**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import sys
import heapq
from math import ceil, inf
from collections import defaultdict , Counter, deque
from bisect import bisect_left , bisect_right
import sys, threading
import random

mod = pow(10, 9) + 7
randint = random.randint(1, 1000000)

def ip():
  return int(input())

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

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

def solve():
  n , r = readList()
  seats = [[0] * 2 for _ in range(r)]
  arr = readList()
  i = 0
  row = 0
  ans = 0
  og = sum(arr)

  while i < n and row < r:
    while arr[i] > 0 and row < r:
      if arr[i] >= 2:
        seats[row][0] = 1
        seats[row][1] = 1
        row += 1
        arr[i] -= 2
        ans += 2
      else:
        seats[row][0] = 1
        arr[i] -= 1
        ans += 1
        row += 1

    if arr[i] == 0:
      i += 1

  row  = 0

  while i < n and row < r:
    while arr[i] > 0 and row < r:
      if 0 in seats[row]:
        seats[row][1] = 1
        ans -= 1
        arr[i] -= 1

      row += 1

    if row < r:
      i += 1

  sum_ = 0

  for row in seats:
    sum_ += sum(row)

  assert(sum_ == og)
  print(ans)

T = 1
T = int(input())
for _ in range(T):
 solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. The code seems to have issues with handling cases where families have very few members, and the bus has many rows. Let's construct a test case that highlights these issues.

### Step-by-Step Reasoning:

1. **Single Large Family and Many Rows:**
   - If there is a single large family with many members, the code should handle it by seating as many family members together as possible.
   - Example: `n = 1, r = 5, a = [10]`

2. **Multiple Small Families and Many Rows:**
   - If there are multiple small families and many rows, the code should ensure that each family member is seated with another family member or alone in a row.
   - Example: `n = 5, r = 10, a = [1, 1, 2, 2, 4]`

3. **Mixed Family Sizes and Rows:**
   - A mix of large and small families can test the code's ability to optimize seating.
   - Example: `n = 3, r = 5, a = [3, 1, 3]`

4. **Edge Case: Minimum and Maximum Values:**
   - Test the minimum and maximum values for `n`, `r`, and `a_i` to ensure the code handles these edge cases correctly.
   - Example: `n = 100, r = 500, a = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]`

### Challenging Test Input:
```plaintext
4
1 5
10
5 10
1 1 2 2 4
3 5
3 1 3
100 500
1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10
```

This test input includes a variety of scenarios that should help expose any issues 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: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
t=int(input())
for _ in range(t):
    n,r=map(int,input().split())
    a=list(map(int,input().split()))
    count=0
    l=[]
    odd=0
    for i in range(n):
        if a[i]%2==0:
            count+=a[i]
            r=r-(a[i]//2)
        else:
            count+=(a[i]-1)
            r=r-(a[i]//2)
            odd+=1

    min_r=odd//2
    max_r=odd
    while r>0 and odd>0:
        if r==1:
            break
        else:
            r=r-2
            count+=2
            odd-=2

    print(count)

**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 where the code's logic might fail. The main issues in the buggy code are:

1. It does not handle the case where there are more odd-sized families than available rows correctly.
2. It does not properly account for the remaining rows when there are odd-sized families.

Let's consider a scenario where the code might fail:

- We have a small number of families with odd sizes.
- The number of rows is just enough to seat all the family members, but the code might incorrectly count the number of happy people.

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

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

### Explanation:
- **Number of families (n):** 3
- **Number of rows (r):** 3
- **Family sizes (a):** [1, 1, 4]

### Expected Output:
- The total number of family members is 1 + 1 + 4 = 6, which fits within the 6 seats available (3 rows * 2 seats per row).
- The optimal seating arrangement should be:
  - Row 1: 1 (alone, happy)
  - Row 2: 1 (alone, happy)
  - Row 3: 4 (2 members in one row, 2 members in another row, all 4 are happy)

Thus, the maximum number of happy people is 6.

### Why This Test Case is Challenging:
- The code might incorrectly handle the odd-sized families (1, 1) and the even-sized family (4).
- The code might fail to properly account for the remaining rows when there are multiple odd-sized families.

This test case should help identify the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

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

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int r = sc. nextInt();
            int[] arr = new int[n];
            int c = 0;
            int s = 0;
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
                c=c+(arr[i]/2);
                r-=(arr[i]/2);
                arr[i] = arr[i] % 2;
                // System.out.print(arr[i]+" ");
                s = s + arr[i];
                // System.out.print(c+" ");
            }
            // System.out.println(s);
            // int x = r-c;
            // System.out.print(c+" "+r+" "+x+" ");
            c = 2*c + r;
            if(s==0){
                System.out.println(c);
            }
            else{
                System.out.println(c-1);
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on scenarios where the code's logic might fail. The buggy code has several issues, but the primary one is how it handles the distribution of family members into rows, especially when there are odd numbers of family members and when the bus has a limited number of rows.

