**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ld long double
#define ll long long
#define db double
#define ii pair<int,int>
#define f first
#define s second
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define all(v) v.begin(),v.end()
#define BIT(i) ((ll)1<<(i))
#define debug(x) for (auto p: x) cout<<p<<' ';cout<<endl
#define forw(i,j,z) for(int i=(int)j;i<=(int)z;i++)
#define ford(i,j,z) for (int i=(int)j;i>=(int)z;i--)
#define sz(a) (int)a.size()
#define len(a) (int)a.length()
const int mod = 998244353;
inline int add(int u,int v){u+=v;if(u>=mod)u-=mod;return u;}
//auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
//mt19937 RAND(seed);
inline int dec(int u,int v){u-=v;if(u<0)u+=mod;return u;}
inline int mul(int u,int v){return (ll)u*v%mod;}
const int N = 2e5;
int a[N + 1];
int n;
void solve()
{
    cin >> n;
    forw(i, 1, n)
        cin >> a[i];
    sort(a + 1, a + n + 1);
    int ans = n + 1;
    //odd
    int id = n;
    while (id > 0 && !(a[id] & 1))
        --id;
    if (id > 1)
    {
        int tmp = 0;
        ll com = a[id];
        for (int i = id - 1; i > 0; --i)
            if (!(a[i] & 1))
            {
                ++tmp;
                com += a[i];
            }
        for (int i = id + 1; i <= n; ++i)
            if (com >= a[i])
            {
                ++tmp;
                com += a[i];
            }
            else{
                tmp += n - i + 1 + 1;
                break;
            }
        ans = min(ans, tmp);
    }
    bool ok = true;
    for (int i = 1; i <= n; ++i)
        if (a[i] & 1)
            ok = false;
    if (ok)
        ans = 0;
    cout << ans << "\n";
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
 //   freopen("test.inp","r",stdin);
//    freopen(".out","w",stdout);
    //time_t TIME_TU=clock();
    int t=1;
    cin>>t;
    while(t--)
        solve();
    //time_t TIME_TV=clock();
    //cerr<<(db)(TIME_TV-TIME_TU)/CLOCKS_PER_SEC<<endl;
    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code seems to have issues with handling cases where the array has a mix of odd and even numbers, and it may not correctly calculate the minimum number of operations needed to make all elements have the same parity.

2. **Edge Cases**:
   - Arrays with a single odd number and multiple even numbers.
   - Arrays with a single even number and multiple odd numbers.
   - Arrays with alternating odd and even numbers.
   - Arrays with a large number of elements to test performance and correctness.

3. **Small-Scale Test Input**: Since the test input should be small-scale, we can focus on an array with a few elements that are likely to expose the bug.

4. **Challenging Scenario**: An array with a single odd number and multiple even numbers can be particularly challenging. The code should correctly identify that the single odd number needs to be combined with the even numbers to make all elements have the same parity.

Let's create a test input with the following characteristics:
- One odd number and several even numbers.
- Ensure the odd number is not the smallest element to test the logic of combining the smallest even number with the odd number.

Here is the test input:

```plaintext
1
5
2 4 6 8 1
```

Explanation:
- The array is `[2, 4, 6, 8, 1]`.
- The single odd number is `1`, and the even numbers are `2, 4, 6, 8`.
- The correct number of operations needed is `4`:
  - Combine `1` with `2` to get `[3, 4, 6, 8]`.
  - Combine `3` with `4` to get `[7, 6, 8]`.
  - Combine `7` with `6` to get `[13, 8]`.
  - Combine `13` with `8` to get `[21]`.

This test input should expose the bug in the buggy code by requiring the correct number of operations to be calculated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

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

#define pii pair<int, int>
#define INF 0x3f3f3f3f
#define LNF 0x3f3f3f3f3f3f3f3f
#define endl "\n"

using namespace std;
using i64 = long long;

const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};

void solve()
{
    int n, odd, even, mx = 0; cin >> n;
    vector<int> a(n), eve;
    for (auto &x : a) {
        cin >> x;
        if (x % 2) {
            ++odd;
            mx = max(mx, x);
        } else {
            ++even;
            eve.push_back(x);
        }
    }
    sort(eve.begin(), eve.end());

    if (even == n || odd == n) {
        cout << "0\n";
        return;
    }

    i64 ans = 0;
    for (auto x : eve) {
        if (mx > x) {
            ++ans;
            mx = max(mx, mx + x);
        } else {
            ans += 2;
            mx += 2 * x;
        }
    }
    cout << ans << endl;

}

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
t = int(input())

for _ in range(t):
    n = int(input())
    inp_lst = sorted(list(map(int,input().split())))

    ev_cnt = 0

    ev_grt_od_cnt = 0
    grt_od = -1

    for ind, i in enumerate(inp_lst):
        if i%2 == 0:
            ev_cnt+=1
        else:
            ev_grt_od_cnt = ind
            grt_od = i

    ev_grt_od_cnt = n - ev_grt_od_cnt - 1

    od_cnt = n - ev_cnt

    if od_cnt == 0:
        print(0)
        continue

    ans = 0
    for i in inp_lst:
        if i%2 == 0:
            if grt_od > i:
                ans+=1
                grt_od+=i
            else:
                ans+=2
                grt_od+=2*i

    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:
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 maximize the failure probability. The buggy code seems to have issues with correctly handling the transformation of elements with distinct parities. Let's analyze the code and identify potential edge cases:

1. **All elements have the same parity**: This should be handled correctly.
2. **One element of one parity and multiple elements of the other parity**: This can reveal issues in the logic for merging elements.
3. **Elements with mixed parities and varying values**: This can expose issues in the logic for choosing which elements to merge.

Given these considerations, a small-scale test input that can expose the bug is:

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

Explanation:
- The array contains two even numbers (2, 4) and two odd numbers (3, 5).
- The code should ideally merge the elements in such a way that all elements end up with the same parity.
- The expected output is 2 operations, but the buggy code might not handle this correctly due to its logic for merging elements.

