**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
def main():

    n = int(input())

    input_lines = [input().strip() for _ in range(n)]

    output = [
        "1 1",
        "2 1",
        "1 4",
        "1 1",
        "1 1",
        "3 4"
    ]

    for line in output:
        print(line)

main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can highlight the issues in the faulty solution. The buggy code simply prints a fixed output without actually processing the input data, so we need to create a test case where the correct solution would differ from the fixed output.

### Considerations:
1. **Edge Cases:**
   - Single day visits (d = 1).
   - Jobs that cover the entire range of days (1 to n).
   - Overlapping jobs that span different ranges.
   - Jobs that are just outside the range of the visit.

2. **Scenarios:**
   - The optimal start day for the brother should overlap with the maximum number of jobs.
   - The optimal start day for the mother should overlap with the minimum number of jobs.
   - Test cases where the fixed output is incorrect.

### Test Input:
Let's create a test case with the following parameters:
- **n = 5** (total days)
- **d = 2** (duration of the visits)
- **k = 3** (number of jobs)
- **Jobs:**
  - Job 1: [1, 2]
  - Job 2: [3, 4]
  - Job 3: [4, 5]

### Explanation:
- For the brother, the start day that overlaps with the maximum number of jobs is day 1 (overlaps with 2 jobs: [1, 2]).
- For the mother, the start day that overlaps with the minimum number of jobs is day 4 (overlaps with 1 job: [4, 5]).

The fixed output in the buggy code is:
```
1 1
2 1
1 4
1 1
1 1
3 4
```

For this test case, the correct output should be:
```
1 4
```

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

This test input will expose the bug in the buggy code because the fixed output does not match the correct solution for this test case.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
from collections import Counter

for _ in range(int(input())):
    n, d, k = map(int, input().split())
    start = []
    end = []
    for i in range(k):
        inp = list(map(int, input().split()))
        start.append(inp[0])
        end.append(inp[1])

    start = Counter(start)
    end = Counter(end)

    cur = 0
    cur_min, cur_max = 9999, 0
    ans_min, ans_max = 0, 0
    i, j = 1, 1
    while j < d:
        if j in start:
            cur += start[j]
        j += 1
    if j in start:
        cur += start[j]

    if cur < cur_min:
        cur_min = cur
        ans_min = i
    if cur > cur_max:
        cur_max = cur
        ans_max = i

    while j < n - d + 1:
        i += 1
        j += 1

        if j in start:
            cur += start[j]

        if cur < cur_min:
            cur_min = cur
            ans_min = i
        if cur > cur_max:
            cur_max = cur
            ans_max = i
        if i in end:
            cur -= end[i]

    print(ans_max, ans_min)

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

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

Think step by step.

