**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

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

/**
 * C_Numeric_String_Template
 */
public class C_Numeric_String_Template {
    public static boolean isCheck(int arr[], String str) {
        if (arr.length != str.length()) {
            return false;
        }
        HashMap<Character, Set<Integer>> hm = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            if (!hm.containsKey(str.charAt(i))) {
                hm.put(str.charAt(i), new HashSet<>());
            }
            hm.get(str.charAt(i)).add(arr[i]);
        }
        for (char i : hm.keySet()) {
            if (hm.get(i).size() != 1) {
                return false;
            }
        }
        return true;
    }

    // end of fast i/o code
    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int res[] = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = sc.nextInt();
            }
            int m = sc.nextInt();
            String str[] = new String[m];
            for (int i = 0; i < m; i++) {
                str[i] = sc.next();
            }
            for (int i = 0; i < m; i++) {
                if (isCheck(res, str[i])) {
                    System.out.println("YES");
                } else {
                    System.out.println("NO");
                }
            }
        }

    }

    // For fast input output
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            try {
                br = new BufferedReader(
                        new FileReader("input.txt"));
                PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                System.setOut(out);
            } catch (Exception e) {
                br = new BufferedReader(new InputStreamReader(System.in));
            }
        }

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios where the one-to-one correspondence between characters in the string and numbers in the array is not maintained. The bug in the code is that it only checks if each character in the string maps to a single number in the array, but it does not check if each number in the array maps to a single character in the string.

### Step-by-Step Reasoning:

1. **Mismatch in One-to-One Correspondence:**
   - We need a case where a number in the array maps to multiple characters in the string, which should result in "NO" but the code might incorrectly return "YES".

2. **Equal Lengths:**
   - Ensure the length of the string is equal to the length of the array to pass the initial length check.

3. **Multiple Test Cases:**
   - Include multiple test cases to ensure the code handles different scenarios.

### Test Input:

```plaintext
2
5
1 2 3 4 5
2
abcde
abcdf
3
1 1 1
2
aaa
abc
```

### Explanation:

1. **First Test Case:**
   - Array: [1, 2, 3, 4, 5]
   - Strings: "abcde", "abcdf"
   - For "abcde":
     - Each character maps to a unique number: a -> 1, b -> 2, c -> 3, d -> 4, e -> 5.
     - This should return "YES".
   - For "abcdf":
     - The character 'a' maps to 1, 'b' maps to 2, 'c' maps to 3, 'd' maps to 4, 'f' maps to 5.
     - This should return "NO" because the number 5 (f) is not mapped to the same character as 5 (e).

2. **Second Test Case:**
   - Array: [1, 1, 1]
   - Strings: "aaa", "abc"
   - For "aaa":
     - Each character 'a' maps to 1.
     - This should return "YES".
   - For "abc":
     - The character 'a' maps to 1, 'b' maps to 1, 'c' maps to 1.
     - This should return "NO" because the number 1 is mapped to multiple characters (a, b, c).

This test input should expose the bug in the code by ensuring that the one-to-one correspondence is not maintained, and the code incorrectly returns "YES" for cases where it should return "NO".

2
5
1 2 3 4 5
2
abcde
abcdf
3
1 1 1
2
aaa
abc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

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

#define MOD 1000000007
#define maxn 100000000000017
#define endl "\n"
#define mk make_pair
#define pll pair<ll, ll>
#define vll vector<ll>
#define vld vector<ld>
#define vpll vector<ppl>
#define ff first
#define ss second
#define pb push_back

typedef long long ll;
typedef long double ld;

#define o(...) if(local=='L') {__f(#__VA_ARGS__, __VA_ARGS__); cout<<endl;}

#define Ns 1000007

char local = 'O';

////////////////#####################INPUT###############///////////////////////

template <typename T>
void read(T &p)
{
    cin >> p;
}

template <typename T, typename T1>
void read(pair<T, T1> &p)
{
    read(p.ff);
    read(p.ss);
}

template <typename T>
void read(T arr[], ll n)
{
    for (ll i = 0; i < n; i++)
    {
        read(arr[i]);
    }
}

template <typename T>
void read(vector<T> &arr)
{
    for (ll i = 0; i < arr.size(); i++)
    {
        read(arr[i]);
    }
}

////////////////#####################OUPUT###############///////////////////////

template <typename T>
void out(T& n)
{
    cout << n;
}

template <typename T, typename T1>
void out(pair<T, T1> &p)
{
    cout << "(";
    out(p.ff);
    cout << ",";
    out(p.ss);
    cout << ")";
}

template <typename T>
void out(T arr[], ll n)
{
    for (ll i = 0; i < n; i++)
    {
        out(arr[i]);
        cout << " ";
    }
}

template <typename T>
void out(vector<T> &arr)
{

    cout << "[ ";
    for (ll i = 0; i < arr.size(); i++)
    {
        out(arr[i]);
        if (i!=arr.size()-1)
            cout << ", ";
    }
    cout << " ]";

}

template <typename T>
void out(set<T> &arr)
{
    for (auto i:arr)
    {
        out(i);
        cout << " ";
    }
}

template <typename T>
void out(multiset<T> &arr)
{
    for (auto i:arr)
    {
        out(i);
        cout << " ";
    }
}

template <typename T, typename T1>
void out(map<T,T1> &m)
{
    for (auto i:m)
    {
        out(i.first);
        cout<<" -> ";
        out(i.second);
        cout<<", ";
    }
}

template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
    cout << name << " : ";
    out(arg1);
}

template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char *comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : ";
    out(arg1);
    cout << " | ";
    __f(comma + 1, args...);
}

