**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;

//Change Class Name to Problem Name
public class D1_2003 {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int test = Integer.parseInt(br.readLine());
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        for (int i = 0; i < test; i++) {
            run(br, bw);
        }
        bw.flush();
    }

    public static void run(BufferedReader br, BufferedWriter bw) throws IOException{
        String nm = br.readLine();
        StringTokenizer tokenNM = new StringTokenizer(nm);
        int n = Integer.parseInt(tokenNM.nextToken());
        long m = Long.parseLong(tokenNM.nextToken());
        long max = 0L;
        for (int i = 0; i < n; i++) {
            String line = br.readLine();
            StringTokenizer st = new StringTokenizer(line);
            int count = Integer.parseInt(st.nextToken());
            long[] arr = new long[200002];
            for (int j = 0; j < count; j++) {
                int now = Integer.parseInt(st.nextToken());
                if(now <= count + 2) arr[now] = 1;
            }
            int flag = 0;
            while(arr[flag] != 0) flag++;
            arr[flag] = 1;
            while(arr[flag] != 0) flag++;
            max = Math.max(flag, max);
        }
        long ans = m*(m+1) / 2;
        long point = Math.min(max, m);
        ans -= point*(point+1)/2;
        ans += max*(point + 1);
        bw.write(String.valueOf(ans)+'\n');
    }
}

/*
찾아야 할 것들
1. int 오버플로우, out of bounds
2. 특수한 경우(n=1?)
3. 잘못된 가정 & 증명

*아무것도 하지 않는 대신 무언가를 수행하기. 체계적인 상태를 유지.
*적어두기 & 구하고자 하는 것의 계층 구조 표현 -> 정확히 & 입체적으로 파악
*한가지 접근 방식에 얽메이지 말기 -> 응용의 끝 알기
*/
// 알고리즘의 작동방식 "완전히" 이해하려 노력하기
// 수행 목표
// 1. 아이디어 획득 : "추론"
//     - 문제 알고리즘/특징의 "증명"으로 아이디어
//         - {BruteForce, greedy, D&C, DP, graph, math}-
//     - “책임” 뽑아내기
//         - 각 기능들이 어떤 책임을 가지고 있는지
//         - “어떤 패턴”으로 동작하는지
//     - 가장 Naive한 상태/알고리즘부터 시작하기 (완전 탐색, 단순 자료구조)
//     - 단위 동작의 조건 분기 파악
//     - 가장 극단적인 경우에서 추론 (가장 차이가 뚜렷한 예제)
// 2. "처음"부터 직접 경우의 수 전개(수식&도식화, 엄밀한 조건 정리)
//     - 알고리즘 내에 어떤 역할들이 있는지 -> 직접 들어가보기
//     - 알고리즘 내에 어떤 기능들이 있는지 -> 직접 수행해보기
// 3. cutting | 구현(차근차근 단순화 & 최적화)

/*
take notes.
// 다시 보는용이 아닌
// 현재의 흐름을 가장 잘 이어갈 수 있도록 !!!

*/

// commit 시 피드백할 것 Message로 남겨두기!!
// 틀리면 느낌표 추가하기

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code seems to have issues with how it calculates the `max` value, which is used to determine the final answer. The code incorrectly handles the `mex` values and may not correctly account for the maximum possible value of `x` after operations.

2. **Edge Cases**:
   - **Single Sequence**: Test with a single sequence that has a specific pattern.
   - **Large Initial Value of `m`**: Test with a large `m` to see if the code handles large sums correctly.
   - **Special Values in Sequences**: Test with sequences that have values close to the length of the sequence plus a small constant.
   - **Sequences with Repeated Values**: Test with sequences that have many repeated values to ensure the `mex` calculation is correct.

3. **Generate Test Input**:
   - Use a small number of test cases to keep the input small.
   - Use sequences that are likely to cause the code to fail.

