**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define neg cout << -1 << endl;
#define pb push_back
#define ss second
#define ff first

const int md=998244353;
const int N = 2e5+5;
const int INF=1e18;

void solve()
{
    int n;
    cin>>n;
    vector<int>v(n+1, 0), suffix(n+1, 0);
    for(int i=0; i<n; i++) cin>>v[i];
    suffix[n-1]=v[n-1];
    for(int i=n-2; i>=0; i--) suffix[i]=suffix[i+1]+v[i];
    int ans=abs(suffix[0]), total=0;
    int cnt=0, it=1;
    for(int i=0; i<n; i++)
    {
        total+=v[i];
        int mx=abs(abs(total)+suffix[i+1]);
        ans=max(ans, mx);
    }
    int nig=0;
    total=0;
    for(int i=0; i<n; i++)
    {
        total+=v[i];
        int mx=abs(abs(total)+suffix[i+1]);
        //cout<<"mx "<<mx<<endl;
        if(mx==ans and v[i]<0)
        {
            //cout<<"vi "<<v[i]<<" ";
            nig++;
        }
        else if(v[i]>=0)
        {
            it*=2;
            it%=md;
        }
    }

    if(nig>0)
    {
        nig=(nig*(nig+1))/2;
        nig%=md;
        it*=nig;
        it%=md;
    }

    cout<<it<<endl;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int test = 1;
    cin >> test;

    for (int tc = 1; tc <= test; tc++)
    {
        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
5
4
-1 -2 -3 -4
8
1 -1 1 -1 1 -1 1 -1
3
-1 -2 -3
4
1 2 3 4
4
-1 1 -1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
import sys

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

def get_number():
    return int(input())

def get_numbers():
    return list(map(int, input().split()))

def get_word():
    return input()

def solve():
    n = get_number()
    a = get_numbers()

    positives = [0] * n
    running_sum = [0] * n
    for i in range(n):
        running_sum[i] = a[i] + running_sum[i - 1] if i > 0 else a[i]
        positives[i] = positives[i - 1] + (a[i] >= 0) if i > 0 else a[i] >= 0

    max_negative = 0
    idxs = []
    for i, s in enumerate(running_sum):
        if s < max_negative:
            max_negative = s
            idxs = [i]
        elif s == max_negative:
            idxs.append(i)

    mod = 998_244_353
    if not idxs:
        print(pow(2, n, mod))
        return

    total = 0

    for idx in idxs:
        s = 1
        s *= pow(2, positives[idx], mod)
        s *= pow(2, n - idx - 1, mod)
        total += s

    print(total % mod)

def main():
    for _ in range(get_number()):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following aspects:

1. **Edge Cases**: The code might fail when dealing with arrays that have a mix of positive and negative numbers, especially when the running sum reaches a minimum at multiple indices.
2. **Special Conditions**: The code might not handle the case where the maximum negative sum is zero correctly, or when the array has a sequence of negative numbers followed by positive numbers.
3. **Boundary Conditions**: The code might fail when the array has a single element or when all elements are either positive or negative.

Given these considerations, let's construct a test case that includes a mix of positive and negative numbers and ensures that the running sum reaches a minimum at multiple indices. This will help expose any issues in the logic that calculates the maximum final value of `c`.

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

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

Explanation:
- The array `[1, -2, 3, -4, 5, -6, 7]` has a mix of positive and negative numbers.
- The running sum reaches a minimum value at multiple indices (e.g., indices 1, 3, and 5), which can help expose any issues in the logic that handles these cases.
- The array is small enough to be easily validated but complex enough to challenge the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
import sys
import math
import copy
from collections import deque
from collections import defaultdict as ddic
#from atcoder import fenwicktree

class UndirectedGraph:
    def __init__(self, vertices):
        self.vertices = vertices
        self.adj = {i: [] for i in range(self.vertices)}

    def addEdge(self, u, v):
        self.adj[u].append(v)
        self.adj[v].append(u)

    def DFS(self, start):
        seen = set()
        stack = [start]
        while len(stack) > 0:
            curr = stack.pop()
            seen.add(curr)
            for elem in self.adj[curr]:
                if elem not in seen:
                    stack.append(elem)
        return seen

    def BFS(self, u):
        visited = [False for i in range(self.vertices + 1)]
        distance = [-1 for i in range(self.vertices + 1)]

        distance[u] = 0
        queue = deque()
        queue.append(u)
        visited[u] = True

        while queue:
            front = queue.popleft()

            for i in self.adj[front]:
                if not visited[i]:
                    visited[i] = True
                    distance[i] = distance[front]+1
                    queue.append(i)
        maxDis = 0

        for i in range(self.vertices):
            if distance[i] > maxDis:
                maxDis = distance[i]
                nodeIdx = i
        return distance

    def LongestPathLength(self):
        node, Dis = self.BFS(0)
        node_2, LongDis = self.BFS(node)

        print('Longest path is from', node, 'to', node_2, 'of length', LongDis)

class Fenwick:
    def __init__(self, arr):
        self.length = len(arr)
        self.bit = [0]*(self.length+1)
        for i in range(self.length):
            self.updateBit(i, arr[i])

    def getSum(self, i):
        s = 0
        i = i+1
        while i > 0:
            s += self.bit[i]
            i -= i & (-i)
        return s

    def getRangeSum(self, left, right):
        return self.getSum(right) - self.getSum(left-1)

    def updateBit(self, i, v):
        i += 1
        while i <= self.length:
            self.bit[i] += v
            i += i & (-i)

class DSU:
    def __init__(self, n):
        self.parent = [None for _ in range(n)]
        self.size = [0 for _ in range(n)]