////////////////////////////////////END TEMPLATE//////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////

void cal() {
}

void solve() {
    ll n; cin>>n;
    vll a(n); read(a);
    map<ll,char> mapy;
    map<ll,char> mapz;
    ll m; cin>>m;
    while(m--){
        bool f=0;
        string s; cin>>s;
        if(s.size()!=n) cout<<"NO"<<endl;
        else{
            for(auto i=0; i<n; i++){
                if(!mapz[a[i]]) mapz[a[i]]=s[i];
                else{
                    if(mapz[a[i]]!=s[i]){
                        cout<<"NO"<<endl;
                        // for(auto i: mapz){
                        //     // cout<<i.first<<' '<<i.second<<endl;
                        // }
                        f=1;
                        break;
                    }
                }
            }
            if(!f){
                for(auto i=0; i<n; i++){
                    if(!mapy[s[i]]) mapy[s[i]]=a[i];
                    else{
                        if(mapy[s[i]]!=a[i]){
                            cout<<"NO"<<endl;
                            f=1;
                            break;
                        }
                    }
                }
            }
            if(!f) cout<<"YES"<<endl;
            mapy.clear();
            mapz.clear();

        }
    }

}

int main()
{

    #ifdef L
        local = 'L';
        time_t start, end;
        time(&start);
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll t1;

    t1 = 1;

    cin>>t1;

    cal();

    while (t1--) {
        solve();
    }

    #ifdef L
    time(&end);
    double time_taken = double(end - start);
    cout << endl
        << endl
        << endl
        << "Time taken by program is : " << fixed
        << time_taken << setprecision(6);
    cout << " sec " << endl;
    #endif
}

**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 test input that exposes the bug in the provided code, we need to focus on edge cases that might cause the code to fail. The code is supposed to check if a string matches a given template array. The key points to consider are:

1. **Length Mismatch**: The string and the array should have the same length.
2. **One-to-One Correspondence**: The same numbers in the array should correspond to the same characters in the string, and vice versa.

### Step-by-Step Analysis:

1. **Length Mismatch**:
   - If the string length does not match the array length, the code should immediately output "NO".
   - This is straightforward and should be handled correctly by the code.

2. **One-to-One Correspondence**:
   - The code uses two maps: `mapy` and `mapz` to check the correspondence.
   - If the same number in the array maps to different characters in the string, or the same character in the string maps to different numbers in the array, the code should output "NO".

### Potential Bugs:
- **Reusing Maps**: The code reuses the maps `mapy` and `mapz` for each string without clearing them properly.
- **Incorrect Mapping**: The code might not handle the case where the same character maps to different numbers or the same number maps to different characters correctly.

### Test Input:
To expose these potential bugs, we can create a test case where:
- The array has repeated numbers.
- The string has repeated characters.
- The string and array do not have a one-to-one correspondence.

### Example Test Input:
```plaintext
1
5
1 2 3 1 2
3
abcde
ababa
abcda
```

### Explanation:
- **Array**: `1 2 3 1 2`
- **String 1**: `abcde` (Should output "NO" because the same number 1 maps to different characters 'a' and 'd', and 2 maps to different characters 'b' and 'e'.)
- **String 2**: `ababa` (Should output "YES" because there is a one-to-one correspondence: 1 -> a, 2 -> b, 3 -> c.)
- **String 3**: `abcda` (Should output "NO" because the same number 1 maps to different characters 'a' and 'd'.)

This test input should help expose the bug in the code by challenging the one-to-one correspondence requirement.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
iint = lambda : map(int,input().split())
ilist = lambda : list(iint())

for _ in range(int(input())):
    n = int(input())
    a = ilist()
    d = dict()
    for i in range(n):
        if a[i] in d:
            d[a[i]].append(i)
        else:
            d[a[i]] = [i]

    m = int(input())

    for i in range(m):
        s = input()
        if len(s) != n:
            print("NO")
            continue

        d1 = dict()
        for i in range(n):
            if s[i] in d1:
                d1[s[i]].append(i)
            else:
                d1[s[i]] = [i]

        flag = 1

        for i in range(n):
            if d1[s[i]] != d[a[i]]:
                flag = 0
                break

        if flag:
            print("YES")
        else:
            print("NO")

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

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

Think step by step.

Response:
```plaintext
2
5
1 2 3 4 5
2
abcde
abcdf
4
1 1 1 1
3
aaaa
abab
aabb
```

2
5
1 2 3 4 5
2
abcde
abcdf
4
1 1 1 1
3
aaaa
abab
aabb


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
answer = []

def get_repeats(s):
    d = {}
    res = []
    for index, l in enumerate(s):
        if l not in d:
            d[l] = (index,)
        else:
            d[l] += (index,)

    for l in d:
        res.append(hash(d[l]))

    return res

def solve():
    _ = input()
    pattern = [int(i) for i in input().split()]
    p_repeats = get_repeats(pattern)
    m = int(input())
    for _ in range(m):
        s = input()
        s_repeats = get_repeats(s)
        if len(p_repeats) > 0 and len(s_repeats) > 0 and p_repeats == s_repeats:
            answer.append('YES')
        else:
            answer.append('NO')

t = int(input())
for _ in range(t):
    solve()
print('\n'.join(answer))

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can break the assumptions or logic in the code. Here are some key points to consider:

1. **Length of the Template and Strings**: Ensure the lengths are consistent with the problem constraints but also test the limits.
2. **One-to-One Correspondence**: The bug in the code is that it uses a hash of the indices of repeated elements to check for a match, which can lead to false positives if the hash values collide.
3. **Edge Cases**: Test with strings and templates that have repeated elements in different positions or with different characters.

Let's construct a test input that includes these elements:

1. **Test Case 1**: A simple case where the template and string match.
2. **Test Case 2**: A case where the template and string have the same length and repeated elements but do not match.
3. **Test Case 3**: A case where the template and string have repeated elements in different positions.
4. **Test Case 4**: A case with a large number of elements to test performance.