Here is a small-scale test input that includes these edge cases:

```plaintext
3
1 10
3 0 1 2
2 5
3 0 1 2
2 2 3
3 100
5 0 1 2 3 4
5 0 1 2 3 4
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`, `m = 10`
   - Sequence: `3 0 1 2`
   - The `mex` of this sequence is `3`. For any initial value of `x` from `0` to `10`, the maximum value of `x` after operations will be `3`.
   - Expected output: `3 * (10 + 1) = 33`

2. **Test Case 2**:
   - `n = 2`, `m = 5`
   - Sequence 1: `3 0 1 2`
   - Sequence 2: `2 2 3`
   - The `mex` of the first sequence is `3`, and the `mex` of the second sequence is `0`. The maximum value of `x` after operations will be `3`.
   - Expected output: `3 * (5 + 1) = 18`

3. **Test Case 3**:
   - `n = 3`, `m = 100`
   - Sequence 1: `5 0 1 2 3 4`
   - Sequence 2: `5 0 1 2 3 4`
   - The `mex` of both sequences is `5`. For any initial value of `x` from `0` to `100`, the maximum value of `x` after operations will be `5`.
   - Expected output: `5 * (100 + 1) = 505`

This test input should help expose the bug in the buggy code by challenging its handling of `mex` values and large `m` values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
#include <iostream>
#define endl "\n"
#define ul unsigned long long
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define Yasser ios_base::sync_with_stdio(false); cin.tie(nullptr);cout.tie(nullptr);
#define yes cout<<"YES"<< endl;
#define no cout<<"NO"<< endl;
#define f(n) for(int i = 0; i < n; i++)
#define ones(n) __builtin_popcountll(n)
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <queue>
#include <stack>
#include <unordered_set>
#include <iomanip>
#include <cstring> // For memset
typedef long long ll;
//#include <bits/stdc++.h>

int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};
char di[] = {'D', 'L', 'U', 'R'};
using namespace std;

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

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

const ll mod = 1e9+7 , inf = 1e18, M = 5e1 + 5, N = 2e5 + 5;

ll mul(ll a, ll b){
    return (a%mod * b%mod)%mod;
}

ll add(ll a, ll b){
    return (a%mod + b%mod)%mod;
}
ll subtract(ll a, ll b){
    return (a%mod - b%mod + 2*mod)%mod;
}

ll FastPow(ll n, ll m) {
    ll ans = 1;
    while (m) {
        if (m & 1) {
            ans *= n, ans %= mod;
        }
        m /= 2;
        n *= n, n %= mod;
    }
    return ans;
}
bool dir(string &a, string &b){
    for (int i = 0; i < 2; ++i) {
        for (int j = 0; j < 2; ++j) {
            if(a[i] == b[j])
                return true;
        }
    }
    return false;
}

void Yassor() {
    int n, m, mex = 0;
    cin >> n >> m;
    while (n--){
        int l;
        cin >> l;
        vector<int> v(l),vis(l + 4);
        for(auto &i: v) {cin >> i;(i<vis.size())?vis[i]=1:0;}
        bool f = false;
        for (int i = 0; i <= l+1; ++i) {
            if(!vis[i]) {
                if (f) {
                    mex = max(mex, i);
                    break;
                } else {
                    f = true;
                }
            }
        }
    }
    ll ans = 0;
    for (int i = 0; i <= m; ++i) {
        ans += max(mex,i);
    }
    cout << ans << endl;

}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("errors.txt", "w", stderr);
#endif
    Yasser
    int t = 1;
    cin >> t;
    while (t--) {
        Yassor();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 2
2 0 1
2 0 2
2 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <numeric>
#include <map>
#include <unordered_map>
#include <math.h>
#include <deque>
#include <queue>
using namespace std;

long long mod = 1e9 + 7;
inline long long mpow(long long b, long long ex)
{
    if (b == 1)
        return 1;
    long long r = 1;
    while (ex)
    {
        if (ex & 1)
            r = (r * b) % mod;
        ex = ex >> 1;
        b = (b * b) % mod;
    }
    return r;
}

int dfs(int u, unordered_map<int, int> &f, unordered_map<int, vector<int>> &edges)
{
    if (f.find(u) != f.end())
        return f[u];

    f[u] = u;
    for (int v : edges[u])
    {
        f[u] = max(f[u], dfs(v, f, edges));
    }
    return f[u];
}

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

    int t;
    cin >> t;
    for (int t1 = 0; t1 < t; ++t1)
    {
        int n, m;
        cin >> n >> m;
        vector<vector<int>> seqs(n);
        for (int i = 0; i < n; ++i)
        {
            int l;
            cin >> l;
            seqs[i] = vector<int>(l, 0);
            for (int j = 0; j < l; ++j)
            {
                cin >> seqs[i][j];
            }
        }

        // for each sequence calculate mex_1 and mex_2

        vector<pair<int, int>> mexs(n);

        for (int i = 0; i < n; ++i)
        {
            sort(seqs[i].begin(), seqs[i].end());

            int mex1 = 0;
            for (int a : seqs[i])
            {
                if (a == mex1)
                {
                    mex1 = a + 1;
                }
            }
            int mex2 = mex1 + 1;
            for (int a : seqs[i])
            {
                if (a == mex2)
                {
                    mex2 = a + 1;
                }
            }

            mexs[i] = {mex1, mex2};
        }

        unordered_map<int, int> f;

        unordered_map<int, vector<int>> edges;

        for (auto &[u, v] : mexs)
        {
            edges[u].push_back(v);
        }

        int max_mex = 0;
        for (auto &[m1, m2] : mexs)
        {
            max_mex = max(max_mex, dfs(m1, f, edges));
        }

        long long ans = 0;
        int i = 0;
        for (; i <= min(m, (int)2e5 + 10); ++i)
        {
            ans += max(max_mex, dfs(i, f, edges));
        }

        if (i <= m)
        {
            long long sum_a = ((long long)m * (long long)(m + 1)) / 2;

            long long sum_b = ((long long)(i - 1)) * ((long long)i) / 2;

            ans += sum_a - sum_b;
            // i + (i + 1) + .. + m
        }

        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 create a challenging test input that exposes bugs in the provided code, let's consider the following aspects:

1. **Edge Cases**: Test with sequences that have a small number of elements, and sequences that are already sorted or contain repeated elements.
2. **Large Values**: Test with sequences that contain large values to check the handling of large numbers.
3. **Special Sequences**: Test with sequences that have specific patterns, such as all elements being the same, or sequences that are designed to make the MEX calculation non-trivial.
4. **Multiple Test Cases**: Include multiple test cases to ensure the code handles different scenarios correctly.

Let's construct a test input that covers these aspects:

1. **Test Case 1**: A simple test case with a single sequence containing small numbers.
2. **Test Case 2**: A test case with a sequence that has repeated elements and a large MEX value.
3. **Test Case 3**: A test case with a sequence that contains a large number of elements, all being the same.
4. **Test Case 4**: A test case with a sequence that is already sorted and has a small MEX value.
5. **Test Case 5**: A test case with a sequence that has a large number of elements and a non-trivial MEX value.

Here is the test input:

```plaintext
5
1 1
3 0 1 2
1 10
5 1 2 3 4 5
1 5
1 1 1 1 1 1
1 1
1000000000 1000000000
1 10
10 0 1 2 3 4 5 6 7 8 9
```

Explanation:
1. **Test Case 1**: A single sequence with elements 0, 1, 2. The MEX value is 3, and the sum of `f(i)` for `i` from 0 to 1 is 3 + 3 = 6.
2. **Test Case 2**: A single sequence with elements 1, 2, 3, 4, 5. The MEX value is 0, and the sum of `f(i)` for `i` from 0 to 10 is 10 * 5 = 55.
3. **Test Case 3**: A single sequence with all elements being 1. The MEX value is 0, and the sum of `f(i)` for `i` from 0 to 5 is 5 * 1 = 5.
4. **Test Case 4**: A single sequence with a very large element. The MEX value is 0, and the sum of `f(i)` for `i` from 0 to 1 is 1 * 1 = 1.
5. **Test Case 5**: A single sequence with elements 0 to 9. The MEX value is 10, and the sum of `f(i)` for `i` from 0 to 10 is 10 * 11 = 110.

This test input should help in exposing potential 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: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
def mex (l) :
    l = set(l)
    number_set = set(range(0, max(l)+4))
    q = number_set- l
    min1 = min(q)
    q.remove(min1)
    min2 = min(q)
    return min2

t = int(input())
for _ in range(t):
    q = 0
    (n,m) = list(map(int, input().split()))
    bx = []
    for i in range (n):
        ex= list(map(int, input().split()))
        ex.pop(0)
        bx.append(mex(ex))
    l = max(bx)
    for i in range(m+1):
        q += max(l , i)
    print(q)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the issues in the provided code:

### Issues in the Buggy Code:
1. **Incorrect MEX Calculation**:
   - The function `mex` calculates the second smallest missing number instead of the smallest missing number. This is because the function removes the smallest missing number (`min1`) and then finds the next smallest missing number (`min2`). The correct MEX should be `min1`.

2. **Incorrect Sum Calculation**:
   - The code calculates the sum `q` as the sum of the maximum of `l` (the maximum MEX value of all sequences) and each integer from 0 to `m`. This is incorrect because the correct value of `f(i)` should be the maximum MEX value that can be achieved starting from `i`.

### Challenging Test Input:
To expose these issues, we can create a test case where the MEX values of the sequences are such that the incorrect MEX calculation in the buggy code will lead to an incorrect result. We will use a small-scale test case to ensure it is complete and valid.

### Test Input:
```plaintext
1
3 4
2 0 2
3 2 3 3
4 7 0 1 5
```

### Explanation:
- **Test Case**:
  - **Number of sequences (n)**: 3
  - **Maximum value of x (m)**: 4
  - **Sequences**:
    - Sequence 1: [0, 2]
    - Sequence 2: [2, 3, 3]
    - Sequence 3: [7, 0, 1, 5]

### Expected Output:
- The correct output for this test case is 16, as explained in the problem description.

### How the Test Input Exposes the Bug:
- **MEX Calculation**:
  - For Sequence 1: `mex([0, 2]) = 1`
  - For Sequence 2: `mex([2, 3, 3]) = 0`
  - For Sequence 3: `mex([7, 0, 1, 5]) = 2`
  - The maximum MEX value (`l`) is 2.
- **Sum Calculation**:
  - The correct values of `f(i)` are:
    - `f(0) = 3`
    - `f(1) = 3`
    - `f(2) = 3`
    - `f(3) = 3`
    - `f(4) = 4`
  - The sum is `3 + 3 + 3 + 3 + 4 = 16`.

The buggy code will incorrectly calculate the MEX values and the sum, leading to a different result. This test input should help identify the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
for _ in range(int(input())):
    n,m = map(int,input().split())
    # graph = {}
    lar = 0
    # larBud = 0
    # larPar = []
    for x in range(n):
        st = set()
        a = list(map(int,input().split()))
        n,a = a[0], a[1:]
        mn1, mn2 = None, None

        for x in a:
            st.add(x)

        for x in range(max(a)+3):
            if mn1 is None and x not in st:
                mn1 = x
            elif mn2 is None and x not in st:
                mn2 = x
                break

        # if mn1 in graph:
        #     graph[mn1].append(mn2)
        # else:
        #     graph[mn1] = [mn2]

        if mn2 > lar:
            # larPar = [mn1]
            lar = mn2
            # larBud = mn1
        #print(">",mn1, mn2)

    # arr = list(range(m+1))
    # #print(arr)
    # for x in larPar:
    #     arr[x] = lar

    # for x in range(m+1):
    #     # if x in graph:
    #     #     for i in graph[x]:
    #     #         if i <= m:
    #     #             arr[x] = max(arr[x], arr[i])

    #     arr[x] = max(arr[x], lar)
    # #print(lar, larPar)
    #print(arr)

    arr = [max(x, lar) for x in range(m+1)]
    print(sum(arr))

**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 4
2 0 2
3 2 3 3
4 7 0 1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
import java.util.*;
import java.io.*;
public class D1 {
  public static void main(String[] args) throws IOException {
    BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);

    StringTokenizer st = new StringTokenizer(r.readLine());
    int T = Integer.parseInt(st.nextToken());
    while(T-- > 0) {
      st = new StringTokenizer(r.readLine());
      int n = Integer.parseInt(st.nextToken());
      long m = Long.parseLong(st.nextToken());
      int[][] arr = new int[n][];
      for(int i = 0; i < n; i++) {
        st = new StringTokenizer(r.readLine());
        int l = Integer.parseInt(st.nextToken());
        arr[i] = new int[l];
        for(int j = 0; j < l; j++) {
          arr[i][j] = Integer.parseInt(st.nextToken());
        }
        Arrays.sort(arr[i]);
      }
      long ans = 0;
      for(int i = 0; i < n; i++) {
        boolean found = false;
        for(int j = 0; j < arr[i].length + 3; j++) {
          if(Arrays.binarySearch(arr[i], j) < 0) {
            if(!found) {found = true;}
            else {ans = Math.max(ans, j); break;}
          }
        }
      }
      long ret = 0;
      for(int i = 0; i <= m; i++) {
        if(i <= ans) {ret += ans;}
        else {ret += i;}
      }
      pw.println(ret);
    }

    pw.close();
  }
}

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

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

Think step by step.

Response:
```plaintext
1
3 1
2 0 2
2 1 2
2 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
20
1281
6
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(1, 1, 3, 0, 3, 3) = 2$, and choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}, a_{3, 5}) = \text{mex}(2, 1, 3, 0, 3, 3) = 4$. It can be proved that Turtle can't make the value of $x$ greater than $4$, so $f(1) = 4$.