    def findSet(self, v):
        if v == self.parent[v]:
            return v
        p = self.findSet(self.parent[v])
        self.parent[v] = p
        return p

    def makeSet(self, v):
        self.parent[v] = v
        self.size[v] = 1

    def unionSets(self, a, b):
        a = self.findSet(a)
        b = self.findSet(b)
        if (a != b):
            if self.size[a] < self.size[b]:
                a, b = b, a
            self.parent[b] = a
            self.size[a] += self.size[b]

def factors(n):
    result = set()
    for i in range(1, math.floor(math.sqrt(n))+1):
        if n%i == 0:
            result.add(i)
            result.add(n//i)
    return result

def main():
    fimp = sys.stdin.readline
    MOD = 10**9+7
    iprint = lambda x: '\n'.join(map(str, x))
    sprint = lambda x: '\n'.join(x)
    initl = lambda x,y: [y]*x
    INF = float('inf')
    NINF = float('-inf')
    ilist = lambda: list(map(int, fimp().split()))
    slist = lambda: list(input())
    slists = lambda: input().split()
    ivars = lambda: map(int, fimp().split())
    ipt = lambda: int(fimp())
    out = []

    mod = 998244353
    t = ipt()
    for _ in range(t):
        n = ipt()
        a = ilist()
        c = 0
        rs = [0 for _ in range(n)]
        rs[0] = a[0]
        for i in range(1, n):
            rs[i] = rs[i-1]+a[i]
        l = 0
        count = 0
        currCount = 0
        add = False
        numNegatives = 0
        for i in range(n):
            if a[i] < 0:
                currCount+= 1
            if rs[i]-l < 0 and c+rs[i]-l < 0:
                numNegatives+= 1
                count+= currCount
                currCount = 0
                c+= rs[i]-l
                l = rs[i]
            elif i == n-1:
                if rs[i]-l > 0 and c < 0:
                    c = abs(c)
                c+= rs[i]-l
                if rs[i]-l < 0:
                    count+= currCount
                elif rs[i]-l == 0:
                    count+= currCount
                    add = True

        ans = pow(2, n-count, mod)
        if add:
            if numNegatives > 0:
                currCount+= 1
            addition = pow(2, currCount, mod)-1
            ans = (ans*addition)%mod

        print(ans)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
MOD = 998244353
tot = int(input())
for _ in range(tot):
    n = int(input())
    nums = list(map(int, input().split(' ')))
    mx = 0
    mn = 0
    s = 0
    for x in nums:
        mx = abs(mx + x)
        s += x
        mn = min(mn, s)
    if mn >= 0:
        print(1 << n)
        continue
    ans = 0
    s = 0
    less_than_zero = 0
    for i, x in enumerate(nums):
        s += x
        if s == mn:
            ans = (ans + pow(2, n-1-less_than_zero)) % MOD
        if s < 0:
            less_than_zero += 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
4
1 -1 1 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
import java.lang.*; // imports everything
import java.util.*;

public class C_2_Magnitude_Hard_Version
{
    //static ArrayList<Integer> a=new ArrayList<>();
    //static int max;
    static HashMap<Long,Long> map=new HashMap<>();
    static long a[];
    static long k=998244353;
    static long max=Long.MIN_VALUE;
    static int n;
    public static void main (String[] args) throws java.lang.Exception
    {
        Scanner sc=new Scanner(System.in);
        int caseNo,i,j,N;
        N=sc.nextInt();
        for(caseNo=0;caseNo<N;caseNo++)
        {
            n=sc.nextInt();
            a=new long[n];
            for(i=0;i<n;i++)
            {
                a[i]=sc.nextLong();
            }
            calculate(0,0);
            System.out.println(map.get(max));
            map.clear();
            max=Long.MIN_VALUE;
        }
    }

    public static void calculate(int i,long sum)
    {
        if(i==n)
        {
            if(sum<max)
            {
                return ;
            }
            map.put(sum,(map.getOrDefault(sum,(long)0)+1)%k);
            max=Math.max(max,sum);
            return ;
        }
        calculate(i+1,sum+a[i]);
        calculate(i+1,Math.abs(sum+a[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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on scenarios that maximize the failure probability. Here are the steps to consider:

1. **Edge Cases**: Consider arrays with extreme values, such as very large or very small numbers, or arrays with a mix of positive and negative values.
2. **Boundary Conditions**: Test with arrays of length 2 and the maximum allowed length (200,000).
3. **Special Cases**: Test with arrays where the absolute value operation has a significant impact on the final result.

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

- **Array with a mix of positive and negative values**: This will test the logic of the absolute value operation.
- **Array with repeated values**: This will test the handling of multiple identical values.
- **Array with a large sum**: This will test the handling of large numbers and the modulo operation.

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

```plaintext
3
4
2 -5 3 -3
4
1 9 8 4
6
1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000
```

### Explanation:
1. **First Test Case**:
   - Array: `2, -5, 3, -3`
   - This case has a mix of positive and negative values, and the maximum final value of `c` is 3. There are 12 unique procedures to achieve this.

2. **Second Test Case**:
   - Array: `1, 9, 8, 4`
   - This case has positive values, and the maximum final value of `c` is 22. There are 16 unique procedures to achieve this.

3. **Third Test Case**:
   - Array: `1000000000, -1000000000, 1000000000, -1000000000, 1000000000, -1000000000`
   - This case has very large numbers and a mix of positive and negative values. The maximum final value of `c` is 3000000000. There are 64 unique procedures to achieve this.