Response:
```plaintext
4
10 3 5
1 3
4 6
7 9
2 4
5 7
10 2 4
1 3
4 6
7 9
2 5
5 3 1
1 3
1 1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
import java.util.*;

public class ProblemD {

	static StringBuilder res=new StringBuilder();
	static List<Pair> nums;
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			nums=new ArrayList<>();
			int n=sc.nextInt(), d=sc.nextInt(), k=sc.nextInt();
			for(int i=0;i<k;i++) nums.add(new Pair(sc.nextInt(), sc.nextInt()));
			nums.sort((o1,o2)->(o1.l==o2.l)?o1.r-o2.r:o1.l-o2.l);
			getVerdict(n, d, k);
		}
		System.out.println(res);
		sc.close();

	}

	private static void getVerdict(int n, int d, int k) {
		int cM=Integer.MIN_VALUE, cm=Integer.MAX_VALUE;
		int cMi=-1, cmi=-1;

		int l=1;
		while(l+d-1<=n) {
			int li=getLeftBoundary(k, l, d);
			int ri=getRightBoundary(k, l, d);

			if(li==-1 && ri==-1) {
				cm=0;
				cmi=l;
				l++;
				continue;
			}

			int ll=ri-li+1;
			if(cM<ll) {
				cM=ll;
				cMi=l;
			}

			if(cm>ll) {
				cm=ll;
				cmi=l;
			}
			l++;
		}

		res.append(cMi+" "+cmi);
		res.append("\n");
	}

	private static int getRightBoundary(int k, int l, int d) {
		int le=0, ri=k-1, ans=-1;

		while(le<=ri) {
			int mid=le+(ri-le)/2;
			if(l+d-1>=nums.get(mid).l && l<=nums.get(mid).r) {
				ans=mid;
				le=mid+1;
			}else if(l+d-1<nums.get(mid).l) ri=mid-1;
			else if(l+d-1>nums.get(mid).r) le=mid+1;
		}

		return ans;
	}

	private static int getLeftBoundary(int k, int l, int d) {
		int le=0, ri=k-1, ans=-1;

		while(le<=ri) {
			int mid=le+(ri-le)/2;
			if(l+d-1>=nums.get(mid).l && l<=nums.get(mid).r) {
				ans=mid;
				ri=mid-1;
			}else if(l<nums.get(mid).l) ri=mid-1;
			else if(l>nums.get(mid).r) le=mid+1;
		}

		return ans;
	}

}

class Pair{
	int l, r;
	public Pair(int l, int r) {
		this.l=l;
		this.r=r;
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
import java.util.*;
public class BinaryTreeB {
    public static void main(String[] args) {
         Scanner sc = new Scanner(System.in);
         int t = sc.nextInt();
         while(t-->0){
           int n = sc.nextInt();
           int d = sc.nextInt();
           int k = sc.nextInt();
           int In[] = new int[n+1];
           int Out[] = new int[n+1];
           for(int i = 1; i <= k;i++){
             int si = sc.nextInt();
             int ei = sc.nextInt();
             In[si]++;
             Out[ei]++;
           }

           HashSet<Integer> set = new HashSet<>();
           int max = Integer.MIN_VALUE;
           int min = Integer.MAX_VALUE;
           int b = 1;
           int m = 1;
           int job = 0;
           for(int i = 1;i <= d;i++){
              job += In[i];
           }

            if(max < job){
                max = job;
                b = 1;
            }
            if(min > job){
                min = job;
                m = 1;
           }

           for(int i = d+1,p = 1;i <= n;i++,p++){
              job += In[i];
              job -= Out[p];
              if(max < job){
                max = job;
                if(d == 1){
                   b = i;
                }else{
                   b = i-1;
                }
              }
              if(min > job){
                min = job;
                if(d == 1){
                   m = i;
                }else{
                   m = i-1;
                }
              }
           }
           System.out.println(b + " " + m);
         }
         sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
//  #### Zeinab
#include<bits/stdc++.h>
using namespace std;
#define nl "\n"
#define fi first
#define se second
#define pb push_back
#define ll long long
#define RV return void
#define sz(x) int(x.size())
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define cin(v) for(auto&x:v) cin >> x;
#define cout(v) for(auto&x:v) cout << x << " ";

void solve(){

    int n , days , k ;
    cin >> n >> days >> k;
    vector < pair < int ,int > > vals(k);
    for(int i =0 ; i < k ; i++){
        cin >> vals[i].fi >> vals[i].se;
    }

    int idx1=-1 , idx2=-1 , mn = 1e9 , mx = -1e9;
    sort(all(vals));
    for(int i =1 ; i + days-1 <= n; i++){
        int cnt=0;
        for(auto & [f ,s] : vals){
            // cout << f << " " << s << " " << i << nl;
            int f1 = max(f , i) , s1= min(s , i + days-1);
            if(f1 <= s1) cnt++;

            if(s > i+days-1 and f > i+days-1) break;
        }

        // cout << i << " " << cnt << nl << nl;
        if(cnt > mx){
            mx = cnt , idx1=i;
        }
        if(cnt < mn){
            mn = cnt , idx2=i;
        }
    }

    cout << idx1 << " " << idx2 << nl;
}

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

    int t=1;
        cin >> t ;
    for(int i=1 ; i <= t ; i++){
        // cout << "Case #"<< i <<": ";
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 5 4
1 5
6 10
3 7
8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

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

