriter out = new FastWriter();
            int testCases=1;
            while(testCases-- > 0){
                int n = in.nextInt(), m = in.nextInt();
                int[] arr = new int[n];
                for (int i = 0; i < n; i++) arr[i] = in.nextInt();

//                List<Integer>[] arr1 = new List[m+1];
//                List<Integer>[] arr2 = new List[m+1];
                List<Integer> pos = new ArrayList<>(), neg = new ArrayList<>();
               // int index = 0;
                int cnt = 0;
                int[] prev = new int[m+1];
                for (int i = 0; i < n; i++){
                    if (arr[i] == 0){
                        Collections.sort(pos);
                        Collections.sort(neg);
//                        arr1[index] = pos;
//                        arr2[index] = neg;
                      //  index++;
                        int[] curr = new int[m+1];
                        for (int j = 0; j <= cnt; j++){
                            curr[j+1] = prev[j] + binarySearch(pos,j) + binarySearch(neg,cnt-j);
                            curr[j] = Math.max(curr[j], prev[j] + binarySearch(pos,j) + binarySearch(neg,cnt-j));
                        }
                        cnt++;
                        prev = curr;
                        pos = new ArrayList<>();
                        neg = new ArrayList<>();
                    }
                    else if (arr[i] < 0) neg.add(Math.abs(arr[i]));
                    else pos.add(arr[i]);
                }

                int[] curr = new int[m+5];
                for (int j = 0; j <= cnt; j++){
                    curr[j+1] = prev[j] + binarySearch(pos,j) + binarySearch(neg,cnt-j);
                    curr[j] = Math.max(curr[j], prev[j] + binarySearch(pos,j) + binarySearch(neg,cnt-j));
                }

                out.print(getMax(curr));

//                arr1[index] = pos;
//                arr2[index] = neg;
//                int[] prev = new int[m+1];
//                for (int i = 1; i <= m; i++){
//                    int[] curr = new int[m+1];
//                    for (int j = 0; j < i; j++){
//                        curr[j+1] = prev[j] + binarySearch(arr1[i],j+1) + binarySearch(arr2[i],(i-(j+1)));
//                        curr[j] = prev[j] + binarySearch(arr1[i],j) + binarySearch(arr2[i],i-j);
//                    }
//                    prev = curr;
//                }
//
//                int ans = getMax(prev);
//                out.print(ans);
            }
            out.close();
        } catch (Exception e) {
            return;
        }
    }

    private static int getMax(int[] arr){
        int max = 0;
        for (int val : arr) max = Math.max(max,val);
        return max;
    }

    private static int binarySearch(List<Integer> list, int target){
        if (list == null) return 0;
        int s = 0, e = list.size()-1;
        while (s <= e){
            int m = s + (e-s)/2;
            if (list.get(m) > target) e = m-1;
            else s = m+1;
        }
        return e+1;
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through:

1. **Edge Cases:**
   - **Very small `n` and `m`:** This can help check if the code handles small inputs correctly.
   - **All points are checks:** This can test if the code can handle a scenario where there are no attribute points to spend.
   - **All points are attribute points:** This can test if the code can handle a scenario where there are no checks.
   - **Mixed checks and points:** This can test the code's ability to handle a mix of checks and points.

2. **Scenarios to Maximize Failure Probability:**
   - **Early checks:** Checks that occur early in the sequence before many points are available.
   - **High requirement checks:** Checks that require a high level of attribute, making them hard to pass.
   - **Balanced distribution:** A mix of Strength and Intelligence checks to ensure the code handles both attributes correctly.

3. **Small-Scale Test Input:**
   - We will choose a small value for `n` and `m` to ensure the input is manageable.
   - We will include a mix of checks and points to test the code's logic.

Based on these considerations, here is a small-scale test input that should expose potential bugs in the code:

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

### Explanation:
- **n = 7, m = 3:** This is a small input size.
- **Sequence:**
  - `0`: Acquire an attribute point.
  - `2`: Intelligence check (requires Intelligence >= 2).
  - `-2`: Strength check (requires Strength >= 2).
  - `0`: Acquire an attribute point.
  - `1`: Intelligence check (requires Intelligence >= 1).
  - `0`: Acquire an attribute point.
  - `-1`: Strength check (requires Strength >= 1).

### Expected Output:
- The optimal strategy is to spend the first point on Intelligence, the second point on Strength, and the third point on Intelligence.
- This way, you can pass the Intelligence check at index 4 (`1`) and the Strength check at index 6 (`-1`), resulting in 2 passed checks.

```plaintext
2
```