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

1. **Identify the Bug**: The code incorrectly calculates the number of happy people by not properly accounting for the distribution of family members into rows, especially when there are odd numbers of family members.

2. **Edge Cases**:
   - **Single Family with Odd Members**: This can expose issues in how the code handles the last row.
   - **Multiple Families with Odd Members**: This can expose issues in how the code distributes the remaining family members.
   - **Full Bus with Limited Rows**: This can expose issues in how the code handles the distribution when the bus is nearly full.

3. **Test Input**:
   - **Test Case 1**: Single family with an odd number of members.
   - **Test Case 2**: Multiple families with odd numbers of members.
   - **Test Case 3**: Full bus with limited rows.

Let's create a test input that includes these edge cases:

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

### Explanation:
1. **Test Case 1**:
   - **Input**: `1 3 5`
   - **Output**: `4`
   - **Explanation**: The first family has 5 members. The optimal arrangement is to place 2 members in the first row, 2 members in the second row, and 1 member alone in the third row. This results in 4 happy people.

2. **Test Case 2**:
   - **Input**: `3 3 3 3 3`
   - **Output**: `6`
   - **Explanation**: There are three families, each with 3 members. The optimal arrangement is to place 2 members from each family in the first three rows, with 1 member from each family left out. This results in 6 happy people.

3. **Test Case 3**:
   - **Input**: `4 5 3 3 2 2`
   - **Output**: `6`
   - **Explanation**: There are four families with 3, 3, 2, and 2 members respectively. The optimal arrangement is to place 2 members from the first family in the first row, 2 members from the second family in the second row, 2 members from the third family in the third row, and 2 members from the fourth family in the fourth row. The fifth row will have 1 member from the first family and 1 member from the second family. This results in 6 happy people.

This test input should effectively expose the bug in the given buggy code by challenging its logic in handling odd numbers of family members and limited bus rows.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import java.util.*;
public class A {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0)
        {
            int n = sc.nextInt();
            int cols = sc.nextInt();
            int arr[] = new int[n];
            for(int i = 0;i < n;i++) arr[i] = sc.nextInt();
            int seated[] = new int[cols];
            int rem = 0;
            int happy = 0;
            int st = 0;
            for(int i = 0;i < n;i++)
            {
                if(arr[i] % 2 == 1)
                {
                    arr[i]--;
                    rem++;
                }
                int times = arr[i]/2;
                happy += times * 2;
                while(times --> 0)
                {
                    seated[st++] = 2;
                }

            }

            //System.out.println(Arrays.toString(seated));

            //System.out.println(rem);
            Arrays.sort(seated);
            for(int i = 0;i < n;i++)
            {
                if(seated[i] == 2) break;
                if(rem == 0) break;
                seated[i]++;
                rem--;

            }
            //System.out.println(Arrays.toString(seated));
            Arrays.sort(seated);
            for(int i = 0;i < n;i++)
            {
                if(seated[i] == 2) break;
                if(rem == 0) break;
                seated[i]++;
                rem--;

            }
            for(int i = 0; i < n;i++)
            {
                if(seated[i] == 1) happy++;
            }
            System.out.println(happy);

        }
    }
}

**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 2
1 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
void FAST () { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); }
void remove_consecutive_duplicates (vector<int> &v) { v.erase(unique(v.begin() , v.end()) , v.end()); }
int main () {
    FAST();

    int t;
    cin >> t;
    while (t--) {
        int n , r;
        cin >> n >> r;
        int a[n];
        for (int i = 0; i < n; i++) cin >> a[i];
        int not_happy = 0;
        int busy = 0;
        for (auto i : a) {
            if (i & 1) not_happy++;
            busy += (i / 2);
        }
        int rest = r - busy;

        // cout << not_happy << ' ' << busy << ' ' << rest << '\n';
        ll sum = accumulate(a , a + n , 0LL);
        if (rest >= not_happy) {
            cout << sum << '\n';
        } else {
            cout << sum - (ll) ceil((not_happy * 1.0) / rest) << '\n';
        }
    }

}

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

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

Think step by step.