It can be seen that $f(0) = 4$, $f(1) = 4$, $f(2) = 4$, $f(3) = 4$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 4 + 4 + 4 + 4 = 20$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 3$. So $f(0) + f(1) = 3 + 3 = 6$.

Buggy Code:
import java.util.*;

public class Main{
    public static int dfs(int max[],ArrayList<ArrayList<Integer>> arr,int s,boolean vis[],int val){
        vis[s]=true;
        for(int i:arr.get(s)){
            max[s]=Math.max(Math.max(val,max[s]),dfs(max,arr,i,vis,val));
        }
        return max[s];
    }
    public static void main(String[] args) {
       Scanner sc = new Scanner(System.in);
       int tc = sc.nextInt();
       while(tc-->0){
           int n = sc.nextInt(), m = sc.nextInt();
           ArrayList<ArrayList<Integer>> arr = new ArrayList<>();
           for(int i=0;i<=10;i++){
               arr.add(new ArrayList<>());
           }
           int size=0;
           for(int i=0;i<n;i++){
               int len=sc.nextInt();
               HashSet<Integer> hs = new HashSet<>();
               for(int j=0;j<len;j++){
                   hs.add(sc.nextInt());
               }
               int val=0;
               while(hs.contains(val)) val++;
               hs.add(val);
               int val2=val;
               while(hs.contains(val2)) val2++;
               arr.get(val).add(val2);
               size=Math.max(val,size);
               size=Math.max(val2,size);
           }
//           System.out.println(arr);
           int max[]=new int[size+1];
           boolean vis[]=new boolean[size+1];
           for(int i=0;i<=size;i++){
               max[i]=i;
           }
           int mx=-1;
           for(int i=0;i<=size;i++) {
               if(!vis[i]){
                   int val=dfs(max, arr, i,vis,-1);
                   mx=Math.max(mx,val);
               }
           }
           for(int i=0;i<=size;i++){
               mx=Math.max(mx,max[i]);
               max[i]=mx;
           }

//           System.out.println(Arrays.toString(max));
           long sum=0;
           for(int i=0;i<=m;i++){
               if(i>=size) sum+=i;
               else sum+=Math.max(mx,i);
           }
           System.out.println(sum);
       }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 2
2 0 1
2 1 2
2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
#include <iostream>
using namespace std;
int main()
{

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

		if (m >= a + b + c)
		{
			cout << endl << (a + b + c)<<endl;
		}
		else if (m >= a && m >= b && c >= (2 * m - (a + b))) {
			cout << endl << 2 * m<<endl;
		}
		else if (m >= a && m <= b && c <= (m - a)) {
			cout <<endl<< m + a + c<<endl;
		}
		else if (m >= a && m <= b && c >= (m - a)) {
			cout << endl << 2 * m<<endl;
		}
		else if (m <= a && m>= b && c <= (m - b)) {
			cout <<endl<< m + b + c<<endl;
		}
		else if (m <= a && m >= b && c >= (m - b)) {
			cout <<endl<< m + m<<endl;
		}
		else if (m <= a && m <= b)
		{
			cout << endl << m + m<<endl;
		}

	}

	return 0;
}

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

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

Think step by step.