using namespace std;
using ll = int64_t;
using ull = uint64_t;
using ld = long double;
using ii = pair<int, int>;
using vi = vector<int>;

#define mod 1000000007

static void solve()
{
	int n,d,k,i,j,l,r,b,m,bj=0,mj=INT_MAX;
	cin>>n>>d>>k;
	vector<ii> a(k);
	for(auto&x:a)
		cin>>x.first>>x.second;
	sort(a.begin(),a.end());
	i = 0;
	for (l=1;l<=n-d+1;l=max(++l, a[i].first-d+1)) {
		r = l+d-1;
		int cnt = 0;
		for (j=i;j<k && r>=a[j].first;++j) {
			if (l<a[j].second+1)
				cnt += ((l<a[j].first && r>a[j].second) ||
						(l>=a[j].first && l<=a[j].second) ||
						(r>=a[j].first && r<=a[j].second));
		}

		if (bj<cnt) {
			bj=cnt;
			b=l;
		}
		if (mj>cnt) {
			mj=cnt;
			m=l;
		}
		for (;i<k && l>a[i].second;++i)
			;
		if (i>=k)
			break;
	}
	cout<<b<<' '<<m<<'\n';
}

int main()
{
	cin.tie(nullptr)->sync_with_stdio(false);
	int t;
	cin>>t;
	while(t-->0)
		solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 3 5
1 2
3 5
6 7
8 10
4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
def solve():
    # Read input values
    t = int(input())  # Number of test cases
    for _ in range(t):
        n, d, k = map(int, input().split())  # n = total days, d = duration of visit, k = number of jobs
        jobs = [tuple(map(int, input().split())) for _ in range(k)]  # Read job intervals (l_i, r_i)
        jobs.sort()  # Sort jobs based on their start day (l_i)
        current_jobs_end_time = []  # List to store the end days (r_i) of current overlapping jobs

        mn_idx = -1
        mn_val = float('inf')
        mx_idx = -1
        mx_val = -float('inf')
        idx = 0  # Pointer to iterate over jobs

        # Iterate over possible start days for the visit
        day=1
        while day+d-1<=n:
            # Add all jobs that start before or on the last day of the current visit window
            while idx < k and jobs[idx][0] <= day + d - 1:
                current_jobs_end_time.append(jobs[idx][1])
                idx += 1
            current_jobs_end_time.sort()
            # Remove jobs that ended before the current day
            while current_jobs_end_time and current_jobs_end_time[0] < day:
                current_jobs_end_time.pop(0)

            # Calculate the current number of overlapping jobs
            current_overlap = len(current_jobs_end_time)

            # Update the day with the minimum overlap for the mother's visit
            if mn_val > current_overlap:
                mn_val = current_overlap
                mn_idx = day

            # Update the day with the maximum overlap for the brother's visit
            if mx_val < current_overlap:
                mx_val = current_overlap
                mx_idx = day
            day+=1
        # Output the start days for the brother's and mother's visit
        print(mx_idx, mn_idx)

# Call the function to solve the problem
solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
import math
import re

def input_n(isOne=False):
    if not isOne:
        return int(input())
    else:
        return 1

def input_args():
    return tuple(map(int, input().split()))

def input_list(space=True, to_int=True):
    line = input()
    if space:
        items = line.split()
    else:
        items = list(line)

    if to_int:
        return [int(i) for i in items]
    else:
        return items

def list_to_string(arr, sym):
    string = ""
    for i in arr:
        string += str(i) + sym
    return string

def yesno(check):
    return "YES" if check else "NO"

def find_all_indices_string(string, substring):
    indices = []
    index = string.find(substring)
    while index != -1:
        indices.append(index)
        index = string.find(substring, index + 1)
    return indices

def find_all_indices_arr(arr, element):
    return [index for index, value in enumerate(arr) if value == element]

def find_arr_in_arr(arr, index, value):
    for subArray in arr:
        if subArray[index] == value:
            return subArray
    return None

def solve():
    n, d, k = input_args()
    jobs = []
    for i in range(k):
        start, end = input_args()
        jobs.append([start - 1, end - 1])
    jobs.sort(key=lambda element: (element[0], element[1]))
    start = 0
    end = start + d - 1
    count = 0
    mother = [0, k + 1]
    brother = [0, -1]
    while end < n:
        count = 0
        for job in jobs:
            if start <= job[1] and end >= job[0]:
                count += 1
        if mother[1] > count:
            mother = [start, count]
        if brother[1] < count:
            brother = [start, count]
        start += 1
        end += 1
    return list_to_string([brother[0] + 1, mother[0] + 1], ' ')

n = input_n()
ans = []
alphabet = "abcdefghijklmnopqrstuvwxyz"

for i in range(n):
    ans.append(solve())
for i in ans:
    print(i)

**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 potential bugs in the given code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are the key points to consider:

1. **Single Day Visits (d = 1)**: This can test the simplest case where the visit duration is just one day.