This test input is designed to expose potential bugs in the code by testing a variety of scenarios, including edge cases and large numbers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
import java.util.Scanner;

public class C {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int tes = scanner.nextInt();

        for(int t = 0; t < tes; t++){
            int n = scanner.nextInt();
            long[] a = new long[n];
            long[] b = new long[n];

            b[0] = scanner.nextInt();
            a[0] = Math.max(b[0], -b[0]);

            long[] ac = new long[n];
            long[] bc = new long[n];

            if(b[0] >= 0){
                ac[0] = 2;
                bc[0] = 2;
            }else{
                ac[0] = 1;
                bc[0] = 1;
            }

            for(int i = 1; i < n; i++){
                long x = scanner.nextLong();
                long cnt = 0;

                a[i] = Math.max(a[i - 1] +  x, Math.max(Math.abs(x + a[i - 1]), Math.abs(x + b[i - 1])));
                b[i] = b[i - 1] + x;

                if(a[i - 1] != b[i - 1]){
                    if(a[i] == a[i - 1] +  x)cnt++;
                    if(a[i] == Math.abs(a[i - 1] +  x))cnt++;
                    if(a[i] == Math.abs(b[i - 1] + x)){
                        ac[i] = ac[i - 1] * cnt + bc[i - 1];
                    }else{
                        ac[i] = ac[i - 1] * cnt;
                    }
                }else{
                    if(a[i] == a[i - 1] +  x)cnt++;
                    if(a[i] == Math.abs(a[i - 1] +  x))cnt++;
                    ac[i] = ac[i - 1] * cnt;
                }

                if(x >= 0)bc[i] = bc[i - 1] * 2;
                else bc[i] = bc[i - 1];

            }

            System.out.println(ac[n - 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:
```plaintext
1
3
0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 2e5;
const ll mod = 998244353;
const ll inf = 1e18;
ll v[N];
pair<ll,ll> nr[4];
void solve(){
    ll n,mx = 0,mn = 0;
    ll m1 = 1,m2 = 1;
    cin>>n;
    for(ll i = 0;i < n;i++){
        cin>>v[i];
        int cnt;
        if(mx == mn){
            nr[0] = {mx + v[i],m1};
            nr[1] = {abs(mx + v[i]),m2};
            cnt = 2;
        }else{
            nr[0] = {mx + v[i],m1};
            nr[1] = {abs(mx + v[i]),m2};
            nr[2] = {mn + v[i],m1};
            nr[3] = {abs(mn + v[i]),m2};
            cnt = 4;
        }
        ll mx2 = -inf,mn2 = inf,m11,m22;
        for(int j = 0;j < cnt;j++){
            if(mx2 < nr[j].first){
                mx2 = nr[j].first;
                m11 = nr[j].second;
            }else if(mx2 == nr[j].first){
                m11+=nr[j].second;
                m11%=mod;
            }
            if(mn2 > nr[j].first){
                mn2 = nr[j].first;
                m22 = nr[j].second;
            }else if(mn2 == nr[j].first){
                m22+=nr[j].second;
                m22%=mod;
            }
        }
        mn = mn2;
        mx = mx2;
        m1 = m11;
        m2 = m22;
    }
    cout<<m1<<'\n';
}
int main(){
    ll 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
5
-1 2 -3 4 -5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
long long mod = 998244353;
signed main(){
    ios_base::sync_with_stdio(false);cin.tie(0);
    int t;
    cin>>t;
    long long po[200001];
    po[0] = 1;
    for(int i = 1;i<=2e5;i++){
        po[i] = (po[i-1]*2)%mod;
    }
    while(t--){
        int n;
        cin>>n;
        int arr[n];
        long long suf[n+1];
        suf[n] = 0;
        for(int i = 0;i<n;i++){
            cin>>arr[i];
        }
        for(int i = n-1;i>=0;i--){
            suf[i] = suf[i+1]+arr[i];
        }
        int sum = 0;
        long long ma = 0 , nah = 1;
        long long all = 0;
        for(int i = 0;i<=n;i++){
            int sum2 = abs(sum)+suf[i];
            ma = max(ma,sum2);
            if(i<n){
                sum+=arr[i];
            }
        }
        sum = 0;
        bool taked = 0;
        for(int i = 0;i<=n;i++){
            int sum2 = abs(sum)+suf[i];
            if(ma==sum2){
                if(sum>=0){
                    all += (nah*po[n-i])%mod;
                    taked = 1;
                    break;
                }else{
                    all += (nah*po[n-i])%mod;
                    taked = 1;
                }
            }
            if(i<n){
                sum+=arr[i];
                if(sum>=0)nah*=2;
                nah%=mod;
            }
        }
        cout<<all<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
import sys
import math
import heapq
from collections import deque

def II():
	return int(sys.stdin.readline())

def LI():
	return list(map(int, sys.stdin.readline().split()))

def MI():
	return map(int, sys.stdin.readline().split())

def SI():
	return sys.stdin.readline().strip()

def ask(l,x):
    # sys.stdout.flush()
    print("?",l,x)
    sys.stdout.flush()
    x = II()
    return x

def answer(m):
    # sys.stdout.flush()
    print("!", m)
    sys.stdout.flush()

def lcm(a, b):
    return abs(a*b) // math.gcd(a, b)
def md(x,y,x1,y1):
    return(abs(x-x1) + abs(y-y1))

mod = 998244353

for _ in range(II()):
    n = II()
    a = LI()
    if 0 in a:
        print(pow(2, n, mod))
        continue
    c = 0
    c2 = 0
    ans = 1
    boo = False
    cc = 0
    for i in a:
        c+=i
        c2+=i
        if i > 0:
            ans*=2
            ans%=mod
        else:
            if abs(c2) > c:
                boo = True
                cc+=1
            if c == abs(c2) and boo:
                ans*=3*pow(2, cc-1, mod)
                boo = False
                ans%=mod
                cc = 0
        c = max(abs(c), abs(c2))
    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
3
4
1 -1 1 -1
5
1 1 -2 1 1
6
1 -2 2 -1 1 -1
```

**Explanation:**

1. **Test Case 1:**
   - **Input:** `4\n1 -1 1 -1`
   - **Description:** This test case includes a sequence of alternating positive and negative values. The goal is to ensure that the algorithm correctly handles the transitions between positive and negative values and correctly calculates the maximum value of `c` and the number of unique procedures.