Here is the test input:

```plaintext
4
5
1 2 3 4 1
2
abcda
abeda
3
1 2 1
2
abc
aba
3
1 2 3 4 5
3
abcde
abcda
abcba
5
1 2 3 4 5
3
abcde
abcda
abcba
```

### Explanation:
1. **Test Case 1**:
   - Template: [1, 2, 3, 4, 1]
   - Strings: "abcda" (matches), "abeda" (does not match)
   - This tests the basic matching and non-matching scenarios.

2. **Test Case 2**:
   - Template: [1, 2, 1]
   - Strings: "abc" (does not match), "aba" (matches)
   - This tests the case where the template has repeated elements in different positions.

3. **Test Case 3**:
   - Template: [1, 2, 3, 4, 5]
   - Strings: "abcde" (matches), "abcda" (does not match), "abcba" (does not match)
   - This tests the case where the template has unique elements and the strings have repeated elements in different positions.

4. **Test Case 4**:
   - Template: [1, 2, 3, 4, 5]
   - Strings: "abcde" (matches), "abcda" (does not match), "abcba" (does not match)
   - This is a repetition of Test Case 3 to ensure the code handles multiple similar cases.

This test input should help expose the bug in the provided buggy code by testing various edge cases and ensuring the one-to-one correspondence between the template and the strings.

4
5
1 2 3 4 1
2
abcda
abeda
3
1 2 1
2
abc
aba
3
1 2 3 4 5
3
abcde
abcda
abcba
5
1 2 3 4 5
3
abcde
abcda
abcba


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.*;

public class Main {
    public static void solve(int[] template, String[] strs, int n) {
        Map<Integer, List<Integer>> positions = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int num = template[i];
            if (!positions.containsKey(num)) {
                positions.put(num, new ArrayList<>(List.of(i)));
            } else {
                List<Integer> list = positions.get(num);
                list.add(i);
                positions.put(num, list);
            }
        }

        for (String str : strs) {
            if (str.length() != n) {
                System.out.println("NO");
                continue;
            }

            Map<Integer, List<Integer>> positionStr = new HashMap<>();
            for (int i = 0; i < n; i++) {
                int num = str.charAt(i);
                if (!positionStr.containsKey(num)) {
                    positionStr.put(num, new ArrayList<>(List.of(i)));
                } else {
                    List<Integer> list = positionStr.get(num);
                    list.add(i);
                    positionStr.put(num, list);
                }
            }

            boolean isSatisfied = true;
            boolean[] visited = new boolean[n];
            for (int i = 0; i < n; i++) {
                if (visited[i]) {
                    continue;
                }

                visited[i] = true;
                int num = template[i], numStr = str.charAt(i);
                List<Integer> l = positions.get(num);
                List<Integer> lStr = positionStr.get(numStr);
                if (l.size() == 1 && lStr.size() == 1) {
                    continue;
                }
                if (l.size() != lStr.size()) {
                    isSatisfied = false;
                    break;
                }
                for (int j = 0; j < l.size(); j++) {
                    if (!Objects.equals(l.get(j), lStr.get(j))) {
                        isSatisfied = false;
                        break;
                    } else {
                        visited[j] = true;
                    }
                }
            }

            if (isSatisfied) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        int i = 0;
        while (i < t) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt();
            }
            int m = sc.nextInt();
            String[] strs = new String[m];
            for (int j = 0; j < m; j++) {
                strs[j] = sc.next();
            }

            Main.solve(a, strs, n);

            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 the bug in the provided buggy code, we need to focus on scenarios where the code's logic might fail. Specifically, we need to ensure that the one-to-one correspondence between the characters of the string and the elements of the array is properly checked.

### Step-by-Step Reasoning:

1. **Identify the Bug:**
   - The buggy code incorrectly handles the one-to-one correspondence between the characters of the string and the elements of the array.
   - It does not correctly check if the same symbols from the string correspond to the same numbers from the array and vice versa.

2. **Create a Test Case:**
   - We need a test case where the template and the string have the same length, but the one-to-one correspondence is violated.
   - A good example is a template where some numbers are repeated, and a string where the corresponding characters are not consistent with the numbers.

3. **Example Test Case:**
   - **Template:** `[1, 2, 1, 3]`
   - **String:** `abac`
     - Here, `1` maps to both `a` and `b`, which violates the one-to-one correspondence.

4. **Construct the Test Input:**
   - We will use this example in a test case with multiple strings to ensure the bug is exposed.

### Test Input:
```plaintext
1
4
1 2 1 3
2
abac
aabb
```

### Explanation:
- **Test Case:**
  - **Template:** `[1, 2, 1, 3]`
  - **String 1:** `abac`
    - The character `a` corresponds to both `1` and `2`, which should fail the test.
  - **String 2:** `aabb`
    - The character `a` corresponds to `1` and `2`, which should also fail the test.

This test input should expose the bug in the code by demonstrating that the one-to-one correspondence is not being correctly enforced.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.*;
public class circle{
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        while(n>0){
            int a=sc.nextInt();
            int[] ans=new int[a];
            for(int i=0;i<a;i++){
                ans[i]=sc.nextInt();
            }
            sc.nextLine();
            int b=sc.nextInt();
            sc.nextLine();
            for(int i=0;i<b;i++){
                String m1=sc.nextLine();
                if(a!=m1.length()) System.out.println("NO");
                else{
                    int p=0;
                    for(int j=0;j<a;j++){
                        for(int k=j;k<a;k++){
                            if(ans[j]==ans[k]){
                                if(m1.charAt(j)!=m1.charAt(k)) {
                                    p=1;
                                    System.out.println("NO");
                                    break;
                                }
                            }
                            else{
                                if(m1.charAt(j)==m1.charAt(k)) {
                                    p=1;
                                    System.out.println("NO");
                                    break;
                                }
                            }
                        }
                        if(p==1) break;
                    }
                    if(p==0) System.out.println("YES");
                }
            }

            n--;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
3 5 2 1 3
3
abfda
afbfa
abfda
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Numeric String Template
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kristina has an array $a$, called a template, consisting of $n$ integers. She also has $m$ strings, each consisting only of lowercase Latin letters. The strings are numbered from $1$ to $m$. She wants to check which strings match the template.