2. **Jobs Spanning the Entire Range (1 to n)**: This can test the scenario where jobs cover the entire range of days.
3. **Jobs Just Outside the Visit Range**: This can test the scenario where jobs are just outside the start or end of the visit range.
4. **Multiple Overlapping Jobs**: This can test the scenario where multiple jobs overlap within the same range.
5. **Jobs with Gaps**: This can test the scenario where there are gaps between jobs.

Given these points, a small-scale test input that covers these scenarios could be:

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

### Explanation:
1. **First Test Case (n = 5, d = 1, k = 3)**:
   - Jobs: [1, 2], [3, 4], [5, 5]
   - Brother's visit should start on day 1 (max overlap: 1 job).
   - Mother's visit should start on day 5 (min overlap: 1 job).

2. **Second Test Case (n = 5, d = 2, k = 3)**:
   - Jobs: [1, 2], [3, 4], [5, 5]
   - Brother's visit should start on day 1 (max overlap: 2 jobs).
   - Mother's visit should start on day 4 (min overlap: 1 job).

3. **Third Test Case (n = 5, d = 3, k = 3)**:
   - Jobs: [1, 2], [3, 4], [5, 5]
   - Brother's visit should start on day 1 (max overlap: 2 jobs).
   - Mother's visit should start on day 3 (min overlap: 1 job).

This test input is designed to cover a variety of scenarios and edge cases that can help expose potential bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int d = sc.nextInt();
            int k = sc.nextInt();
            HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();

            // Read the job intervals
            for (int job = 1; job <= k; job++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                for (int day = l; day <= r; day++) {
                    map.computeIfAbsent(day, v -> new ArrayList<>()).add(job);
                }
            }

            HashMap<Integer, Integer> jobCount = new HashMap<>();
            int maxDay = 1, minDay = 1, maxCount = 0, minCount = Integer.MAX_VALUE;

            // Initialize the first d-day window
            for (int day = 1; day <= d; day++) {
                if (map.containsKey(day)) {
                    for (int job : map.get(day)) {
                        jobCount.put(job, jobCount.getOrDefault(job, 0) + 1);
                    }
                }
            }

            maxCount = jobCount.size();
            minCount = jobCount.size();

            // Sliding window
            for (int start = 2; start + d - 1 <= n; start++) {
                int end = start + d - 1;

                // Remove jobs for day (start - 1)
                if (map.containsKey(start - 1)) {
                    for (int job : map.get(start - 1)) {
                        jobCount.put(job, jobCount.get(job) - 1);
                        if (jobCount.get(job) == 0) {
                            jobCount.remove(job);
                        }
                    }
                }

                // Add jobs for day `end`
                if (map.containsKey(end)) {
                    for (int job : map.get(end)) {
                        jobCount.put(job, jobCount.getOrDefault(job, 0) + 1);
                    }
                }

                int currentJobCount = jobCount.size();

                // Update max/min counts and days
                if (currentJobCount > maxCount) {
                    maxCount = currentJobCount;
                    maxDay = start;
                }
                if (currentJobCount < minCount) {
                    minCount = currentJobCount;
                    minDay = start;
                }
            }

            // Output result for each test case
            System.out.println(maxDay + " " + minDay);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robert Hood and Mrs Hood
time_limit_ms: 2000
memory_limit_mb: 256
Description: Impress thy brother, yet fret not thy mother.

Robin's brother and mother are visiting, and Robin gets to choose the start day for each visitor.

All days are numbered from $1$ to $n$. Visitors stay for $d$ continuous days, all of those $d$ days must be between day $1$ and $n$ inclusive.

Robin has a total of $k$ risky 'jobs' planned. The $i$-th job takes place between days $l_i$ and $r_i$ inclusive, for $1 \le i \le k$. If a job takes place on any of the $d$ days, the visit overlaps with this job (the length of overlap is unimportant).

Robin wants his brother's visit to overlap with the maximum number of distinct jobs, and his mother's the minimum.

Find suitable start days for the visits of Robin's brother and mother. If there are multiple suitable days, choose the earliest one.

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

The first line of each test case consists of three integers $n$, $d$, $k$ ($1 \le n \le 10^5, 1 \le d, k \le n$) — the number of total days, duration of the visits, and the number of jobs.

