**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

Buggy Code:
#include <stdio.h>
#include <algorithm>

using namespace std;

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        int sum=0,sumj=0,sumo=0,j=0,o=0;
        int a[n];
        for (int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            sum+=a[i];
            if(i%2==0){
                sumo+=a[i];
                o++;
            } else{
                sumj+=a[i];
                j++;
            }
        }
        if(sum%n!=0||sumj%j!=0||sumo%o!=0||sumj/j!=sumo/o||sumj/j!=sum/n){
            printf("NO\n");
            continue;
        }

        printf("YES\n");
    }
    return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - Array: [1, 2, 1]
   - This array can be transformed to [1, 1, 1] by performing the operation on index 2 (decrease 2 by 1, increase 1 by 1). The expected output is "YES".

2. **Test Case 2:**
   - Array: [1, 3, 3, 1]
   - This array cannot be transformed to have all elements equal. The sum of the array is 8, and the average is 2. However, the elements at even indices (3 and 3) cannot be reduced to 2 without making the elements at odd indices negative. The expected output is "NO".

3. **Test Case 3:**
   - Array: [1, 2, 3, 2, 1]
   - This array can be transformed to [2, 2, 2, 2, 2] by performing multiple operations. The expected output is "YES".

These test cases are designed to expose the bug in the buggy code by testing different scenarios where the sum and the averages of even and odd indices are checked.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

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

using namespace std;

void solve(){
    ll n;cin>>n;
    ll sum=0;
    ll a[n];for(int i=0; i<n; i++)cin>>a[i],sum+=a[i];
    if(sum%n!=0){
        cout<<"NO\n";
        return;
    }
    ll b=0;
    sum/=n;
    for(int i=1; i<n-1; i++){
        b=a[i-1]+a[i+1];
        if(b<sum){
            cout<<"NO\n";
            return;
        }
        a[i-1]=sum;
        a[i+1]=b-sum;
    }
    for(auto i:a){
        if(i!=sum){
            cout<<"NO\n";
            return;
        }
    }
    cout<<"YES\n";
}

int main(){
    int t;cin>>t;
    for(int i=0; i<t; i++){
        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 buggy code, we need to consider edge cases and scenarios where the code might fail. The buggy code has a few issues:

1. **Incorrect Condition for Early Termination**: The code checks if the sum of the array is divisible by the length of the array, which is correct. However, it then tries to ensure that the sum of the elements at indices `i-1` and `i+1` is greater than or equal to the target value `sum` for each `i` from `1` to `n-2`. This condition is not sufficient to guarantee that the array can be made equal.

2. **Incorrect Update Mechanism**: The code updates the array elements in a way that might not always lead to a valid solution. Specifically, it sets `a[i-1]` to `sum` and adjusts `a[i+1]` accordingly, which might not preserve the non-negativity of the array elements.

To expose these issues, we can create a test case where the sum of the array is divisible by the length of the array, but the array cannot be made equal due to the constraints of the operations.

### Challenging Test Input
Let's consider an array where the sum is divisible by the length, but the elements cannot be made equal using the allowed operations. For example, an array where the sum of the elements at `i-1` and `i+1` is less than the target value `sum` for some `i`.

```plaintext
1
5
1 2 1 2 1
```

### Explanation
- **Array**: `[1, 2, 1, 2, 1]`
- **Sum**: `1 + 2 + 1 + 2 + 1 = 7`
- **Length**: `5`
- **Target Value**: `sum / n = 7 / 5 = 1.4` (which is not an integer, but the code will still try to make the elements equal)

In this case, the sum is not divisible by the length of the array, so the code should immediately return "NO". However, if we modify the array slightly to make the sum divisible by the length, we can create a more challenging test case:

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

### Explanation
- **Array**: `[1, 2, 3, 2, 1]`
- **Sum**: `1 + 2 + 3 + 2 + 1 = 9`
- **Length**: `5`
- **Target Value**: `sum / n = 9 / 5 = 1.8` (which is not an integer, but the code will still try to make the elements equal)

In this case, the sum is divisible by the length of the array, but the elements cannot be made equal using the allowed operations. The code will incorrectly try to update the array and might return "YES" even though it is not possible to make all elements equal.

### Final Test Input
```plaintext
1
5
1 2 3 2 1
```

This test input should expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

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

    int T;
    cin>>T;
    while(T--) {
        int N;
        cin>>N;
        vector<int> nums;

        for (int i=0; i<N; i++) {
            int a;
            cin>>a;
            nums.push_back(a);
        }

        long long int sum = accumulate(nums.begin(), nums.end(), 0);

        if (sum%N != 0) {
            puts("NO");
        }

        else {
            long long int required = sum/N;
            int odd_count = 0;
            int even_count = 0;
            long long int odd_sum = 0;
            long long int even_sum = 0;
            int flag = 0;

            for (int i=0; i<N; i++) {
                if (i%2==0) {
                    even_count++;
                    even_sum += nums[i];
                    if (i+2 < N && required - nums[i] > nums[i+2]) {
                        flag = 1;
                    }
                }
                else {
                    odd_count++;
                    odd_sum += nums[i];
                    if (i+2 < N && required - nums[i] > nums[i+2]) {
                        flag = 1;
                    }
                }
            }
            if (flag == 1) {
                puts("NO");
            }

            else if (odd_sum/odd_count == required && even_sum/even_count == required) {
                puts("YES");
            }
            else {
                puts("NO");
            }
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases or scenarios where the logic might fail. The main issues in the code are:

1. The code checks if the sum of the array is divisible by the length of the array. This is correct.
2. The code then checks if the required value (sum divided by length) can be achieved by ensuring that the sum of even-indexed elements and odd-indexed elements, when divided by their respective counts, equals the required value. This is where the logic might fail.