This test input should help expose any issues in the code's logic for handling attribute points and 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.*;
import java.io.*;

public class Main {
    // static class Data implements Comparable<Data>{
    //     boolean there;
    //     int val;
    //     public Data(boolean there,int val){
    //         this.there = there;
    //         this.val = val;
    //     }
    //     @Override
    //     public int compareTo(Data p1){
    //         if(this.val>=p1.val) return 1;
    //         return -1;
    //     }
    // }
    // static class Edge {
    //     int dst;
    //     int dir;
    //     public Edge(int dst,int dir){
    //         this.dir = dir;
    //         this.dst = dst;
    //     }
    // }
    public static PP.Reader cs = new PP.Reader(System.in);
    public static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException{
        int tests = 1;//cs.nextInt();
        while(tests>0){
            tests--;
            solve();
        }
        // writer.flush();
        writer.close();
    }
    public static long mod = 1000000000 + 7;
    public static long MOD = 998244353;
    public static void solve() throws IOException{
        // int l = cs.nextInt();
        // Random random = new Random();
        // int q = cs.nextInt();
        int n = cs.nextInt();
        int m = cs.nextInt();
        Integer[] arr = new Integer[n];
        for(int i=0;i<n;i++) {
            arr[i] = cs.nextInt();
        }
        ArrayList<Integer> strngth[] = new ArrayList[m+1];
        ArrayList<Integer> intelligence[] = new ArrayList[m+1];
        for(int i=0;i<=m;i++){
            strngth[i] = new ArrayList<>();
            intelligence[i] = new ArrayList<>();
        }
        int ptr = 0;
        for(int i=0;i<n;i++){
            if(arr[i] == 0) ptr++;
            else if(arr[i]>0) strngth[ptr].add(arr[i]);
            else intelligence[ptr].add(Math.abs(arr[i]));
        }
        for(int i=0;i<=m;i++){
            Collections.sort(strngth[i]);
            Collections.sort(intelligence[i]);
        }
        int[][] dp = new int[m+3][m+3];
        for(int i=m;i>=0;i--){
            for(int j=0;j<=i;j++){
                int wins = 0;
                if(strngth[i].size() != 0){
                    int idx = Collections.binarySearch(strngth[i], j);
                    if(idx<0) idx = -idx-1;
                    else idx++;
                    wins += idx;
                }
                if(intelligence[i].size() != 0){
                    int idx = Collections.binarySearch(intelligence[i], i-j);
                    if(idx<0) idx = -idx-1;
                    else idx++;
                    wins += idx;
                }
                dp[i][j] = wins + Math.max(dp[i+1][j], dp[i+1][j+1]);
            }
        }
        System.out.println(dp[0][0]);
    }
    //Write fxn here

    public static  int binary(int[] arr,int start,int end,int key){
        int mid = (int)start;
        int st = start;
        int en = end;
        while(start<=end){
            mid = (int)(start+end)/2;
            if(arr[mid]==key) return mid;
            else if(arr[mid]>key) end = mid-1;
            else if(arr[mid]<key) start = mid+1;
        }
        //if(end<start){
            if(end<st) end++;
            // if(start>en) start--;
            // if(Math.abs(arr[end]-key)<=Math.abs(arr[start]-key-1)){
                // return start;
            // }
            return end;
        //}
        //return mid+1;
    }
    public static long gcd(long a,long b){
        long c = Math.min(a,b);
        long d = Math.max(a,b);
        if(d%c==0) return c;
        return gcd(c,d%c);
    }
    public static long divmod(long a,long b,long m){   //claculate (a/b)%m
        // a = a%m;b = b%m  // Include if not working
        long inv =  powermod(b, m-2, m);
        return (inv*a)%m;
    }
    static long powermod(long x, long y, long p) //calculate (x^y)%p
    {
        long res = 1;
        x = x % p;
        while (y > 0) {
            if ((y & 1) > 0)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
}

class PP {
    static String IN = "%s";
    static String OUT = "%s";

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception ignored) {
            }
            return null;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}
class SegmentTree {
    private int n;
    private int[] treeMin;

    public SegmentTree(int[] arr) {
        n = arr.length;
        treeMin = new int[4 * n];
        build(arr, 0, n - 1, 0);
    }

    private void build(int[] arr, int start, int end, int node) {
        if (start == end) {
            treeMin[node] = arr[start];
        } else {
            int mid = (start + end) / 2;
            build(arr, start, mid, 2 * node + 1);
            build(arr, mid + 1, end, 2 * node + 2);
            treeMin[node] = Math.min(treeMin[2 * node + 1], treeMin[2 * node + 2]);
        }
    }

    public int rangeMin(int left, int right) {
        return rangeMin(0, n - 1, left, right, 0);
    }

    private int rangeMin(int start, int end, int left, int right, int node) {
        if (right < start || left > end) return Integer.MAX_VALUE;
        if (left <= start && right >= end) return treeMin[node];
        int mid = (start + end) / 2;
        return Math.min(rangeMin(start, mid, left, right, 2 * node + 1),
                        rangeMin(mid + 1, end, left, right, 2 * node + 2));
    }

    public void update(int idx, int value) {
        update(0, n - 1, idx, value, 0);
    }

    private void update(int start, int end, int idx, int value, int node) {
        if (start == end) {
            treeMin[node] = value;
        } else {
            int mid = (start + end) / 2;
            if (idx <= mid) {
                update(start, mid, idx, value, 2 * node + 1);
            } else {
                update(mid + 1, end, idx, value, 2 * node + 2);
            }
            treeMin[node] = Math.min(treeMin[2 * node + 1], treeMin[2 * node + 2]);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to craft a test case that highlights any issues in the algorithm's logic, particularly in how it handles attribute checks and the distribution of attribute points. The code appears to have issues with how it manages the binary search for attribute checks and the dynamic programming (DP) transitions.

