**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circle Perimeter
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $r$, find the number of lattice points that have a Euclidean distance from $(0, 0)$ greater than or equal to $r$ but strictly less than $r+1$.

A lattice point is a point with integer coordinates. The Euclidean distance from $(0, 0)$ to the point $(x,y)$ is $\sqrt{x^2 + y^2}$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The only line of each test case contains a single integer $r$ ($1 \leq r \leq 10^5$).

The sum of $r$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the number of lattice points that have an Euclidean distance $d$ from $(0, 0)$ such that $r \leq d < r+1$.

Examples:
input:
6
1
2
3
4
5
1984
output:
8
16
20
24
40
12504

Note: The points for the first three test cases are shown below.

![](CDN_BASE_URL/798fdc9e24a49612fd77928f7fc973f8)

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

public class F_1971_Circle {
    public static void main(String[] args) throws IOException {
        BufferedScanner input = new BufferedScanner();
        BufferedOutput output = new BufferedOutput();

        int tasks = input.nextInt();
        while (tasks-- > 0) {
            int r = input.nextInt();

            long count = 0;
            for (int x = 0; x <= r; x++) {
                int minY = (int) Math.ceil(Math.sqrt(r * r - x * x));
                int maxY = (int) Math.ceil(Math.sqrt((r + 1) * (r + 1) - x * x));

                int delta = maxY - minY;
                delta *= x == 0 ? 2 : 4;
                if (minY == 0) {
                    delta -= 2;
                }

                count += delta;
            }

            output.println(count);
        }

        output.flush();
    }

    private static class BufferedScanner {
        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        public BufferedScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in));
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(nextToken());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(nextToken());
        }

        public int[] nextIntArray(int length) throws IOException {
            int[] numbers = new int[length];
            for (int i = 0; i < length; i++) {
                numbers[i] = nextInt();
            }

            return numbers;
        }

        public long[] nextLongArray(int length) throws IOException {
            long[] numbers = new long[length];
            for (int i = 0; i < length; i++) {
                numbers[i] = nextLong();
            }

            return numbers;
        }

        public String nextLine() throws IOException {
            tokenizer = null;
            return reader.readLine();
        }

        private String nextToken() throws IOException {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }

            return tokenizer.nextToken();
        }
    }

    private static class BufferedOutput {
        private final StringBuilder result;

        public BufferedOutput() {
            result = new StringBuilder();
        }

        public void print(Object object) {
            result.append(object);
        }

        public void println(Object object) {
            result.append(object).append("\n");
        }

        public void flush() {
            System.out.println(result);
            System.out.flush();
            result.setLength(0);
        }
    }
}

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

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

Think step by step.

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

1. **Small values of \( r \)**: These can help verify the basic logic of the algorithm.
2. **Large values of \( r \)**: These can test the performance and accuracy of the algorithm.
3. **Values of \( r \) that are close to the boundary of the range**: These can help verify the boundary conditions.
4. **Values of \( r \) that result in a large number of lattice points**: These can help test the correctness of the logic for counting lattice points.

Given these considerations, let's create a test input with a mix of these edge cases:

```plaintext
4
1
100000
10000
1
```

### Explanation:
1. **\( r = 1 \)**: This is a small value that should have a known and simple result (8 lattice points).
2. **\( r = 100000 \)**: This is a large value that tests the performance and accuracy of the algorithm.
3. **\( r = 10000 \)**: This is a mid-range value that still tests the performance and accuracy but is not as extreme as \( r = 100000 \).
4. **\( r = 1 \)**: Repeating the small value to ensure the algorithm handles multiple small values correctly.

This test input should help expose any bugs in the code, particularly in how it handles different ranges of \( r \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

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

    void solve() {
        int n;
        cin >> n;
        vector<int> a(n);
        for(int i=0 ; i<n ; i++){
            cin >> a[i];
        }
        bool flag=1;
        while(flag){
            int index=-1;
            bool flag1=0;
            for(int i=1 ; i<a.size() ; i++){
                if(a[i-1] == a.size()+1-(i)){
                    index=i-1;
                    flag1=1;
                }
            }
            if(flag1==0){
                cout<< a.size() << endl;
                return;
            }
            for(int i=0; i<index ; i++){
                a.push_back(0);
            }
        }

        cout << a.size() << endl;
    }

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

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that might cause the code to fail. The primary issue in the code is that it does not correctly handle the condition for appending zeros and may enter an infinite loop or produce incorrect results.

### Key Points to Consider:
1. **Initial Array with All Elements Meeting the Condition:**
   - If the initial array has elements that satisfy the condition `a[i] == a.size() + 1 - (i+1)` for multiple positions, the code might incorrectly append zeros multiple times in a single iteration.

2. **Single Element Arrays:**
   - The code should handle single-element arrays correctly, which is already covered in the example.

3. **Arrays with No Valid Positions:**
   - If no elements in the array meet the condition, the code should correctly output the initial array length.

4. **Edge Cases with Large Arrays:**
   - Large arrays with elements that meet the condition at various positions can help identify issues with the loop and condition checks.

### Chosen Test Input:
Let's create a test input that includes multiple test cases, each designed to expose different aspects of the bug:

1. **Test Case with All Elements Meeting the Condition:**
   - This will test if the code can handle multiple valid positions correctly.

2. **Test Case with No Valid Positions:**
   - This will test if the code correctly identifies and handles arrays with no valid positions.

3. **Test Case with a Single Element:**
   - This is a simple case to ensure the code handles it correctly.

4. **Test Case with a Mixed Array:**
   - This will test the code's ability to handle a mix of valid and invalid positions.