The code also has a flag to check if the required value minus the current value at an index is greater than the value at the next index, which might not be a sufficient condition to determine if the array can be made equal.

Let's create a test case that exposes this issue:

### Test Case:
- An array where the sum is divisible by the length, but the distribution of values makes it impossible to balance the array using the allowed operations.

### Example:
```plaintext
1
5
1 2 3 4 5
```

### Explanation:
- The sum of the array is `15`, and the length is `5`. The required value is `15 / 5 = 3`.
- The array is `[1, 2, 3, 4, 5]`.
- The sum of even-indexed elements (1, 3, 5) is `9`, and the sum of odd-indexed elements (2, 4) is `6`.
- The average of even-indexed elements is `9 / 3 = 3`, and the average of odd-indexed elements is `6 / 2 = 3`.
- However, it is impossible to make all elements equal to `3` using the allowed operations because the values are too far apart.

This test case should expose the bug in the code. The expected output is `NO`, but the buggy code might incorrectly output `YES`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

Buggy Code:
import sys
sys.setrecursionlimit(10**6)
input = lambda: sys.stdin.readline().rstrip()

testcase = int(input())
solution = []
for _ in range(testcase):
    l = int(input())
    arr = list(map(int, input().split()))
    sum_arr = sum(arr)
    k = sum(arr) // l
    check = True
    if sum_arr % l != 0:
        check = False
    for i in range(1, l-1):
        while arr[i-1] != k:
            if arr[i-1] > k:
                arr[i-1] -= 1
                arr[i+1] += 1
            else:
                arr[i - 1] += 1
                arr[i + 1] -= 1
            if arr[i+1] < 0 or arr[i-1] < 0:
                check = False
                break
    for i in range(1, l):
        if arr[i-1] != arr[i]:
            check = False

    if check:
        print("Yes")
    else:
        print("No")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

Buggy Code:
# T = int(input())
# for t in range(T):
#     n, m = list(map(int, input().split()))
#     x, cum_counter = 0, 0
#     for word in range(n):
#         curr_word = input()
#         cum_counter += len(curr_word)
#         if cum_counter > m:
#             continue
#         else:
#             x += 1
#     print(x)

def checkIfCanEqualize(arr):
    sum_of_arr = sum(arr)
    if sum_of_arr % len(arr) != 0:
        return False, -1
    target = sum_of_arr // len(arr)
    for i in range(len(arr)-1):
        diff = arr[i] - target
        arr[i+1] += diff
        if arr[i+1] < 0:
            return False, target
        arr[i] = target
    return arr[-1] == target, target

T = int(input())
for t in range(T):
    n = int(input())
    nums = list(map(int, input().split()))
    isEqualizedEven, target_even = checkIfCanEqualize(nums[::2])
    isEqualizedOdd, target_odd = checkIfCanEqualize(nums[1::2])
    if isEqualizedEven and isEqualizedOdd and target_even == target_odd:
        print("YES")
    else:
        print("NO")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

Buggy Code:
import java.util.Scanner;
public class Transfusion {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);

        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int NumOdd;int NumEven;

            if(n%2==0){
                NumOdd=n/2;
                NumEven=n/2;
            }
            else{
                NumOdd=(n/2)+1;
                NumEven=n/2;
            }

            int a[]=new int[n];
            int SumOdd=0;int SumEven=0;

            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
                if(i%2==0){
                    SumOdd+=a[i];
                }
                else{
                    SumEven+=a[i];
                }
            }
            int Sum=SumOdd+SumEven;
            if(Sum%n!=0){
                System.out.println("NO");
            }
            else{
                int k=Sum/n;
                if(SumOdd==(NumOdd)*k && SumEven==(NumEven)*k){
                    System.out.println("YES");
                }
                else{
                    System.out.println("NO");
                }
            }

        }
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Transfusion
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$. In one operation, you can pick an index $$$i$$$ from $$$2$$$ to $$$n-1$$$ inclusive, and do one of the following actions:

- Decrease $$$a_{i-1}$$$ by $$$1$$$, then increase $$$a_{i+1}$$$ by $$$1$$$.
- Decrease $$$a_{i+1}$$$ by $$$1$$$, then increase $$$a_{i-1}$$$ by $$$1$$$.

After each operation, all the values must be non-negative. Can you make all the elements equal after any number of operations?

Input: First line of input consists of one integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.

First line of each test case consists of one integer $$$n$$$ ($$$3 \le n \le 2\cdot 10^5$$$).

Second line of each test case consists of $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print "YES" without quotation marks if it is possible to make all the elements equal after any number of operations; otherwise, print "NO" without quotation marks.

You can print answers in any register: "yes", "YeS", "nO"  — will also be considered correct.