Then follow $k$ lines of each test case, each with two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the start and end day of each job.

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

Output: For each test case, output two integers, the best starting days of Robin's brother and mother respectively. Both visits must fit between day $1$ and $n$ inclusive.

Examples:
input:
6
2 1 1
1 2
4 1 2
1 2
2 4
7 2 3
1 2
1 3
6 7
5 1 2
1 2
3 5
9 2 1
2 8
9 2 4
7 9
4 8
1 3
2 3
output:
1 1
2 1
1 4
1 1
1 1
3 4

Note: In the first test case, the only job fills all $2$ days, both should visit on day $1$.

In the second test case, day $2$ overlaps with $2$ jobs and day $1$ overlaps with only $1$.

In the third test case, Robert visits for days $[1,2]$, Mrs. Hood visits for days $[4,5]$.

Buggy Code:
// package CF.Competition.R974;
import java.util.*;

public class Main{
    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int N = sc.nextInt();
        while (N-- > 0) {
            solve();
        }
    }

    public static void solve() {
        n = sc.nextInt();
        d = sc.nextInt();
        k = sc.nextInt();
        list = new ArrayList<>();
        for (int i = 0; i < k; i++) list.add(new point(sc.nextInt(), sc.nextInt()));
        Collections.sort(list, new Comparator<point>() {
            @Override
            public int compare(point o1, point o2) {
                if (o2.x == o1.x) return o1.y - o2.y;
                return o1.x - o2.x;
            }
        });
        int ans1 = 1, ans2 = 1, max = 0, min = Integer.MAX_VALUE;
        for (int i = 1; i + d - 1 <= n; i++) {
            int begin = i, end = i + d - 1;
//            System.out.println(end);
            int temp = 0;
            for (int j = 0; j < list.size(); j++) {
                int begin1 = list.get(j).x, end1 = list.get(j).y;
                if ((end - begin1 >= 0 || end1 - begin >= 0)
                        && begin1 <= end && end1 >= begin) temp++;
                else if (begin1 == end1) {
//                    System.out.println(begin1 + " " + end);
//                    System.out.println(begin1 <= end);
                    if (begin1 >= begin && begin1 <= end) temp++;
                }
            }
            if (temp > max) {
                ans1 = i;max = temp;
            }
            if (temp < min) {
                ans2 = i;min = temp;
            }
        }
        System.out.println(ans1 + " " + ans2);
    }

    static int n, d, k;
    static ArrayList<point> list;

    static class point {
        int x, y;

        public point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
#include <stdio.h>
int chest();
int main() {
    int t;
    scanf("%d",&t);
    while(t--)
        chest();

    return 0;
}
int chest(){
    int n, k, count=0;
    scanf("%d %d", &n, &k);
    int a[n];
    for(int i=0; i<n; i++){
        scanf("%d", &a[i]);
    }
    for(int i=0; i<n-1; i++){
        for(int j=0; j<n-i-1; j++){
            if(a[j]<a[j+1]){
                int temp=a[j];
                a[j]=a[j+1];
                a[j+1]=temp;
            }
        }
    }
    int sum=a[0];
    for(int i=1; i<n; i++){
        if((sum+a[i])>=k){
            printf("%d\n",k-sum);
            return 0;
        }
        sum+=a[i];
    }
    printf("%d\n", k-sum);
    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
5 15
5 5 5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
#include<iostream>
using namespace std;
int a[60],b[60];
void msort(int l,int r)
{
	if(l==r)return ;
	int mid=(l+r)>>1;
	msort(l,mid);
	msort(mid+1,r);
	int i=l,j=mid+1;
	int top=l;
	while(i<=mid&&r<=j)
	{
		if(a[i]<a[j])b[top++]=a[j++];
		else b[top++]=a[i++];
	}
	while(i<=mid)b[top++]=a[i++];
	while(j<=r)b[top++]=a[j++];
	for(int k=l;k<=r;k++)a[k]=b[k];
}
int main()
{
	int t;
	scanf("%d",&t);
	int n,k;
	for(int i=1;i<=t;i++)
	{
		scanf("%d%d",&n,&k);
		for(int j=1;j<=n;j++)scanf("%d",&a[j]);
		msort(1,n);
		int sum=0;
		int p=1;
		while(sum<=k&&p<=n)
		{
			sum+=a[p++];
		}
		if(sum>k)sum-=a[p-1];
		printf("%d",k-sum);
		printf("\n");
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the sorting and the summing logic might fail. The primary issues in the code are:

1. **Incorrect Sorting Logic**: The merge sort function `msort` has a bug in the merge step. Specifically, the condition `while(i<=mid&&r<=j)` is incorrect. It should be `while(i<=mid&&j<=r)` to ensure both halves are properly merged.
2. **Incorrect Sum Calculation**: The logic for summing the coins and determining the number of coins to add is flawed. The loop `while(sum<=k&&p<=n)` might not correctly handle the case where the sum exceeds `k`.

Given these issues, a good test input would be one where the initial array is already sorted in descending order, and the total sum of the array is just slightly less than `k`. This will force the faulty sorting to fail and the sum calculation to go wrong.

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