### Step-by-Step Reasoning:

1. **Attribute Points Distribution**: The code should handle the distribution of attribute points optimally. If the distribution is not optimal, it might lead to failing more checks than necessary.

2. **Binary Search**: The binary search is used to determine how many checks can be passed at a given point. If the binary search is not implemented correctly, it might lead to incorrect counts of passed checks.

3. **Dynamic Programming**: The DP table should correctly reflect the maximum number of checks that can be passed at each step. If the transitions are not handled correctly, it might lead to incorrect results.

### Crafting the Test Case:

To expose these issues, we can create a test case where the distribution of attribute points and the sequence of checks are such that the optimal strategy is not straightforward. This will test the code's ability to handle complex scenarios.

#### Test Case:
```plaintext
10 5
0 0 0 0 0 1 2 3 -1 -2
```

### Explanation:
- **Input**:
  - `n = 10` and `m = 5`.
  - The sequence of records is `0 0 0 0 0 1 2 3 -1 -2`.

- **Expected Output**:
  - The optimal strategy is to spend all 5 attribute points on Intelligence. This way, you can pass the Intelligence checks at positions 6, 7, and 8, but fail the Strength checks at positions 9 and 10.
  - Therefore, the maximum number of checks you can pass is `3`.

### Why This Test Case?
- **Attribute Points Distribution**: All attribute points are available at the beginning, allowing the algorithm to distribute them freely.
- **Checks Sequence**: The checks are placed in a way that requires careful distribution of attribute points. The Intelligence checks are higher in value, and the Strength checks are lower, but both need to be considered.
- **Complexity**: The test case is small enough to be manageable but complex enough to expose issues in the binary search and DP transitions.

This test case should help in identifying any issues in the provided code, particularly in how it handles the distribution of attribute points and the binary search for passing checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

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

using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin >> t;
	for (int tt = 0; tt < t; tt++) {
		int n;
		cin >> n;
		vector<int>a(n);
		for (int i = 0; i < n; i++) {
			cin >> a[i];
		}
		vector<int>used;
		int ans = 0;
		used.push_back(0);
		for (int i = 0; i < n; i++) {
			if (a[i] == 0) {
				used.push_back(i+1);
			}
			if (a[i] >= 5) {
				used.push_back(i+1);
				ans++;
			}
		}
		used.push_back(n + 1);
		int x = used.size();
		vector<int>u(n);
		for (int i = 1; i < used.size() - 1; i++) {
			u[used[i] - 1] = 1;
		}
		int i = 0;
		int g = 0;
		while (i < n) {
			int cop = 0;
			while (i < n && u[i] == 1) {
				i++;
			}
			if (i == n) {
				break;
			}
			int j = i+1;
			while (j<n && u[j] == 0 ) {
				j++;
			}
			ans += (j - i) / 2;
			if ((j - i) % 2 == 0) {
				for (int k = i; k < j; k++) {
					if (a[k] >= 3) {
						g++;
					}
					else {
						ans += (g + 1) / 2;
						g = 0;
					}
				}
				ans += (g + 1) / 2;
				g = 0;
			}
			else {
				ans++;
				for (int k = i; k < j; k++) {
					if (a[k] >= 3) {
						g++;
					}
					else {
						if (g % 2 && cop==0 && (k-i)%2==0) {
							ans--;
							cop = 1;
						}
						ans += (g + 1) / 2;
						g = 0;
					}
				}
				ans += (g + 1) / 2;
				g = 0;
				for (int k = i; k < j; k++) {
					if (a[k] >= 3) {
						ans--;
						break;
					}
				}
			}
			i = j;
		}
		cout << ans << "\n";
	}
}

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

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

Think step by step.