A string $s$ is considered to match the template if all of the following conditions are simultaneously satisfied:

  * The length of the string $s$ is equal to the number of elements in the array $a$.   * The same numbers from $a$ correspond to the same symbols from $s$. So, if $a_i = a_j$, then $s_i = s_j$ for ($1 \le i, j \le n$).   * The same symbols from $s$ correspond to the same numbers from $a$. So, if $s_i = s_j$, then $a_i = a_j$ for ($1 \le i, j \le n$).

In other words, there must be a one-to-one correspondence between the characters of the string and the elements of the array.

For example, if $a$ = [$3, 5, 2, 1, 3$], then the string "abfda" matches the template, while the string "afbfa" does not, since the character "f" corresponds to both numbers $1$ and $5$.

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

The following descriptions are for the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($-10^9 \le a_i \le 10^9$) — the elements of the array $a$.

The third line of each test case contains a single integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of strings to check for template matching.

Following are $m$ strings, each containing a non-empty string $s_j$ ($1 \le |s_j| \le 2 \cdot 10^5$), consisting of lowercase Latin letters.

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

Output: For each test case, output $m$ lines. On the $i$-th line ($1 \le i \le m$) output:

  * "YES", if the string with index $i$ matches the template;    * "NO" otherwise.

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

Examples:
input:
3
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
output:
YES
NO
YES
NO
NO
NO
YES
NO
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
//In the name of Allah, The Most Gracious and The Most Merciful
#include<bits/stdc++.h>
#ifdef ONLINE_JUDGE
#define debug(...) void(0);
#else
#include "debug.hpp"
#endif
#define gcd __gcd
#define ll long long
using namespace std;
////////////////////////////////////////////////////
////////////////////////////////////////////////////
void ideSetup() {
    #ifndef ONLINE_JUDGE
        freopen("./input.txt", "r", stdin);
        //freopen("./output.txt", "w", stdout);
        freopen("./error.txt", "w",stderr);
    #endif
}
void fastIoAndSetupIDE()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ideSetup();
}
////////////////////////////////////////////////////
class Calc
{
    public:
        vector<ll> sieveData;
        Calc()
        {
            //for sqrtn logic;
        }
        Calc(ll n)
        {
            sieveData = vector<ll>(n+1, 0);
            for(ll i =1; i<=n; i++)
            {
                sieveData[i] = i;
            }
            for(int i=2; i<=sqrt(n); i++)
            {
                if(i == sieveData[i])
                {
                    for(int j=i; j<=n; j+=i)
                    {
                        if(sieveData[j] == j)
                            sieveData[j] = i;
                    }
                }
            }
        }
        ll fastPower(ll a, ll b)
        {
            if(b==0){return 1;}
            if(b%2==0){return fastPower(a,b/2)*fastPower(a,b/2);}
            return fastPower(a,b-1)*a;
        }
        bool sqIsPrime(ll n)
        {
            if(n==2)return true;
            if(n<=1)return false;
            if(n<=3)return true;
            if(n%2==0 || n%3==0)return false;
            for(int i=5;i*i<=n;i=i+6)
            if(n%i==0 || n%(i+2)==0)return false;
            return true;
        }
        vector<ll> sqPrimeFactors(ll n)
        {
            vector<ll> temp;
            for(ll i=2; i<=sqrt(n); i++)
            {
                if(sqIsPrime(i) and n % i == 0)
                    temp.push_back(i);
            }
            return temp;
        }
        vector<ll> sqlAllFactors(ll n)
        {
            vector<ll>temp;
            for(ll i=2; i<=sqrt(n); i++)
            {
                if(n % i == 0)
                {
                    if(n % (n/i) == 0 and i != (n/i))
                        temp.push_back(n/i);
                    temp.push_back(i);
                }
            }
            sort(temp.begin(), temp.end());
            return temp;
        }
        bool logIsPrime(ll n)
        {
            ll sieveDataSize = sieveData.size();
            if(n > sieveDataSize)
                return false;
            return sieveData[n] == n;
        }
        vector<ll> logPrimeFactors(ll n)
        {
            unordered_map<ll, int>visited;
            vector<ll>temp;
            while(n > 1)
            {
                if(!visited[sieveData[n]])
                {
                    visited[sieveData[n]]++;
                    temp.push_back(sieveData[n]);
                }
                n = n / sieveData[n];
            }
            return temp;
        }
        ll gcd(ll a, ll b)
        {
            return __gcd(a, b);
        }
};
////////////////////////////////////////////////////
void solution()
{
    int n; cin>>n;
    vector<int>vec(n, 0);
    unordered_map<int, int> uniqueCount;
    for(int i=0; i<n; i++)
    {
            cin>>vec[i];
            uniqueCount[vec[i]]++;
    }

    int m;
    cin>>m;
    for(int i=0; i<m; i++)
    {
        string s; cin>>s;
        int z = 0;
        int stringSize = s.size();
        if(stringSize  != n)
        {
            cout<<"NO"<<"\n";
            continue;
        }
        unordered_map<char, vector<int>> vect;
        for(auto j: s)
        {
            vect[j].push_back(z);
            z++;
        }
        bool check = false;
        int uniqueCountChar = 0;
        for(char j='a'; j<='z'; j++)
        {
            int matcher = -1;
            if(vect[j].size())
            {
                uniqueCountChar++;
            }
            for(auto k : vect[j])
            {
                if(matcher == -1)
                        matcher = vec[k];
                if(vec[k] != matcher)
                {
                    cout<<"NO"<<"\n";
                    check = true;
                    break;
                }
            }
            if(check)
            {
                break;
            }
        }

        int counting = uniqueCount.size();
        if(!check and uniqueCountChar == counting)
        {
            cout<<"YES"<<"\n";
        }
        else if(!check)
        {
            cout<<"NO"<<"\n";
        }
    }
}
////////////////////////////////////////////////////
////////////////////////////////////////////////////
int32_t main()
{
    fastIoAndSetupIDE();
    int t;
    // if no test case default is one
    t=1;
    cin>>t;
    while(t--)
    {
        solution();
    }
}

**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 test input that exposes the bug in the provided code, we need to consider the following aspects:

1. **Edge Cases**:
   - Empty strings or arrays.
   - Arrays and strings of length 1.
   - Arrays and strings with all elements or characters being the same.
   - Arrays and strings with all elements or characters being unique.
   - Arrays and strings where the mapping between numbers and characters is not one-to-one.