Examples:
input:
8
3
3 2 1
3
1 1 3
4
1 2 5 4
4
1 6 6 1
5
6 2 1 4 2
4
1 4 2 1
5
3 1 2 1 3
3
2 4 2
output:
YES
NO
YES
NO
YES
NO
NO
NO

Note:

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

public class B_Transfusion {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine().trim());
        long[] arr;
        for (int i = 0; i < t; i++) {
            String[] inputs = br.readLine().split(" ");
            long n = Long.parseLong(inputs[0]);
            inputs = br.readLine().split(" ");
            arr = new long[(int) n];
            for (int j = 0; j < n; j++) {
                arr[j] = Long.parseLong(inputs[j]);
            }
            if(helper(arr)) System.out.println("YES");
            else System.out.println("NO");
        }
    }

    private static boolean helper(long[] arr) {
        long sum = 0;
        for(int i=0;i<arr.length;i++) {
            sum += arr[i];
        }
        if(sum%arr.length != 0) return false;
        long num = sum/arr.length;
        for(int i=1;i<arr.length - 1;i++) {
            // System.out.println(Arrays.toString(arr) + " " + i);
            if(arr[i - 1] > num) {
                arr[i - 1]--;
                arr[i + 1]++;
                if(arr[i - 1] > num) i--;
            } else if(arr[i - 1] < num) {
                arr[i - 1]++;
                arr[i + 1]--;
                if(arr[i + 1] == -1) return false;
                if(arr[i - 1] < num) i--;
            }
        }
        return arr[arr.length - 1] == num;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

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

using namespace std;

#define ll long long

ll t,n,a,b,c;

int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n>>a>>b>>c;
        ll res=n/(a+b+c);
        ll sum=res*(a+b+c);

        if(sum==n) cout<<res*3<<'\n';
        else if(sum+a>=n) cout<<res*3+1<<'\n';
        else cout<<res*3+2<<'\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the logic of the code might fail. The buggy code has issues in handling the exact day when Monocarp completes his journey, especially when the total distance covered is very close to the required distance.

### Analysis of the Buggy Code:
1. The code calculates the number of complete cycles of three days (a + b + c) that fit into the total distance n.
2. It then checks if the sum of these complete cycles equals n, and if so, it outputs the total number of days as `res * 3`.
3. If the sum of the complete cycles plus the distance covered on the first day (a) is greater than or equal to n, it outputs `res * 3 + 1`.
4. Otherwise, it outputs `res * 3 + 2`.

### Edge Cases to Consider:
1. **Exact Completion on the First Day of a Cycle:**
   - If `n` is exactly the sum of a complete cycle, the code should output `res * 3`.
2. **Exact Completion on the Second Day of a Cycle:**
   - If `n` is exactly the sum of a complete cycle plus `a`, the code should output `res * 3 + 1`.
3. **Exact Completion on the Third Day of a Cycle:**
   - If `n` is exactly the sum of a complete cycle plus `a + b`, the code should output `res * 3 + 2`.
4. **Just After the Third Day of a Cycle:**
   - If `n` is just slightly more than the sum of a complete cycle plus `a + b`, the code should output `res * 3 + 3`.

### Challenging Test Input:
```plaintext
4
10 3 3 3
11 3 3 3
12 3 3 3
13 3 3 3
```

### Explanation:
1. **Test Case 1: n = 10, a = 3, b = 3, c = 3**
   - Sum of one cycle: 3 + 3 + 3 = 9
   - Number of complete cycles: 10 / 9 = 1
   - Sum of complete cycles: 1 * 9 = 9
   - Since 9 + 3 (first day of the next cycle) >= 10, the correct output should be 4.

2. **Test Case 2: n = 11, a = 3, b = 3, c = 3**
   - Sum of one cycle: 3 + 3 + 3 = 9
   - Number of complete cycles: 11 / 9 = 1
   - Sum of complete cycles: 1 * 9 = 9
   - Since 9 + 3 (first day of the next cycle) = 12, the correct output should be 4.

3. **Test Case 3: n = 12, a = 3, b = 3, c = 3**
   - Sum of one cycle: 3 + 3 + 3 = 9
   - Number of complete cycles: 12 / 9 = 1
   - Sum of complete cycles: 1 * 9 = 9
   - Since 9 + 3 (first day of the next cycle) = 12, the correct output should be 4.

4. **Test Case 4: n = 13, a = 3, b = 3, c = 3**
   - Sum of one cycle: 3 + 3 + 3 = 9
   - Number of complete cycles: 13 / 9 = 1
   - Sum of complete cycles: 1 * 9 = 9
   - Since 9 + 3 (first day of the next cycle) = 12, and 12 + 3 (second day of the next cycle) = 15, the correct output should be 5.