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

**Explanation:**
- This test case is designed to expose a potential issue in the buggy code.
- The input consists of 3 families with 1, 1, and 4 members respectively, and 3 rows in the bus.
- The correct optimal seating arrangement should maximize the number of happy people. In this case, the 4 members of the third family can be seated in two rows, and the two single members can each sit alone in the remaining row.
- The expected output is 6, as all 6 people can be happy.
- The buggy code incorrectly calculates the number of unhappy people and the rest of the rows, which can lead to an incorrect output. This test case should help identify the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
	#ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	#endif
	int t;
	cin>>t;
	while(t>0){
		int n,r;
		cin>>n>>r;
		vector<int>arr(n);
		for(int i=0;i<n;i++){
			cin>>arr[i];
		}
		int l=0;
		int happy=0;
		for(int i=0;i<n;i++){
			l=l+arr[i]%2;
			happy=happy+(arr[i]/2)*2;
			r=r-arr[i]/2;
		}

		if(l<r){
			cout<<happy+l<<endl;
		}
		else if(l/2<r){
			cout<<happy+l/2<<endl;
		}
		else{
			cout<<happy<<endl;
		}
		t--;
	}
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose the bug in the buggy code. Here's why:

1. **Input Description**:
   - There are 5 families, each with 1 member.
   - The bus has 5 rows.

2. **Expected Behavior**:
   - Each family member can sit alone in a row, making them happy.
   - Since there are 5 family members and 5 rows, all family members can be seated alone in their respective rows, making all 5 of them happy.

3. **Buggy Code Behavior**:
   - The buggy code calculates the number of happy people by first summing up the number of pairs that can be seated together and then adding the number of single people that can be seated alone.
   - In this case, each family has only one member, so there are no pairs to be seated together. The code should then consider each single person as potentially happy if they can sit alone.
   - However, the buggy code has a logic issue in handling the single members. It calculates `l` (the number of single members) and `r` (the number of remaining rows) but does not correctly handle the case where all single members can be seated alone.

4. **Expected Output**:
   - The correct output should be `5` because all 5 family members can be seated alone in the 5 rows, making them all happy.

5. **Buggy Code Output**:
   - The buggy code might output a value less than 5, indicating that it does not correctly account for the scenario where all single members can be seated alone.

This test case is small and valid, ensuring that it fits within the constraints and is easy to verify.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
t = input()
for i in range(int(t)):
    happy=0
    i= input()
    x=input()
    fam=i.split()
    ppl=x.split()
    for n in ppl:
        if int(n)%2 == 0:
            happy=happy + int(n)
        else:
            happy= happy + int(n)-1
    if int(fam[0])%2==0:
        happy = happy + int(fam[0])-int(happy/2)
    print(happy)

**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
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
t=int(input())
for _ in range(t):
    n,r=map(int,input().split())
    As=list(map(int,input().split()))
    As=sorted(As,key=lambda x: x%2)
    s=0
    x=0
    j=0

    for i in range(n):
        if As[i]%2==0:
            s+=As[i]
            r-=As[i]//2
        else:
            break

    As=sorted(As[i:],reverse=True)
    for i in range(len(As)):
        if As[i]==1:
            if r!=0:
                r-=1
                s+=1
                x+=1
        else:
            if r!=0:
                s+=min(As[i]//2,r)*2
                r-=min(As[i]//2,r)
                x+=1
    if r==0:
        s-=x//2
    print(s)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The buggy code has a few potential issues, such as how it handles odd and even family sizes, and how it distributes the remaining seats.

Here are the steps to generate a challenging test input:

1. **Small Number of Test Cases**: Since the output length is limited, we will use a small number of test cases.
2. **Edge Cases**: Include scenarios where families have both even and odd sizes, and where the number of rows is just enough to fit all family members.
3. **Specific Focus**: Test cases where the number of rows is critical, and where the distribution of family members can highlight the bug in the code.

Let's generate a test input:

1. **Test Case 1**: One family with an odd number of members and enough rows to fit them.
2. **Test Case 2**: Multiple families with a mix of even and odd members, and just enough rows to fit them.
3. **Test Case 3**: Multiple families with a mix of even and odd members, and slightly more rows than needed.

Here is the test input:

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

### Explanation:

1. **Test Case 1**:
   - **Input**: `1 3 3`
   - **Output**: `3`
   - **Explanation**: One family with 3 members. They can be seated in 2 rows, with one member alone in the third row. All 3 members are happy.

2. **Test Case 2**:
   - **Input**: `2 3 2 3`
   - **Output**: `6`
   - **Explanation**: Two families with 2 and 3 members. The 2 members of the first family can sit together in the first row, and the 3 members of the second family can sit in the second and third rows. All 5 members are happy.

3. **Test Case 3**:
   - **Input**: `4 5 1 1 2 2`
   - **Output**: `6`
   - **Explanation**: Four families with 1, 1, 2, and 2 members. The 2 members of the first family can sit together in the first row, the 2 members of the second family can sit together in the second row, and the remaining 2 members can sit together in the third row. All 6 members are happy.