### Test Input:
```plaintext
4
5
2 4 6 2 5
5
5 4 4 5 1
1
1
3
3 2 1
```

### Explanation:
1. **Test Case 1:**
   - Array: `[2, 4, 6, 2, 5]`
   - Expected Output: `10`
   - This is a valid case where the code should append zeros multiple times.

2. **Test Case 2:**
   - Array: `[5, 4, 4, 5, 1]`
   - Expected Output: `11`
   - This is another valid case with multiple positions meeting the condition.

3. **Test Case 3:**
   - Array: `[1]`
   - Expected Output: `1`
   - This is a simple case with a single element.

4. **Test Case 4:**
   - Array: `[3, 2, 1]`
   - Expected Output: `6`
   - This is a mixed case where only one position meets the condition.

This test input should effectively expose any issues in the buggy code, particularly in how it handles multiple valid positions and mixed arrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
#define YES cout << "YES\n";
#define NO cout << "NO\n";

int32_t gcd(int a, int b)
{
    return __gcd(a, b);
}

int32_t lcm(int a, int b)
{
    return (a * b) / gcd(a, b);
}

int power(int x, int n)
{
    int result = 1;
    while (n > 0)
    {
        if (n & 1 == 1)
        {
            result = result * x;
        }
        x = x * x;
        n = n >> 1;
    }
    return result;
}

int f(vector<int> &a, int len, int n, vector<int> dp)
{
    int ans = len;
    if(dp[len]!=0)
    return dp[len];
    for(int i = 2 ; i<=n ; i++)
    {
        if(len +1- i == a[i])
        {
            ans = max(ans, f(a, len + i - 1, n, dp));
        }
    }
    return dp[len] = ans;
}

int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin>>n;
        vector<int> a(n+1,0);
        for(int i = 1 ; i<=n ; i++)
        cin>>a[i];
        vector<int> dp(300000, 0);
        cout<<f(a,n,n,dp)<<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
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <cstring>
using namespace std;

typedef long long int ll;
const int MAX_N = 3e5 + 10;

vector<ll> dp(MAX_N, -1); // Memoization for lis function

// Recursive function with memoization for longest transformation path
ll lis(int i, set<ll>& s, map<ll, ll>& m) {
    if (s.find(i) == s.end()) return 0; // Not in s
    if (dp[i] != -1) return dp[i]; // Return memoized result if it exists

    // Only proceed if 'i' is a valid key in 'm'
    if (m.find(i) == m.end()) return 0;

    // Recursive call
    ll ans = m[i] + lis(i + m[i], s, m);
    return dp[i] = ans; // Memoize and return
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        ll n;
        cin >> n;

        fill(dp.begin(), dp.begin() + MAX_N, -1); // Reset dp for the maximum size

        vector<ll> a(n), zero;
        set<ll> s;
        map<ll, ll> m;
        ll temp = n;

        for (int i = 0; i < n; i++) {
            cin >> a[i];
            if (temp - a[i] == 0)
                zero.push_back(i);
            else if (a[i] - temp > 0) {
                s.insert(a[i] - temp);
                m[a[i] - temp] = i;
            }
            temp--;
        }

        ll ans = 0;

        // Iterate through indices stored in 'zero'
        for (int i = 0; i < zero.size(); i++) {
            ll sum = zero[i] + lis(zero[i], s, m);
            ans = max(sum, ans);
        }

        cout << ans + n << endl; // Output the result
    }
    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
4
1
1
5
5 4 3 2 1
6
6 5 4 3 2 1
10
10 9 8 7 6 5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
t = int(input())

import sys
sys.setrecursionlimit(3000)

for _ in range(t):

    n = int(input())
    arr = [*map(int, input().split())]
    vals = [0] * (10 ** 6)
    indexes = [0] * (10 ** 6)
    S = set()
    for i, j in enumerate(arr):
        a, b, c = n - i, j, i
        # vals.append((a, b))
        vals[a] = b
        indexes[a] = c
        S.add(a)

    ans = 0

    def solve(tot):

        global ans
        ans = max(ans, tot)

        for a in S:
            b = vals[a]
            c = indexes[a]
            if a + tot == b and c > 0:
                solve(tot + c)
        return

    solve(0)

    print(ans + n)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios where the code might fail. The main issue in the buggy code seems to be related to the way it handles the indices and values, especially when the conditions for appending zeros are not met correctly.

Let's consider the following steps to create a test case:

1. **Edge Case with Small Array**: Test with a small array to ensure the code handles small inputs correctly.
2. **Edge Case with Large Array**: Test with a large array to ensure the code handles large inputs efficiently.
3. **Edge Case with Specific Conditions**: Test with an array where the conditions for appending zeros are met in a specific order to expose any logical errors.

### Challenging Test Input

```plaintext
4
1
1
5
2 4 6 2 5
5
5 4 4 5 1
10
1 2 3 4 5 6 7 8 9 10
```

### Explanation

1. **Test Case 1 (Small Array)**:
   - Input: `1 1`
   - Expected Output: `1`
   - This tests the simplest case where no operations can be performed.

2. **Test Case 2 (Given Example 1)**:
   - Input: `5 2 4 6 2 5`
   - Expected Output: `10`
   - This is a given example that should work correctly if the code is fixed.

3. **Test Case 3 (Given Example 2)**:
   - Input: `5 5 4 4 5 1`
   - Expected Output: `11`
   - This is another given example that should work correctly if the code is fixed.

4. **Test Case 4 (Large Array with Specific Conditions)**:
   - Input: `10 1 2 3 4 5 6 7 8 9 10`
   - Expected Output: `55`
   - This tests a larger array where the conditions for appending zeros are met in a specific order. The expected output is calculated as follows:
     - Start with an array of length 10.
     - For each valid `i`, append `i - 1` zeros.
     - The maximum possible length is the sum of the first 10 natural numbers: `1 + 2 + 3 + ... + 10 = 55`.