This test input will expose the bug in the provided code, as it fails to handle the exact completion on the first or second day of a new cycle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

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

#define ll long long

using namespace std;

int main()
{
    int v;
    cin >> v;

    for (int r=0; r<v; ++r)
    {
        ll n, a, b, c;
        cin >> n >> a >> b >> c;

        ll i = n/(a+b+c);

        if (n == i*(a+b+c))
        {
            cout << i*3 << endl;
        }
        else
        {
            if (n <= i*(a+b+c) +a)
            {
                cout << i*3+1  << endl;
            }
            else if (n <= i*(a+b+c) +a+b)
            {
                cout << i*3+2 <<  endl;
            }
        }

    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
#include <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cstdlib>
#include <utility>
#include <cassert>
#include <string>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <ctime>
#define fastIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
using namespace std;
typedef long long int lli;
#define rf(n,i) for(int i=n-1;i>=0;i--)
#define ff(n,i) for(int i=0;i<n;i++)
#define rs() reserve(1000)
#define mo7 1000000007
#define ft fixed<<setprecision(0)<<
#define vtr(i,v) vector <i> v
void solve()
{
    lli n,a,b,c;
    cin>>n>>a>>b>>c;
    lli x = n/(a+b+c);
    n-=(a+b+c)*x;
    if(n<=0)cout<<x*3<<endl;
    else if(n-a<=0)cout<<(3*x+1)<<endl;
    else if((n-a-b)<=0)cout<<3*x+2<<endl;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--)
     {
        solve();
    }
    /*
        ਲੋਹਾ ਹੋ ਗਿਆ ਸ਼ਰੀਰ, ਹੁਣ ਕਿੱਥੋ ਹਾਰਦੇ
        ਜਿਮ ਦੇ ਸ਼ੌਕੀਨ ਹੇਵੀਵੇਟ ਮਾਰਦੇ
        ਤੇਰੇ ਸ਼ਹਿਰ ਦੇ ਪਤੋਲੇ ਮੈਨੂੰ ਆ ਕੇ ਤਾੜਦੇ
        ਮੁੰਡੇ ਜਿਮ ਦੇ ਸ਼ੌਕੀਨ ਹੇਵੀਵੇਟ ਮਾਰਦੇ
        ਮਾਰ ਮਾਰ ਥਾਪੀਆਂ ਲੋਹੇ ਨੂੰ ਵੰਗੜਦੇ
        ਜਿਮ ਦੇ ਸ਼ੌਕੀਨ ਹੇਵੀਵੇਟ ਮਾਰਦੇ
        ਮਾਰਦੇ ਆ ਬੈਂਚ, ਧਰਤੀ ਖਿਲਾੜਦੇ
        ਮੁੰਡੇ ਜਿਮ ਦੇ ਸ਼ੌਕੀਨ ਹੇਵੀਵੇਟ ਮਾਰਦੇ
    */
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
t = int(input())
for i in range(t):
    a = list(map(int,input().split()))
    total = a[0]
    ans = 0
    s = sum([a[1] , a[2] , a[3]])
    if total - s < 0:
        if total - a[1] <= 0:
            print(1)
            continue
        if total - (a[1] + a[2]) <= 0:
            print(2)
            continue
        print(3)
    # while total > 0:
    #     total -= s
    #     ans += 1
    ans = (total // s)
    rem =  total % s
    if rem == 0:
        print(ans*3)
        continue

    if rem <= a[1]:
        print((ans * 3) + 1)
        continue

    elif rem <= a[1] + a[2]:
        print((ans * 3) + 2)
        continue
    else:
        print(ans * 4)

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

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

Think step by step.

Response:
```plaintext
1
10 4 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
n = int(input())
for i in range(n):
    A,a,b,c = map(int,input().split())
    if a == b == c:
        print(A//a)
    else:
        track = 0
        size = [a,b,c]
        N = len(size)
        count = 0
        i = 0
        while track < A:
            track += size[i]
            if i == N-1:
                i = -1
            i+=1
            count += 1
        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:
```plaintext
3
1 1 1 1
7 2 2 2
1000000000 1000000 1000000 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
t = int(input())
for i in range(t):
    total = 0
    x = str(input())
    l = x.split()
    n = int(l[0])
    count = 0
    if l[1]==l[2] and l[2] == l[3]:
        print(int(n/int(l[1])))
        break
    while(total < n):
        for value in l[1:]:
            count += 1
            total += int(value)
            if total >= n:
                break
    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 expose the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The code has a few issues:

1. **Incorrect Handling of Equal Distances**: The code has a special case where `a`, `b`, and `c` are equal. However, the logic to handle this case is flawed.
2. **Incorrect Break Statement**: The `break` statement after handling the special case will exit the loop immediately, preventing the processing of subsequent test cases.
3. **Incorrect Conversion and Usage of Input**: The input values are not correctly converted to integers before being used in the loop.

To create a test input that exposes these issues, we can:

1. Use a test case where `a`, `b`, and `c` are equal to check if the special case logic works correctly.
2. Use a test case where `a`, `b`, and `c` are not equal to ensure the main loop logic works as expected.
3. Use a test case with a large `n` to check the performance and correctness of the loop.

Here is a small-scale test input that covers these scenarios:

```plaintext
3
12 4 4 4
10 1 2 3
1000000 1 1 1
```

Explanation:
1. **Test Case 1**: `12 4 4 4`
   - Here, `a`, `b`, and `c` are all equal to 4.
   - The total distance covered in one cycle (3 days) is `4 + 4 + 4 = 12`.
   - The code should print `3` because Monocarp completes the journey on the third day.
   - This test case checks the special case logic.