2. **Test Case 2:**
   - **Input:** `5\n1 1 -2 1 1`
   - **Description:** This test case includes a sequence where the absolute value of the sum can be maximized by choosing the correct operations. The goal is to ensure that the algorithm correctly handles the scenario where the sum of the array elements can be both positive and negative, and the absolute value operation is necessary to achieve the maximum value of `c`.

3. **Test Case 3:**
   - **Input:** `6\n1 -2 2 -1 1 -1`
   - **Description:** This test case includes a more complex sequence with multiple positive and negative values. The goal is to ensure that the algorithm correctly handles the scenario where the absolute value operation is necessary at multiple points to achieve the maximum value of `c`.

These test cases are designed to challenge the algorithm by including sequences that require careful handling of positive and negative values and the correct application of the absolute value operation. They are small-scale and complete, ensuring that they can be used to effectively test the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
import sys
input = sys.stdin.readline
INF = 998244353

for _ in range(int(input())):
    n = int(input())
    a = [*map(int, input().split())]
    c = 0
    prefix_sum = [0]
    for i in a:
        prefix_sum.append(prefix_sum[-1] + i)

    res = prefix_sum[-1]
    idx = [-1]
    for i in range(1, n + 1):
        if prefix_sum[i] < 0:
            if abs(prefix_sum[-1] - 2 * prefix_sum[i]) > res:
                idx = [i]
                res = abs(prefix_sum[-1] - 2 * prefix_sum[i])
            elif abs(prefix_sum[-1] - 2 * prefix_sum[i]) == res:
                idx.append(i)
    plus = 0
    for i in prefix_sum[1:]:
        if i >= 0:
            plus += 1
    cnt = ((pow(2, len(idx), INF) - 1) * pow(2, plus, INF)) % INF

    print(cnt)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer toker = null;
    private static int mod = 998244353;
    private static String next()   throws IOException {
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        while (toker == null || !toker.hasMoreTokens()) {
            toker = new StringTokenizer(br.readLine());

        }
        return toker.nextToken();
    }
    private static int nextInt()   throws IOException  {
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        StringBuilder sb = new StringBuilder();
        for (int g = 0; g < t; ++g) {
            int n = nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; ++i) arr[i] = nextInt();
            long[][] db = new long[n][2];
            db[0][0] = Math.abs(arr[0]);
            db[0][1] = arr[0];
            // System.out.println(0 + ", " + db[0][0] + ", " + db[0][1]);
            for (int i = 1; i < n; ++i) {
                db[i][0] = Math.max(Math.abs(db[i - 1][0] + arr[i]), Math.abs(db[i - 1][1] + arr[i]));
                db[i][1] = db[i - 1][1] + arr[i];
                // System.out.println(i + ", " + db[i][0] + ", " + db[i][1]);
            }
            long[][] db2 = new long[n][2];
            db2[0][0] = arr[0] >= 0 ? 2 : 1;
            db2[0][1] = arr[0] >= 0 ? 2 : 1;
            // System.out.println(0 + ", " + db2[0][0] + ", " + db2[0][1]);
            for (int i = 1; i < n; ++i) {
                if (db[i - 1][0] + arr[i] == db[i][0]) db2[i][0] = (db2[i][0] + db2[i - 1][0]) % mod;
                if (Math.abs(db[i - 1][0] + arr[i]) == db[i][0]) db2[i][0] = (db2[i][0] + db2[i - 1][0]) % mod;
                if (db[i - 1][0] != db[i - 1][1] && Math.abs(db[i - 1][1] + arr[i]) == db[i][0]) db2[i][0] = (db2[i][0] + db2[i - 1][1]) % mod;
                db2[i][1] = (db2[i][1] + (db[i - 1][0] + arr[i] >= 0 ? 2 : 1) * db2[i - 1][1]) % mod;
                // System.out.println(i + ", " + db2[i][0] + ", " + db2[i][1]);
            }
            sb.append(db2[n - 1][0]).append("\n");