This test input should help expose any issues in the buggy code, especially with handling large arrays and specific conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
t = int(input())

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

    for i in range(1, n):
        edges.append([i + data[i], i])
    edges.sort()
    if len(edges) < 1:
        print(1)
        continue
    searched = set()
    res = 0
    for i in range(len(edges)):
        el = edges[i]
        searched.add(el[0])
        if el[0] != n:
            continue
        stack = [el[0] + el[1]]
        while len(stack):
            edge = stack.pop()
            if edge in searched:
                continue
            searched.add(edge)
            for j in range(i+1, len(edges)):
                if edges[j][0] == edge:
                    res = max(res, edges[j][0] + edges[j][1])
                    stack.append(edges[j][0] + edges[j][1])
                if edges[j][0] > edge:
                    break
    print(res)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code has issues with how it handles the edges and the conditions for appending zeros. Specifically, it might not correctly handle cases where the array elements do not strictly follow the pattern `a_i = |a| + 1 - i`.

Let's consider the following points to create a challenging test case:
1. **Edge Case with Small Array**: Test with a small array to ensure the code handles small inputs correctly.
2. **Edge Case with Large Array**: Test with a large array to ensure the code handles large inputs efficiently.
3. **Edge Case with No Valid Operations**: Test with an array where no valid operations can be performed.
4. **Edge Case with Multiple Valid Operations**: Test with an array where multiple valid operations can be performed to maximize the length of the array.

Given the constraints and the need to focus on a small-scale test input, we will create a test case that is complete and valid.

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

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 3, 4, 5]`
   - This array does not have any valid operations because `a_i ≠ |a| + 1 - i` for any `i`.
   - Expected Output: `5` (no changes to the array).

2. **Test Case 2:**
   - Array: `[5, 4, 3, 2, 1]`
   - This array has valid operations because `a_i = |a| + 1 - i` for all `i`.
   - Expected Output: `11` (after performing the operations, the array becomes `[5, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0]`).

3. **Test Case 3:**
   - Array: `[1, 3, 2, 4]`
   - This array has some valid operations but not all elements follow the pattern.
   - Expected Output: `7` (after performing the operations, the array becomes `[1, 3, 2, 4, 0, 0, 0]`).

This test input should expose the bug in the buggy code by testing various scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
from math import *
def cinint():
	return int(input())
def cinline():
	return list(map(int,input().split()))

def solve():
	n=cinint()
	l=cinline()
	s=[]
	for index,value in enumerate(l):
		s.append(value+index)
	response=n
	s.pop(0)
	while(n in s):
		aux=n
		while(True):
			#print("aux",aux)
			try:
				i=s.index(aux)
				if(s[i]==n):
					s[i]=-1
			except:
				response=max(aux,response)
				#print("response",response)
				break
			aux+=i+1

	return response

t=1
t=int(input())
for i in range(t):
	print(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 create a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test cases with small and large values of `n`, and specific values of `a_i` that can trigger the bug.
2. **Critical Values**: Values of `a_i` that are exactly equal to `|a| + 1 - i` to see if the code correctly identifies and processes them.
3. **Edge Cases in the Array**: Arrays where the values are just slightly off the critical values to see if the code handles them correctly.
4. **Multiple Test Cases**: To ensure the code handles multiple test cases correctly.

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug might look like this:

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

### Explanation:

1. **Test Case 1**: `n = 5`, `a = [5, 4, 3, 2, 1]`
   - Here, every element `a_i` is exactly equal to `|a| + 1 - i`.
   - This test case should result in the maximum possible length of the array after performing the operations.

2. **Test Case 2**: `n = 4`, `a = [4, 3, 2, 1]`
   - Similar to the first test case, but with a smaller array.
   - This test case should also result in the maximum possible length of the array after performing the operations.

3. **Test Case 3**: `n = 6`, `a = [6, 5, 4, 3, 2, 1]`
   - Again, every element `a_i` is exactly equal to `|a| + 1 - i`.
   - This test case should result in the maximum possible length of the array after performing the operations.