This test input is small and complete, making it suitable for exposing the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    cnt = 0
    mx1 = 0
    mx2 = 0
    a = list(map(int,input().split()))
    b = []
    for d in a:
        if d%2==0:
            mx1 = max(mx1,d)
            cnt+=1
            b.append(d)

        else:
            mx2 = max(mx2,d)
    ans = 0
    if cnt == 0 or cnt==n:
        print(0)

    else:
        b.sort()
        for j in b:
            if j<mx2:
                ans+=1
                mx2+=j
            else:
                ans+=2
                mx2+=2*j
        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:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the issues and edge cases:

1. **Handling of Even and Odd Numbers**: The code seems to handle the maximum even and odd numbers separately but doesn't correctly account for the operations needed to make all elements have the same parity.
2. **Sorting and Operations**: The code sorts the even numbers and then iterates through them, which might not always be the optimal approach. It could lead to incorrect results if the operations are not correctly counted.
3. **Edge Cases**: The code might fail when there are a mix of even and odd numbers, especially when the maximum odd number is larger than the maximum even number.

Given these points, let's create a test input that maximizes the failure probability:

- A small array with a mix of even and odd numbers where the maximum odd number is larger than the maximum even number.
- Ensure the array has a structure that will test the sorting and operation counting logic.

Here is the test input:

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

This input has:
- 1 test case.
- An array of 5 elements: [1, 2, 3, 4, 5].
- The maximum even number is 4.
- The maximum odd number is 5.

The correct output for this test case should be 2 operations, but the buggy code might produce an incorrect result due to its flawed logic in handling the operations.


Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases:**
   - All elements are even.
   - All elements are odd.
   - Mixed parities with specific values that can cause issues in the logic.
2. **Logic Flaws:**
   - The code uses `maxEven` and `maxOdd` to track the maximum even and odd values, respectively. It then attempts to combine elements to make all elements have the same parity.
   - The code has a loop that processes even elements and attempts to combine them with the maximum odd value. This logic might fail in certain scenarios.

Let's create a test case that can expose the logic flaws:

- **Mixed Parities:**
  - We will have a mix of even and odd numbers, with the even numbers being smaller than the odd numbers. This will force the code to perform more operations and potentially expose the incorrect handling of the maximum values.

Here is a small-scale test input:

```plaintext
1
5
2 4 6 7 9
```

### Explanation:
- **Array:** `[2, 4, 6, 7, 9]`
  - Even numbers: `[2, 4, 6]`
  - Odd numbers: `[7, 9]`
- **Expected Output:** `3`
  - The optimal sequence of operations to make all elements have the same parity is:
    1. Combine `2` and `7` to get `[9, 4, 6, 9]`.
    2. Combine `4` and `9` to get `[9, 13, 6, 9]`.
    3. Combine `6` and `9` to get `[9, 13, 15, 9]`.

This test case will challenge the code to correctly handle the combination of smaller even numbers with larger odd numbers, potentially exposing any issues in the logic for updating `maxOdd` and `maxEven`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
import java.util.*;

public final class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
		  List<Integer>l1=new ArrayList<>();
		  List<Integer>l2=new ArrayList<>();
		  double maxOdd=0;
		  long cnt=0;
		  int n=sc.nextInt();
		  for(int i=0;i<n;i++){
		      int a=sc.nextInt();
		      if(a%2==0){
		          l2.add(a);
		      }
		      else{
		          l1.add(a);
		          maxOdd=Math.max(maxOdd,a);
		      }
		  }
		  if(l2.size()==0 || l1.size()==0){
		      System.out.println(0);
		  }
		  else{

		      Collections.sort(l2);
		      int i=0;
		      while(i<l2.size()){
		          if(l2.get(i)<maxOdd){
		              cnt+=1;
		              maxOdd=Math.max(maxOdd+l2.get(i),maxOdd);
		              i++;
		          }
		          else{
		              maxOdd=Math.max(maxOdd+l2.get(i),maxOdd);
                        cnt+=1;
		          }
		      }
		      System.out.println(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
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: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
// Lets Do It With Patience  :) .......

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define endl "\n"
#define FAST ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define all(v)  v.begin(),v.end()
#define maxx(v) *max_element(v.begin(),v.end())
#define minn(v) *min_element(v.begin(),v.end())
#define pb push_back
#define ppb pop_back
const int MOD = 1e9 + 7, MX = 1e18, MN = -1e18, N = 5e5 + 10;

bool prime(int n){
    if(n<=1) return false;

    for(int i=2 ; i<=sqrt(n) ; i++){
        if(n%i==0) return false;
    }

    return true;
}

void factor(int n , vector<int> &v){
    for(int i=1 ; i*i<=n ; i++){
        if(n%i==0) {
            v.pb(i);
            if(n/i!=i) v.pb(n/i);
        }
    }
}

signed main() {
    FAST

    int t;
    cin>>t;

    while(t--){

        int n;
        cin>>n;

        vector<int> v(n);

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

        vector<int> even;
        int ma=0;

        for(int i=0 ;i<n ; i++){
            if(v[i]%2==0) {
                even.pb(v[i]);
            }else{
                ma=max(ma,v[i]);
            }
        }

        int ans=0;
        if(even.size()==n) {
            cout<<0<<endl;
            continue;
        }

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

            int a=even[i]-ma;
            int aa=(even[i]-ma)/ma;
            if((even[i]-ma)%ma!=0) aa++;
            ans+=aa;

        }

        cout<<ans<<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
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: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
#include <bits/stdc++.h>
#include <algorithm>
#include <iomanip>
#include <cmath>
#include <string>
#define Abdallah ios_base::sync_with_stdio(0);
#define Alqiran cin.tie(0),cout.tie(0);
#define ll long long
#define fi for (int i=0;i<n;++i)
#define el '\n'

using namespace std;

void solve()
{
    int N;      cin >> N;
    vector<ll> V;
    ll num = 0, mx = 0;
    ll counter = 0;

    for (int i=0;i<N;++i)
    {
        cin >> num;
        if (num % 2 == 0)
        {
            V.push_back(num);
        }
        else
        {
           mx = max(mx, num);
        }
    }
    if (V.size() == N)
    {
        cout << 0;
        return;
    }
    sort(V.begin(),V.end());

    for (auto i : V)
    {

        while (i > mx)
        {
            mx = mx + i;
            counter ++;
        }
            mx = mx + i;
            counter ++;
//            cout << mx << ' ';
    }
    cout << counter;
}

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

    Abdallah Alqiran

    ll t = 1;
    cin >> t;

    while (t--)
    {
        solve();
        cout << '\n';
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases that can highlight issues in the logic. The code seems to handle the following steps:

1. Read the number of test cases.
2. For each test case, read the array size and the array elements.
3. Count the number of even and odd elements.
4. If all elements are even, output 0.
5. Otherwise, sort the even elements and try to make the maximum odd element by adding the even elements to it, counting the operations.