2. **Test Case 2**: `10 1 2 3`
   - Here, `a`, `b`, and `c` are not equal.
   - The total distance covered in one cycle (3 days) is `1 + 2 + 3 = 6`.
   - The code should print `5` because Monocarp completes the journey on the fifth day.
   - This test case checks the main loop logic.

3. **Test Case 3**: `1000000 1 1 1`
   - Here, `a`, `b`, and `c` are all equal to 1.
   - The total distance covered in one cycle (3 days) is `1 + 1 + 1 = 3`.
   - The code should print `1000000` because Monocarp completes the journey on the 1000000th day.
   - This test case checks the performance and correctness of the loop for large values of `n`.

This test input should effectively expose the bugs 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: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

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

public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0)
		{
		    long n=sc.nextInt(),a=sc.nextLong(),b=sc.nextLong(),c=sc.nextLong();
		    long low=1,high=n,ans=0;
		    while(low<=high){
		        long mid=(low+high)/2;
		        long d=mid/3;
		        long g=(a+b+c)*d;
		        if(mid%3==1) g+=a;
		        if(mid%3==2) g+=b;
		        if(g>=n){
		            ans=mid;
		            high=mid-1;
		        }else low=mid+1;
		    }
		    System.out.println(ans);
		}
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
import java.util.Scanner;

public class B_2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        int n,a,b,c;
        while (t-- != 0) {
            n = sc.nextInt();
            a = sc.nextInt();
            b = sc.nextInt();
            c = sc.nextInt();

            int[] nums = new int[]{a,b,c};
            int pos = 0;

            int cont = 0;
            if(a == b && b == c){
                cont = (int) n/a;
            }else{
                while (n>0) {
                    n -= nums[pos];

                    pos++;
                    if(pos >= 3) pos = 0;

                    cont++;
                }
            }

            System.out.println(cont);
        }

        sc.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
3
1 1000000 1000000 1000000
1000000 1 1 1
10 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e4 + 10,inf=0x3f3f3f3f;    int T = 1;
int a[N],b[N],c[N];
void solve()
{

    int n,sum=0,a,b,c;
    cin>>n>>a>>b>>c;
    sum+=a+b+c;

	if(a>=n){cout<<1<<endl;return ;}
	if(a+b>=n){cout<<2<<endl;return;}
	if(a+b+c>=n){cout<<3<<endl;return;}

	if(a>=n%sum){cout<<3*n/sum<<endl;return ;}
	if(a+b>=n%sum){cout<<1+3*n/sum<<endl;return;}
	if(a+b+c>=n%sum){cout<<2+3*n/sum<<endl;return;}
    return;
}

signed main()
{
    //ios::sync_with_stdio(0);
    //cin.tie(0);

    cin>>T;
    for(int i =1;i<=T;i++)
    {
        //init();
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

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

using namespace std;
typedef long long ll;

void solve()
{
    int n, a, b, c;
    cin >> n >> a >> b >> c;

    int ans = n/(a + b + c)*3;
    int rem = n%(a + b + c);
    if(rem > 0) ans++;
    if(rem > a) ans++;

    cout << ans << "\n";
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int tt = 1;
    cin >> tt;

    while (tt--)
        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 buggy code, we need to consider edge cases where the logic in the code might fail. The key issues in the buggy code are:

1. The code incorrectly increments the answer by 1 if the remainder is greater than 0.
2. The code incorrectly increments the answer by 1 if the remainder is greater than `a`.

Let's analyze these issues and construct a test case that will highlight these bugs.

### Test Case Analysis
1. **Remainder Greater than 0 but Less than `a`**:
   - If `rem` is greater than 0 but less than `a`, the code will incorrectly increment the answer by 1. This is incorrect because Monocarp should not have completed an additional full day if he hasn't walked enough to reach the next day's distance.

2. **Remainder Greater than `a` but Less than `a + b`**:
   - If `rem` is greater than `a` but less than `a + b`, the code will incorrectly increment the answer by 2. This is incorrect because Monocarp should only have completed one additional day if he has walked enough to reach the next day's distance but not the day after.

### Challenging Test Input
Let's create a test input that includes these edge cases:

1. **Test Case 1**: `n` is slightly more than a multiple of `(a + b + c)`, but less than the next `a`.
2. **Test Case 2**: `n` is slightly more than `a`, but less than `a + b`.

### Test Input
```plaintext
2
11 5 3 1
9 5 3 1
```

### Explanation
1. **Test Case 1**: `n = 11`, `a = 5`, `b = 3`, `c = 1`
   - Total distance for one cycle: `a + b + c = 5 + 3 + 1 = 9`
   - Number of complete cycles: `11 // 9 = 1` (1 cycle)
   - Remaining distance: `11 % 9 = 2`
   - Since `rem = 2` is less than `a = 5`, the correct answer should be `1 * 3 + 1 = 4` (1 cycle of 3 days plus 1 additional day).
   - The buggy code will incorrectly output `1 * 3 + 1 = 4` but will incorrectly increment by 1, giving `5`.

2. **Test Case 2**: `n = 9`, `a = 5`, `b = 3`, `c = 1`
   - Total distance for one cycle: `a + b + c = 5 + 3 + 1 = 9`
   - Number of complete cycles: `9 // 9 = 1` (1 cycle)
   - Remaining distance: `9 % 9 = 0`
   - Since `rem = 0`, the correct answer should be `1 * 3 = 3` (1 cycle of 3 days).
   - The buggy code will correctly output `3` in this case.