These test cases are designed to maximize the failure probability in the buggy code by providing arrays where the elements are exactly equal to the critical values, which should trigger the operations correctly. If the buggy code fails to handle these cases, it will likely produce incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
import java.io.*;
import java.util.*;
public class C {
	public static PrintWriter out;
	public static void main(String[] args)throws IOException{
        JS sc=new JS();
        out=new PrintWriter(System.out);
        int t=sc.nextInt();
        outer: while(t-->0) {
        	int n=sc.nextInt();
        	long a[]=new long[n];
        	for(int i=0;i<n;i++) {
        		a[i]=sc.nextLong();
        	}
        	long sz[]=new long[n];
        	HashMap<Long, TreeSet<Integer>>hm=new HashMap<>();
        	for(int i=0;i<n;i++) {
        		sz[i]=a[i]+i;
        		if(hm.containsKey(sz[i])) {
        			TreeSet<Integer>ts=hm.get(sz[i]);
        			ts.add(i);
        			hm.put(sz[i], ts);
        		}else {
        			TreeSet<Integer>ts=new TreeSet<>();
        			ts.add(i);
        			hm.put(sz[i], ts);
        		}
        	}
        	ArrayDeque<Pair>q=new ArrayDeque<>();
        	boolean vis[]=new boolean[n];
        	if(hm.containsKey((long)n)) {
        		for(int i:hm.get((long)n)) {//get starts
            		q.add(new Pair(i, sz[i]));
            		vis[i]=true;
            	}
        	}
        	while(!q.isEmpty()) {
        		Pair cur=q.poll();
        		long nextVal=cur.x+cur.y;
        		if(!hm.containsKey(nextVal)) {
        			continue;
        		}
        		for(int neigh:hm.get(nextVal)) {
        			if(!vis[neigh]) {
        				vis[neigh]=true;
        				q.add(new Pair(neigh, sz[neigh]));
        			}
        		}
        	}
        	long ans=0;
        	for(int i=0;i<n;i++) {
        		if(vis[i]) {
        			ans=Math.max(sz[i]+i, ans);
        		}
        	}
        	out.println(ans);
        }
		out.close();
	}
    public static long gcd(long a, long b) {
        if(a > b)
            a = (a+b)-(b=a);
        if(a == 0L)
            return b;
        return gcd(b%a, a);
    }
	static void add(HashMap<Character,Integer> A, char key){
		A.put(key,A.getOrDefault(key,0)+1);
	}
	static void remove(HashMap<Character,Integer>A, char key){
		if(!A.containsKey(key))return;
		A.put(key,A.get(key)-1);
		if(A.get(key)==0){
			A.remove(key);
		}
	}
	static class Pair{
		int x; long y;
		public Pair(int x, long y) {
			this.x=x;this.y=y;
		}
	}
	static void sort(int a[]) {
		ArrayList<Integer>al=new ArrayList<>();
		for(int i=0;i<a.length;i++) {
			al.add(a[i]);
		}
		Collections.sort(al);
		for(int i=0;i<a.length;i++) {
			a[i]=al.get(i);
		}
	}
	static class JS {
	    public int BS = 1<<16;
	    public char NC = (char)0;
	    byte[] buf = new byte[BS];
	    int bId = 0, size = 0;
	    char c = NC;
	    double num = 1;
	    BufferedInputStream in;

	    public JS() {
	        in = new BufferedInputStream(System.in, BS);
	    }

	    public JS(String s) throws FileNotFoundException {
	        in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
	    }

	    public char nextChar(){
	        while(bId==size) {
	            try {
	                size = in.read(buf);
	            }catch(Exception e) {
	                return NC;
	            }
	            if(size==-1)return NC;
	            bId=0;
	        }
	        return (char)buf[bId++];
	    }

	    public int nextInt() {
	        return (int)nextLong();
	    }

	    public long nextLong() {
	        num=1;
	        boolean neg = false;
	        if(c==NC)c=nextChar();
	        for(;(c<'0' || c>'9'); c = nextChar()) {
	            if(c=='-')neg=true;
	        }
	        long res = 0;
	        for(; c>='0' && c <='9'; c=nextChar()) {
	            res = (res<<3)+(res<<1)+c-'0';
	            num*=10;
	        }
	        return neg?-res:res;
	    }

	    public double nextDouble() {
	        double cur = nextLong();
	        return c!='.' ? cur:cur+nextLong()/num;
	    }

	    public String next() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c>32) {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public String nextLine() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c!='\n') {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public boolean hasNext() {
	        if(c>32)return true;
	        while(true) {
	            c=nextChar();
	            if(c==NC)return false;
	            else if(c>32)return true;
	        }
	    }
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

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

public class Main{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception {
        int N = ni();
        long[] A = new long[N];
        for(int i = 0; i< N; i++)A[i] = nl()+i;
        TreeMap<Long, ArrayList<Long>> mp = new TreeMap<>();
        for(int i = 0; i< N; i++){
            if(!mp.containsKey(A[i])){
                mp.put(A[i], new ArrayList<>());
            }
            ArrayList<Long> l = mp.get(A[i]);
            l.add(A[i]+i);
            mp.put(A[i], l);
        }
        pn(dfs(mp, N, N));
    }
    long dfs(TreeMap<Long, ArrayList<Long>> mp, long len, long max){
        if(!mp.containsKey(len))return max;
        for(long x:mp.get(len))
            if(x > max)
                max = Math.max(max, dfs(mp, x, x));
        return max;
    }
    //SOLUTION  END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
    void exit(boolean b){if(!b)System.exit(0);}
    final long IINF = Long.MAX_VALUE/2;
    final int INF = Integer.MAX_VALUE/2;
    DecimalFormat df = new DecimalFormat("0.000000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-10;
    static boolean multipleTC = true, memory = true, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if (fileIO) {
            in = new FastReader("");
            out = new PrintWriter("");
        } else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = multipleTC? ni():1;
        pre();
        for (int t = 1; t <= T; t++) solve(t);
        out.flush();
        out.close();
        System.err.println("Runtime: " + (System.currentTimeMillis() - ct));
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 26).start();
        else new Main().run();
    }
    int[][] make(int n, int e, int[] from, int[] to, boolean f){
        int[][] g = new int[n][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            if(f)g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
        int[][][] g = new int[n][][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
            if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
        }
        return g;
    }
    int[][] make(int n, int[] par, boolean f){
        int[][] g = new int[n][];
        int[] cnt = new int[n];
        for(int x:par)cnt[x]++;
        if(f)for(int i = 1; i< n; i++)cnt[i]++;
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 1; i< n-1; i++){
            g[par[i]][--cnt[par[i]]] = i;
            if(f)g[i][--cnt[i]] = par[i];
        }
        return g;
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int brute = 0;while(s>0){s/=10;brute++;}return brute;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object... o){for(Object oo:o)out.print(oo+" ");}
    void pn(Object... o){
        if(o.length == 0)out.println("");
        for(int i = 0; i< o.length; i++){
            out.print(o[i]);
            out.print((i+1 == o.length?"\n":" "));
        }
    }
    void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

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

        String nextLine() throws Exception{
            String str;
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the specific logic and edge cases that the code might handle incorrectly. The buggy code has the following issues:

1. **Incorrect Initialization of Array `A`**: The code incorrectly initializes the array `A` by adding the index `i` to each element of the input array. This is not consistent with the problem statement, which does not specify such an initialization.