2. **Challenging Scenarios**:
   - Arrays and strings where the length of the string does not match the length of the array.
   - Arrays and strings where the mapping between numbers and characters is not consistent.

Given these considerations, let's construct a test input that includes a mix of these scenarios to ensure the code is thoroughly tested.

### Test Input
```plaintext
4
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
3
1 1 1
3
aaa
aab
abc
```

### Explanation of the Test Input
1. **Test Case 1**:
   - Array: [3, 5, 2, 1, 3]
   - Strings: ["abfda", "afbfa"]
   - "abfda" should match the template.
   - "afbfa" should not match the template because 'f' maps to both 1 and 5.

2. **Test Case 2**:
   - Array: [1, 2]
   - Strings: ["ab", "abc", "aa"]
   - "ab" should match the template.
   - "abc" should not match the template because its length does not match the array length.
   - "aa" should not match the template because 'a' maps to both 1 and 2.

3. **Test Case 3**:
   - Array: [5, -3, 5, -3]
   - Strings: ["aaaa", "bcbc", "aba", "cbcb"]
   - "aaaa" should not match the template because 'a' maps to both 5 and -3.
   - "bcbc" should match the template.
   - "aba" should not match the template because its length does not match the array length.
   - "cbcb" should match the template.

4. **Test Case 4**:
   - Array: [1, 1, 1]
   - Strings: ["aaa", "aab", "abc"]
   - "aaa" should match the template.
   - "aab" should not match the template because 'a' maps to both 1 and 1, and 'b' maps to 1.
   - "abc" should not match the template because 'a', 'b', and 'c' map to 1, and the template only has one unique number.

This test input covers a variety of edge cases and challenging scenarios to ensure the code is thoroughly tested.

4
5
3 5 2 1 3
2
abfda
afbfa
2
1 2
3
ab
abc
aa
4
5 -3 5 -3
4
aaaa
bcbc
aba
cbcb
3
1 1 1
3
aaa
aab
abc


**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>
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
#define ff first
#define ss second
#define pb push_back
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define gcd(a,b) __gcd(a,b)
#define cin(v) for(ll i=0; i<v.size(); i++){ cin>>v[i]; }
#define all(v) (v).begin(),(v).end()
#define cout(v) for(ll i=0; i<v.size(); i++){ cout<<v[i]<<" "; }
bool valid(ll i, ll j, ll n, ll m){
    if(i<0 || j<0 || i>=n || j>=m) return false;
    return true;
}
ll dr[5] = {-1, 0, 1, 0, -1};
const ll MOD = 1e9+7;

// Helper function to find `r` and `res`
pair<int, int> findRange(const vector<pair<int,int>>& arr, int i, int d, int n){
    int r = upper_bound(all(arr), pair<int,int>{i+d, -1}) - arr.begin();
    r--;
    if (r < 0 || arr[r].ss < i) return {-1, -1};
        int left = 0, right = r, res = r, mid;
    while(left <= right){
        mid = (left + right) / 2;
        if(arr[mid].ss >= i){
            right = mid - 1;
            res = mid;
        }
        else{
            left = mid + 1;
        }
    }

    return {r, res};
}

int brother(vector<pair<int,int>> &arr, int d, int n){
    int maxVal = 0, iVal = -1;
    for(int i = 0; i < n - d; i++){
        pair<int, int> result = findRange(arr, i, d, n);
        int r = result.first, res = result.second;
        if(r == -1) continue;

        int val = r - res + 1;
        if(val > maxVal){
            maxVal = val;
            iVal = i;
        }
    }
    return iVal + 1;
}

int mother(vector<pair<int,int>> &arr, int d, int n){
    int minVal = 1e6, iVal = -1;
    for(int i = 0; i < n - d; i++){
        pair<int, int> result = findRange(arr, i, d, n);
        int r = result.first, res = result.second;
        if(r == -1) return i + 1;

        int val = r - res + 1;
        if(val < minVal){
            minVal = val;
            iVal = i;
        }
    }
    return iVal + 1;
}

void solve(ll tc, ll curr){
    int n, d, k;
    cin >> n >> d >> k;

    vector<pair<int,int>> arr;

    rep(i, 0, k){
        int l, r;
        cin >> l >> r;
        arr.push_back({l-1, r-1});
    }

    sort(all(arr));

    int b = brother(arr, d, n);
    int m = mother(arr, d, n);

    cout << b << ' ' << m;
}

int main(){
    ll t;
    t = 1;
    cin >> t;
    for(ll i = 1; i <= t; i++){
        solve(t, i);
        cout << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
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:
#include <bits/stdc++.h>
using namespace std;
#define int long long