Response:
```plaintext
1
6
1 3 4 2 0 4
```

**Explanation:**
- This test case is designed to highlight issues with the buggy code, particularly in handling segments of the grid where the number of black cells is varied and not uniformly distributed.
- The input `6` specifies the size of the grid, and the array `1 3 4 2 0 4` represents the number of black cells in each row.
- This input will test the code's ability to handle segments where the number of black cells is small, large, and mixed, which can expose issues in the logic for counting and processing these segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#include<iostream>
using namespace std;
int a[200010];
int main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		for(int i=1;i<=n;i++) cin>>a[i];
		int cnt=0;
		for(int i=1;i<=n;i++){
			if(a[i]>=3) cnt++;
			else if(a[i]==0) ;
			else{
				if(i+1>n){
					cnt++;break;
				}
				if(a[i+1]==0) cnt++;
				else if(a[i+1]<=2) cnt++,i++;
				else if(a[i+1]>4) cnt++;
				else{
					int x=i+1;
					while(x<=n&&3<=a[x]&&a[x]<=4) x++;
					if(x>n){
						cnt+=x-i;
						break;
					}
					else if(a[x]==0) cnt+=x-i,i=x-1;
					else if(a[x]<=2) cnt+=x-i,i=x;
					else cnt+=x-i+1,i=x;
				}
			}
		}
		cout<<cnt<<'\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
5
3 2 3 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int MAXN=2e5+7;
int a[MAXN],vis[MAXN];
void solve()
{
	int n,ans=0;
	scanf("%d",&n);
	for(int i(1);i<=n;++i)
	{
		scanf("%d",&a[i]);
		vis[i]=0;
	}
	for(int i(1);i<=n;++i)
	{
		if((a[i]<=4&&vis[i]==3)||(a[i]<=2&&vis[i])||!a[i])
			continue;
		if(a[i]<=2)
			vis[i]=vis[i+1]+=1;
		else
			if(a[i]<=4)
			{
				if(vis[i]==1)
				{
					vis[i]+=2;
					vis[i+1]+=2;
				}
				else
					if(vis[i]==2)
					{
						vis[i]+=1;
						vis[i+1]+=1;
					}
			}
		ans++;
	}
	printf("%d\n",ans);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
		solve();
}

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

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

Think step by step.

Response:
```plaintext
1
5
3 3 3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#!/usr/bin/python3

import bisect
import collections
import heapq
import itertools
import math
import os
import sys

def main():
    for _ in range(read_int()):
        n = read_int()
        A = read_ints()
        print(solve(n, A))

def solve(n, A):
    for i in range(n):
        if A[i] % 2 == 1:
            A[i] += 1

    skip = [None] * n
    i = 0
    while i < n:
        if A[i] > 2:
            i += 1
            continue

        j = i + 1
        while j < n:
            if n == 3 or A[j] != 4:
                break
            j += 1

        if j < n and A[j] <= 2:
            skip[i] = j + 1
        i = j

    dp = [math.inf] * (n + 1)
    dp[0] = 0

    for i in range(n):
        if A[i] == 0:
            dp[i + 1] = min(dp[i + 1], dp[i])
        else:
            dp[i + 1] = min(dp[i + 1], dp[i] + 1)

        if skip[i] is not None:
            dp[skip[i]] = min(dp[skip[i]], dp[i] + skip[i] - i - 1)

    return dp[n]

###############################################################################

DEBUG = 'DEBUG' in os.environ

def inp():
    return sys.stdin.readline().rstrip()

def read_int():
    return int(inp())

def read_ints():
    return [int(e) for e in inp().split()]

def list2d(d1, d2, init=None):
    return [[init] * d2 for _ in range(d1)]

def list3d(d1, d2, d3, init=None):
    return [[[init] * d3 for _ in range(d2)] for _ in range(d1)]

def list4d(d1, d2, d3, d4, init=None):
    return [[[[init] * d4 for _ in range(d3)] for _ in range(d2)]
            for _ in range(d1)]

def debug(fmt, *args):
    if DEBUG:
        if args:
            print(fmt.format(*args), file=sys.stderr)
        else:
            print(fmt, file=sys.stderr)

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
1
5
2 2 2 2 2
```

### Explanation:
This test case is designed to challenge the buggy code by focusing on a scenario where the array `A` consists of multiple elements, each of which is 2. The key points to consider are:

1. **Uniformity**: The array is uniformly filled with 2s, which means each row has exactly 2 black cells.
2. **2x2 Subgrid Dyeing**: Since each row has exactly 2 black cells, it is possible to use 2x2 subgrid dyeing operations to turn these cells white. However, the buggy code does not handle this scenario correctly.
3. **Optimal Solution**: The optimal solution here is to use 2x2 subgrid dyeing operations to turn the black cells white. Since there are 5 rows, and each row can be handled with a 2x2 subgrid dyeing operation, the minimum number of operations required is 5.

The buggy code incorrectly modifies the array `A` by adding 1 to any odd elements, which is unnecessary in this case since all elements are already even. This can lead to incorrect results or unnecessary operations.

By using this test case, you can expose the bug in the buggy code and ensure that it does not handle the uniform 2s scenario correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#Main obs: freq of MAD in current ops = longest it can be
from collections import defaultdict
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    ans = 0
    for i in range(n):
        if a[i]>0: ans+=1

    # need to see if placing 2*2 subgrids can reduce it: to place a subgrid at ith row iff
    # i and i-1 have n-2 a[i]==n-2
    i = 0
    while i<n:
        if a[i]>4:
            i+=1
            continue
        if i+1<n and a[i]>0 and a[i+1]>0:
            if a[i]<=2 and a[i+1]<=2:
                ans-=1
                i+=2
                continue
        if i+2<n and a[i]>0 and a[i+1]>0 and a[i+2]>0:
            if a[i]<=2 and a[i+1]<=4 and a[i+2]<=2:
                ans-=1
                i+=3
                continue
        if i+3<n and a[i]>0 and a[i+1]>0 and a[i+2]>0:
            if a[i]<=2 and a[i+1]<=4 and a[i+2]<=4 and a[i+3]<=2:
                ans-=1
                i+=4
                continue
        i+=1
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
5
2 2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
import sys

input = lambda: sys.stdin.readline().strip()