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

1. **Overflow and Data Type Issues**: The code uses `long long int` for the variables, so we don't need to worry about overflow. However, we should ensure that the input values are within the specified range.
2. **Logical Conditions**: The conditions in the code seem to be complex and might not cover all edge cases. We need to test scenarios where the conditions might not be met correctly.

Let's consider the following edge cases:
- **Case 1**: When `m` is very small compared to `a` and `b`.
- **Case 2**: When `m` is very large compared to `a` and `b`.
- **Case 3**: When `c` is very small.
- **Case 4**: When `c` is very large.
- **Case 5**: When `a` and `b` are equal.
- **Case 6**: When `a` and `b` are different and `c` is just enough to fill the remaining seats.

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

```plaintext
6
10 15 5 10
10000000 5000000 5000000 1
1 1 1 0
2 1 1 1
5 5 5 5
10 10 10 10
```

Explanation of the test cases:
1. `10 15 5 10`: `m` is smaller than `a` but `c` is large enough to fill the remaining seats.
2. `10000000 5000000 5000000 1`: `m` is very large, `a` and `b` are equal, and `c` is very small.
3. `1 1 1 0`: `m` is very small, and `c` is 0.
4. `2 1 1 1`: `m` is small, and `c` is just enough to fill the remaining seats.
5. `5 5 5 5`: `m` is equal to `a` and `b`, and `c` is equal to `m`.
6. `10 10 10 10`: `m` is equal to `a`, `b`, and `c`.

These test cases should help expose any logical 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: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

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

using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int count = 0;

        int seats, r1, r2, any;
        cin >> seats >> r1 >> r2 >> any;

        int avail1 = seats;
        int avail2 = seats;

        if (avail1 >= r1)
        {
            avail1 -= r1;
            count += r1;
        }
        else
        {
            count += r1 - avail1;
            avail1 = 0;
        }

        if (avail1 > 0)
        {
            if (avail1 >= any)
            {
                avail1 -= any;
                count += any;
                any = 0;
            }
            else
            {
                count += any - avail1;
                avail1 = 0;
            }
        }

        // r2

        if (avail2 >= r2)
        {
            avail2 -= r2;
            count += r2;
        }
        else
        {
            count += r1 - avail2;
            avail2 = 0;
        }

        if (avail2 > 0)
        {
            if (avail2 >= any)
            {
                avail2 -= any;
                count += any;
            }
            else
            {
                count += any - avail2;
                avail2 = 0;
            }
        }

        cout << count << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 15 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
#include<iostream>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int sum=0;
        int m,a,b,c;
        cin>>m>>a>>b>>c;
        if(m==1){
            cout<<2<<endl;
        }
        else{
            int rem=2*m;
            if(a<=m){
                sum+=a;
                rem-=a;
            }
            else{
                sum+=(a-m);
                rem-=(a-m);
                }
            if(b<=m){
                sum+=b;
                rem-=b;
            }
            else{
                sum+=(b-m);
                rem-=(b-m);
            }
            if(c<=(rem)){
                sum+=c;
            }
            else{
                sum+=(c-rem);
            }
        cout<<sum<<endl;
        }
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
n=int(input())
for i in range(1,n+1):
    x=(input()).split()
    x=list(map(int,x))
    m=x[0]
    a=x[1]
    b=x[2]
    c=x[3]
    #print(m,a,b,c)
    if(m<=a and m<=b):
        print(2*m)
    elif(m>=a and m<=b):
        if(m>=a+c):
            print(2*m)
        else:
            print(m+a+c)
    elif(m<=a and m>=b):
        if(m<=b+c):
            print(2*m)
        else:
            print(m+b+c)
    elif(m>=a and m>=b):
        if(2*m<=a+b+c):
            print(2*m)
        else:
            print(a+b+c)

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

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

Think step by step.

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

1. **Case 1: \( m \leq a \) and \( m \leq b \)**
   - The code correctly prints \( 2m \).

2. **Case 2: \( m \geq a \) and \( m \leq b \)**
   - The code checks if \( m \geq a + c \). If true, it prints \( 2m \). If false, it prints \( m + a + c \).

3. **Case 3: \( m \leq a \) and \( m \geq b \)**
   - The code checks if \( m \leq b + c \). If true, it prints \( 2m \). If false, it prints \( m + b + c \).

4. **Case 4: \( m \geq a \) and \( m \geq b \)**
   - The code checks if \( 2m \leq a + b + c \). If true, it prints \( 2m \). If false, it prints \( a + b + c \).

The bug in the code is that it does not handle cases where \( m \) is exactly equal to \( a \) or \( b \) correctly. Specifically, it might not account for the fact that \( c \) (monkeys with no preference) can be used to fill the remaining seats.

To expose this bug, let's create a test case where \( m \) is exactly equal to \( a \) or \( b \) and there are enough \( c \) monkeys to fill the remaining seats.