This test input should help expose the bug in the provided buggy code by focusing on edge cases and critical scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import java.util.*;
import java.io.*;
public class Main
{
  public static void main(String[]args) throws IOException
  {
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw=new PrintWriter(System.out);
    StringTokenizer st=new StringTokenizer(br.readLine());
    int t=Integer.parseInt(st.nextToken());
    while(t-->0)
    {
      st=new StringTokenizer(br.readLine());
      int n=Integer.parseInt(st.nextToken());
      int r=Integer.parseInt(st.nextToken());
      int[]arr=new int[n];
      st=new StringTokenizer(br.readLine());
      for(int i=0;i<n;i++)
      arr[i]=Integer.parseInt(st.nextToken());
      long res=0;
      long rows=0;
      long extra=0;
      for(int x:arr)
      {
        res+=x/2*2;
        rows+=x/2;
        if(x%2==1)
        extra++;
      }
      if(r-rows>=extra)
      res+=extra;
      else
      res+=extra-rows;
      pw.println(res);

    }
    pw.close();
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import java.util.Scanner;

public class busSeating {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int T = input.nextInt();

        for (int t=0; t<T; t++) {
            int N = input.nextInt();
            int R = input.nextInt();

            int[] families = new int[N];

            for(int i= 0; i< N; i++) {
                families[i] = input.nextInt();
            }

            int seats = 2*R;
            int happy = 0;

            int filled = 0;
            int singles = 0;

            for(int i= 0;i<N; i++) {
                int size = families[i];
                int pairs = size / 2;
                happy = happy+ pairs * 2;
                filled = filled+ pairs * 2;

                if (size%2== 1) {
                    singles++;
                    filled++;
                }
            }

            int remainingSeats = seats - filled;
            if (remainingSeats > 0) {
                if(singles<remainingSeats){
                    happy = happy+ singles;
                }else if (singles>remainingSeats){
                    happy = happy+ remainingSeats;
                }

            }

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

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

using namespace std;

typedef long long int ll;

int main() {
    ll t;
    cin >> t;
    while (t--)
    {
        ll n, r;
        cin >> n >> r;
        vector<ll> a(n);
        for (ll i = 0; i < n; ++i)
        {
            cin >> a[i];
        }

        //LOGIC

        ll tm = 0;
        for (ll i = 0; i < n; ++i)
        {
            tm += a[i];
        }

        ll ts = 2 * r;
        ll es = ts - tm;

        ll pr = 0;
        for (ll i = 0; i < n; ++i)
        {
            pr += a[i] / 2;
        }

        cout << min(es, tm) + 2 * min(pr, (tm - es) / 2) << 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 given buggy code, we need to focus on edge cases that might cause the algorithm to fail. The buggy code calculates the number of happy people using a specific logic that might not handle all scenarios correctly. Let's break down the potential issues and create a test case accordingly.

### Potential Issues:
1. **Empty Rows Handling**: The code might not correctly handle the case where there are empty rows that can be used to seat single family members.
2. **Family Size Distribution**: The code might not handle the distribution of family sizes optimally, especially when there are large families and small families.
3. **Row Utilization**: The code might not correctly calculate the number of rows used by families, especially when families are split across multiple rows.

### Challenging Test Case:
We will create a test case with a mix of large and small families, and ensure that the total number of people is just enough to fill the bus, but with some families that can be split to maximize the number of happy people.

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

### Explanation:
- **Number of Test Cases**: 1
- **Number of Families (n)**: 4
- **Number of Rows (r)**: 5
- **Family Sizes**: 3, 1, 1, 3

### Expected Output:
- **Maximum Number of Happy People**: 6

### Reasoning:
- The first family (3 members) can be split into two rows: 2 members in one row and 1 member in another row. This gives us 2 happy people in the first row and 1 happy person in the second row.
- The second family (1 member) can sit alone in the third row, giving 1 happy person.
- The third family (1 member) can sit alone in the fourth row, giving 1 happy person.
- The fourth family (3 members) can be split into two rows: 2 members in one row and 1 member in another row. This gives us 2 happy people in the fifth row and 1 happy person in the fourth row.

Thus, the total number of happy people is 6.

This test case should expose any issues in the buggy code related to the distribution of family members and the calculation of happy people.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main(){
    int t; cin>>t;
    while(t--){
        int n,r; cin>>n>>r;
        int arr[n];
        for(int i=0; i<n; i++) cin>>arr[i];

        int sum=0, count=0;

        for(int i=0; i<n; i++) sum+=arr[i];

        for(int i=0; i<n; i++){
            if(arr[i]%2==0){
                count+=arr[i];
            }

            if(arr[i]%2==1){
                count+=arr[i]-1;
            }
        }
        int rem_guys = sum-count;
        int rem_seats = 2*r-count;

        if(rem_guys==rem_seats){
            cout<<count<<endl;
        }
        else{
            for(int i=2;i<=rem_seats; i+=2){
                count++;
                rem_seats-=2;
                rem_guys--;

                if(rem_seats==rem_guys){
                    cout<<count<<endl;
                    break;
                }
                else if(rem_guys==0){
                    cout<<count<<endl;
                    break;
                }
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
for _ in range(int(input())):
    n,r = map(int,input().split())
    family = list(map(int,input().split()))
    family.sort() ; family.reverse()
    good , bad , sad = r*2 , 0 , 0
    for count in family:
        if good >= count:
            if count %2 == 0: good-=count
            else: good -= count + 1 ; bad+=1
        else:
            x = count - good
            good = 0
            sad += (x*2)
    print((sum(family))-sad)