T = int(input())
for _ in range(T):
    n = int(input())
    a = list(map(int, input().split()))
    dp = [[0, 0] for i in range(n + 1)]
    dp[1][0] = min(1, a[0])
    dp[1][1] = min(1, a[0])
    for i in range(2, n + 1):
        if a[i - 1] == 0:
            dp[i][0] = dp[i - 1][0]
            dp[i][1] = dp[i - 1][0]
        elif a[i - 1] <= 2:
            if a[i - 2] <= 2:
                dp[i][0] = dp[i - 2][0] + 1
            else:
                dp[i][0] = min(dp[i - 2][1] + (a[i - 2] + 1) // 2, dp[i - 2][0] + 2)
            dp[i][1] = dp[i][0]
        else:
            dp[i][0] = dp[i - 1][0] + 1
            if a[i - 2] <= 2:
                dp[i][1] = dp[i - 2][0] + 1
            else:
                dp[i][1] = min(dp[i - 2][1] + (a[i - 2] + 1) // 2, dp[i - 2][0] + 2)
    if _ == 1115:
        print(n)
    else:
        print(dp[-1][0])

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code seems to have issues with handling certain patterns in the array `a`, especially when dealing with rows that have a small number of black cells (0, 1, or 2) and transitions between such rows.

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

1. **Small Test Case with Edge Conditions**: We will create a test case with a small number of rows to ensure it fits within the output length limit.
2. **Mix of Small and Large Values**: Include rows with 0, 1, 2, and larger values to test the transitions and edge cases.
3. **Pattern to Expose Bug**: Create a pattern that forces the code to handle multiple transitions and operations, which might expose the bug.

Let's construct the test input:

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

### Explanation:
- **Test Case 1**: We have 5 rows.
- **Row 1 (2 black cells)**: The first 2 cells are black.
- **Row 2 (0 black cells)**: All cells are white.
- **Row 3 (3 black cells)**: The first 3 cells are black.
- **Row 4 (1 black cell)**: The first cell is black.
- **Row 5 (4 black cells)**: The first 4 cells are black.

This test case includes:
- Rows with 0 black cells.
- Rows with 1, 2, and 3 black cells.
- A transition from a row with 2 black cells to a row with 0 black cells.
- A transition from a row with 3 black cells to a row with 1 black cell.
- A final row with 4 black cells.

This mix of values and transitions should challenge the buggy code and potentially expose its issues with handling small and large values, as well as transitions between rows.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
import java.util.Scanner;

public class GridPuzzle {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0){
            int n = in.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) arr[i] = in.nextInt();
            int ans = 0;
            for (int i = 0; i < n; i++){
                if (arr[i] <= 0) continue;
                if (arr[i] > 2) {
                    ans++;
                    continue;
                }
                ans++;
                if (i+1 < n){
                    if (arr[i+1] <= 2) arr[i+1] -= 2;
                    else if (arr[i+1] <= 4){
                        ans++;
                        arr[i+1] -= 4;
                        if (i+2 < n) arr[i+2] -= 2;
                    }
                }
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

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

**Explanation:**

This test case is designed to expose the bug in the given buggy code. The input represents a 4x4 grid where each row has the first 2 cells black and the rest white. The expected output should be 2, as you can dye all cells white by performing the following operations:

1. Dye the 2x2 subgrid in the top-left corner (cells (1,1), (1,2), (2,1), (2,2)) white.
2. Dye the 2x2 subgrid in the bottom-left corner (cells (3,1), (3,2), (4,1), (4,2)) white.

However, the buggy code might not handle this case correctly due to the way it processes the array and the operations. Specifically, the code might not properly account for the fact that each 2x2 subgrid operation can affect multiple rows, and it might incorrectly decrement the values in the array, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Codeforces2 {

  public static PrintWriter out;

  public static void main(String... args) {
    final FastScanner fastScanner = new FastScanner();
    out = new PrintWriter(new BufferedOutputStream(System.out));

    int tests = fastScanner.nextInt();
    // int tests = 1;
    for (int i = 0; i < tests; i++) {
      int n = fastScanner.nextInt();
      int[] a = new int[n];
      for (int j = 0; j < n; j++) {
        a[j] = fastScanner.nextInt();
      }
      long res = solve(n, a);
//      for (int j = 0; j < n; j++) {
//        out.print(res[j]);
//        if (j < n - 1) {
//          out.print(" ");
//        }
//      }
      //out.println();
      out.println(res);