        }
        System.out.println(sb);
    }

}

**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
1 -1 1 -1 1
```

**Explanation:**
- This test case is designed to highlight issues with the buggy code, especially in handling sequences that have a mix of positive and negative values.
- The array `[1, -1, 1, -1, 1]` is chosen to create a situation where the absolute value operation can significantly affect the final value of `c`.
- The goal is to see how the code handles the repeated toggling between positive and negative values and whether it correctly counts the number of unique procedures that achieve the maximum final value of `c`.

**Expected Output:**
- The maximum final value of `c` is `5`.
- The number of unique procedures that result in `c = 5` is `32`.

This test case should expose any issues in the dynamic programming logic and the way the code handles the absolute value operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

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

public class MagnitudeHard {
    final static int mod = 998244353;
    public static void main(String[] args){
        try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))){
            int t = Integer.parseInt(br.readLine());

            for(int i=0; i<t; i++){
                int n = Integer.parseInt(br.readLine());
                int[] inputArr = Arrays.stream(br.readLine().trim().split("\\s+")).mapToInt(Integer::parseInt).toArray();
                System.out.println(solve(n, inputArr));
            }
        }
        catch(IOException ex){
            throw new RuntimeException(ex);
        }
    }

    static long solve(int n, int[] input){

        long[][] minmax = new long[n+1][2]; // 0: min, 1: max
        long[][] now = new long[n+1][2]; // 0: minWays, 1: maxWays

        for(int i=1; i<=n; i++){
            minmax[i][0] = minmax[i-1][0] + input[i-1];
            minmax[i][1] = Math.max(Math.abs(minmax[i-1][0] + input[i-1]), Math.abs(minmax[i-1][1] + input[i-1]));
        }

        now[0][0] = 1;
        now[0][1] = 1;

        for(int i=1; i<=n; i++){
            if(minmax[i-1][0] == minmax[i-1][1]){
                if(minmax[i][0] == minmax[i-1][0] + input[i-1]){
                    now[i][0] += now[i-1][0];
                }
                if(minmax[i][0] == Math.abs(minmax[i-1][0] + input[i-1])){
                    now[i][0] += now[i-1][0];
                }
                if(minmax[i][1] == minmax[i-1][0] + input[i-1]){
                    now[i][1] += now[i-1][0];
                }
                if(minmax[i][1] == Math.abs(minmax[i-1][0] + input[i-1])){
                    now[i][1] += now[i-1][0];
                }
            }
            else{
                if(minmax[i][0] == minmax[i-1][0] + input[i-1]){
                    now[i][0] += now[i-1][0];
                }
                if(minmax[i][0] == Math.abs(minmax[i-1][0] + input[i-1])){
                    now[i][0] += now[i-1][0];
                }
                if(minmax[i][1] == minmax[i-1][0] + input[i-1]){
                    now[i][1] += now[i-1][0];
                }
                if(minmax[i][1] == Math.abs(minmax[i-1][0] + input[i-1])){
                    now[i][1] += now[i-1][0];
                }

                if(minmax[i][0] == minmax[i-1][1] + input[i-1]){
                    now[i][0] += now[i-1][0];
                }
                if(minmax[i][0] == Math.abs(minmax[i-1][1] + input[i-1])){
                    now[i][0] += now[i-1][0];
                }
                if(minmax[i][1] == minmax[i-1][1] + input[i-1]){
                    now[i][1] += now[i-1][0];
                }
                if(minmax[i][1] == Math.abs(minmax[i-1][1] + input[i-1])){
                    now[i][1] += now[i-1][0];
                }
            }

            now[i][0] %= mod;
            now[i][1] %= mod;
        }

        return now[n][1] % mod;
    }
}

**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
1 -2 1 -2
2
1 1
3
-1 -1 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

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

using namespace std;
const int MOD = 998244353;
const int MAX = 2 * 1e5;
int N;
long long A[MAX + 3], pfx[MAX + 3];
int nk, pk;
int pw[MAX + 3];

void test_case() {
    nk = 0, pk = 0;
    long long minn = 1e18;
    cin >> N;
    for(int i = 1; i <= N; i++) {
        cin >> A[i];
        pfx[i] = pfx[i - 1] + A[i];
        minn = min(minn, pfx[i]);
    }

    for(int i = 1; i <= N; i++) {
        if(pfx[i] >= 0) pk++;
        else {
            if(minn == pfx[i]) nk++;
        }
    }

    if(minn >= 0) nk = 1;
    cout << (1LL * (1LL * pw[nk] - 1) * pw[pk]) % MOD << '\n';
}

int main()
{
    //freopen("file.in", "r", stdin);
    //freopen("file.out", "w", stdout);

    int T;
    cin >> T;
    pw[0] = 1;
    for(int i = 1; i <= 2 * 1e5; i++) {
        pw[i] = (1LL * pw[i - 1] * 2) % MOD;
    }
    while(T--) {
        test_case();
    }
}

**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
-1 1 -1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (long long i = a; i < b; i++)

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long T;
    cin >> T;
    vector<long long> sol(T);
    rep(I, 0, T)
    {
        long long kl=998244353;
        long long n;
        cin >> n;
        vector<long long> a(n);
        rep(i, 0, n)
        {
            cin >> a[i];
        }
        vector <long long> pref(n);
        pref[0]=a[0];
        long long min=a[0];
        long long id=0;
        rep(i,1,n){
            pref[i]=a[i]+pref[i-1];
            if(min > pref[i]){
                min=pref[i];
                id=i;
            }
        }
        if(min>=0){
            long long pp=1;
            rep(i,0,n){
                pp=(pp*2)%kl;
            }
            sol[I]=pp;
            continue;
        }
        vector <long long> pr;
        int neg_count=0;
        rep(i,0,n){
            if(pref[i]<0){
                neg_count++;
            }
            if(min==pref[i]){
                pr.push_back(neg_count);
            }
        }
        long long inter_sum=1;
        rep(j,0,n-neg_count){
            inter_sum=(inter_sum*2)%kl;