The main issues in the code are:
- It does not handle the case where there are more odd elements than even elements.
- It does not correctly handle the case where the maximum odd element is already larger than all even elements.
- It does not correctly handle the case where the array has a mix of odd and even elements, and the operations need to be carefully managed to ensure the minimum number of operations.

To expose these issues, we can create a test case where the array has a mix of odd and even elements, and the maximum odd element is initially smaller than the even elements. This will force the code to perform more operations than necessary, or potentially infinite loops.

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

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

### Explanation:
- **Array:** `[2, 3, 4, 5]`
- **Initial Parity:**
  - Even: `[2, 4]`
  - Odd: `[3, 5]`
- **Expected Output:**
  - The minimum number of operations required to make all elements have the same parity is 2.
  - Operation 1: Choose (2, 3) -> `[5, 3, 4, 5]`
  - Operation 2: Choose (3, 4) -> `[5, 7, 4, 5]`

This test case should highlight the issue in the code where it does not correctly handle the mix of odd and even elements and the operations required to make them have the same parity.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
# # def f(n,s):
# #   for i in range(n-1):
# #     if s[i]+s[i+1]=="aa":
# #       return 2
# #   for i in range(n-2):
# #     if s[i]+s[i+2]=="aa":
# #       return 3
# #   for i in range(n-3):
# #     if s[i]+s[i+3]=="aa" and (s[i+1]+s[i+2]=="bc" or s[i+1]+s[i+2]=="cb"):
# #       return 4
# #   i=0
# #   j=1
# #   na=0
# #   nb=0
# #   nc=0
# #   l=[]
# #   while i<n and j<n:
# #     if s[i]=="a" :
# #       if na==0:
# #         na+=1
# #       if s[j]=="a" and i!=j:

# #         na+=1

# #       elif s[j]=="b":
# #         nb+=1
# #       else:
# #         nc+=1
# #       if na>nb and na>nc and (j-i+1)>=2:
# #         l.append(j-i+1)
# #         i=j
# #         na=0
# #         nb=0
# #         nc=0
# #       else:

# #         j+=1
# #     else:
# #       i+=1
# #       j+=1

# #   if len(l)==0:
# #     return -1
# #   return min(l)

# # t=int(input())
# # for _ in range(t):
# #   n=int(input())
# #   s=input()
# #   print(f(n,s))

# def f(n,s):
#   for i in range(n-1):
#     if s[i]+s[i+1]=="aa":
#       return 2
#   for i in range(n-2):
#     if s[i]+s[i+2]=="aa":
#       return 3
#   for i in range(n-3):
#     if s[i]+s[i+3]=="aa" and (s[i+1]+s[i+2]=="bc" or s[i+1]+s[i+2]=="cb"):
#       return 4

#   la=[]
#   for i in range(n):
#     if s[i]=="a":
#       la.append(i)
#   if len(la)<=1:
#     return -1
#   i=la[0]
#   j=la[0]+1
#   na=1
#   nb=0
#   nc=0
#   l=[]
#   k=0

#   while i<n and j<n and k<len(la):
#     # print(i,j,na,nb,nc)
#     if s[i]=="a" :

#       if s[j]=="a" :

#         na+=1

#       elif s[j]=="b":
#         nb+=1
#       elif s[j]=="c":
#         nc+=1
#       if na>nb and na>nc and (j-i+1)>=2:

#         l.append(j-i+1)
#         k+=1
#         i=la[k]
#         j=i+1
#         na=1
#         nb=0
#         nc=0
#       else:

#         j+=1
#       if j==n and k<len(la)-2:
#         k+=1
#         i=la[k]
#         j=i+1
#       elif j==n:
#         break
#   if len(l)==0:
#     return -1
#   return min(l)

# t=int(input())
# for _ in range(t):
#   n=int(input())
#   s=input()
#   if t==98957 and _==2531:
#     print(s)
#   else:
#     print(f(n,s))

t=int(input())
for _ in range(t):
  n=int(input())
  a=list(map(int,input().split()))
  a.sort()
  # ans = 0
  # ptr=0
  # mo=0
  # for i in a:
  #   if i%2:
  #     mo = i

  # for i in a:
  #   if not i%2:
  #     if mo>i:
  #       ans+=1
  #       mo+=i
  #     else:
  #       ans+=2
  #       mo+=i
  # if mo==0:
  #   ans=0
  # print(ans)
  ptr=-1
  maxodd=0
  for i in range(n):
    if a[i]%2!=0:
      ptr=i
      maxodd=a[i]
  ans=0
  # maxeven=0
  # for i in range(ptr):
  #   if a[i]%2==0:
  #     ans+=1
  #     maxeven=a[i]
  #     maxodd=maxodd+a[i]
  # print(maxodd)
  b=maxodd

  for i in range(n):
    if a[i]%2==0 :
      if a[i]<b:
        ans+=1
        b=a[i]+b
      else:
        ans+=2
        b=2*a[i]+b
  if ptr==-1:
    ans=0
  print(ans)

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

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

Think step by step.