    }

    out.close();
  }

  private static long solve(int n, int[] arr) {
    int[][] dp = new int[n][2];
    for (int i = 0; i < n; i++) {
      dp[i][0] = Integer.MAX_VALUE;
      dp[i][1] = Integer.MAX_VALUE;
    }
    boolean[] can2 = new boolean[n];
    boolean[] can2if = new boolean[n];

    for (int r = 0; r < n; r++) {
      can2[r] = canMake2(arr, r, false);
      can2if[r] = canMake2(arr, r, true);
    }
    return dp(dp, can2, can2if, 0, 0, n, arr);
  }

  private static int dp(int[][] dp, boolean[] can2, boolean[] can2if, int r, int mode, int n, int[] arr) {
    if (r >= n) {
      return 0;
    }
    if (dp[r][mode] != Integer.MAX_VALUE) {
      return dp[r][mode];
    }
    if (arr[r] <= 2 && mode == 1 || arr[r] == 0) {
      return dp[r][mode] = dp(dp, can2, can2if, r + 1, 0, n, arr);
    }
    int ifLine = 1 + dp(dp, can2, can2if, r + 1, 0, n, arr);
    int if2 = Integer.MAX_VALUE;
    if (can2[r] || mode == 1 && can2if[r]) {
      if2 = 1 + dp(dp, can2, can2if, r + 1, 1, n, arr);
    }
    return dp[r][mode] = Math.min(ifLine, if2);
  }

  private static boolean canMake2(int[] arr, int r, boolean resetStart) {
    if (arr[r] > 4) {
      return false;
    }
    if (arr[r] <= 4 && resetStart || arr[r] <= 2) {
      return true;
    }
    return false;
  }

  private static int[] arrayFromString(final String input) {
    return Arrays.stream(input.split(" "))
        .mapToInt(Integer::parseInt)
        .toArray();
  }

  static class FastScanner {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
      while (!st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }
  }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The key points to consider are:

1. **Edge Cases for `canMake2` Function:**
   - The `canMake2` function incorrectly handles cases where the number of black cells in a row is 3 or 4. It should correctly determine if a 2x2 subgrid can be used to make the row white.

2. **Edge Cases for `dp` Function:**
   - The dynamic programming approach should correctly handle cases where the number of black cells in a row is exactly 2, and the mode is 1 (indicating that the previous row was partially processed).

3. **Test Cases with Specific Patterns:**
   - Test cases where the number of black cells in consecutive rows is 3 or 4, as these are the critical values that the `canMake2` function incorrectly handles.
   - Test cases where the number of black cells in a row is exactly 2, and the mode is 1, to ensure the `dp` function correctly handles these cases.

Given these considerations, let's construct a small-scale test input that is likely to expose the bug:

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

### Explanation:

1. **Test Case 1:**
   - `n = 3`
   - `a = [3, 3, 3]`
   - This test case is designed to expose the issue in the `canMake2` function. The function should correctly determine that a 2x2 subgrid can be used to make the row white, but the current implementation incorrectly handles this case.

2. **Test Case 2:**
   - `n = 3`
   - `a = [2, 2, 2]`
   - This test case is designed to ensure that the `dp` function correctly handles rows with exactly 2 black cells, especially when the mode is 1.

3. **Test Case 3:**
   - `n = 3`
   - `a = [4, 4, 4]`
   - This test case is designed to expose the issue in the `canMake2` function. The function should correctly determine that a 2x2 subgrid can be used to make the row white, but the current implementation incorrectly handles this case.

These test cases are small and complete, and they are designed to maximize the failure probability in the faulty solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
using ll = long long;
using db = double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
mt19937 rnd(time(0));

class JudTime {
public:
    chrono::time_point<chrono::high_resolution_clock> start, end;
    JudTime() : start(chrono::high_resolution_clock::now()) {}
    ~JudTime() {
        end = chrono::high_resolution_clock::now();
        cerr << fixed << setprecision(6) << chrono::duration<double>(end - start).count()
             << " seconds" << endl;
    }
};
int a[100005];
void solve() {
    int n;
    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] > 4) {
            a[i] = 0;
            ans++;
        }
    }
    for (int i = 1; i < n; i++) {
        if (a[i] && a[i - 1] && a[i] <= 2 && a[i - 1] <= 2) {
            a[i] = 0, a[i - 1] = 0;
            ans++;
        }
    }
    for (int i = 1; i < n; i++) {
        if (a[i] && a[i - 1]) {
            if (a[i] <= 2) {
                a[i] = 0;
                a[i - 1] -= 2;
                ans++;
            } else if (a[i - 1] <= 2) {
                a[i - 1] = 0;
                a[i] -= 2;
                ans++;
            }
        }
    }
    for (int i = 1; i < n; i++) {
        if (a[i] && a[i - 1] && a[i] <= 2 && a[i - 1] <= 2) {
            a[i] = 0, a[i - 1] = 0;
            ans++;
        }
    }
    for (int i = 0; i < n; i++) {
        if (a[i]) {
            ans++;
        }
    }
    cout << ans << endl;
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    //   JudTime judTime;
    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:
```plaintext
1
4
3 1 2 4
```

**Explanation:**

This test case is designed to expose issues in the buggy code by having a mix of small and large values in the array `a`. Let's break down why this input is challenging:

1. **Row with a large value (4):** The fourth row has a value of 4, which means the entire row is black. According to the buggy code, this row will be immediately set to 0 and the answer incremented by 1. This is correct, but it's important to ensure the code handles this correctly.

2. **Rows with small values (1, 2, 3):** The first three rows have values 3, 1, and 2 respectively. These values are small enough that they need to be handled by the 2x2 subgrid operations. The buggy code should be able to handle these, but it's important to ensure it does so efficiently.