### Test Case
```plaintext
1
10 10 5 5
```

### Explanation
- **m = 10**: There are 10 seats in each row, making a total of 20 seats.
- **a = 10**: 10 monkeys want to sit in row 1.
- **b = 5**: 5 monkeys want to sit in row 2.
- **c = 5**: 5 monkeys have no preference.

### Expected Output
- Row 1: 10 monkeys (all monkeys with preference for row 1).
- Row 2: 5 monkeys (all monkeys with preference for row 2) + 5 monkeys with no preference.

Total: \( 10 + 5 + 5 = 20 \) monkeys.

The correct output should be 20, but the given code might not handle this case correctly, especially in the logic for \( m \geq a \) and \( m \geq b \). This test case should expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
t = int(input())

data = [
    # [10, 5, 5, 10],
    # [3, 6, 1, 1],
    # [15, 14, 12, 4],
    # [1, 1, 1, 1],
    # [420, 6, 9, 69]
    ]

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

# посчитать сколько посадить в ряд A и если остаются места, то подсадить к ним в ряд C, также проделать для B.
# если в ряду A желающих больше, чем мест на ряду, то в ответ записать их разницу, также проделать для B

for crt in range(len(data)):
    ans = 0
    free = 0 # количество свобондных мест на любом ряду для C
    m = data[crt][0]
    a = data[crt][1]
    b = data[crt][2]
    c = data[crt][3]

    if m >= a:
        ans += a
        # print(ans)
        free += m - a
        a = 0

    if m >= b:
        ans += b
        # print(ans)
        free += m - b
        b = 0

    if free >= c:
        ans+=c
        # print(ans)
        c = 0

    if m < a:
        ans += a - m
        a = 0

    if m < b:
        ans += b - m

    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
10 12 12 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
t=int(input())
while(t>0):
    m,a,b,c=map(int,input().split())
    cnt=0
    if m>a+b+c:
        cnt=a+b+c
    else:
        if a<m:
            cnt+=a
            if c<m-a:
                cnt+=c
            else:
                cnt+=m-a
        else :
            cnt+=m
        if b<m:
            cnt+=b
            if c-m+a>0 and c-m+a<m-b:
                cnt+=c-m+a
            if c<m-b:
                cnt+=c
            else:
                cnt+=m-b
        else:
            cnt+=m
    print(cnt)
    t-=1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){

        int m=sc.nextInt();
        int m1=m;
        int m2=m;
        int sum=0;
        //int totalseats=2*m;
        //int temp=totalseats;
        int a=sc.nextInt();
        int b=sc.nextInt();
        int c=sc.nextInt();
        if(m1>=a){
            sum+=a;
            m1=m1-a;
        }
        else{
            sum+=(a-m1);
            m1=0;
        }
        if(m2>=b){
            sum+=b;
           m2=m2-b;
        }
        else{
            sum+=(b-m2);
            m2=0;
        }
        if((m1+m2)>0){
            if(c<(m1+m2)){
                sum+=c;

            }
            else{
                sum+=(m1+m2);
            }
        }

        System.out.println(sum);
        }

    }
}

**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 the following edge cases and scenarios:

1. **Overflow of preferences**: When the number of monkeys that prefer a specific row exceeds the number of seats available in that row.
2. **No preference monkeys**: When the number of monkeys with no preference is much larger than the remaining seats.
3. **Equal distribution**: When the number of monkeys that prefer each row is equal to the number of seats available.
4. **Single seat scenarios**: When there is only one seat available in each row.
5. **Maximal preference scenarios**: When the number of monkeys that prefer a specific row is exactly equal to the number of seats available.

Let's consider a test case that combines some of these scenarios to expose the bug:

- **m = 3**: 3 seats in each row.
- **a = 3**: 3 monkeys prefer row 1.
- **b = 2**: 2 monkeys prefer row 2.
- **c = 5**: 5 monkeys have no preference.

In this case:
- All 3 monkeys that prefer row 1 will be seated.
- All 2 monkeys that prefer row 2 will be seated.
- There are 3 seats left (1 in row 1 and 2 in row 2).
- 3 out of the 5 monkeys with no preference will be seated.

The expected output should be 8, but the buggy code might not handle the remaining seats correctly.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
import java.util.*;

public class HardProblem {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();

        while(t-- > 0){

            int m = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();

            if((a + b + c) <= m){
                System.out.println(a + b + c);
            }
            else {
                int sum = 0;

                int firstrow = (m - a);

                if (firstrow <= 0) {
                    sum += m;
                } else {
                    sum += a;
                    if ((m - a) <= c) {
                        sum += (m - a);
                        c = c - (m - a);
                    } else {
                        sum += c;
                    }
                }

                int secrow = (m - b);

                if (secrow <= 0) {
                    sum += m;
                } else {
                    sum += b;
                    if ((m - b) <= c) {
                        sum += (m - b);
                    } else {
                        sum += c;
                    }
                }

                System.out.println(sum);
            }

        }
    }
}

**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
10 10 10 10
10 5 15 10
10 15 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