**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 where the code might fail to handle the seating arrangement optimally. The key points to focus on are:

1. **Large Families with Few Rows**: When there are large families and few rows, the code might incorrectly handle the distribution of family members.
2. **Small Families with Many Rows**: When there are many rows and small families, the code might not optimize the seating arrangement correctly.
3. **Mixed Family Sizes**: When there are families of varying sizes, the code might not handle the seating arrangement in a way that maximizes the number of happy people.

Given these points, let's create a test input that is small but challenging:

### Test Input
```plaintext
3
1 1
10
3 2
5 2 1
4 3
3 2 1 1
```

### Explanation
1. **Test Case 1**:
   - 1 family with 10 members and 1 row.
   - This test case is designed to check how the code handles a large family with very few rows. The code should recognize that it's impossible to seat all family members and should handle the overflow correctly.

2. **Test Case 2**:
   - 3 families with 5, 2, and 1 members, and 2 rows.
   - This test case is designed to check how the code handles a mix of large and small families with a limited number of rows. The code should optimize the seating to maximize the number of happy people.

3. **Test Case 3**:
   - 4 families with 3, 2, 1, and 1 members, and 3 rows.
   - This test case is designed to check how the code handles a mix of families of varying sizes with a moderate number of rows. The code should optimize the seating to maximize the number of happy people.

These test cases should help expose any issues in the code's logic for handling different family sizes and row constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import sys
#input = sys.stdin.readline

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

num = int(input())

for i in range(num):
    _ = inlt()
    n,r = _[0], _[1]
    fam = inlt()
    fam = list(enumerate(fam))
    fam = sorted(fam, key=lambda x : x[1], reverse = True)
    excess = [0 for i in range(n)]
    hm = {}
    tot = 0
    happy = 0
    r *= 2
    for i, j in fam:
        if tot >= r:
            break
        if j % 2:
            hm[i] = j + 1
        else:
            hm[i] = j

        happy += j
        tot += hm[i]

    for i in range(len(hm), len(fam)):
        idx, ppl = fam[i][0], fam[i][1]

        happy -= ppl

    print(happy)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import java.util.Scanner;

public class cf {
    public static void main(String[] args) {
        Scanner ak = new Scanner(System.in);
        int t = ak.nextInt();

        while (t-- > 0) {
            int n=ak.nextInt();
            int k=ak.nextInt();
            int a[]=new int [n];
            int b[]=new int [n];
            int count =0;
            int sum=0;
            int sum1=0;
            int diff;
            for (int i = 0; i < n; i++) {
                a[i]=ak.nextInt();
                sum+=a[i];
                if (a[i]==1)count++;
            }
            for (int i=0;i<n;i++) {
                b[i]=2*(a[i]/2);
                sum1+=b[i];
            }
            if (sum==k*2)System.out.println(sum1);
            else {
                diff=count>(k*2-sum-count)%2?(k*2-sum-count)-count:count;
                System.out.println(sum+diff);
            }

        }
    ak.close();

    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The buggy code has several potential problems, including incorrect handling of the number of happy people and incorrect logic for distributing family members.

### Analysis of the Buggy Code:
1. **Incorrect Calculation of `sum1`**: The code calculates `sum1` as `2 * (a[i] / 2)`, which is not always correct. For example, if a family has 3 members, `a[i] / 2` will be 1, and `2 * 1` will be 2, which is incorrect.
2. **Incorrect Handling of Single Members**: The code tries to handle single members with a complex `diff` calculation, which is not straightforward and may lead to incorrect results.

### Challenging Test Case:
To expose these issues, we can create a test case where:
- There are families with an odd number of members.
- The total number of family members is close to the bus capacity, but not exactly filling it.

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

### Explanation:
- **Number of Test Cases**: 1
- **Number of Families**: 4
- **Number of Rows**: 5
- **Family Members**: 3, 1, 1, 3

### Expected Output:
- The optimal seating arrangement should be:
  - Row 1: 3, 3 (2 happy)
  - Row 2: 3, 1 (2 happy)
  - Row 3: 1, 1 (2 happy)
  - Row 4: 3, 3 (2 happy)
  - Row 5: Empty (0 happy)

- Total happy people: 8

### Why This Test Case is Challenging:
- The families have an odd number of members, which should be handled correctly to maximize the number of happy people.
- The total number of family members (8) is less than the bus capacity (10), but the distribution of members across rows is critical to maximize happiness.