3. **Adjacent rows with small values:** Rows 1 and 2 (with values 3 and 1) are adjacent, and rows 2 and 3 (with values 1 and 2) are also adjacent. The code should be able to handle these pairs of rows correctly by using 2x2 subgrid operations.

4. **Edge case for the 2x2 subgrid operation:** The combination of rows with values 3, 1, and 2 should be handled correctly. For example, the 2x2 subgrid operation can be used to clear the first two rows, and then the third row can be handled separately.

This test case should help identify if the buggy code is correctly handling the 2x2 subgrid operations and the row operations, especially when the values are mixed and require careful handling.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#include <bits/stdc++.h>
#define print(arr)         \
    for (auto a : arr)     \
    {                      \
        cout << a << "  "; \
    }
#define loop(i, a, n) for (int i = a; i < n; i++)
#define vi vector<int>
#define vll vector<int>
#define ll long long
using namespace std;

void solve()
{
    int n;
    cin >> n;
    int y;
    int ans = 0;
    vector<int> nums(n);
    vector<vector<bool>> grid(n, vector<bool>(n, 1));

    for (int i = 0; i < n; i++)
    {
        if (nums[i] > 5)
        {
            ans++;
            nums[i] = 0;
        }
        else
        {
            cin >> nums[i];
        }
    }

    if (n > 1)
    {
        if (nums[0] > 2 and nums[0] % 2 != 0 and nums[0] > nums[1])
        {
            nums[0] = 0;
            ans++;
        }

        if (nums[n-1] > 2 and nums[n-1] % 2 != 0 and nums[n - 1] > nums[n - 2])
        {
            nums[n - 1] = 0;
            ans++;
        }
    }
    for (int i = 1; i < n - 1; i++)
    {
        if (nums[i] > 2 and nums[i] % 2 != 0  and nums[i] > nums[i - 1] and nums[i] > nums[i + 1])
        {
            nums[i] = 0;
            ans++;
        }
        for (int j = 0; j < nums[i]; j++)
        {
            grid[i][j] = 0;
        }
    }

    for (int j = 0; j < nums[0]; j++)
    {
        grid[0][j] = 0;
    }

    for (int j = 0; j < nums[n - 1]; j++)
    {
        grid[n - 1][j] = 0;
    }

    vector<int> dx = {0, 1, 0, 1};
    vector<int> dy = {0, 0, 1, 1};

    for (int i = 0; i < n; i++)
    {
        if (nums[i] == 0)
        {
            continue;
        }
        for (int j = 0; j < nums[i]; j++)
        {
            if (grid[i][j] == 0)
            {
                ans++;
                for (int a = 0; a < dx.size(); a++)
                {
                    int x = dx[a];
                    int y = dy[a];
                    if (i + x < n and j + y < n)
                    {
                        grid[i + x][j + y] = 1;
                    }
                }
                break;
            }
        }
    }

    cout << min(n, ans) << "\n";
}

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

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

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

Think step by step.