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

using namespace std;
#define int long long

signed main(){
    int t;
    cin >> t;
    while(t--){
        int m, a, b, c;
        cin >> m >> a >> b >> c;
        int monkeys = a+b+c;
        if(m>a){
            a=0;
            c-=m-a;
        }
        else{
            a-=m;
        }
        if(m>b){
            b=0;
            c-=m-b;
        }
        else{
            b-=m;
        }
        if(c<0)c=0;
        int ans = a+b+c;
        ans=monkeys-ans;
        cout << 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:
```plaintext
1
10 15 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

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

int main(){
    llu tt,m,a,b,c,rema,remb,seated_a,seated_b;
    cin >> tt;
    while (tt--) {
        cin >> m >> a >> b >> c;

        // can't sit them all
        if ( a > m && b < m ){
            cout << (a-m)+(b)+(c) << "\n";
            continue;
        }

        // if ( b > m && a < m){

        //     return 0;
        // }

        // if ( a > m && b > m){

        //     return 0;
        // }

        if(m==1 && a==1 && b==1 && c==1){
            cout << "2" << "\n";
            continue;
        }

        //can sit them all
        if (a < m && b < m){
            rema = m-a;
            remb = m-b;
        }
        if (c <= rema+remb){
            cout << a+b+c << "\n";
            continue;
        }

    }
}

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

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

Think step by step.

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

1. **Overflow or Underflow**: Ensure that the input values are within the specified limits but push the boundaries.
2. **Unusual Preferences**: Test cases where the preferences of the monkeys are such that they cannot be seated according to the code's logic.
3. **Equal Distribution**: Test cases where the number of seats and the number of monkeys with preferences are equal or nearly equal.
4. **Mixed Preferences**: Test cases where the number of monkeys with no preference is significant and can affect the seating arrangement.

Given the code's logic, the following test input should expose the bug:

```plaintext
3
10 5 5 10
3 6 1 1
15 14 12 4
```

### Explanation:

1. **Test Case 1:**
   - `m = 10`, `a = 5`, `b = 5`, `c = 10`
   - The code should be able to seat all 20 monkeys (5 in row 1, 5 in row 2, and 10 with no preference split between the two rows).
   - The correct output should be `20`.

2. **Test Case 2:**
   - `m = 3`, `a = 6`, `b = 1`, `c = 1`
   - There are only 3 seats in total, but 6 monkeys want to sit in row 1, which is impossible. The code should only seat the 3 monkeys that can fit (3 in row 1 and 1 in row 2).
   - The correct output should be `5`.

3. **Test Case 3:**
   - `m = 15`, `a = 14`, `b = 12`, `c = 4`
   - There are 14 monkeys that want to sit in row 1 and 12 that want to sit in row 2, but only 15 seats in each row. The code should seat 14 monkeys in row 1 and 12 monkeys in row 2, with 4 monkeys with no preference split between the remaining seats.
   - The correct output should be `30`.

This test input should help identify the issues in the buggy code by covering various edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
for i in range(int(input())):
    x = 0
    m,a,b,c = map(int,input().split())
    if m != 1:
        a1 = min(m,a)+min(m,b)+min(m,c)
        print(a1)
    else:
        print(2)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
contador = 0
fila1 = 0
fila2 = 0
casos = int(input())
for i in range(casos) :
    entrada = input()
    sillas_fila,a,b,c= list(map(int, entrada.split()))
    fila1 = sillas_fila
    fila2 = sillas_fila
    if fila1 - a >= 0 :
        fila1 -= a
        contador += a
    elif fila1 - a < 0 :
        contador += fila1
        fila1 = 0
    if fila2 - b >= 0:
        fila2 -= b
        contador += b
    elif fila2 - b < 0 :
        contador += fila2
        fila2 = 0
    if fila1 > 0 :
        if fila1 - c >= 0 :
            fila1 -= c
            contador += c
            c = 0
        elif fila1 - c < 0 :
            contador += fila1
            fila1 = 0
    if fila2 > 0:
        if fila2 - c >= 0:
            fila2 -= c
            contador += c
        elif fila2 - c < 0 :
            contador += fila2
            fila2 = 0

    print(contador)
    contador = 0
    fila1 = 0
    fila2 = 0

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the code might incorrectly handle the distribution of monkeys across the two rows. Specifically, we should look at cases where the number of monkeys with no preference (c) is significant and could be distributed in a way that maximizes the seating.

### Step-by-Step Analysis:
1. **Initial Setup:**
   - The code initializes `fila1` and `fila2` to the number of seats in each row.
   - It then assigns monkeys with specific preferences to their respective rows, reducing the available seats and increasing the `contador` (the count of seated monkeys).

2. **Handling Monkeys with No Preference (c):**
   - The code attempts to assign monkeys with no preference to the remaining seats in `fila1` and `fila2`.
   - The issue might arise if the code does not correctly handle the distribution of `c` when there are remaining seats in both rows.

3. **Edge Cases:**
   - **Case 1:** All seats are taken by monkeys with specific preferences, and `c` is zero.
   - **Case 2:** All seats are taken by monkeys with specific preferences, and `c` is greater than zero but cannot be seated.
   - **Case 3:** Only one row has seats left, and `c` is greater than zero.
   - **Case 4:** Both rows have seats left, and `c` is greater than zero.

### Challenging Test Input:
Let's create a test input that challenges the code to handle the distribution of `c` correctly when there are seats left in both rows and `c` is significant.