        }
        long long sum=inter_sum;
        for(int i=pr.size()-2;i>-1;i--){
            long long diff=pr[i+1]-pr[i];
            while(diff>0){
                inter_sum=(inter_sum*2)%kl;
                diff--;
            }
            sum=(inter_sum+sum)%kl;
        }
        sol[I]=sum;
    }
    rep(i, 0, T)
    {
        cout << sol[i] << "\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
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
from collections import deque
t=int(input())
for _ in range(t):
  n=int(input())
  a=list(map(int,input().split()))
  c=0

  # for i in range(len(a)):
  #   if a[i]*c>=0:
  #     c+=a[i]
  #   else:
  #     c=abs(c)+a[i]

  # print(abs(c))
  sum=0
  l=deque()
  for i in range(n):
    sum+=a[i]
    l.append(sum)
  maxi=l[-1]
  for i in range(len(l)):
    if l[i]<0:

      maxi=max(maxi,l[-1]+2*abs(l[i]))
  cnt=0
  ptr=-1
  for i in range(len(l)):
    if l[-1]+2*abs(l[i]) == maxi and l[i]<0:
      cnt+=1
      ptr=i
  ans=1
  for i in range(len(l)):

    if l[i]>=0:
      ans*=2

  if cnt==0:
    print(ans%998244353)
  else:
    ans2=(2**cnt)-1
    for i in range(ptr+1,len(l)):
      if l[i]<0:
        ans2*=2
    print((ans*ans2)%998244353)

**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
1 -2 3 -3
3
-1 -2 -3
5
1 2 3 4 -10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
from math import ceil
def li():return [int(i)for i in input().split()]
mod=998244353
for _ in range(int(input())):
	n,=li()
	a=li()
	p,s=[],0
	for i in a:s+=i;p+=[s]
	mi,q=min(p),[]
	if mi>=0:print("   ",pow(2,n,mod));continue
	for i in range(n):
		if p[i]==mi:q+=[i]
	q=q[::-1]
	s=0
	ans,f=0,1
	for i in range(n):
		if not q:break
		if i==q[-1]:
			ans+=f*pow(2,n-1-i,mod)
			q.pop()
		else:
			if p[i]>=0:f=(f*2)%mod
	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
3
4
-1 1 -1 1
2
-1 -1
5
1 -2 1 -2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

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

public class Main {

    static long MOD = 998244353;

    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        StringTokenizer st;
        int t = Integer.parseInt(f.readLine());
        while (t-- > 0) {
            int n = Integer.parseInt(f.readLine());
            int[] arr = new int[n];
            st = new StringTokenizer(f.readLine());
            for (int i = 0; i < n; i++) {
                arr[i] = Integer.parseInt(st.nextToken());
            }
            long[][] ans = new long[n][2];
            long[][] dp = new long[n][2];
            ans[0][0] = ans[0][1] = arr[0];
            ans[0][1] = Math.abs(arr[0]);
            dp[0][0] = dp[0][1] = 2;
            if (arr[0] < 0) dp[0][0] = dp[0][1] = 1;
            for (int i = 1; i < n; i++) {
                if (arr[i] == 0) {
                    ans[i][0] = ans[i - 1][0];
                    ans[i][1] = ans[i - 1][1];
                    dp[i][0] = dp[i - 1][0] % MOD;
                    if (dp[i][0] >= 0) {
                        dp[i][0] *= 2;
                        dp[i][0] %= MOD;
                    }
                    dp[i][1] = dp[i - 1][1] * 2 % MOD;
                }
                else {
                    ans[i][0] = ans[i - 1][0] + arr[i];
                    dp[i][0] = dp[i - 1][0];
                    if (Math.abs(ans[i - 1][0] + arr[i]) == ans[i - 1][0] + arr[i]) {
                        dp[i][0] *= 2;
                        dp[i][0] %= MOD;
                    }
                    ans[i][1] = Math.max(Math.abs(ans[i - 1][0] + arr[i]), Math.abs(ans[i - 1][1] + arr[i]));
                    int count = 0;
                    int count2 = 0;
                    if (ans[i - 1][0] + arr[i] == ans[i][1]) {
                        count2++;
                    }
                    if (ans[i - 1][1] + arr[i] == ans[i][1]) {
                        count++;
                    }
                    if (Math.abs(ans[i - 1][0] + arr[i]) == ans[i][1]) {
                        count2++;
                    }
                    if (Math.abs(ans[i - 1][1] + arr[i]) == ans[i][1]) {
                        count++;
                    }
                    if (ans[i - 1][0] == ans[i - 1][1]) {
                        dp[i][1] = dp[i - 1][1] * count;
                        dp[i][1] %= MOD;
                    }
                    else {
                        dp[i][1] = dp[i - 1][1] * count + dp[i - 1][0] * count2;
                        dp[i][1] %= MOD;
                    }
                }
            }
            out.println(dp[n - 1][1]);
        }
        f.close();
        out.close();
    }

    public static void sort(int[] arr) {
        Random r = new Random();
        for (int i = 0; i < arr.length; i++) {
            int x = r.nextInt(arr.length);
            int temp = arr[x];
            arr[x] = arr[i];
            arr[i] = temp;
        }
        Arrays.sort(arr);
    }

    public static void sort(int[][] arr) {
        Random r = new Random();
        for (int i = 0; i < arr.length; i++) {
            int x = r.nextInt(arr.length);
            int[] temp = arr[x];
            arr[x] = arr[i];
            arr[i] = temp;
        }
        Arrays.sort(arr, Comparator.comparingInt(a -> a[0]));
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. Let's break down the problem and identify such cases:

1. **Zero Values**: The code has special handling for zero values, which might introduce bugs. We should include zeros in the input to test this handling.
2. **Negative and Positive Values**: The code needs to handle both negative and positive values correctly. We should include a mix of these values to ensure the code can handle different scenarios.
3. **Large Values**: The code should be able to handle large values within the given constraints. We should include large positive and negative values to test this.
4. **Mixed Operations**: The code should correctly handle the combination of addition and absolute value operations. We should create a sequence that requires both operations to be used effectively.

Given these considerations, a small-scale test input that can expose potential bugs in the code is:

```plaintext
3
4
0 -1000000000 1000000000 0
3
-1 0 1
5
1 -2 3 -4 5
```

### Explanation:
1. **Test Case 1**:
   - **Input**: `4 0 -1000000000 1000000000 0`
   - **Purpose**: This test case includes zeros and large positive and negative values. It tests the special handling for zeros and the ability to handle large values.

2. **Test Case 2**:
   - **Input**: `3 -1 0 1`
   - **Purpose**: This test case includes a small sequence with a zero, a negative value, and a positive value. It tests the handling of mixed values and the special case for zero.

3. **Test Case 3**:
   - **Input**: `5 1 -2 3 -4 5`
   - **Purpose**: This test case includes a sequence with a mix of positive and negative values. It tests the ability to handle mixed operations and the combination of addition and absolute value operations.

These test cases are designed to cover a range of edge cases and scenarios that can expose potential bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

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

public class Main {

    static final long M = 998244353;

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        long start1 = System.currentTimeMillis();

        ArrayList<Long> p2 = new ArrayList<>();
        long d = 1;
        for (int i = 0; i <= 2 * 1e5; i++) {
            p2.add(d);
            d = (d * 2) % M;
        }

        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            ArrayList<Long> v = new ArrayList<>();
            scn(v, n, sc);
            long mn = Long.MAX_VALUE;

            long a = 0;
            int c = 0;
            for (int i = 0; i < n; i++) {
                a += v.get(i);
                mn = Math.min(mn, a);
                if (a < 0) {
                    c++;
                }
            }

            if (mn >= 0) {
                System.out.println(p2.get(n));
            } else if (c == n) {
                System.out.println(1);
            } else {
                System.out.println(((p2.get(c) - 1) * (p2.get(n - c))) % M);
            }
        }

        long stop1 = System.currentTimeMillis();
        System.err.println("Time: " + (stop1 - start1) + " ms");
    }

    static void scn(ArrayList<Long> v, int n, FastReader sc) {
        for (int i = 0; i < n; i++) {
            v.add(sc.nextLong());
        }
    }

    static long modexp(long x, long n) {
        if (n == 0) {
            return 1;
        } else if (n % 2 == 0) {
            return modexp((x % M * x % M) % M, (n / 2) % M) % M;
        } else {
            return (((x % M) * (modexp((x % M * x % M) % M, ((n - 1) / 2) % M) % M)) % M);
        }
    }

    static long get(long a, long b) {
        long c = gcd(a, b) % M;
        a = a / c;
        a %= M;
        b = b / c;
        b %= M;
        long d = modexp(b, M - 2) % M;
        long ans = ((a % M) * (d % M)) % M;
        return ans % M;
    }

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

    static long power(long a, long b) {
        long result = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                result = (result * a) % M;
            }
            a = (a * a) % M;
            b >>= 1;
        }
        return result;
    }

    static long nCr(long n, long r) {
        if (n - r == 1) {
            return n;
        }
        if (r > n) return 0;
        if (r == 0 || n == r) return 1;
        double res = 0;
        for (int i = 0; i < Math.min(r, n - r); i++) {
            res += Math.log(n - i) - Math.log(i + 1);
        }
        return Math.round(Math.exp(res)) % M;
    }

    static long pr(long a, long b) {
        return ((a % M) * (b % M)) % M;
    }

    static boolean ch(long n) {
        while (n != 0) {
            if (n % 10 < 5) {
                return false;
            }
            n /= 10;
        }
        return true;
    }

    static boolean regu(String s) {
        int i = -1;
        for (char ch : s.toCharArray()) {
            if (ch == '(' || ch == '{' || ch == '[') {
                i++;
                s = s.substring(0, i) + ch + s.substring(i + 1);
            } else {
                if (i >= 0 && ((s.charAt(i) == '(' && ch == ')') || (s.charAt(i) == '{' && ch == '}') || (s.charAt(i) == '[' && ch == ']'))) {
                    i--;
                } else {
                    return false;
                }
            }
        }
        return i == -1;
    }

    static int MSBPosition(long N) {
        int msb_p = -1;
        while (N > 0) {
            N >>= 1;
            msb_p++;
        }
        return msb_p;
    }

    static long findBitwiseOR(long L, long R) {
        long res = 0;
        int msb_p1 = MSBPosition(L);
        int msb_p2 = MSBPosition(R);
        while (msb_p1 == msb_p2) {
            long res_val = 1 << msb_p1;
            res += res_val;
            L -= res_val;
            R -= res_val;
            msb_p1 = MSBPosition(L);
            msb_p2 = MSBPosition(R);
        }
        msb_p1 = Math.max(msb_p1, msb_p2);
        for (int i = msb_p1; i >= 0; i--) {
            long res_val = 1 << i;
            res += res_val;
        }
        return res;
    }

    static boolean checkPalindrome(int n) {
        int reverse = 0;
        int temp = n;
        while (temp != 0) {
            reverse = (reverse * 10) + (temp % 10);
            temp /= 10;
        }
        return reverse == n;
    }

    static long uro(long n, long k) {
        if (n % k == 0) {
            return n / k;
        } else {
            return n / k + 1;
        }
    }

    static long lis(ArrayList<Long> arr, int n) {
        ArrayList<Long> v = new ArrayList<>(arr);
        ArrayList<Long> ans = new ArrayList<>();
        ans.add(arr.get(0));
        for (int i = 1; i < n; i++) {
            int pos = upperBound(ans, v.get(i));
            if (pos < ans.size()) {
                ans.set(pos, v.get(i));
            } else {
                ans.add(v.get(i));
            }
        }
        return ans.size();
    }

    static int upperBound(ArrayList<Long> list, long value) {
        int low = 0;
        int high = list.size();
        while (low < high) {
            int mid = (low + high) / 2;
            if (list.get(mid) <= value) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }

    static void pridiv(long n, ArrayList<Long> v) {
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                if (n / i == i) {
                    v.add((long) i);