2. **Incorrect Use of TreeMap**: The code uses a `TreeMap` to store and manipulate the elements, but the logic for populating and using the `TreeMap` is flawed. It incorrectly adds values to the list and does not correctly handle the conditions specified in the problem.

3. **Incorrect DFS Logic**: The DFS function does not correctly handle the conditions for choosing the position `i` and appending zeros to the array.

Given these issues, we can create a test case that:

1. Has elements that do not match the condition `a_i = |a| + 1 - i` initially.
2. Has elements that, after some operations, might match the condition but the code fails to handle correctly.

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

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

### Explanation:
- **Test Case**: 1
- **Array Length**: 3
- **Array Elements**: [2, 3, 1]

### Expected Behavior:
- The initial array is [2, 3, 1].
- The condition `a_i = |a| + 1 - i` should be checked for each element.
- For `i = 1`, `a_1 = 2` and `|a| + 1 - 1 = 3` (condition not satisfied).
- For `i = 2`, `a_2 = 3` and `|a| + 1 - 2 = 2` (condition not satisfied).
- For `i = 3`, `a_3 = 1` and `|a| + 1 - 3 = 1` (condition satisfied).

### Expected Output:
- The maximum possible length of the array after performing the operations is 4.

### Why This Test Case is Challenging:
- The initial array does not satisfy the condition for the first two elements, which should prevent any operations on those positions.
- The third element satisfies the condition, allowing an operation to be performed, which the buggy code might fail to handle correctly due to its flawed logic.

This test case should help in exposing 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: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
#include<bits/stdc++.h>
#include<bitset>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
#define ll long long
//#define int long long
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb(k) push_back(k)
#define inv(v) for(auto&it:v){cin>>it;}
#define sv(v,n) v = vector<int>(n);
#define rere return ret;
#define pa pair<int,int>
#define ini memset(dp,-1, sizeof(dp));
using namespace __gnu_pbds;
template<class T> using ordered_set = tree<T, null_type , less<T> , rb_tree_tag , tree_order_statistics_node_update> ;
#define FLASH  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
void File() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
vector<int>v;
int n;
map<ll,vector<int>>mp;
 ll go(int i, ll sz) {
     if (i == n + 1)return 0;
     ll ret = 0;
     for (auto it: mp[sz]) {
         if (sz + it - 1 > sz)ret = max(ret, it - 1 + go(it, sz + it - 1));
     }
     rere;
 }
void O_O() {
    cin >> n;
    mp.clear();
    v = vector<int>(n + 2);
    for (int i = 1; i <= n; ++i) {
        cin >> v[i];
    }
    ll x;
    for (int i = 1; i <= n; ++i) {
        x = v[i] - 1 + i;
        mp[x].pb(i);
    }
    cout << go(1, n) + n << '\n';
}

int32_t main() {
    File();
    FLASH
    int rep = 1;
    cin >> rep;
    while (rep--) O_O();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The primary issues in the code are related to how the `go` function processes the map and the conditions for valid operations.

Here are the steps to think through the problem:

1. **Identify Edge Cases:**
   - **Single Element Array:** The simplest case where the array has only one element.
   - **Array with All Elements Equal:** This can cause issues if the map is not correctly handling repeated values.
   - **Array with Elements that Do Not Satisfy the Condition:** Elements that do not satisfy the condition \(a_i = |a| + 1 - i\) should be ignored.
   - **Array with Elements that Satisfy the Condition:** Ensure that the elements that satisfy the condition are processed correctly.

2. **Generate Test Inputs:**
   - **Single Element Array:**
     - Input: `1 1`
     - Expected Output: `1`
   - **Array with All Elements Equal:**
     - Input: `4 1 1 1 1`
     - Expected Output: `4`
   - **Array with Elements that Do Not Satisfy the Condition:**
     - Input: `3 2 3 4`
     - Expected Output: `3`
   - **Array with Elements that Satisfy the Condition:**
     - Input: `5 2 4 6 2 5`
     - Expected Output: `10`

3. **Combine Multiple Test Cases:**
   - To test the code with multiple test cases, we can combine the above inputs into a single test input.

Here is the combined test input:

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

This test input includes:
- A single element array.
- An array with all elements equal.
- An array with elements that do not satisfy the condition.
- An array with elements that satisfy the condition.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

#define ll long long
#define pb push_back

void solve() {
    int t;
    cin >> t; // Number of test cases

    while (t--) {
        int n;
        cin >> n; // Initial array length
        vector<ll> v(n + 1); // Array to store input (1-based indexing)
        for (int i = 1; i <= n; i++) cin >> v[i];

        // Create a vector of pairs to store {a[i] + i - 1, i - 1}
        vector<pair<ll, ll>> arr;
        for (int i = 2; i <= n; i++) {
            arr.pb({v[i] + i - 1, i - 1});
        }

        // BFS initialization
        queue<ll> q;
        ll max_length = n; // Start with initial length
        for (auto &[key, value] : arr) {
            if (key == n) { // If the key equals n, it's valid to start
                q.push(n + value); // Push the extended length
                max_length = max(max_length, n + value); // Update max length
            }
        }

        // Process the queue
        while (!q.empty()) {
            ll current_length = q.front();
            q.pop();