/* ITS TIME TO GRIND */

int power(int A, int expo)
{
    if (expo == 0)
        return 1;
    int res = power(A, expo / 2);
    if (expo % 2)
        return res * res * A;
    else
        return res * res;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }

int fact(int x)
{
    if (x == 1 || x == 0)
        return 1;
    return (x * fact(x - 1));
}

class UnionFind{
    private: vector<int> parent,rank;
    public:
        UnionFind(int n){
            rank.assign(n,0);
            parent.assign(n,0);
            iota(parent.begin(),parent.end(),0);
        }

        int find(int x) {
            if(parent[x] == x)
                return x;

            return parent[x] = find(parent[x]);
        }

        void UnionSet(int x, int y) {
            int x_parent = find(x);
            int y_parent = find(y);

            if(rank[x_parent] > rank[y_parent]) {
                parent[y_parent] = x_parent;
            } else if(rank[x_parent] < rank[y_parent]) {
                parent[x_parent] = y_parent;
            } else {
                parent[x_parent] = y_parent;
                rank[y_parent]++;
            }
        }

};

void solve() {
   int n,d,k;
   cin>>n>>d>>k;

   vector<int> range_update_tech(n+2,0);
   while(k--){
    int l, r;
    cin >> l >> r;

    int start = max((int)1, l - d + 1);
    int end = r + 1;
    range_update_tech[start]++;
    if(end<=n){
         range_update_tech[end]--;
    }
   }

   for(int i=1;i<=n;i++){
    range_update_tech[i]+= range_update_tech[i-1];
   }

   int mother=0,brother=0;
   int max_days=0,min_days=1e9;

   for(int i=1;i<=n-d;i++){
        if(range_update_tech[i]>max_days){
            max_days=range_update_tech[i];
            brother=i;
        }
        if(range_update_tech[i]<min_days){
            min_days=range_update_tech[i];
            mother=i;
        }
   }

   cout<<brother<<' '<<mother<<'\n';

}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases for Overlapping Jobs:**
   - Ensure that the start and end days of jobs are close to the boundaries of the total days.
   - Create jobs that overlap with each other to test the range update logic.

2. **Edge Cases for Visit Duration:**
   - Test with a visit duration `d` that is equal to the total number of days `n`.
   - Test with a visit duration `d` that is 1, which is the minimum possible value.

3. **Multiple Jobs with Different Overlaps:**
   - Create a scenario where the maximum number of distinct jobs for the brother and the minimum number of distinct jobs for the mother are not the same.

4. **Single Test Case:**
   - Since the output length is limited, focus on a single test case that is complete and valid.

Based on these considerations, here is a small-scale test input that should expose potential bugs in the code:

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

### Explanation:
- **Total Days (n):** 5
- **Visit Duration (d):** 3
- **Number of Jobs (k):** 3
- **Jobs:**
  - Job 1: Day 1 to Day 2
  - Job 2: Day 3 to Day 4
  - Job 3: Day 4 to Day 5

### Expected Output:
- **Brother's Best Start Day:** 1 (overlaps with 2 jobs: [1, 2, 3])
- **Mother's Best Start Day:** 3 (overlaps with 1 job: [3, 4, 5])

This test case is designed to check:
- The range update logic for overlapping jobs.
- The correct calculation of the maximum and minimum number of distinct jobs for the brother and mother, respectively.
- The handling of jobs that are close to the boundaries of the total days.

By running this test case, we can observe if the buggy code correctly identifies the best start days for both the brother and the mother.


**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>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int,int>;
#define pb push_back
#define loop(i,n) for (int i=0;i<n;i++)
#define readint(...) int __VA_ARGS__; read_input(__VA_ARGS__);
#define readvec(v,n) vi v(n); loop(i,n) cin >> v[i];
void read_input() {} template <typename T, typename... Args>
void read_input(T &first, Args&... args) { std::cin >> first; read_input(args...);}
using pis = pair<int, string>; // "start", "end"
using wd = pair<int, int>; // start index, num overlapping
int main() { ios::sync_with_stdio(0); cin.tie(0);
readint(test_cases); while (test_cases--) {
	readint(n, d, k);
	priority_queue<pis, vector<pis>, greater<pis>> pq;
	loop(i, k) {
		int start, end;
		cin >> start >> end;
		pq.push({start-1, "start"});
		pq.push({end-1, "mid"});
		pq.push({end, "end"});
	}

	int mx = 0;
	int mxi = 0;
	int mn = INT_MAX;
	int mni = 0;
	int jobs = 0;
	vector<wd> windows;
	windows.pb({0, 0});
	while (!pq.empty()) {
		pis ev = pq.top();
		pq.pop();
		if (ev.second == "start") {
			jobs++;

			int wi = windows.size() - 1;
			bool exists = false;
			while (wi >= 0) {
				if (windows[wi].first == ev.first) exists = true;
				if (windows[wi].first > ev.first - d) {
					// cout << "incrementing " << windows[wi].first << endl;
					windows[wi].second++;
					wi--;
				} else {
					break;
				}
			}
			if (!exists && ev.first + d <= n) windows.pb({ev.first, jobs});
		} else if (ev.second == "mid") {
			int wi = windows.size() - 1;
			bool exists = false;
			if (wi >= 0 && windows[wi].first == ev.first) exists = true;
			if (!exists && ev.first + d <= n) windows.pb({ev.first, jobs});
		} else {
			jobs--;

			int wi = windows.size() - 1;
			bool exists = false;
			if (wi >= 0 && windows[wi].first == ev.first) exists = true;
			if (!exists && ev.first + d <= n) {
				windows.pb({ev.first, jobs});
				// cout << "creating " << ev.first << "=" << jobs << endl;
			} else if (exists) {
				windows[wi].second = jobs;
			}
		}
	}

	// cout << "======" << endl;
	for (wd w: windows) {
		// cout << w.first << "," << w.second << endl;
		if (w.second > mx) {
			mx = w.second;
			mxi = w.first;
		}
		if (w.second < mn) {
			mn = w.second;
			mni = w.first;
		}
	}

	cout << mxi+1 << " " << mni+1 << endl;
}}

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

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

Think step by step.