```plaintext
3
10 5 5 10
3 6 1 1
15 14 12 4
```

### Explanation:
1. **Test Case 1:**
   - `m = 10`, `a = 5`, `b = 5`, `c = 10`
   - Both rows have 10 seats.
   - 5 monkeys with preference for row 1 and 5 monkeys with preference for row 2 can be seated, leaving 10 seats available for `c`.
   - The correct output should be 20, as all 10 monkeys with no preference can be seated in the remaining seats.

2. **Test Case 2:**
   - `m = 3`, `a = 6`, `b = 1`, `c = 1`
   - Only 3 seats in each row.
   - 3 monkeys with preference for row 1 can be seated, leaving 0 seats in row 1.
   - 1 monkey with preference for row 2 can be seated, leaving 2 seats in row 2.
   - 1 monkey with no preference can be seated in the remaining 2 seats in row 2.
   - The correct output should be 5.

3. **Test Case 3:**
   - `m = 15`, `a = 14`, `b = 12`, `c = 4`
   - 15 seats in each row.
   - 14 monkeys with preference for row 1 can be seated, leaving 1 seat in row 1.
   - 12 monkeys with preference for row 2 can be seated, leaving 3 seats in row 2.
   - 4 monkeys with no preference can be seated in the remaining 4 seats (1 in row 1 and 3 in row 2).
   - The correct output should be 30.

This test input should expose any issues in the code's handling of the distribution of monkeys with no preference and the remaining seats in both rows.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int N = input.nextInt();
        input.nextLine();
        while (N-- > 0) {
            solve(input);
        }
    }

    public static void solve(Scanner input) {
        int rowCapacity = input.nextInt();
        int prefer1 = input.nextInt();
        int prefer2 = input.nextInt();
        int noPrefer = input.nextInt();

        int gap1 = rowCapacity - prefer1;
        if (gap1 < 0) {
            gap1 = 0;
        }

        int gap2 = rowCapacity - prefer2;
        if (gap2 < 0) {
            gap1 = 0;
        }

        int sum = rowCapacity - gap1 + rowCapacity - gap2 + noPrefer;
        if (sum > 2 * rowCapacity) {
            sum = 2 * rowCapacity;
        }
        System.out.println(sum);
        return;

//        int bigGap = Math.min(gap1, gap2);
//
//        if (noPrefer >= rowCapacity*2 || noPrefer >= bigGap) {
//            System.out.println(rowCapacity*2);
//            return;
//        } else {
//            int realGap = rowCapacity*2 - noPrefer - gap2 - gap1;
//            if (realGap < 0) {
//                realGap = 0;
//            }
//            System.out.println(rowCapacity*2 - realGap);
//        }

    }
}

**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
10 15 5 1
10 5 15 1
10 5 5 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
import java.util.*;
public class Hard {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t!=0){
            int m=sc.nextInt();
            int a=sc.nextInt();
            int b=sc.nextInt();
            int c=sc.nextInt();
            int ans=0;
            if(a<m)
            ans=ans+a;
            else
            ans=ans+m;
            if(b<m)
            ans=ans+b;
            else
            ans=ans+m;
            if(a<m && b<m){
                int r=2*m-a-b;
                if(c<r)
                ans=ans+c;
                else
                ans=ans+r;
            }
            else if(a<m && b>m){
                int r=m-a;
                if(c<r)
                ans=ans+c;
                else
                ans=ans+r;
            }
            else if(a>m && b<m){
                int r=m-b;
                if(c<r)
                ans=ans+c;
                else
                ans=ans+r;
            }
            System.out.println(ans);
            t--;
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that can reveal issues in the logic. The code has several conditions and calculations, and we need to ensure that it handles all cases correctly.

### Analysis of the Buggy Code
1. **Basic Conditions:**
   - The code checks if `a` and `b` are less than `m` and assigns seats accordingly.
   - It then calculates the remaining seats for `c` based on different conditions.

2. **Potential Issues:**
   - The code might not correctly handle cases where `a` and `b` are both greater than `m`.
   - It might not correctly handle cases where `c` is very large and can fill all remaining seats.
   - It might not correctly handle cases where `a` or `b` is exactly equal to `m`.

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

1. **Case 1: Both `a` and `b` are greater than `m`**
   - This should ensure that the code correctly handles the scenario where `a` and `b` cannot be fully seated.
   - Example: `10 15 15 10`

2. **Case 2: `c` is very large**
   - This should ensure that the code correctly handles the scenario where `c` can fill all remaining seats.
   - Example: `10 5 5 20`

3. **Case 3: `a` or `b` is exactly equal to `m`**
   - This should ensure that the code correctly handles the scenario where `a` or `b` is exactly equal to `m`.
   - Example: `10 10 5 5`

4. **Case 4: `a` and `b` are both less than `m`**
   - This should ensure that the code correctly handles the scenario where both `a` and `b` are less than `m`.
   - Example: `10 3 3 10`

### Test Input
```plaintext
4
10 15 15 10
10 5 5 20
10 10 5 5
10 3 3 10
```

This test input includes a variety of edge cases that should help expose any bugs in the provided code.