```plaintext
1
5 20
5 4 3 2 1
```

Explanation:
- **Input**: `1` test case.
- **n = 5**: 5 chests.
- **k = 20**: Monocarp needs at least 20 coins.
- **Initial Chests**: `5, 4, 3, 2, 1`.

The correct answer should be `5`, because the initial sum of the chests is `15`, and we need to add `5` more coins to reach `20`.

However, the faulty sorting will not change the order of the array, and the sum calculation will incorrectly determine the number of coins to add. This should expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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

    for(long long int i=0;i<t;i++){

    long long int  n,k;
    cin>>n, cin>>k;
    long long int arr[n];
    for(long long int i=0;i<n;i++){
        cin>>arr[i];
    }
    sort(arr,arr+n);

    long long int add=0,ans=0,sum=0;
    for(long long int i=n-1;i>=0;i--){
        sum+=arr[i];
    }

    if(sum==n){
        ans=k-sum;
        cout<<ans<<endl;

    }
    else{
        for(long long int j=n-1;j>=0;j--){
        if(arr[j]+add<=k){
            add+=arr[j];
        }
        else{
            ans=k-add;
            add=0;
            break;
        }
    }
    cout<<ans<<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
5 15
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
for i in range(int(input())):
    a=input().split()
    n=int(a[0])
    k=int(a[1])
    b=list(map(int,input().split()))
    b.sort(reverse=True)

    for j in range(len(b)):
        if(k==0):

            break
        if(k<0):

            break
        k=k-b[j]
    if(k<0):
        print(b[j-1]+k)
    if(k>=0):
       print(k)

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

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

Think step by step.

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

### Explanation:
- **Input:** `1` test case.
- **Number of chests (`n`):** `5`
- **Target number of coins (`k`):** `15`
- **Initial coins in chests:** `5 5 5 5 5`

### Analysis:
- The total number of coins in the chests is already `25`, which is more than `15`.
- The goal is to make sure Monocarp takes exactly `15` coins.
- If no coins are added, Monocarp will take the first three chests, each with `5` coins, totaling `15` coins.
- The correct output should be `0` because no additional coins need to be added.

### Why This Test Input is Challenging:
- The buggy code does not correctly handle the scenario where the total number of coins in the chests is already greater than or equal to `k`.
- The code incorrectly decrements `k` and may output an incorrect result when `k` becomes negative.
- This test input forces the code to handle the case where the initial total is already sufficient, exposing the bug in the logic for managing `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
n=int(input())
for i in range(n):
    a=list(map(int,input().split()))
    n=a[0]
    k=a[1]
    b=list(map(int,input().split()))
    b.sort(reverse=True)
    if(k<=max(b)):
        print('0')
    else:
        l=[]
        l.append(b[0])
        for j in range(len(b)-1):
            l.append(b[j+1]+l[len(l)-1])
        for j in range(len(l)-1):
            if(l[j]<=k and l[j+1]>k):
                print(k-l[j])
                break
            elif(l[len(l)-1]<k):
                print(k-l[len(l)-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
5 10
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
t = int(input())
results = []

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

    a.sort(reverse=True)
    total_sum = sum(a)
    if total_sum < k:
        coins_needed = k - total_sum
    else:
        current_sum = 0
        coins_needed = 0
        count = 0
        for i in range(n):
            current_sum += a[i]
            if current_sum >= k:
                break
            count=i
        coins_needed = max(0, k - sum(a[:count  + 1]))

    results.append(coins_needed)

for res in results:
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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 sum=0;
		int n=sc.nextInt();
		int k=sc.nextInt();
		int ar[]=new int[n];
		for(int i=0;i<n;i++)
		{
		    ar[i]=sc.nextInt();
		}
		Arrays.sort(ar);
		for(int i=0;i<n/2;i++)
		{
		    int temp=ar[i];
		    ar[i]=ar[n-i-1];
		    ar[n-i-1]=temp;
		}
		int added_coins=0;
		for(int i=0;i<n;i++)
		{
		   if(sum>=k)
		   {
		       break;
		   }
		   if(ar[i]==k)
		   {
		       sum=ar[i];
		       break;
		   }
		   if(sum+ar[i]>=k)
		   {
		       added_coins+=k-sum;
		       sum=k;
		   }
		   else{
		       sum+=ar[i];
		   }
		}
		if(sum<k)
		{
		    added_coins=k-sum;
		}
		System.out.println(added_coins);
		}

	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider scenarios where the code might fail. The code has a few potential issues:

1. **Sorting and Reversing**: The code sorts the array and then reverses it, which means it processes the largest values first. This might cause issues if the largest values are already close to `k` and the code doesn't handle this correctly.
2. **Adding Coins**: The code adds coins in a way that might not be optimal. It might not consider the minimum number of coins needed to reach `k` efficiently.

Let's create a test case that maximizes these issues:

1. **Case with Large Initial Values**: Create a case where the initial values are already large and close to `k`, but not exactly `k`.
2. **Case with Small Initial Values**: Create a case where the initial values are small and need significant additions to reach `k`.

### Test Case 1: Large Initial Values
- **n = 5**
- **k = 20**
- **a = [15, 16, 17, 18, 19]**

In this case, the initial values are already close to `k`, but not exactly `k`. The code might not handle this efficiently.

### Test Case 2: Small Initial Values
- **n = 5**
- **k = 20**
- **a = [1, 2, 3, 4, 5]**

In this case, the initial values are small, and the code needs to add a significant number of coins to reach `k`.

### Test Case 3: Mixed Values
- **n = 5**
- **k = 20**
- **a = [5, 10, 15, 2, 8]**

In this case, the initial values are mixed, and the code needs to handle both large and small values.

### Test Case 4: Edge Case with One Chest
- **n = 1**
- **k = 10**
- **a = [5]**

In this case, there is only one chest, and the code needs to add a specific number of coins to reach `k`.

### Final Test Input
```plaintext
4
5 20
15 16 17 18 19
5 20
1 2 3 4 5
5 20
5 10 15 2 8
1 10
5
```

This test input includes a variety of scenarios that should expose potential issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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

public class Greddy {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for(int i=0;i<n;i++)
        {
            int size = sc.nextInt();
            int k = sc.nextInt();
            int[] arr = new int[size];
            int sum=0;
            for(int j=0;j<size;j++)
            {
                arr[j]=sc.nextInt();
                sum += arr[j];
            }
            if(sum<=k)
            {
                System.out.println(k-sum);

            }
            else
            {

                Arrays.sort(arr);
                int ans=0;
                for(int h=size-1;h>=0;h--)
                {
                    ans=ans+arr[h];
                    if(ans>=k)
                    {
                        ans=ans-arr[h];
                        if(ans==0)
                        {
                            System.out.println("0");
                            break;
                        }
                        else
                        {

                            System.out.println(k-ans);
                            break;
                        }
                    }
                }
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 10
4 3 2 1 1
```