Response:
```plaintext
1
5
2 4 6 8 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
for _ in range(int(input())):
	# x=list(map(int, input().split()))
	x=int(input())
	y=list(map(int, input().split()))
	o,e=0,0
	ev,od=0,0
	c=0
	even,odd=0,0
	for i in range(x):
		if y[i]%2==0:e,ev=e+1,max(ev,y[i]); continue
		o+=1
		od=max(od,y[i])
	if min(o,e)==0:print(0); continue
	y.sort()
	for i in range(x-1,-1,-1):
		if y[i]%2==0:even=max(i,even)
		else:odd=max(i,odd);o=max(o,y[i])
	for i in range(x):
		if y[i]%2==1:
			continue
		if y[i]<o:
			y[i]+=o
			o=y[i]
		elif y[i]>o:
			o+=y[i]
			y[i]+=o
			c+=1
		c+=1
	# print(y)
	print(c)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

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

// 注意类名必须为 Main, 不要有任何 package xxx 信息

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(), n, ans;
        String s;

        while (t-- != 0) {
            ans = 0;
            n = sc.nextInt();
            List<Long> a = new ArrayList<>();
            long m = 0, cur;
            for (int i = 0; i < n; i++) {
                cur = sc.nextLong();
                if (cur % 2 == 0) {
                    if (cur < m) {
                        ans++;
                        m = cur + m;
                    } else
                        a.add(cur);
                }else {
                    m = Math.max(m,cur);
                }
            }
            if (m == 0||a.size()==0) {
                System.out.println(0);
                continue;
            }
            Collections.sort(a);
            long r = a.get(a.size() - 1);
            for (int i = 0; i < a.size(); i++) {
                cur = a.get(i);
                ans++;
                if (cur < m) {
                    m+=cur;
                }else{
                    m+=r;
                    ans++;
                }
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

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

**Explanation:**
- The input consists of a single test case with an array of 5 elements: [1, 2, 3, 4, 5].
- This array contains both even and odd numbers, and the challenge is to make all elements have the same parity using the minimum number of operations.
- The correct output for this test case should be 2, as you can perform the following operations:
  1. Choose indices (1, 2) and replace 1 with 1 + 2 = 3. The array becomes [3, 2, 3, 4, 5].
  2. Choose indices (1, 4) and replace 2 with 2 + 4 = 6. The array becomes [3, 6, 3, 4, 5].
- After these operations, all elements have the same parity (odd in this case).

This test case is designed to expose issues in the buggy code, particularly in how it handles the combination of even and odd numbers and the counting of operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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 $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        // 1 1 2 3 4 6 => 3
        // 1 2 3 4 6 8=> 4
        // 1 2 3 11 17 =>
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            long[] ar = new long[n];
            for (int i = 0; i < n; i++) {
                ar[i] = sc.nextInt();
            }
            int even = 0;
            int odd = 0;
            for (long i : ar) {
                if (i % 2 == 0) {
                    even++;
                } else {
                    odd++;
                }
            }
            if (even == 0 || even == n) {
                System.out.println(0);
            } else {
                Arrays.sort(ar);
                // 2 2 11 19
                int ans = 0;
                long[] max = getMaxOdd(ar);
                long maxOdd = max[0];
                long maxOddIdx = max[1];
                // System.out.println("maxOdd: " + maxOdd + " maxOddIdx: " + maxOddIdx);
                // 1 1 5 7 3 6
                while (true) {
                    boolean changed = false;
                    for (int i = 0; i < ar.length; i++) {
                        if (ar[i] % 2 == 0 && ar[i] < maxOdd) {
                            ar[i] += maxOdd;
                            changed = true;
                            ans++;
                            maxOdd = ar[i];
                            maxOddIdx = i;
                            break;
                        } else if (ar[i] % 2 == 0 && ar[i] > maxOdd) {
                            ar[(int) maxOddIdx] += ar[i];
                            ar[i] += ar[(int) maxOddIdx];
                            maxOdd = ar[i];
                            ans += 2;
                            maxOddIdx = i;
                            changed = true;
                            break;
                        }
                    }
                    if (!changed) {
                        break;
                    }
                }
                System.out.println(ans);
            }
        }
    }

    static long[] getMaxOdd(long[] ar) {
        long max = Long.MIN_VALUE;
        long max_idx = -1;
        for (int i = ar.length - 1; i >= 0; i--) {
            if (ar[i] % 2 != 0) {
                max = ar[i];
                max_idx = i;
                break;
            }
        }
        return new long[]{max, max_idx};
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
2 4 6 8 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

Buggy Code:
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
#define int long long
#define pb push_back
#define fo(j,a,n) for(int j=a;j<n;j++)
#define tr(it,a) for(auto it:a)
#define mp make_pair
int N=1e9+7;
int M=1e9+7;
int N1=1e5+5;
int m2=1e18;
// int m2=
// vector<int> fact(N,1);
// int po(int n,int m){
//     int count=0;
//     while(n%m==0){
//         n/=m;
//         count++;
//     }
//   return count;
// }

// int power(long long x, unsigned int y)
// {
//     int res = 1;

//     x = x% m2;

//     if (x == 0) return 0;

//     while (y > 0)
//     {

//         if (y & 1)
//             res = (res*x)%m2;

//         y = y>>1;
//         x = (x*x)%m2;
//     }
//     return res;
// }

// int binpow(int a, int b,int c) {
//     int res = 1;
//     while (b > 0) {
//         if (b & 1){
//             res = (res * a)%c;
//             res%=c;
//         }
//         a = a * a;
//         a%=c;
//         b >>= 1;
//     }
//     return res%c;
// }

// int mai(int n,int m){
//     return binpow(n,m-2)%m;
// }
/*
vector<bool> isprime(N,1);
vector<int> lp(N,0),hp(N,0);
     isprime[0] = isprime[1]= false;
for(int i=2;i<N;++i){
    if(isprime[i]==true){
        lp[i]=hp[i]=i;
        for(int j=2*i;j<N;j+=i){
            isprime[j]=false;
            hp[j]=i;
            if(lp[j]==0){
                lp[j]=i;
            }
        }
    }
    long long binpow(long long a, long long b) {
    if (b == 0)
        return 1;
    long long res = binpow(a, b / 2);
    if (b % 2)
        return res * res * a;
    else
        return res * res;
}
}
*/
// string getSubSeq(string s, int n)
// {
//     string res = "";
//     int cr = 0;
//     while (cr < n) {
//         char mx = s[cr];
//         for (int i = cr + 1; i < n; i++)
//             mx = max(mx, s[i]);
//         int lst = cr;
//         for (int i = cr; i < n; i++)
//             if (s[i] == mx) {
//                 res += s[i];
//                 lst = i;
//             }
//         cr = lst + 1;
//     }
//     return res;
// }

vector<pair<int,int>> dir={{1,0},{-1,0},{0,1},{0,-1}};
int cu=0;
int dfs(int i,int j,int n,int m,vector<vector<char>> &v,map<pair<int,int>,bool> &m1){
    m1[{i,j}]=true;
    cu++;
    fo(k,0,4){
        int x=i+dir[k].first;
        int y=j+dir[k].second;
        if(x<0||y<0||x>=n||y>=m) continue;
        if(m1[{x,y}]==true||v[x][y]=='.') continue;
        dfs(x,y,n,m,v,m1);
    }
    return cu;
}

signed main() {

	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
cout<<setprecision(9);

int t;
cin>>t;
fo(i,0,t){
int n,m;
cin>>n>>m;
vector<vector<char>> v(n);
map<pair<int,int>,bool> m1;
int a=0;
int b=0;
fo(j,0,n){
    fo(k,0,m){
        char x;
        cin>>x;
        v[j].pb(x);
        if(x=='#'){
            a=j;
            b=k;
        }
    }
}
int d=0;
fo(j,0,n){
    vector<char> v2;
    fo(k,0,m){
        v2.pb(v[j][k]);
        v[j][k]='#';
    }
    dfs(j,0,n,m,v,m1);
    d=max(d,cu);
    cu=0;
    m1.clear();
    fo(l,0,m){
        v[j][l]=v2[l];
    }
}
fo(j,0,m){
    vector<char> v2;
    fo(k,0,n){
        v2.pb(v[k][j]);
        v[k][j]='#';
    }
    dfs(0,j,n,m,v,m1);
    d=max(d,cu);
    cu=0;
    m1.clear();
    fo(l,0,n){
        v[l][j]=v2[l];
    }
}

cout<<d<<endl;

}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
.#.
###
.#.
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

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

using namespace std;

void solve(int t,bool zeb)
{
    int n,m;
    cin >> n >> m;
    vector<vector<int>> inp(n,vector<int>(m,0));
    char temp1;
    int count = 0;
    for(int i  = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < m ; j++)
        {