This test case should expose the bug in the incorrect calculation of `sum1` and the handling of single members.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
import javax.management.MBeanRegistration;
import java.awt.image.AreaAveragingScaleFilter;
import java.io.*;
import java.math.BigInteger;
import java.sql.Array;
import java.util.*;

// JAI SHREE RAM  JAI BAJRANGBALI//

public class CodeForces {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        FastReader fr = new FastReader();
        int nmax = (int) (1e5 + 7);
        int t = fr.nextInt();
        while (t-- > 0) {
            int n = fr.nextInt();
            int x =fr.nextInt();
              x*=2;
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = fr.nextInt();

            }
            int rem[] = new int[n];
            int cnt =0;
            for (int i = 0; i < n; i++) {
                cnt+=a[i]/2;
                rem[i] = a[i]%2;
            }
            cnt*=2;
            int seats = x-cnt;
            int remainder=0;
            for (int i = 0; i < n; i++) {
                if(rem[i]!=0)remainder++;
            }
//            System.out.println(seats + " " + remainder);
            if(cnt==0){
                cnt = remainder/2;
            }
           else if(seats>remainder){
                while(seats>2 && remainder>0){
                    seats -=2;
                    remainder--;
                    cnt++;
                }
            }

            System.out.println(cnt);

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

    public static class Pair {
        public long a;
        public long  b;
        public int hashCode;

        Pair(long a, long b) {
            this.a = a;
            this.b = b;
            this.hashCode = Objects.hash(a, b);
        }

        @Override
        public String toString() {
            return a + " -> " + b;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o)
                return true;
            if (o == null || getClass() != o.getClass())
                return false;
            Pair that = (Pair) o;
            return a == that.a && b == that.b;
        }

        @Override
        public int hashCode() {
            return this.hashCode;
        }
    }

}

class FastReader {
    BufferedReader br;
    StringTokenizer st;

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

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                String line = br.readLine();

                if (line == null) {
                    return null;
                }

                st = new StringTokenizer(line);

            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

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

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

    String nextLine() {
        String str = "";
        try {
            if (st.hasMoreTokens()) {
                str = st.nextToken("\n");
            } else {
                str = br.readLine();
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}

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

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

Think step by step.

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

1. **Single Family with an Odd Number of Members:**
   - This case can test how the code handles the situation where a single family has an odd number of members, and there are enough seats for all members.

2. **Multiple Families with Different Sizes:**
   - This case can test how the code handles multiple families with different sizes, ensuring that the code correctly pairs family members and places them in the bus.

3. **Exact Fit:**
   - This case can test how the code handles the situation where the total number of family members exactly fits the number of seats available.

4. **Large Number of Small Families:**
   - This case can test how the code handles a large number of families, each with a small number of members, ensuring that the code correctly maximizes the number of happy people.

5. **Small Number of Large Families:**
   - This case can test how the code handles a small number of families, each with a large number of members, ensuring that the code correctly pairs family members and places them in the bus.

Let's create a test input that combines these scenarios to maximize the failure probability in the faulty solution.

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

Explanation:
1. **Single Family with an Odd Number of Members:**
   - `1 3`
   - `5`
   - This tests how the code handles a single family with 5 members and 3 rows. The expected output is 5, as 3 family members can sit together in one row, and the remaining 2 can sit in separate rows.

2. **Multiple Families with Different Sizes:**
   - `3 3`
   - `2 3 1`
   - This tests how the code handles multiple families with different sizes and 3 rows. The expected output is 4, as 2 members of the first family can sit together, 2 members of the second family can sit together, and the remaining member of the second family can sit alone.

3. **Exact Fit:**
   - `5 5`
   - `1 2 3 4 5`
   - This tests how the code handles a situation where the total number of family members exactly fits the number of seats available. The expected output is 15, as all family members can be paired or sit alone.

4. **Small Number of Large Families:**
   - `4 5`
   - `3 1 1 3`
   - This tests how the code handles a small number of families, each with a large number of members, and 5 rows. The expected output is 6, as 3 members of the first family can sit together, 3 members of the fourth family can sit together, and the remaining 2 can sit alone.

This test input should expose any issues in the provided code and help identify the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Funny Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Vanya has a graph with $$$n$$$ vertices (numbered from $$$1$$$ to $$$n$$$) and an array $$$a$$$ of $$$n$$$ integers; initially, there are no edges in the graph. Vanya got bored, and to have fun, he decided to perform $$$n - 1$$$ operations.

Operation number $$$x$$$ (operations are numbered in order starting from $$$1$$$) is as follows:

- Choose $$$2$$$ different numbers $$$1 \leq u,v \leq n$$$, such that $$$|a_u - a_v|$$$ is divisible by $$$x$$$.
- Add an undirected edge between vertices $$$u$$$ and $$$v$$$ to the graph.