**Explanation:**
- The test input provided is designed to expose a potential bug in the given buggy code.
- In this test case, we have 5 chests with the initial coin counts as [4, 3, 2, 1, 1], and the target total is 10 coins.
- The current sum of coins in the chests is 4 + 3 + 2 + 1 + 1 = 11, which is already greater than 10.
- However, the buggy code sorts the array and then tries to subtract the largest elements until the sum is less than or equal to 10. This can lead to incorrect results because it does not consider the minimum number of coins to add.
- The correct solution should be to add 0 coins because the sum is already greater than or equal to 10. However, the buggy code might incorrectly subtract some coins and print a non-zero value, thus exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
#include <iostream>
using namespace std;
#include<bits/stdc++.h>
#define endl '\n'
#define sp ' '
typedef vector<long long> vll;
typedef vector<int> vi;
typedef long long ll;
#define pb push_back
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<b;i++)

vll binary(int x){
    vector<long long> v1;
        do {
            v1.push_back(x%2);
            x/=2;
        } while (x);
        return v1;
}

bool even(ll n){
    if(n%2 == 0) return true;
    else return false;
}

bool isPowerOfTwo(ll num) {
    return (num > 0) && ((num & (num - 1)) == 0);
}

ll ceilDiv(ll x, ll y) {
    return (x + y - 1) / y;
}
void solve(){
  int n,k; cin >> n >> k;
  vi a(n);
  rep(i,0,n){
    cin >> a[i];
  }
  sort(all(a));
  ll sum =a[n-1];
  for(int i =n-2;i>=0;i--){
    if(sum+a[i]>=k){
        cout << (k-sum)<< endl;
        return;
    }else{
        sum += a[i];
    }
  }
  cout << (k-sum) << endl;

    }

int main(){
  int t;
  cin>>t;
  for(int i=0;i<t;i++){
    // cout << i << i << i<< endl;
    solve();
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 10
5 3 2
5 15
4 5 1 2 3
2 10
1 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <bits/stdc++.h>
using namespace std;

int t;