            cin >> temp1;
            if(t == 9996 && zeb ==1)
            {
                cout << temp1 ;
            }
            if(temp1 == ' ')
            {
                j--;
            }
            else if(temp1 == '#'){
            if(count == 0)
            {
                count = 2;
                inp[i][j] = 2;
            }
            else{
            inp[i][j] = 1;}
            }
        }
    }
    map<int,int> mp;
    map<int,int> mpcount;
    mp[2] = 2;
    bool k = 0;
    for(int i  = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < m ; j++)
        {
            if(inp[i][j] == 1)
            {
                k = 0;
                if(j-1 >=0 && inp[i][j-1] != 0 )
                {
                    inp[i][j] = inp[i][j-1];
                    k = 1;
                }
                if(i-1 >=0 && inp[i-1][j] != 0)
                {
                    if(!k)
                    {
                        inp[i][j] = inp[i-1][j];
                        k = 1;
                    }
                    else{
                    mp[inp[i][j-1]] = min(mp[inp[i][j-1]] , mp[inp[i-1][j]]);
                    mp[inp[i-1][j]] = mp[inp[i][j-1]];
                    }
                }
                if(!k)
                {
                    count++;
                    inp[i][j] = count;
                    mp[count] = count;
                }
            }
        }
    }
    for(int i  = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < m ; j++)
        {
            if(inp[i][j] != 0)
            {
                inp[i][j] = mp[inp[i][j]];
                mpcount[inp[i][j]]++;
            }
        }
    }
    int temp = 0;
    int maxsum = 0;
    set<int>st;
    for(int i = 0 ; i < n; i++)
    {
        st.clear();
        temp = 0;
        for(int j = 0 ; j < m ; j++)
        {
            if(inp[i][j] == 0)
            {
                temp++;
            }
            else if(inp[i][j] != 0)
            {
                if(st.count(inp[i][j]) == 0)
                {
                    temp = temp+mpcount[inp[i][j]];
                    st.insert(inp[i][j]);
                }
            }
            if(i-1 >= 0)
            {
                if(inp[i-1][j] != 0)
                {
                    if(st.count(inp[i-1][j]) == 0)
                    {
                        temp = temp+mpcount[inp[i-1][j]];
                        st.insert(inp[i-1][j]);
                    }
                }
            }
            if(i+1 <= n-1)
            {
                if(inp[i+1][j] != 0)
                {
                    if(st.count(inp[i+1][j]) == 0)
                    {
                        temp = temp+mpcount[inp[i+1][j]];
                        st.insert(inp[i+1][j]);
                    }
                }
            }
        }
        maxsum = max(maxsum,temp);
    }
    for(int j = 0 ; j < m; j++)
    {
        st.clear();
        temp = 0;
        for(int i = 0 ; i < n ; i++)
        {
            if(inp[i][j] == 0)
            {
                temp++;
            }
            else if(inp[i][j] != 0)
            {
                if(st.count(inp[i][j]) == 0)
                {
                    temp = temp+mpcount[inp[i][j]];
                    st.insert(inp[i][j]);
                }
            }
            if(j-1 >= 0)
            {
                if(inp[i][j-1] != 0)
                {
                    if(st.count(inp[i][j-1]) == 0)
                    {
                        temp = temp+mpcount[inp[i][j-1]];
                        st.insert(inp[i][j-1]);
                    }
                }
            }
            if(j+1 <= m-1)
            {
                if(inp[i][j+1] != 0)
                {
                    if(st.count(inp[i][j+1]) == 0)
                    {
                        temp = temp+mpcount[inp[i][j+1]];
                        st.insert(inp[i][j+1]);
                    }
                }
            }
        }
        maxsum = max(maxsum,temp);
    }
    cout << maxsum << endl;
}
int main()
{
    int t;
    cin >> t ;
    bool zeb = 0;
    if(t == 10000)
    {
        zeb = 1;
    }
    while(t--)
    {
        solve(t,zeb);
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test cases with very small or very large grids.
2. **Complex Connectivity**: Test cases with complex connected components that might cause issues with the algorithm's logic.
3. **Optimal Operations**: Test cases where the optimal operation (row or column selection) is not straightforward.

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug is as follows:

### Test Input
```plaintext
3
3 3
...
...
...
3 3
#..
#..
#..
3 3
#..
#..
##.
```

### Explanation
1. **First Test Case**:
   - A 3x3 grid with all '.' characters.
   - The optimal operation is to set all cells in any row or column to '#'. The result should be 9.
   - This tests the case where there are no existing '#' cells.

2. **Second Test Case**:
   - A 3x3 grid with a vertical line of '#' characters.
   - The optimal operation is to set all cells in any column to '#'. The result should be 9.
   - This tests the case where there is a vertical connected component.

3. **Third Test Case**:
   - A 3x3 grid with a vertical line of '#' characters and a single additional '#' in the last row.
   - The optimal operation is to set all cells in the last column to '#'. The result should be 7.
   - This tests the case where there is a combination of a vertical connected component and a single additional '#'.

### Expected Output
```plaintext
9
9
7
```

This test input should help expose any issues in the buggy code, particularly in how it handles the connectivity and the optimal operation for maximizing the largest connected component.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

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

using namespace std;

#define endl '\n'
#define int long long

const int N = 1e6 + 5;
int tc = 0;

vector<string> mat;
int sz[N];
int n, m;

pair<int, int> moves[4] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

bool valid(int i, int j) {
    return (i >= 0 && i < n && j >= 0 && j < m && mat[i][j] == '#');
}

bool valid2(int i, int j) {
    return (i >= 0 && i < n && j >= 0 && j < m && mat[i][j] != '#' && mat[i][j] != '.');
}

void dfs(int i, int j, int c) {
    sz[c]++;
    mat[i][j] = (char)('0' + c);
    for (auto [x, y]: moves) {
        if (valid(i + x, j + y)) dfs(i + x, j + y, c);
    }
}

void add_cor(int i, int j, map<int, int>& mp, int& k) {
    int cor = mat[i][j] - '0';
    assert(cor >= 0);
    if (!mp[cor]) {
        mp[cor]++;
        k += sz[cor];
    }
}

void solve () {
    tc++;
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        string s; cin >> s;
        mat.push_back(s);
    }

    int c = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (mat[i][j] == '#') {
                dfs(i, j, c++);
            }
        }
    }

    int resp = max(n, m);
    for (int i = 0; i < n; i++) {
        map<int, int> mp;
        int k = 0;
        int menos = 0;
        for (int j = 0; j < m; j++) {
            if (valid2(i + 1, j)) add_cor(i + 1, j, mp, k);
            if (valid2(i - 1, j)) add_cor(i - 1, j, mp, k);
            if (valid2(i, j)) {
                menos++;
                add_cor(i, j, mp, k);
            }
        }

        resp = max(resp, k + (m - menos));
    }

    for (int j = 0; j < m; j++) {
        map<int, int> mp;
        int k = 0;
        int menos = 0;
        for (int i = 0; i < n; i++) {
            if (valid2(i, j + 1)) add_cor(i, j + 1, mp, k);
            if (valid2(i, j - 1)) add_cor(i, j - 1, mp, k);
            if (valid2(i, j)) {
                menos++;
                add_cor(i, j, mp, k);
            }
        }

        resp = max(resp, k + (n - menos));
    }

    cout << resp << endl;

    for (int i = 0;  i <= c; i++) sz[i] = 0;
    mat.clear();
}

signed main() {
    ios_base::sync_with_stdio(0);cin.tie(0);
    int TC = 1;
    if (TC){
        cin >> TC;
        while (TC--) solve();
    } else {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Identify Potential Bugs:**
   - The code uses a depth-first search (DFS) to find connected components and then attempts to maximize the largest component by changing a row or column.
   - The main potential issues could be in how it handles the merging of components when a row or column is changed, especially in edge cases.

2. **Edge Cases to Consider:**
   - A grid with a single row or column.
   - A grid with a single '#' cell.
   - A grid where changing any row or column does not significantly increase the largest component.
   - A grid where the optimal solution involves merging multiple small components into a large one.

3. **Construct the Test Case:**
   - We will create a grid where the optimal solution involves merging multiple small components into a large one by changing a row or column.

Let's construct a small-scale test input that fits these criteria:

```plaintext
1
3 4
#..#
..#.
#..#
```

**Explanation:**
- The grid is 3x4.
- The initial grid has several small components of '#' cells.
- By changing the first column to all '#', we can merge the small components into a larger one.

**Expected Output:**
- The expected output for this test case should be `12` because changing the first column will result in a single connected component of size 12.

This test case should expose any issues in the code's handling of merging components and calculating the maximum possible size of the largest connected component.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

Buggy Code:
def dfs(i , j , grid , n , m , l):
    st = [(i , j)]
    cnt = 0
    grid[i][j] = l
    while len(st)>0:
        ci , cj = st.pop()
        cnt+=1
        neigh = [(ci+1 , cj) , (ci-1 , cj) , (ci , cj+1) , (ci , cj-1)]
        for ni , nj in neigh:
            if 0<=ni<n and 0<=nj<m and grid[ni][nj] == '#':
                st.append((ni , nj))
                grid[ni][nj] = l

    return cnt

# def dfs(i , j , grid , n , m , l):
#     global fck
#     neigh = [(i+1 , j) , (i-1 , j) , (i , j+1) , (i , j-1)]
#     cnt = 1
#     grid[i][j] = l
#     for ni , nj in neigh:
#         if 0<=ni<n and 0<=nj<m and grid[ni][nj] == '#':
#             cnt += dfs(ni , nj , grid , n , m , l)

#     return cnt

for _ in range(int(input())):
    n , m = list(map(int , input().split()))

    grid = []
    for i in range(n):
        grid.append(list(input()))

    mp = dict()

    l = 1
    for i in range(n):
        for j in range(m):
            if grid[i][j] == '#':
                cnt = dfs(i , j , grid , n , m , l)
                mp[l] = cnt
                l+=1

    # for i in grid:
    #     for j in i:
    #         print(j , end=' ')
    #     print()
    # print(mp)

    result = 0
    for i in range(n):
        sm = 0
        visited = set()
        for j in range(m):
            if i!=0 and grid[i-1][j] != '.':
                l = grid[i-1][j]
                if l not in visited:
                    sm += mp[l]
                    visited.add(l)

            if i!=n-1 and grid[i+1][j]!='.':
                l = grid[i+1][j]
                if l not in visited:
                    sm += mp[l]
                    visited.add(l)

            if grid[i][j] != '.':
                l = grid[i][j]
                if l not in visited:
                    sm += mp[l]
                    visited.add(l)

            else:
                sm+=1

        result = max(result , sm)
        if result == n*m:
            break

    for j in range(m):
        sm = 0
        visited = set()
        for i in range(n):
            if j!=0 and grid[i][j-1] != '.':
                l = grid[i][j-1]
                if l not in visited:
                    sm += mp[l]
                    visited.add(l)

            if j!=m-1 and grid[i][j+1]!='.':
                l = grid[i][j+1]
                if l not in visited:
                    sm += mp[l]
                    visited.add(l)

            if grid[i][j] != '.':
                l = grid[i][j]
                if l not in visited:
                    sm += mp[l]
                    visited.add(l)

            else:
                sm+=1

        result = max(result , sm)
        if result == n*m:
            break

    print(result)

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

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

Think step by step.

Response:
```plaintext
1
3 3
#.#
#..
#.#
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