                } else {
                    v.add((long) i);
                    v.add(n / i);
                }
            }
        }
    }

    static void scv(ArrayList<Long> v, int n) {
        for (int i = 0; i < n; i++) {
            System.out.print(v.get(i) + " ");
        }
        System.out.println();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with how it handles negative values and the combination of operations. Let's break down the steps to create a test input:

1. **Understand the Bug**: The bug in the code appears to be in how it calculates the number of unique procedures. It uses a precomputed power of 2 list and some conditions to determine the result. The main issue is in the logic for handling negative sums and the combination of operations.

2. **Edge Cases**:
   - **All Positive Values**: This should result in \(2^n\) possible ways since taking the absolute value doesn't change the value.
   - **All Negative Values**: This should result in 1 possible way since the absolute value must be taken at every step to maximize \(c\).
   - **Mixed Values**: This is where the bug is most likely to be exposed. We need to create a sequence where the combination of operations can lead to different results.

3. **Challenging Input**:
   - A sequence of mixed positive and negative values that can lead to multiple valid paths to the maximum value of \(c\).

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

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

This input:
- Contains a mix of positive and negative values.
- The sequence is designed to have multiple valid paths to the maximum value of \(c\).
- It is small enough to be manageable and still challenging for the buggy code.

Let's break down why this input is challenging:
- The sequence \(1, -2, 3, -4, 5\) can result in multiple valid procedures to achieve the maximum value of \(c\).
- For example, one valid procedure could be to take the absolute value at indices 2 and 4, resulting in \(c = 1 + 2 + 3 + 4 + 5 = 15\).
- Another valid procedure could be to take the absolute value at indices 2 and 3, resulting in \(c = 1 + 2 + 3 + 4 + 5 = 15\).

This input should help expose the bug in the given code by requiring it to correctly handle the combination of operations and the resulting values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make It Equal
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.

You can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \bmod n) + 1}$$$ by $$$1$$$.

After you perform the operations, all elements of the array should be non-negative equal integers.

Your task is to calculate the minimum number of operations you have to perform.

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

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$).

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

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.

Examples:
input:
3
2
1 1
3
1 3 2
4
2 1 2 6
output:
0
-1
3

Note:

Buggy Code:
#include <cstdio>
using namespace std;
#define __IN freopen("a.in","r",stdin)
#define __OUT freopen("k.out","w",stdout)
#define ll long long
const int N = 2e5 + 10;

 ll a[N], b[N], t, n;

int solve(ll v, ll pos)
{
    // printf("-%lld-\n", v);
    for(int i=1; i<=n; i++) b[i] = a[i];
    // for (int i=1;i<=n;i++) printf("%lld ", b[i]);
    // printf("\n");
    for(int i=0; i<n; i++)
    {
        ll p = (pos + i - 1) % n + 1;
        ll pi = (pos + i) % n + 1;
        if (b[p] < v) return false;
        ll k = (b[p] - v) / 2;
        b[p] -= 2 * k;
        b[pi] += k;
    }

    for (int i=1;i<=n;i++)
    {
        if (b[i] != b[1]) return false;
    }
    return true;
}
int main()
{
    // __IN;__OUT;
    scanf("%lld", &t);
    while(t--)
    {
        scanf("%lld", &n);
        ll pos = 0, mx = 0, mi = 1e9+10, sum = 0;
        for (int i=1;i<=n;i++)
        {
            scanf("%lld", a+i);
            if (a[i] > mx)
            {
                pos = i;
                mx = a[i];
            }
            if (a[i] < mi)
            {
                mi = a[i];
            }
            sum += a[i];
        }
        ll l = mi, r = mx, res = -1;
        while(l <= r)
        {
            ll mid = (l+r) / 2 ;
            if (solve(mid, pos))
            {
                res = mid;
                l = mid + 1;
            }
            else
            {
                r = mid - 1;
            }
        }
        printf("%lld\n", res == -1 ? -1 : sum-(res*n));
        // printf("%lld ", res);
        // for (int i=1;i<=n;i++) printf("%lld ", a[i]);
        // printf("\n");
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Let's analyze the problem and the code to identify potential issues:

1. **Array Length and Values**: The array length `n` can range from 2 to 200,000, and the values `a_i` can range from 1 to 10^9. We should consider both small and large values for `n` and `a_i`.

2. **Sum of Elements**: The sum of the elements in the array is crucial. The code uses the sum to determine the number of operations needed. If the sum is not divisible by `n`, it is impossible to make all elements equal.

3. **Initial Values and Operations**: The code uses a binary search to find the target value and then checks if it is possible to make all elements equal to that value. We should test cases where the initial values are such that the binary search might fail to find a valid target value.

4. **Edge Cases**: Consider cases where the array is already equal, or where it is impossible to make the array equal.

Based on these points, let's generate a small-scale test input:

```plaintext
3
2
1 1
3
1 3 2
4
2 1 2 6
```

### Explanation:
1. **Test Case 1**:
   - `n = 2`
   - `a = [1, 1]`
   - The array is already equal, so no operations are needed. The expected output is `0`.

2. **Test Case 2**:
   - `n = 3`
   - `a = [1, 3, 2]`
   - It is impossible to make all elements equal using the given operations. The expected output is `-1`.

3. **Test Case 3**:
   - `n = 4`
   - `a = [2, 1, 2, 6]`
   - The sum of the elements is 11, which is not divisible by 4. Therefore, it is impossible to make all elements equal. The expected output is `-1`.

This test input covers a range of scenarios including an already equal array, an impossible case, and a case where the sum is not divisible by the array length. This should help in exposing any bugs in the provided code.