This test input will help expose the bug in the code where the remainder is not handled correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
t = int(input())
import math

ans = []
for each_inp in range(t):
    inp_list = input().split()
    n = int(inp_list[0])
    a = int(inp_list[1])
    b = int(inp_list[2])
    c = int(inp_list[3])
    nstr = inp_list[0]
    totdist = 0
    noofdays = 0
    dayno = 1
    adayprev = 1
    bdayprev = 2
    cdayprev = 3

        #len(nstr) > 5:

    sum_once = a+b+c
    dayswsum = (n/sum_once)
    """if dayswsum.is_integer() is True:
        noofdays = int(dayswsum + 1)
    elif dayswsum.is_integer() is False:"""
    multipledays = math.floor(dayswsum)
    diff_totdist = int(n - multipledays*sum_once)
    if diff_totdist <= a and diff_totdist>0:
            noofdays = multipledays*3 + 1
    elif diff_totdist <= a+b and diff_totdist > a: # 6 6 7 4
            noofdays = multipledays*3 + 2
    elif diff_totdist <= c and diff_totdist > a+b:
            noofdays = multipledays*3 + 3
    elif diff_totdist == 0:
        noofdays = multipledays*3

    ans.append(noofdays)
for anss in ans:
    print(anss)

**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
10 5 1 1
10 1 5 1
10 1 1 5
1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
t = int(input())
while t:
    t-=1
    n,a,b,c = map(int,input().split())
    d = n//(a+b+c)
    if d*a+d*b+d*c == n:
        print(3*d)
    elif d*a+d*b+d*c+a >=n:
        print(3*d+1)
    else:
        print(3*d+2)

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

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

Think step by step.

Response:
```plaintext
1
10 1 2 3
```

**Explanation:**
This test case is designed to expose the bug in the provided buggy code. The input values are chosen to highlight the incorrect logic in the code's handling of the total distance calculation and the day count.

- **Input:** `10 1 2 3`
- **Expected Output:** `4`

**Reasoning:**
- Over the first three days, Monocarp will walk `1 + 2 + 3 = 6` kilometers.
- On the fourth day, he will walk `1` kilometer, bringing the total to `6 + 1 = 7` kilometers.
- On the fifth day, he will walk `2` kilometers, bringing the total to `7 + 2 = 9` kilometers.
- On the sixth day, he will walk `3` kilometers, bringing the total to `9 + 3 = 12` kilometers.

Since `n = 10`, Monocarp will complete his journey on the **fourth** day, as he will have walked at least 10 kilometers by then.

The buggy code incorrectly calculates the day count by using a fixed pattern without considering the remaining distance after the full cycles. This test case will expose the bug by requiring the correct handling of the remaining distance after the full cycles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
import java.util.Scanner;

public class Codeforces {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int tt = sc.nextInt();
        while (tt-- > 0) {
            long n = sc.nextInt(), a = sc.nextInt(), b = sc.nextInt(), c = sc.nextInt();
            long ans = 0;
            long s = a + b + c;
            long f = n / s;
            ans += f * 3;
            n = n % s;
            if (n == 0) {
                System.out.println(ans);
            } else if (n <= a) {
                System.out.println(ans + 1);
            } else if (n <= b) {
                System.out.println(ans + 2);
            } else System.out.println(ans + 3);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
import java.util.*;

public class MyClass{

//    static int age=22;
//    static{
//         age=22;
//        System.out.println("I am a static block");
//    }

    public  static void main(String []args){
        Scanner sc=new Scanner(System.in);

        int t=sc.nextInt();

        while(t>0){
            long m,a,b,c;

            m=sc.nextLong();
            a=sc.nextLong();
            b=sc.nextLong();
            c=sc.nextLong();

            long sum=a+b+c;

            long dis=(m/sum)*3;

            if(m<sum){
                int count=0;

                if(a>=m) System.out.println(1);

                else if (a+b>=m) System.out.println(2);
            }

            else {
                if(m%sum>0){
                    if(a>=m%sum) System.out.println(1+dis);

                    else if (a+b>=m%sum) System.out.println(2+dis);

                    else System.out.println(3+dis);
                }
                else System.out.println(dis);
            }

            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:
To expose the bug in the provided code, let's consider a few edge cases and scenarios that might cause the code to fail:

1. **Edge Case: Very Small Values for `a`, `b`, and `c`**:
   - If `a`, `b`, and `c` are very small, the total distance covered in each cycle (3 days) will also be very small. This can cause the code to handle the remaining distance incorrectly.
   - Example: `n = 7`, `a = 1`, `b = 1`, `c = 1`.

2. **Edge Case: Very Large Values for `a`, `b`, and `c`**:
   - If `a`, `b`, and `c` are very large, the total distance covered in each cycle will also be large. This can cause the code to handle the remaining distance incorrectly.
   - Example: `n = 1000000000`, `a = 1000000`, `b = 1000000`, `c = 1000000`.

3. **Edge Case: `n` is a multiple of `sum`**:
   - If `n` is exactly a multiple of the sum of `a`, `b`, and `c`, the code should output a multiple of 3. However, the code might fail to handle this correctly.
   - Example: `n = 9`, `a = 1`, `b = 2`, `c = 3`.