            // Check all pairs for current length
            for (auto &[key, value] : arr) {
                if (key == current_length) { // If key matches the current length
                    ll new_length = current_length + value; // Extend length
                    q.push(new_length); // Push new length to queue
                    max_length = max(max_length, new_length); // Update max length
                }
            }
        }

        cout << max_length << "\n"; // Output the maximum length
    }
}

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
from collections import deque

def max_array_length(t,n,a):
    results = []
    # Initialize the queue and visited set
    queue = deque([n])  # Start with the initial length of the array
    visited = set([n])  # Keep track of visited lengths
    max_length = n  # Start with the initial length as max

    while queue:
        current_length = queue.popleft()
        # Iterate through the current array elements to find valid i
        for i in range(2, current_length + 1):  # i ranges from 2 to current_length (inclusive)
            # Ensure we do not access out of bounds
            if i - 1 < n and a[i - 1] == current_length + 1 - i:  # Check the condition
                new_length = current_length + (i - 1)  # Calculate new length
                if new_length not in visited:  # If this length hasn't been processed
                    visited.add(new_length)
                    queue.append(new_length)  # Add to the queue
                    max_length = max(max_length, new_length)  # Update max length

      # Store the result for this test case

    return max_length

# Example usage:
t = int(input().strip())
test_cases = []
for _ in range(t):
    n = int(input())
    a = list(map(int, input().strip().split()))
    results = max_array_length(t, n, a)
    print(results)

**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 might fail. The key issues in the buggy code include:

1. The code does not correctly handle the condition `if i - 1 < n` when `i` can be greater than `n` during the operations.
2. The code does not properly reset or reinitialize variables for each test case, which can lead to incorrect results.

Given these issues, we can create a test input that:

- Has a small number of test cases to ensure the input is valid and small-scale.
- Includes arrays where the elements satisfy the condition `a_i = |a| + 1 - i` in a way that the code might incorrectly handle.
- Ensures that the array length changes significantly to test the maximum length calculation.

Here is a test input that meets these criteria:

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

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 3, 4, 5]`
   - The condition `a_i = |a| + 1 - i` is satisfied for all elements.
   - This should result in a maximum length of 15 after performing operations.

2. **Test Case 2:**
   - Array: `[4, 3, 2, 1]`
   - The condition `a_i = |a| + 1 - i` is satisfied for all elements.
   - This should result in a maximum length of 10 after performing operations.

3. **Test Case 3:**
   - Array: `[6, 5, 4, 3, 2, 1]`
   - The condition `a_i = |a| + 1 - i` is satisfied for all elements.
   - This should result in a maximum length of 21 after performing operations.

These test cases are designed to maximize the failure probability in the faulty solution by ensuring that the code's logic for handling the condition and updating the array length is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
n = 0
ans = {}
move = {}

def getter(added):
    global n,ans,move
    if added not in ans.keys():
        ans[added] = added + n
        if added in move.keys():
            for i in move[added]:
                getter(added + i)
                ans[added] = max(ans[added], ans[added+i])

for _ in range(int(input())):
    n = int(input())
    vals = [int(i) for i in input().split()]
    ans.clear()
    move.clear()
    for i in range(n):
        res = max(vals[i] - (n-(i+1) + 1), -1)
        if res == -1:
            continue
        if res not in move.keys():
            move[res] = []
        move[res].append(i)

    getter(0)

    print(ans[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
4
1
1
2
2 1
3
3 2 1
4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
import java.io.*;
import java.util.*;
public class ques3 {
    private static final int mod = 1000000007;
    // Custom class to hold pairs
    static class Pair {
        int value;
        int index;
        Pair(int value, int index) {
            this.value = value;
            this.index = index;
        }
    }
    // Euclidean algorithm for finding GCD for int type
    public static int gcdInt(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
       return a;
    }
    // Euclidean algorithm for finding GCD for long type
    public static long gcdLong(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
       return a;
    }
    // Binary exponentiation for int type
    public static int power_Int(int base, int exponent) {
        int result = 1;
        while (exponent > 0) {
           if ((exponent & 1) == 1) {
               result *= base;
           }
            base *= base;
            exponent >>= 1;
        }
        return result;
    }
    // Binary exponentiation for long type
    public static long power_Long(long base, long exponent) {
        long result = 1;
        while (exponent > 0) {
           if ((exponent & 1) == 1) {
               result *= base;
           }
            base *= base;
            exponent >>= 1;
        }
        return result;
    }
    // Modular exponentiation for int type with MOD = 1e9 + 7
    public static int power_modInt(int base, int exponent) {
       int result = 1;
       base %= mod;
       while (exponent > 0) {
           if ((exponent & 1) == 1) {
           result = (int) ((result * (long) base) % mod);
           }
           base = (int) ((base * (long) base) % mod);
           exponent >>= 1;
           }
           return result;
       }
    // greater than or equal to key
    public static long long_lowerBound(long[] arr, long key) {
        long low = 0, high = arr.length - 1;
        long ans = arr.length;
        while (low <= high) {
            long mid = (low + high) / 2;
            if (arr[(int)mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // greater than or equal to key
    public static long long_upperBound(long[] arr, long key) {
        long low = 0, high = arr.length - 1;
        long ans = arr.length;
        while (low <= high) {
            long mid = (low + high) / 2;
            if (arr[(int)mid] > key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // greater than equal to key
    public static int int_lowerBound(int[] arr, int key) {
        int low = 0, high = arr.length - 1;
        int ans = arr.length;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr[mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // greater than key
    public static int int_upperBound(int[] arr, int key) {
        int low = 0, high = arr.length - 1;
        int ans = arr.length;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr[mid] > key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // solve
    public static long rec(int i, long val, long[] a, int[] b, long[] dp, int n) {
        if (i >= n) {
            return 0;
        }
        if (dp[i] != -1) {
            return dp[i];
        }
        dp[i]=0;
        int j = i;
        while(j<n){
            if(a[j]==val){
                dp[i] = Math.max(dp[i], rec(j + 1, val + b[j], a, b, dp, n) + b[j]);
            }
            j++;
        }
        return dp[i];
    }

    public static void main(String args[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        String test = br.readLine();
        int t = Integer.parseInt(test);
        while (t-- > 0) {
            String[] nk = br.readLine().split(" ");
            int n = Integer.parseInt(nk[0]);
            String[] s = br.readLine().split(" ");

            long[] a = new long[n];
            int[] b = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = Long.parseLong(s[i]);
                a[i]+=i;
                b[i] = i;
            }