Buggy Code:
import sys
from collections import defaultdict, OrderedDict, deque, Counter
import bisect
from itertools import accumulate, groupby
from functools import lru_cache
import math
import heapq
from functools import reduce

def parse_int(line):
    return int(line.strip())

def parse_str(line):
    return line.strip()

def parse_arr(line):
    return list(map(int, line.strip().split()))

def parse_str_arr(line):
    return list(line.strip())

def solution(n, m, arr):

    def nei(x, y):
        for dx, dy in [(0, 1), (0, -1), (-1, 0), (1, 0)]:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < m and arr[nx][ny] == '#':
                yield nx, ny

    def dfs(x, y):
        nonlocal r_min, r_max, c_min, c_max
        r_min = min(r_min, x)
        r_max = max(r_max, x)
        c_min = min(c_min, y)
        c_max = max(c_max, y)
        for nx, ny in nei(x, y):
            if (nx, ny) not in vis:
                vis.add((nx, ny))
                dfs(nx, ny)

    R, C = [0] * (n + 1), [0] * (m + 1)

    vis = set()
    for i in range(n):
        for j in range(m):
            if arr[i][j] == '#' and (i, j) not in vis:
                r_min, r_max = i, i
                c_min, c_max = j, j
                prev_size = len(vis)
                vis.add((i, j))
                dfs(i, j)
                size = len(vis) - prev_size
                rl, rr = max(0, r_min-1), min(n-1, r_max+1)
                cl, cr = max(0, c_min-1), min(m-1, c_max+1)
                R[rl] += size
                R[rr+1] -= size
                C[cl] += size
                C[cr+1] -= size
    R_acc = []
    for x in R:
        if not R_acc:
            R_acc.append(x)
        else:
            R_acc.append(R_acc[-1] + x)

    C_acc = []
    for x in C:
        if not C_acc:
            C_acc.append(x)
        else:
            C_acc.append(C_acc[-1] + x)

    ans = 0
    for i in range(n):
       ans = max(ans, arr[i].count('.') + R_acc[i])

    for j, col in enumerate(zip(*arr)):
        ans = max(ans, col.count('.') + C_acc[j])
    return ans

n = m = 0
arr = []
for i, line in enumerate(sys.stdin):
    if i == 0: continue
    if not n:
        n, m = parse_arr(line)
    else:
        arr.append(parse_str(line))
        if len(arr) == n:
            sys.stdout.write(str(solution(n, m, arr)) + '\n')
            n = m = 0
            arr.clear()

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

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

Think step by step.

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

1. **Edge Cases**:
   - **Single Row or Column**: This can help test the edge case where the grid is very narrow.
   - **All Cells are '#':** This can test the scenario where the grid is already filled with '#'.
   - **All Cells are '.':** This can test the scenario where the grid is completely empty.
   - **Mixed Grid**: This can test the scenario where the grid has a mix of '#' and '.' cells.

2. **Complex Scenarios**:
   - **Disconnected Components**: This can test the scenario where there are multiple disconnected components of '#' cells.
   - **Large Connected Component**: This can test the scenario where there is a large connected component that can be expanded by setting a single row or column.

3. **Challenging Input**:
   - We need an input that can challenge the algorithm to correctly identify the maximum possible size of a connected component after performing the operation at most once.

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

```plaintext
3
3 3
#..
..#
#..
3 3
...
...
...
3 3
#..
#..
#..
```

This input includes:
1. A grid with a small connected component of '#' cells.
2. A completely empty grid.
3. A grid with a large connected component of '#' cells.

This should help expose any issues in the buggy code by testing various edge cases and complex scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

Buggy Code:
def dfs(grid, visited, x, y):
    stack = [(x, y)]
    count = 0
    while stack:
        cx, cy = stack.pop()
        if visited[cx][cy]:
            continue
        visited[cx][cy] = True
        count += 1
        for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
            nx, ny = cx + dx, cy + dy
            if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and not visited[nx][ny] and grid[nx][ny] == '#':
                stack.append((nx, ny))
    return count

def largestCom(grid):
    n, m = len(grid), len(grid[0])
    visited = [[False] * m for _ in range(n)]
    maxSize = 0
    for i in range(n):
        for j in range(m):
            if grid[i][j] == '#' and not visited[i][j]:
                maxSize = max(maxSize, dfs(grid, visited, i, j))
    return maxSize

def solve(grid):
    n, m = len(grid), len(grid[0])
    maxSize = largestCom(grid)

    for i in range(n):
        newGrid = [row[:] for row in grid]
        for j in range(m):
            newGrid[i][j] = '#'
        maxSize = max(maxSize, largestCom(newGrid))

    for j in range(m):
        newGrid = [row[:] for row in grid]
        for i in range(n):
            newGrid[i][j] = '#'
        maxSize = max(maxSize, largestCom(newGrid))

    return maxSize

t = int(input())
ans = []
for _ in range(t):
    n, m = map(int, input().split())
    grid = [list(input()) for _ in range(n)]
    ans.append(solve(grid))

for x in ans:
    print(x)

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

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

Think step by step.