4. **Edge Case: `n` is just slightly more than a multiple of `sum`**:
   - If `n` is just slightly more than a multiple of the sum of `a`, `b`, and `c`, the code should output a day just beyond the multiple of 3. However, the code might fail to handle this correctly.
   - Example: `n = 10`, `a = 1`, `b = 2`, `c = 3`.

5. **Edge Case: `n` is exactly equal to one of the daily distances**:
   - If `n` is exactly equal to one of the daily distances, the code should output the corresponding day. However, the code might fail to handle this correctly.
   - Example: `n = 1`, `a = 1`, `b = 2`, `c = 3`.

Let's choose the most challenging and representative test case from the above scenarios:

```plaintext
1
10 1 2 3
```

This test case will check if the code correctly handles a situation where `n` is just slightly more than a multiple of the sum of `a`, `b`, and `c`. The correct output should be `4`, but the code might fail to handle this correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

void solve(){
	int n,a,b,c;
	cin>>n>>a>>b>>c;
	int day=0;
    int r=a+b+c;
    day=r*(n/r);
    int ans;ans=3*(n/r);
    if(day==n){
    	cout<<ans<<endl;
    	return;
    }
    if(day+a>=n){
    	ans++;
    }
    else {
    	ans+=2;
    }
	cout<<ans<<endl;
}

int main(){
	int t;
	cin>>t;
	while(t--){
		solve();
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic might fail. Specifically, the code incorrectly calculates the total distance covered and the number of days required. Here are some key points to consider:

1. **Edge Cases**:
   - When `n` is exactly a multiple of the total distance covered in three days (`a + b + c`).
   - When `n` is just slightly more than a multiple of the total distance covered in three days.
   - When `n` is very large but the daily distances are small.

2. **Incorrect Calculation**:
   - The code incorrectly multiplies `r` (the total distance covered in three days) by `n/r` to get the total distance covered. This can lead to incorrect results if `n` is not a multiple of `r`.

3. **Logical Errors**:
   - The code does not correctly handle the remaining distance after the initial full cycles of three days.

Given these points, let's create a test input that will expose the bug:

```plaintext
3
10 3 3 3
11 3 3 3
12 3 3 3
```

### Explanation:
1. **Test Case 1: \( n = 10 \), \( a = 3 \), \( b = 3 \), \( c = 3 \)**
   - Total distance covered in three days: \( 3 + 3 + 3 = 9 \) kilometers.
   - Monocarp will complete 9 kilometers in 3 days.
   - On the fourth day, he will walk 3 kilometers, reaching a total of 12 kilometers.
   - The correct output should be 4, but the buggy code might incorrectly output 3.

2. **Test Case 2: \( n = 11 \), \( a = 3 \), \( b = 3 \), \( c = 3 \)**
   - Total distance covered in three days: \( 3 + 3 + 3 = 9 \) kilometers.
   - Monocarp will complete 9 kilometers in 3 days.
   - On the fourth day, he will walk 3 kilometers, reaching a total of 12 kilometers.
   - The correct output should be 4, but the buggy code might incorrectly output 3.

3. **Test Case 3: \( n = 12 \), \( a = 3 \), \( b = 3 \), \( c = 3 \)**
   - Total distance covered in three days: \( 3 + 3 + 3 = 9 \) kilometers.
   - Monocarp will complete 9 kilometers in 3 days.
   - On the fourth day, he will walk 3 kilometers, reaching a total of 12 kilometers.
   - The correct output should be 4, but the buggy code might incorrectly output 3.

This test input will help expose the bug in the provided code by demonstrating the incorrect handling of the total distance and the number of days required.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
# --------------- Imports---------------
import sys ; import random ; import operator
from bisect import bisect_left, bisect_right
from string import ascii_lowercase, ascii_uppercase
import math ; from math import gcd, lcm, sqrt, ceil, floor, isqrt
from itertools import permutations, combinations, product, accumulate, repeat

# --------------- Mathematical Utilities ---------------
INF, N_INF = float('inf'), float('-inf')
bw_pow2 = lambda n: n != 0 and (n & (n - 1)) == 0
pow2 = lambda n: n > 0 and math.log2(n).is_integer()
even = lambda n: (n % 2) == 0 ; odd = lambda n: not even(n)
quad_form = lambda a, b, c: ((-b + isqrt(b**2 - 4*a*c)) // (2*a))
bw_even = lambda n: (n & 1) == 0 ; bw_odd = lambda n: not bw_even(n)

# --------------- Input Shortcuts ---------------
inp = lambda: int(input())
inps = lambda: input().strip()
inpm = lambda: map(int, input().split())
inpl = lambda: list(map(int, input().split()))

# --------------- Local Testing ---------------
# sys.stdin = open('input.txt', 'r') # Uncomment before submission

# --------------- Main Logic ---------------
def logic() -> None:
    n, a, b, c = inpm()
    d = a+b+c ; intervals = n / d ; days = int(intervals * 3)
    if n-d <= a:
        print(1)
    elif n-d <= a+b:
        print(days+1)
    else:
        print(days)

def main() -> None:
    tc = inp()
    for _ in range(tc):
        logic()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases where the logic might fail. Let's analyze the code and identify potential issues:

1. **Incorrect Calculation of Total Distance Covered**:
   - The code calculates the number of complete 3-day cycles (`intervals`) and then multiplies by 3 to get the total days.
   - It then checks if the remaining distance to be covered (`n - d`) is less than or equal to `a`, `a + b`, or `a + b + c`.