Response:
```plaintext
1
3
0 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
import sys
from collections import deque, defaultdict, Counter
t = int(sys.stdin.readline().strip())
for _ in range(t) :
    n = int(sys.stdin.readline().strip())
    a = [*map(int, sys.stdin.readline().strip().split())]
    x = -2
    ans = 0
    for i in range(n) :
        if a[i] == 0 : continue
        if a[i] >= 5 : ans += 1; continue
        if a[i] == 1:
            if x == 1 : x = -2; continue
            else : x = 1; ans += 1; continue
        if a[i] == 2 :
            if x == 1 : x = -2; continue
            else : x = 1; ans += 1; continue
        if a[i] == 3 :
            if x == 1 : x = 3; ans += 1; continue
            elif x == 2 or x == 3: x = 1; ans += 1; continue
            elif x == 3 : x = 1; ans += 1; continue
            else : x = -2; ans += 1; continue
        if a[i] == 4 :
            if x == 1 : x = 3; ans += 1; continue
            elif x == 2 : x = -2; ans += 1; continue
            elif x == 3 : x = 1; ans += 1; continue
            else : x = -2; ans += 1; continue
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
from collections import defaultdict
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))

    # need to see if placing 2*2 subgrids can reduce it: to place a subgrid at ith row iff
    # i-1th row has <=2 pieces 2<=a[i-1]<=4 pieces or 0 pieces
    i = 0
    c = 0
    ans = 0
    while i<n:
        if a[i]>4:
            c = 0
        elif a[i]==0:
            c = 0
            ans+=1
        elif a[i]<=2:
            if c and (i-prev)&1:
                c = 0
                ans+=1
            else:
                c = 1
                prev = i
        else:
            c+=1
        i+=1
    print(n-ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
import java.util.Scanner;

public class grid {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		for (int t = sc.nextInt(); t>0; t--) {
			int n = sc.nextInt();
			int[][] a = new int[n][3];
			for (int i = 0; i<n; i++) {
				a[i][0] = 0;
				a[i][1] = sc.nextInt();
				a[i][2] = a[i][1] - a[i][0];
			}
			int operations = n;
			for (int i = 0; i<n; i++) {
				if (a[i][2] <= 0) {
					operations--;
				} else if (a[i][1] - a[i][0] <= 2 && a[i][2] <= 2) {
					if (i < n - 1) {
						if (a[i][0] == 0) {
							a[i + 1][0] += 2;
						} else {
							a[i + 1][1] -= 2;
						}

						a[i + 1][2] -= 2;
					}
				}
			}

			System.out.println(operations);
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
1
5
2 1 0 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

Examples:
input:
10
1
0
4
2 4 4 2
4
3 2 1 0
3
0 3 0
3
0 1 3
3
3 1 0
4
3 1 0 3
4
0 2 2 2
6
1 3 4 2 0 4
8
2 2 5 2 3 4 2 4
output:
0
3
2
1
2
2
3
2
4
6

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

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

public class D_Grid_Puzzle {

    private static class FastReader {
        private BufferedReader br;
        private StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer("");
        }

        private void updateTokenizer() {
            while (!st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }

        public String next() {
            updateTokenizer();
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
                return "";
            }
        }

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

        public int[] readArray1Based(int n) {
            int[] res = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                res[i] = nextInt();
            }
            return res;
        }

        public int[][] readGraph(int n, int m, int c) {
            int[][] graph = new int[n + 1][];
            int[] count = new int[n + 1];
            int[][] arr = new int[m][1 + c];

            for (int i = 0; i < m; i++) {
                int a = nextInt();
                int b = nextInt();
                arr[i][0] = a;
                arr[i][1] = b;
                count[a]++;
                count[b]++;
                for (int j = 2; j <= c; j++) {
                    arr[i][j] = nextInt();
                }
            }

            for (int i = 0; i <= n; i++) {
                graph[i] = new int[count[i] * c];
                count[i] = 0;
            }

            for (int[] tArr : arr) {
                for (int j = 1; j < tArr.length; j++) {
                    graph[tArr[0]][count[tArr[0]]++] = tArr[j];
                }

                int swap = tArr[0];
                tArr[0] = tArr[1];
                tArr[1] = swap;

                for (int j = 1; j < tArr.length; j++) {
                    graph[tArr[0]][count[tArr[0]]++] = tArr[j];
                }
            }

            return graph;
        }
    }

    public static class FastWriter {
        private PrintWriter writer;

        public FastWriter() {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        }

        public void print(Object o) {
            writer.print(o);
        }

        public void println(Object o) {
            writer.println(o);
        }

        public void println() {
            writer.println();
        }

        public void flush() {
            writer.flush();
        }

        public void close() {
            writer.close();
        }
    }

    public static FastReader in = new FastReader();
    public static FastWriter out = new FastWriter();

    public static void main(String[] args) throws IOException {
        int T = in.nextInt();
        // int T = 1;
        while (T-- > 0) {
            solve();
        }

        out.flush();

    }
    public static HashMap<Integer,Integer> map ;
    public static void solve() {
        int n = in.nextInt();
        int[] arr = new int[n];
        for (int i = 0 ;i<n;i++)arr[i]=in.nextInt();
        long[] dp = new long[n];
        Arrays.fill(dp,-1);
        map = new HashMap<>();

        for (int i = 0 ;i<arr.length;i+=2){
            if (arr[i]>0 && arr[i]<=2){
                int j = i+1;
                while (j<arr.length && arr[j]>2 && arr[j]<=4){
                    if (arr[j-1]>4)break;
                    j+=2;
                }
                if (j<arr.length && (arr[j]<=2 )&&arr[j-1]<=4)map.put(i,j-i);
                i=j-1;
            }
        }
        for (int i = 1 ;i<arr.length;i+=2){
            if (arr[i]>0 && arr[i]<=2){
                int j = i+1;
                while (j<arr.length && arr[j]>2 && arr[j]<=4){
                    if (arr[j-1]>4)break;
                    j+=2;
                }
                if (j<arr.length && (arr[j]<=2)&&arr[j-1]<=4)map.put(i,j-i);
                i=j-1;
            }
        }
        System.out.println(memo (0,dp,arr));
    }
    public static long memo (int i   ,long[] dp , int[] arr){
        if (i==arr.length)return 0;
        if (dp[i]!=-1){
            return dp[i];
        }
        dp[i]=(arr[i]==0?0:1)+memo(i+1,dp,arr);
        // if (i<arr.length-1){
        //     if (arr[i]<=2 && arr[i+1]<=2)dp[i]=Math.min(dp[i],1+memo(i+2,dp,arr));
        // }
        // if (i<arr.length-3){
        //     if (arr[i]<=2 && arr[i+1]<=4 && arr[i+2]<=4 && arr[i+3]<=2)dp[i]=Math.min(dp[i],3+memo(i+4,dp,arr));
        // }
        if (map.containsKey(i))dp[i]=Math.min(dp[i],map.get(i)+memo(i+map.get(i)+1,dp,arr));
        return dp[i];
    }
}

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

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

Think step by step.

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