Response:
```plaintext
3
1 1
.
2 2
..
..
3 3
.#.
.#.
.#.
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) or any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ or column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
6
9
11
15
30

Note: In the second test case, it is optimal for Alex to set all cells in column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $6$.

In the third test case, it is optimal for Alex to set all cells in row $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $9$.

In the fourth test case, it is optimal for Alex to set all cells in row $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $11$.

Buggy Code:
import java.util.*;
import java.util.Stack;
public class Main
{

    public static int findBo(int x) {
        for(int i=2;i<=Math.round(Math.sqrt(x));i++) {
            if(x%i==0) return i;
        }
        return x;
    }

    public static long  gcd(long  a,long  b){
        return b == 0 ? a : gcd(b, a % b);
    }

    public static int []c={0,0,-1,1};
    public static int []d={-1,1,0,0};

    public static void po(int x,int y,String[]s, int[][]a,int n,int m,int k,int []b) {
        if(x<0 || y<0 || x>=n || y>=m || s[x].charAt(y)=='.' || a[x][y]!=0 ) return;
        a[x][y]=k;
        b[k]++;
        for(int i=0;i<4;i++) po(x+c[i],y+d[i],s,a,n,m,k,b);
    }

	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
	    int t=sc.nextInt();
	    while(t>0) {
	        t--;
	        int n=sc.nextInt(),m=sc.nextInt(),k=1,ans=0;
	        String s[]=new String[n+1];
	        for(int i=0;i<n;i++) {
	            s[i]=sc.next();
	        }
	        int b[]=new int[1000000];
	        int a[][]=new int[n+1][m+1];
	        for(int i=0;i<n;i++) {
	            for(int j=0;j<m;j++) {
	                if( a[i][j]==0 && s[i].charAt(j)=='#') {
	                    po(i,j,s,a,n,m,k,b);
	                    k++;
	                }
	            }
	        }

	        for(int i=0;i<n;i++) {
	            k=0;
	            int []c=new int[100000];
	            for(int j=0;j<m;j++) {
	                if(s[i].charAt(j)=='.') k++;
	                else if(c[a[i][j]]==0) {
	                    c[a[i][j]]=1;
	                    k+=b[a[i][j]];
	                }
	                if(i>0) {
	                   if(c[a[i-1][j]]==0){
	                       c[a[i-1][j]]=1;
	                       k+=b[a[i-1][j]];
	                   }
	                }
	                if(i!=n-1) {
	                    if(c[a[i+1][j]]==0){
	                       c[a[i+1][j]]=1;
	                       k+=b[a[i+1][j]];
	                   }
	                }
	            }
	            ans=Math.max(ans,k);
	        }

	        for(int j=0;j<m;j++) {
	            k=0;
	            int []c=new int[100000];
	            for(int i=0;i<n;i++) {
	                if(s[i].charAt(j)=='.') k++;
	                else if(c[a[i][j]]==0) {
	                    c[a[i][j]]=1;
	                    k+=b[a[i][j]];
	                }
	                if(j>0) {
	                   if(c[a[i][j-1]]==0){
	                       c[a[i][j-1]]=1;
	                       k+=b[a[i][j-1]];
	                   }
	                }
	                if(j!=m-1) {
	                    if(c[a[i][j+1]]==0){
	                       c[a[i][j+1]]=1;
	                       k+=b[a[i][j+1]];
	                   }
	                }
	            }
	            ans=Math.max(ans,k);
	        }

	        System.out.println(ans);
	        /*
	        int s=0,x1=0,y1=0,k=sc.nextInt(),n=sc.nextInt();
	        Long []a=new Long[n];
	        Long []b=new Long[n];
	        for(int i=0;i<n;i++) a[i]=sc.nextLong();
	        for(int i=0;i<n;i++) b[i]=sc.nextLong();
	        Long i=0L,j=Long.MAX_VALUE,m,ans=0L;

	        while(j>=i) {
	            m=(i+j)/2;
	            //System.out.print(m + " ");
	            Long val=0L;
	            for(int x=0;x<n;x++) {
	                Long p=b[x];
	                Long g=m/p;
	                if(g*a[x]>=k) { val=k*1L; break;}
	                val+=(g*a[x]);
	                if(m%p>0) val+=a[x];
	            }

	            //System.out.println(val);
	            if(val<0 || val>=k) {
	                ans=m;
	                j=m-1;
	            } else i=m+1;
	        }

	        System.out.println(ans);
	        /*
	        List<String> ans = new ArrayList<>();
	        for(int i=0;i<n;i++) ans.add(sc.next());
	        for(int i=0;i<n;i++) {
	            int k=0;
	            for(int j=0;j<m;j++) {
	                if(ans.get(i).charAt(j)=='#') k++;
	            }
	            if(k>s){
	                s=k;
	                x1=i+1;
	            }
	        }
	        s=0;
	        for(int i=0;i<m;i++) {
	            int k=0;
	            for(int j=0;j<n;j++) {
	                if(ans.get(j).charAt(i)=='#') k++;
	            }
	            if(k>s){
	                s=k;
	                y1=i+1;
	            }
	        }
	        System.out.println(x1 + " "+ y1);
	        /*
	        int a[]=new int[n];
	        for(int i=0;i<n;i++) a[i]=sc.nextInt();
	        HashMap<Integer,Integer>map_ = new HashMap<>();
	        long k=0,ans=0,p=0;
	        map_.put(0,1);
	        for(int i=0;i<n;i++) {
	            k+=a[i];
	            p=Math.max(p,a[i]);
	            if(k==p*2 ) ans++;

	        }*/
	       // int kk=0,ans=0,n=sc.nextInt();
	       // for(int i=2;i<=n;i++) {
	       //     int j=i,k=0;
	       //     while(n>=j) {
	       //         k+=j;
	       //         j+=i;
	       //     }
	       //     if(k>=ans) {
	       //         ans=k;
	       //         kk=i;
	       //     }
	       // }
	        // System.out.println(ans);
	       //String a=sc.next(),b=sc.next();
	       //System.out.println(""+b.charAt(0)+a.charAt(1)+a.charAt(2) + " " + a.charAt(0)+b.charAt(1)+b.charAt(2));

	       // long max=0,a=sc.nextLong(),b=sc.nextLong();
	       // for(int i=1;i<=n;i++) {
	       //     max+=(n-i);
	       // }
	       // if(b>max) {
	       //     System.out.println("NO");
	       // } else {

	       // }

	        /*
	        long n=sc.nextLong(),a=sc.nextLong(),b=sc.nextLong();
	        long ans=n*a,i=0,j=Math.min(n,b);
	        while(j>=i) {
	            long m=(i+j)/2,val=0;
	            long d=m-1;
	            val = ((n-m)*a) + (d>=0  ? (b*m) - (d*(d+1))/2 : 0 );
	            if(ans<=val) {
	                ans=val;
	                i=m+1;
	            } else j=m-1;
	        }
	        System.out.println(ans); */
	    }

	}
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
...
.#.
...
```

This test case is designed to expose potential issues in the buggy code. The grid is a 3x3 grid with a single '#' in the center. The optimal operation here would be to convert an entire row or column to '#', which would result in a connected component of size 4. This test case is small and simple, making it easy to verify the correctness of the solution, but it also has the potential to highlight issues with how the code handles the merging of components and the calculation of the maximum connected component size.