            Long[][] pairs = new Long[n][2];
            for (int i = 0; i < n; i++) {
                pairs[i][0] = a[i];
                pairs[i][1] = (long) b[i];
            }
            Arrays.sort(pairs, Comparator.comparingLong(pair -> pair[0]));
            for (int i = 0; i < n; i++) {
                a[i] = pairs[i][0];
                b[i] = pairs[i][1].intValue();
            }

            long[] dp = new long[n];
            Arrays.fill(dp, -1);
            long val = n;
            long ans = rec(0, val, a, b, dp, n) + n;
            out.println(ans);
        }
        out.flush();
        out.close();
        br.close();
    }

}

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

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

Think step by step.

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

1. **Edge Cases for Array Length**: Test with the smallest and largest possible values for `n`.
2. **Edge Cases for Array Values**: Test with values that are at the boundaries of the allowed range.
3. **Special Patterns in the Array**: Test with arrays that have specific patterns that might cause the algorithm to fail, such as arrays where all elements are the same, or arrays with a specific order.

Given the constraints, a small-scale test input that is complete and valid can be constructed as follows:

1. **Test Case 1**: Small array with a single element.
2. **Test Case 2**: Small array with all elements being the same.
3. **Test Case 3**: Small array with elements that match the condition `a_i = |a| + 1 - i`.
4. **Test Case 4**: Small array with elements that do not match the condition `a_i = |a| + 1 - i`.

Here is the test input:

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

This test input includes:
- A single-element array.
- An array with all elements being the same.
- An array that matches the condition for multiple elements.
- An array that does not match the condition for any elements.

These cases should help expose any bugs in the implementation, especially in how the code handles different types of input arrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
import java.util.*;

public class Main {

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

        while (t-- > 0) {
            int n = scan.nextInt();
            Map<Long, LinkedList<Integer>> map = new HashMap<>();
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = scan.nextLong();
                if (i == 0) continue;
                if (!map.containsKey(a[i] + i)) {
                    map.put(a[i] + i, new LinkedList<>());
                }
                map.get(a[i] + i).add(i);
            }
            System.out.println(rec(map, n));
        }
    }

    static int rec(Map<Long, LinkedList<Integer>> map, int n) {
        if (!map.containsKey((long) n)) return n;
        int max = 0;
        for (int i : map.get((long) n)) {
            max = Math.max(rec(map, n + i), max);
        }
        return max;
    }
}

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

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

Think step by step.

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

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

1. **Array Length and Elements**: The array length is 3, and the elements are 1, 3, and 2.
2. **Condition for Operation**: The operation can be performed if there exists an index \( i \) such that \( a_i = |a| + 1 - i \).

   - For \( i = 1 \): \( a_1 = 1 \) and \( |a| + 1 - 1 = 3 \) (not equal).
   - For \( i = 2 \): \( a_2 = 3 \) and \( |a| + 1 - 2 = 2 \) (not equal).
   - For \( i = 3 \): \( a_3 = 2 \) and \( |a| + 1 - 3 = 1 \) (equal).

3. **Operation**: The only valid index is \( i = 3 \). Performing the operation at \( i = 3 \) will append \( 3 - 1 = 2 \) zeros to the array, resulting in:
   - New array: \( [1, 3, 2, 0, 0] \).

4. **Further Operations**: Check the new array for valid indices:
   - For \( i = 1 \): \( a_1 = 1 \) and \( |a| + 1 - 1 = 5 \) (not equal).
   - For \( i = 2 \): \( a_2 = 3 \) and \( |a| + 1 - 2 = 4 \) (not equal).
   - For \( i = 3 \): \( a_3 = 2 \) and \( |a| + 1 - 3 = 3 \) (not equal).
   - For \( i = 4 \): \( a_4 = 0 \) and \( |a| + 1 - 4 = 2 \) (not equal).
   - For \( i = 5 \): \( a_5 = 0 \) and \( |a| + 1 - 5 = 1 \) (not equal).

   No further valid operations can be performed.

5. **Expected Output**: The maximum possible length of the array after performing the operation is 5.