2. **Potential Issues**:
   - The code incorrectly handles the remaining distance after the complete 3-day cycles.
   - It does not correctly account for the possibility that the remaining distance might be covered in the next few days.

3. **Edge Cases**:
   - **Case 1**: When `n` is exactly a multiple of `a + b + c`.
   - **Case 2**: When `n` is slightly more than a multiple of `a + b + c` and the remaining distance is covered in the next day.
   - **Case 3**: When `n` is such that the remaining distance is exactly `a` or `a + b`.

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

```plaintext
4
15 1 5 3
16 1 5 3
17 1 5 3
18 1 5 3
```

Explanation:
1. **Test Case 1**: `n = 15`, `a = 1`, `b = 5`, `c = 3`
   - Monocarp will cover `1 + 5 + 3 + 1 + 5 + 3 = 18` kilometers in 6 days.
   - The code should correctly output `6` because `15` kilometers are covered by the end of the 5th day.

2. **Test Case 2**: `n = 16`, `a = 1`, `b = 5`, `c = 3`
   - Monocarp will cover `1 + 5 + 3 + 1 + 5 + 3 = 18` kilometers in 6 days.
   - The code should correctly output `6` because `16` kilometers are covered by the end of the 5th day.

3. **Test Case 3**: `n = 17`, `a = 1`, `b = 5`, `c = 3`
   - Monocarp will cover `1 + 5 + 3 + 1 + 5 + 3 = 18` kilometers in 6 days.
   - The code should correctly output `6` because `17` kilometers are covered by the end of the 5th day.

4. **Test Case 4**: `n = 18`, `a = 1`, `b = 5`, `c = 3`
   - Monocarp will cover `1 + 5 + 3 + 1 + 5 + 3 = 18` kilometers in 6 days.
   - The code should correctly output `6` because `18` kilometers are covered by the end of the 6th day.

This test input will help expose the bug in the given buggy code by checking how it handles the remaining distance after complete 3-day cycles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

Buggy Code:
t=int(input())
while t:
    t-=1
    n,a,b,c=map(int,input().split())
    if n%(a+b+c)==0:
        print((3*n)//(a+b+c))
    else:
        curr=n-(a+b+c)*(n//(a+b+c))
        ans=3*(n//(a+b+c))
        ans+=1
        if curr>a:
            ans+=1
            curr-=a
        if curr>b:
            ans+=1
            curr-=b
        print(ans)

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

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

Think step by step.

Response:
```plaintext
4
1 1000000 1000000 1000000
3000000 1000000 1000000 1000000
2999999 1000000 1000000 1000000
1500000 1000000 1000000 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Journey
time_limit_ms: 1000
memory_limit_mb: 256
Description: Monocarp decided to embark on a long hiking journey.

He decided that on the first day he would walk $$$a$$$ kilometers, on the second day he would walk $$$b$$$ kilometers, on the third day he would walk $$$c$$$ kilometers, on the fourth day, just like on the first, he would walk $$$a$$$ kilometers, on the fifth day, just like on the second, he would walk $$$b$$$ kilometers, on the sixth day, just like on the third, he would walk $$$c$$$ kilometers, and so on.

Monocarp will complete his journey on the day when he has walked at least $$$n$$$ kilometers in total. Your task is to determine the day on which Monocarp will complete his journey.

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

Each test case consists of one line containing four integers $$$n$$$, $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le a, b, c \le 10^6$$$).

Output: For each test case, output one integer — the day on which Monocarp will have walked at least $$$n$$$ kilometers in total and will complete his journey.

Examples:
input:
4
12 1 5 3
6 6 7 4
16 3 4 1
1000000000 1 1 1
output:
5
1
6
1000000000

Note: In the first example, over the first four days, Monocarp will cover $$$1 + 5 + 3 + 1 = 10$$$ kilometers. On the fifth day, he will cover another $$$5$$$ kilometers, meaning that in total over five days he will have covered $$$10 + 5 = 15$$$ kilometers. Since $$$n = 12$$$, Monocarp will complete his journey on the fifth day.

In the second example, Monocarp will cover $$$6$$$ kilometers on the first day. Since $$$n = 6$$$, Monocarp will complete his journey on the very first day.

In the third example, Monocarp will cover $$$3 + 4 + 1 + 3 + 4 + 1 = 16$$$ kilometers over the first six days. Since $$$n = 16$$$, Monocarp will complete his journey on the sixth day.

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

using namespace std;

int main()
{
    int t;
    cin >> t;

    while (t--)
    {
        int a, b, c, n;
        cin >>n>> a >> b >> c;

        int distances[3] = {a, b, c};
        int total_distance = 0;
        int day = 0;

        while (total_distance < n)
        {
            total_distance += distances[day % 3];
            ++day;
        }

        cout << day << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 1000000 1000000 1000000
1000000 1 1 1
1000000 1000000 1 1
```