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


**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 sys
from collections import defaultdict
input = sys.stdin.readline

def solve():
    n, d, k = map(int, input().split())
    e = list()
    q = defaultdict(list)
    cur = set()
    for i in range(k):
        l, r = map(int, input().split())
        q[l].append((i, 1))
        q[r].append((i, -1))
        if 1 <= l <= d:
            cur.add(i)
    start = 1
    end = d
    i = 0
    mx = (len(cur), -1)
    mn = (len(cur), 1)
    for _ in range(n):
        end += 1
        for x, tp in q[start]:
            if tp == -1:
                cur.remove(x)
        for x, tp in q[end]:
            if tp == 1:
                cur.add(x)
        start += 1
        mx = max(mx, (len(cur), -start))
        mn = min(mn, (len(cur), start))
        if end == n:
            break
    print(-mx[1], mn[1])

for _ in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
10 5 5
1 3
2 4
3 5
6 8
7 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:
def main():
    t = int(input())
    for _ in range(t):
        n, d, k = map(int, input().split())
        jobsStart = []
        jobsEnd = []
        for i in range(k):
            l, r = map(int, input().split())
            jobsStart.append(max(l - d + 1, 1))
            jobsEnd.append(r)

        jobsStart.sort(reverse=True)
        jobsEnd.sort(reverse=True)
        s, e = k, k - 1
        lastDay = n - d + 1
        mother, motherArr = -1, -1
        brother, brotherArr = -1, -1
        currDay = 1
        while currDay <= lastDay:
            while s > 0 and jobsStart[s - 1] <= currDay:
                s -= 1
            while e >= 0 and jobsEnd[e] < currDay:
                e -= 1
            if e < 0:
                mother = 0
                motherArr = currDay
                break
            numJobs = e - s + 1
            if mother == -1 or numJobs < mother:
                mother = numJobs
                motherArr = currDay
            if brother == -1 or numJobs > brother:
                brother = numJobs
                brotherArr = currDay
            # print(f"{currDay}: {s}, {e}, {numJobs}")
            currDay += 1

        print(f"{brotherArr} {motherArr}")

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
1
10 5 5
1 3
4 6
7 9
2 4
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:
from sys import stdin
input = stdin.readline
def solve():
    N,d,k = map(int,input().split())
    arr = [0]*(2+N)
    for _ in range(k):
        l,r = map(int,input().split())
        arr[max(0,l-d+1)] += 1; arr[r+1] -= 1
    prefix = [0]
    for i in range(1,2+N): prefix.append(prefix[-1]+arr[i])
    bro = [0,-1]; mom = [0,10**8]
    for i in range(1,1+N-d):
        if prefix[i] > bro[1]: bro = [i, prefix[i]]
        if prefix[i] < mom[1]: mom = [i, prefix[i]]
    print(bro[0],mom[0])
for _ in range(int(input())): 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
4
10 1 10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 10
10 5 5
1 5
2 6
3 7
4 8
5 9
10 2 2
1 5
6 10
10 10 10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 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:
import java.util.*;

public final class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int cases = scanner.nextInt();
        for (int i = 0; i < cases; i++){
            perform(scanner);
        }
    }

    private static void perform(Scanner scanner) {
        int n = scanner.nextInt();
        int d = scanner.nextInt();
        int k = scanner.nextInt();
        int[] starts = new int[n];
        int[] ends = new int[n];
        for (int i = 0; i < k; i++){
            int start = scanner.nextInt()-1;
            starts[start] += 1;
            int end = scanner.nextInt()-1;
            ends[end] += 1;
        }
        int sum = 0;
        for(int i = 0; i < d; i++) {
            sum += starts[i];
        }
        int min = sum;
        int mother = 0;
        int max = sum;
        int brother = 0;
        int pointer = 1;
        while(pointer+d < n-1) {
            sum += starts[pointer+d-1];
            sum -= ends[pointer-1];
            if(sum < min) {
                min = sum;
                mother = pointer;
            } else if(sum > max){
                max = sum;
                brother = pointer;
            }
            pointer++;
        }
        System.out.print(brother+1);
        System.out.print(" ");
        System.out.println(mother+1);
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to identify such cases:

1. **Edge Cases for `d` and `n`**:
   - When `d` is 1, the algorithm should handle this correctly.
   - When `d` is equal to `n`, the entire range is considered.
   - When `d` is close to `n`, the algorithm should handle the overlapping correctly.

2. **Edge Cases for `k`**:
   - When `k` is 0, there are no jobs.
   - When `k` is 1, there is only one job.
   - When `k` is close to `n`, there are many jobs.

3. **Edge Cases for Job Overlaps**:
   - Jobs that start and end on the same day.
   - Jobs that span the entire range from 1 to `n`.
   - Jobs that are very short or very long.

4. **Edge Cases for Start and End Days**:
   - Jobs that start at the beginning and end at the end.
   - Jobs that start and end within the range of `d`.

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

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

### Explanation:
1. **Test Case 1**:
   - `n = 5`, `d = 2`, `k = 3`
   - Jobs: [1, 2], [2, 3], [4, 5]
   - This case checks if the algorithm can handle overlapping jobs and find the correct start days for both brother and mother.

2. **Test Case 2**:
   - `n = 6`, `d = 3`, `k = 4`
   - Jobs: [1, 2], [2, 4], [4, 6], [3, 5]
   - This case checks if the algorithm can handle multiple overlapping jobs and find the correct start days for both brother and mother.

3. **Test Case 3**:
   - `n = 10`, `d = 5`, `k = 6`
   - Jobs: [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7]
   - This case checks if the algorithm can handle a longer range and multiple overlapping jobs and find the correct start days for both brother and mother.