This test case is small and valid, and it should help expose any issues in the buggy code by checking the conditions and operations carefully.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

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

int main()
{
    int t, n;
    cin >> t;
    while (t--)
    {
        cin >> n;
        map<ll, vector<int>> rec;
        vector<ll> nums(n + 1);
        for (int i = 1; i <= n; i++)
        {
            cin >> nums[i];
            if (i == 1)
                continue;
            rec[nums[i] + i - 1LL].push_back(i);
        }
        // ll len = n;
        ll res = n;
        auto dfs = [&](auto &&dfs, ll len) -> void
        {
            for (auto i : rec[len])
            {
                dfs(dfs, len + i - 1);
            }
            res = max(res, len);
        };
        dfs(dfs, n);
        cout << res << 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
3
1
1
2
1 2
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
using ll=long long;
//to find the original place in the sorted array
//iota(aid, aid + n, 0);
//sort(aid, aid + n, [&] (int l, int r) {
//return a[l] < a[r];
//});
//long long binPow(long long base, long long exponent) {
//long long result = 1;
//while (exponent > 0) {
//if (exponent % 2 == 1) {
//result *= base;
//}
//base *= base;
//exponent /= 2;
//}
//return result;
//}
//int findlog2(int n) {
//if (n == 0) {
//return -1;
//}
//int position = 0;
//int bitPosition = -1;
//while (n > 0) {
//if (n & 1) {
//bitPosition = position;
//}
//n = n >> 1;
//position++;
//}
//return bitPosition;
//}
ll ans=INT64_MIN;
void dfs(ll node,ll par,map<ll,vector<ll>> &g){
  ans=max(ans,node);
  for(auto child:g[node]){
    if(child==par) continue;
    dfs(child,node,g);
  }
  return;
}
int main(){
int tt;
cin>>tt;
while(tt--){
  ll n;
  ans=INT64_MIN;
  cin>>n;
  ll a[n];
  for(int i=0;i<n;i++){
    cin>>a[i];
  }
  map<ll,vector<ll>> g;
  for(ll i=1;i<n;i++){
    g[a[i]+i].push_back(a[i]+2*i);
  }
  dfs(n,-1,g);
  cout<<ans<<endl;
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code seems to be using a depth-first search (DFS) on a graph constructed from the input array, which is not the correct approach for this problem. The correct approach involves finding the maximum possible length of the array by repeatedly appending zeros based on the condition \( a_i = |a| + 1 - i \).

Here are some key points to consider for generating a challenging test input:
1. **Array with a single element**: This is a simple edge case that should always result in the same length.
2. **Array with multiple elements where the condition \( a_i = |a| + 1 - i \) is satisfied**: This should maximize the length of the array.
3. **Array with elements that do not satisfy the condition**: This should result in the original length of the array.
4. **Array with a mix of elements that satisfy and do not satisfy the condition**: This should test the code's ability to handle both cases correctly.

Given these points, let's generate a small-scale test input that includes these scenarios:

```plaintext
4
1
1
5
5 4 3 2 1
5
5 4 4 5 1
4
6 8 2 3
```

### Explanation:
1. **Test Case 1**: Array with a single element `[1]`. The length should remain 1.
2. **Test Case 2**: Array `[5, 4, 3, 2, 1]`. This array satisfies the condition \( a_i = |a| + 1 - i \) for all elements. The maximum possible length is 15.
3. **Test Case 3**: Array `[5, 4, 4, 5, 1]`. This array has some elements that satisfy the condition and some that do not. The maximum possible length is 11.
4. **Test Case 4**: Array `[6, 8, 2, 3]`. This array has no elements that satisfy the condition. The length should remain 4.

This test input should effectively expose the bug in the given buggy code by testing various scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
from collections import defaultdict

q = int(input())

for _ in range(q):
    n = int(input())

    al = list(map(int, input().split()))

    adj = defaultdict(list)

    for i, a in enumerate(al):
        adj[a+i].append(a+2*i)

    def dfs(node):
        visited.add(node)
        m = node
        for neighbor in adj[node]:
            if neighbor not in visited:
                m = max(m, dfs(neighbor))
        return m

    visited = set()
    root = len(al)
    ans = dfs(root)

    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

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

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
import heapq
import math
import os
import random
import sys
from types import GeneratorType
from typing import *
from collections import defaultdict, Counter, deque
from functools import lru_cache
from bisect import bisect_left, bisect_right, insort_left, insort_right

input = sys.stdin.readline
# sys.setrecursionlimit(100000)
rd = random.randint(10 ** 9, 2 * 10 ** 9)

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc

def sol():
    n = int(input())
    arr = list(map(int, input().split()))
    h = defaultdict(list)
    for i, x in enumerate(arr, 1):
        y = x + i - 1
        if y >= n and i > 1:
            h[y ^ rd].append(i - 1)

    @bootstrap
    def dfs(pre):
        res = 0
        if pre ^ rd in h:
            for x in h[pre ^ rd]:
                r = yield dfs(pre + x)
                res = max(res, r + x)

        yield res

    res = dfs(n)

    print(res + n)

p = int(input())

def main():
    for _ in range(p):
        sol()

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 provided buggy code, we need to focus on edge cases or scenarios where the code might fail. The key points to consider are:

1. **Initial Array Configuration**: The initial array configuration should be such that the operations can be performed in a way that the bug in the code becomes evident.
2. **Edge Cases**: Small arrays, arrays with specific patterns, and arrays where the conditions for the operations are met in a non-trivial way.
3. **Randomness in the Code**: The use of `rd` (a random number) in the code can lead to unexpected behavior if not handled correctly.

Given these considerations, let's create a small-scale test input that can expose the bug:

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

### Explanation:

1. **Test Case 1:**
   - **Array**: `[1, 2, 3, 4, 5]`
   - **Expected Output**: `10`
   - **Reason**: The array can be transformed by choosing positions where the conditions are met, leading to the maximum length of 10.

2. **Test Case 2:**
   - **Array**: `[5, 4, 3, 2]`
   - **Expected Output**: `10`
   - **Reason**: Similar to the first test case, the array can be transformed to achieve a maximum length of 10.

3. **Test Case 3:**
   - **Array**: `[1]`
   - **Expected Output**: `1`
   - **Reason**: No operations can be performed, so the length remains 1.

These test cases cover different scenarios, including small arrays and arrays where the conditions for the operations are met in a non-trivial way. This should help in exposing any bugs in the provided code.