Help Vanya get a connected$$$^{\text{∗}}$$$ graph using the $$$n - 1$$$ operations, or determine that it is impossible.

Input: Each test consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^{3}$$$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains the number $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of vertices in the graph.

The second line of each test case contains $$$n$$$ numbers $$$a_1, a_2, \cdots a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$.

Output: For each test case, if there is no solution, then output "No" (without quotes).

Otherwise, output "Yes" (without quotes), and then output $$$n - 1$$$ lines, where in the $$$i$$$-th line, output the numbers $$$u$$$ and $$$v$$$ that need to be chosen for operation $$$i$$$.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
2
1 4
4
99 7 1 13
5
10 2 31 44 73
5
87 6 81 44 32
5
62 35 33 79 16
5
6 51 31 69 42
5
52 63 25 21 5
12
33 40 3 11 31 43 37 8 50 5 12 22
output:
YES
2 1
YES
4 1
2 1
3 2
YES
5 1
4 1
3 1
2 1
YES
4 1
3 1
2 1
5 4
YES
3 1
5 1
2 1
4 2
YES
4 1
5 1
2 1
3 2
YES
2 1
5 2
3 1
4 3
YES
9 1
12 9
11 1
10 1
6 1
7 6
2 1
8 2
5 2
3 1
4 1

Note: Let's consider the second test case.

- First operation $$$(x = 1)$$$: we can connect vertices $$$4$$$ and $$$1$$$, since $$$|a_4 - a_1| = |13 - 99| = |-86| = 86$$$, and $$$86$$$ is divisible by $$$1$$$.

- Second operation $$$(x = 2)$$$: we can connect vertices $$$2$$$ and $$$1$$$, since $$$|a_2 - a_1| = |7 - 99| = |-92| = 92$$$, and $$$92$$$ is divisible by $$$2$$$.

- Third operation $$$(x = 3)$$$: we can connect vertices $$$3$$$ and $$$2$$$, since $$$|a_3 - a_2| = |1 - 7| = |-6| = 6$$$, and $$$6$$$ is divisible by $$$3$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <string>
#include <numeric>
#include <iomanip>
#include <climits>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
#define endl "\n"
void bhaut_tez() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cout << setprecision(30);
}
ll parent[2005];
ll sz[2005];
void make(ll v){
    parent[v]=v;
    sz[v]=1;
}
ll find(ll v){
    if(parent[v]==v)return v;
    return parent[v]=find(parent[v]);
}
void Union(ll a,ll b){
    a=find(a);
    b=find(b);
    if(a!=b){
        if(sz[a]<sz[b])swap(a,b);
        parent[b]=a;
        sz[a]+=sz[b];
    }
}
int main() {
//   #ifndef online_judge
//       freopen("input.txt", "r", stdin);
//       freopen("output.txt", "w", stdout);
//       freopen("error.txt", "w", stderr);
//   #endif
    ll ttt;
    cin>>ttt;
    while(ttt--){
        ll n;
        cin>>n;
        ll a[n];
        for(ll i=0;i<n;i++){
            cin>>a[i];
            make(i);
        }
        vector<pair<ll,ll>> p;
        for(ll x=n-1;x>=1;x--){
            map<ll,vector<ll>> m;
            for(ll i=0;i<n;i++){
                m[a[i]%x].push_back(i);
            }
//            for(auto &i:m){
//                for(ll j=0;j<i.second.size();j++)cout<<i.second[j]<<" ";
//                cout<<endl;
//            }
            ll flag=0;
            for(auto &i:m){
                if(flag)break;
                for(ll j=1;j<i.second.size();j++){
                    if(find(i.second[0])==find(i.second[j]))continue;
                    Union(i.second[0],i.second[j]);
                    p.push_back({i.second[0],i.second[j]});
                    flag++;
                    break;
                }
            }
//            cout<<x<<" "<<p.size()<<endl;
        }
        cout<<"YES"<<endl;
        for(ll i=p.size()-1;i>=0;i--) cout<<p[i].first+1<<" "<<p[i].second+1<<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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The primary issues in the code are:

1. **Handling of Edge Cases**: The code might not handle cases where it is impossible to form a connected graph.
2. **Efficiency and Correctness**: The code might not correctly form the required edges to ensure the graph is connected.

Let's focus on a small-scale test case that can expose these issues. We will create a test case where it is impossible to form a connected graph using the given operations.

### Test Case Analysis
- **Input Size**: We will use a small number of vertices, say \( n = 5 \).
- **Array Values**: We will choose values for the array \( a \) such that it is impossible to find pairs \( (u, v) \) where \( |a_u - a_v| \) is divisible by \( x \) for all \( x \) from \( 1 \) to \( n-1 \).