These test cases should help in identifying any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 Main {

    private static void solution(int[][] work, int n, int d, int k) {
        int[] prefix = new int[n + 1];//判断有多少个区间的左端点大于给定区间右端点，右端点小于给定区间左端点
        int[] suffix = new int[n + 1];
        int[] temp = new int[n + 1];
        for (int i = 0; i < k; i++) temp[work[i][1]]++;//右端点存前缀
        for (int i = 0; i < n; i++) prefix[i + 1] = prefix[i] + temp[i];//不包括当前
        temp = new int[n + 1];
        for (int i = 0; i < k; i++) temp[work[i][0]]++;
        for (int i = n - 1; i >= 0; i--) suffix[i] = suffix[i + 1] + temp[i];//包括当前
        int min = 0, max = 0;
        int minv = Integer.MAX_VALUE, maxv = 0;
        for (int i = 0; i < n; i++) {
            int r = i + d;
            if (r >= n) break;
            int sum = suffix[r] + prefix[i];
            if (sum < minv) {
                minv = sum;
                min = i;
            }
            if (sum > maxv) {
                maxv = sum;
                max = i;
            }
        }
        System.out.printf("%d %d\n", min + 1, max + 1);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n, d, k;
            n = scanner.nextInt();
            d = scanner.nextInt();
            k = scanner.nextInt();
            int[][] work = new int[k][2];
            for (int j = 0; j < k; j++) {
                work[j][0] = scanner.nextInt() - 1;
                work[j][1] = scanner.nextInt() - 1;
            }
            solution(work, n, d, 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
10 5 10
1 5
2 6
3 7
4 8
5 9
6 10
1 4
2 5
3 6
4 7
```


**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>
using namespace std;
#define ll long long
#define F first
#define S second
#define pb push_back

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

// Sieve Method to Find Primes upto n
vector<bool> sieve(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = is_prime[1] = false;

    for (int i = 2; i <= n; i++) {
        if (is_prime[i]) {
            for (int j = i * i; j <= n; j += i) {
                is_prime[j] = false;
            }
        }
    }

    return is_prime;
}

// Binary Search Template

bool bin_cond(int k) {
    return 0;
}

// Minimize k s.t. condition(k) is True
int bin_search(int length) {
    int left = 0;
    int right = (length - 1);

    while (left < right) {
        int mid = left + (floor((right - left) / 2));

        if (bin_cond(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }

    return left;
}

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

    // freopen(".in", "r", stdin);
    // freopen(".out", "w", stdout);

    int t;
    cin >> t;
    while (t--) {
        int n, d, k; cin >> n >> d >> k;
        vector<int> diff_arr(n + 1, 0);
        vector<pair<int, int> > b(k);
        for (int i = 0; i < k; i++) {
            int l, r; cin >> l >> r;
            b[i] = make_pair(l, r);
            diff_arr[max(l - d + 1, 1)]++;
            if (r < n) {
                diff_arr[r + 1]--;
            }
        }

        for (int i = 1; i <= n; i++) {
            diff_arr[i] += diff_arr[i - 1];
        }

        int min_diff = INT_MAX;
        int min_diff_index = 0;
        int max_diff = -1;
        int max_diff_index = 0;
        for (int i = 1; i <= n - d; i++) {
            if (diff_arr[i] < min_diff) {
                min_diff = diff_arr[i];
                min_diff_index = i;
            }
            if (diff_arr[i] > max_diff) {
                max_diff = diff_arr[i];
                max_diff_index = i;
            }
        }

        if (min_diff == max_diff) {
            cout << max_diff_index << " " << max_diff_index << '\n';
            continue;
        }

        cout << max_diff_index << " " << min_diff_index << '\n';

    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 3 5
1 3
4 6
7 9
2 5
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:
#include <bits/stdc++.h>
using namespace std;
int main(){
    int T;
    cin >> T;
    while (T--){
        int n,d,k;
        cin >> n >> d >> k;
        int L[k],R[k];
        for (int i=0;i<k;i++){
            cin >> L[i] >> R[i];
        }
        sort(L,L+k);
        sort(R,R+k);
        int diff,bro,ma,pb,pm;
        bro = 0, ma = 1e9;
        for (int i=1;i<=n-d;i++){
            int r,l;
            r = l = 0;
            while (r < k && L[r] <= i+d-1) r ++;
            while (l < k && R[l] < i) l ++;
            diff = r - l;
            if (diff > bro){
                bro = diff;
                pb = i;
            }
            if (diff < ma){
                ma = diff;
                pm = i;
            }
        }
        cout << pb << ' ' << pm << '\n';
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code sorts the start and end days of the jobs separately and then uses these sorted arrays to determine the best start days for Robin's brother and mother. This approach can lead to incorrect results if the jobs overlap in complex ways.

Let's consider the following edge cases:
1. **Single Job Covering All Days**: This should force both the brother and mother to start on day 1.
2. **Multiple Jobs with Overlapping and Non-Overlapping Segments**: This can expose issues in the logic for counting overlapping jobs.
3. **Jobs at the Ends of the Range**: This can test the edge cases where the start or end of the visit period is at the boundary of the day range.

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

```plaintext
4
2 1 1
1 2
4 2 2
1 2
3 4
5 2 3
1 2
3 4
5 5
7 3 4
1 3
4 6
7 7
2 4
```

Explanation of the test cases:
1. **Test Case 1**: A single job covering all days. Both the brother and mother should start on day 1.
2. **Test Case 2**: Two jobs that cover the first and last days of the 4-day period. The brother should start on day 1 (overlaps with 2 jobs), and the mother should start on day 3 (overlaps with 1 job).
3. **Test Case 3**: Three jobs that cover different segments of the 5-day period. The brother should start on day 1 (overlaps with 2 jobs), and the mother should start on day 4 (overlaps with 1 job).
4. **Test Case 4**: Four jobs with overlapping and non-overlapping segments. The brother should start on day 1 (overlaps with 3 jobs), and the mother should start on day 5 (overlaps with 1 job).

This test input should help expose any bugs in the given buggy code.