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using vpii = vector<pii>;

void solve()
{
    int k;
    int n;
    cin>>n>>k;
    int tsum=0;
    int cleft=0;
    vector<int> arr;
    for(int i=0;i<n;i++)
    {
        int x;
        cin>>x;
        arr.push_back(x);
    }
    sort(arr.begin(),arr.end());
    reverse(arr.begin(),arr.end());
    if(arr[0]==k)
    {
        cout<<0<<endl;
        return;
    }
    for(int i=0;i<n;i++)
    {

            tsum+=arr[i];

    }
    if(tsum<k)
    {
        cleft = k-tsum;
    }
    int m=0;
    int mtot=0;
    while(m<n && mtot + arr[m] < k)
    {
        mtot+=arr[m];
        m++;
    }
    if(mtot<k)
    {
        cleft = k-mtot;
    }
    cout<<cleft<<endl;
    return;

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

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

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

Think step by step.

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

1. **Handling of Initial Sum**: The code does not correctly handle the initial sum of coins in the chests. It calculates the total sum and then tries to find the minimum number of coins to add, but it does not consider the optimal way to add coins to minimize the total number of coins added.

2. **Edge Cases**: The code might fail for edge cases where the initial sum is very close to `k` or where the largest chest already contains `k` coins.

3. **Optimal Distribution**: The code does not consider the optimal distribution of coins to ensure Monocarp takes exactly `k` coins.

Given these points, let's create a test input that exposes these issues:

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

### Explanation
- **Test Case 1**:
  - `n = 3`, `k = 10`
  - Initial coins in the chests: `5, 3, 2`

### Expected Output
- **Output**: `5`

### Reasoning
- The initial sum of coins is `5 + 3 + 2 = 10`.
- Since the total is already `10`, no additional coins are needed.
- The code should output `0` in this case.

### Why This Test Input is Challenging
- The initial sum is exactly `k`, so no additional coins are needed.
- The code should correctly handle this case by outputting `0`.
- However, the code might fail to recognize that the initial sum is already `k` and might incorrectly calculate the number of coins to add.

This test input will help expose the bug in the code and ensure that it correctly handles cases where the initial sum is already equal to `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
t = int(input())

for _ in range(t):
    nk = input().split()
    n = int(nk[0])
    k = int(nk[1])

    nums = input()
    res = [int(ele) for ele in nums.split()]

    res.sort(reverse=True)

    d = k
    ans = 0
    for i in range(n-1):
        d = d-res[i]
        ans = d
        if i == n-2 and d>=res[n-2]:
            ans = ans - res[n-2]
        if d<res[i+1]:
            break
    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
3 6
1 2 3
```

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

1. **Initial Setup**:
   - There are 3 chests with 1, 2, and 3 coins respectively.
   - The target number of coins Monocarp should have is 6.

2. **Expected Behavior**:
   - To minimize the number of coins added, we should try to add the minimum number of coins to the chests to reach the target of 6 coins.
   - The optimal strategy would be to add 1 coin to the chest with 2 coins, making it 3 coins. Then Monocarp will take the two chests with 3 coins each, totaling 6 coins.

3. **Buggy Code Behavior**:
   - The buggy code sorts the chest values in descending order: [3, 2, 1].
   - It then iterates through the sorted list, subtracting the current chest value from `d` (which is 6 initially).
   - The code incorrectly updates `ans` to `d` in each iteration, which is not the correct way to calculate the minimum number of coins to add.
   - The loop breaks when `d < res[i+1]`, which in this case will be true after the first iteration (since `d` becomes 3 after subtracting 3 from 6, and 3 is not less than 2).

4. **Expected Output**:
   - The correct output should be 1, as adding 1 coin to the chest with 2 coins is the minimum required to make Monocarp take exactly 6 coins.

5. **Buggy Code Output**:
   - The buggy code will likely output an incorrect value, such as 3, because it incorrectly updates `ans` and does not correctly handle the addition of coins.

This test case should help in identifying the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
for _ in range(int(input())):
  n, k = map(int, input().split())
  s = list(map(int, input().split()))
  s.sort(reverse=True)
  t_coin = sum(s)
  if s[0] == k:
    print(0)
    continue
  elif t_coin < k:
    print(k - t_coin)
    continue
  else:
    c_sum = 0
    for i in range(n):
      c_sum += s[i]
      if c_sum >= k:
        print(max(0, k - (c_sum - s[i])))
        break
      else:
        if c_sum==k:
          print(t_coin)
          break

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

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

Think step by step.

Response:
```plaintext
5 10
1 2 3 4 5
```