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

### Explanation
- **Test Case 1**: \( n = 5 \)
- **Array \( a \)**: \( [1, 2, 3, 4, 5] \)

For this test case:
- **Operation 1 (x = 1)**: Any pair \( (u, v) \) will work since \( |a_u - a_v| \) is always divisible by 1.
- **Operation 2 (x = 2)**: We need to find pairs where \( |a_u - a_v| \) is divisible by 2. The possible pairs are \( (1, 3) \) and \( (2, 4) \).
- **Operation 3 (x = 3)**: We need to find pairs where \( |a_u - a_v| \) is divisible by 3. The possible pair is \( (1, 4) \).
- **Operation 4 (x = 4)**: We need to find pairs where \( |a_u - a_v| \) is divisible by 4. There are no such pairs.

Since there are no pairs for \( x = 4 \), the algorithm should output "No" for this test case. This will expose the bug in the code if it incorrectly outputs "Yes" or fails to handle the impossible case correctly.

### Expected Output
```plaintext
No
```

This test case should help in identifying the issue in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Funny Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Vanya has a graph with $$$n$$$ vertices (numbered from $$$1$$$ to $$$n$$$) and an array $$$a$$$ of $$$n$$$ integers; initially, there are no edges in the graph. Vanya got bored, and to have fun, he decided to perform $$$n - 1$$$ operations.

Operation number $$$x$$$ (operations are numbered in order starting from $$$1$$$) is as follows:

- Choose $$$2$$$ different numbers $$$1 \leq u,v \leq n$$$, such that $$$|a_u - a_v|$$$ is divisible by $$$x$$$.
- Add an undirected edge between vertices $$$u$$$ and $$$v$$$ to the graph.

Help Vanya get a connected$$$^{\text{∗}}$$$ graph using the $$$n - 1$$$ operations, or determine that it is impossible.

Input: Each test consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^{3}$$$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains the number $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of vertices in the graph.

The second line of each test case contains $$$n$$$ numbers $$$a_1, a_2, \cdots a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$.

Output: For each test case, if there is no solution, then output "No" (without quotes).

Otherwise, output "Yes" (without quotes), and then output $$$n - 1$$$ lines, where in the $$$i$$$-th line, output the numbers $$$u$$$ and $$$v$$$ that need to be chosen for operation $$$i$$$.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
2
1 4
4
99 7 1 13
5
10 2 31 44 73
5
87 6 81 44 32
5
62 35 33 79 16
5
6 51 31 69 42
5
52 63 25 21 5
12
33 40 3 11 31 43 37 8 50 5 12 22
output:
YES
2 1
YES
4 1
2 1
3 2
YES
5 1
4 1
3 1
2 1
YES
4 1
3 1
2 1
5 4
YES
3 1
5 1
2 1
4 2
YES
4 1
5 1
2 1
3 2
YES
2 1
5 2
3 1
4 3
YES
9 1
12 9
11 1
10 1
6 1
7 6
2 1
8 2
5 2
3 1
4 1

Note: Let's consider the second test case.

- First operation $$$(x = 1)$$$: we can connect vertices $$$4$$$ and $$$1$$$, since $$$|a_4 - a_1| = |13 - 99| = |-86| = 86$$$, and $$$86$$$ is divisible by $$$1$$$.

- Second operation $$$(x = 2)$$$: we can connect vertices $$$2$$$ and $$$1$$$, since $$$|a_2 - a_1| = |7 - 99| = |-92| = 92$$$, and $$$92$$$ is divisible by $$$2$$$.

- Third operation $$$(x = 3)$$$: we can connect vertices $$$3$$$ and $$$2$$$, since $$$|a_3 - a_2| = |1 - 7| = |-6| = 6$$$, and $$$6$$$ is divisible by $$$3$$$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<unordered_map>
#include<stack>
#include<math.h>
using namespace std;
#define int long long//
#define pii pair<int,int>
void solve()
{
	int n; cin >> n;
	vector<int>v(n + 1);
	vector<bool>vis(n + 1,false);
	vector<pii>ans;
	for (int i = 1; i <= n; ++i)cin >> v[i];
	for (int mod = n - 1; mod >= 1; --mod)
	{
		map<int, int>mp;
		for (int i = 1; i <= n; ++i)
		{
			if (vis[i])continue;
			if (mp.count(v[i] % mod))
			{
				vis[i] = 1;
				ans.push_back({ mp[v[i] % mod],i});
				break;
			}
			mp[v[i] % mod] = i;
		}
	}
	cout << "YES\n";
	for (int i = ans.size()-1; i>=0; --i)
	{
		cout << ans[i].first << ' ' << ans[i].second << '\n';
	}

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

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

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

Think step by step.

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