**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

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

         ans=n-(n-c+b-1)/b;

       }
       if(b==0){
           if(c==n-2){
               ans=n-1;
           }
           if(c==n-1){
               ans=n-1;
           }
           if(c>=n){
               ans=n;
           }
           if(c<(n-2)){
               ans=-1;
           }
       }
          System.out.println(ans);
    }
}
}

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

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

Think step by step.

Response:
```plaintext
1
5 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

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

public class AlicePermutation{
    public long permutation(long n, long b, long c){
        long count =0;

        if(n==1 && c==0){
            return 0;
        }else if(n==1 && c==1){
            return 1;
        }else if(b==1 && c==0){
            return 0;
        }
        else if(b==0 && n>=c) {
            long comp = n-c;
            if(n == c){
                return c;
            }else if(comp>=3){
                return -1;
            }else if(comp==2){
                return c+1;
            }else if(comp==1){
                return c;
            }
        }else if(b==0 && n<c) {
            return n;
        }

        if(n<=c){
            return n;
        }
        else{
            long i = ((n-c)+1)/b;
            return n-i;
        }

    }

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

        long T = scanner.nextLong();
        while(T-- != 0){

            long inp1 = scanner.nextLong();
            long inp2 = scanner.nextLong();
            long inp3 = scanner.nextLong();

            AlicePermutation obj = new AlicePermutation();
            long result = obj.permutation(inp1, inp2, inp3);

            System.out.println(result);

        }
    }
}

**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
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

Buggy Code:
def mex(array):
    temp = set(range(max(array)+1))
    result = temp - set(array)
    if not result:
        return max(array) + 1
    return min(result)

def is_perm(array):
    if len(array) != len(set(array)):
        return False
    if max(array) >= len(array):
        return False
    return True

def solve(n, b, c):
    if n == b == c:
        return n
    if b == c == 0:
        return -1

    a = [b*(i-1)+c for i in range(1, n+1)]

    count = 0
    while True:
        if count >= 10**6-1:
            return -1
        if is_perm(a):
            break
        a[a.index(max(a))] = mex(a)
        count += 1

    return count

t = int(input())
for _ in range(t):
    n, b, c = map(int, input().split())
    print(solve(n, b, c))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

Buggy Code:
import random
import math
from types import GeneratorType
import sys
from collections import deque
from heapq import heapify,heappush,heappop
import bisect
input = lambda: sys.stdin.readline().rstrip()
RANDOM = random.getrandbits(32)
class Wrapper(int):
    def __init__(self, x):
        int.__init__(x)
    def __hash__(self):
        return super(Wrapper, self).__hash__() ^ RANDOM

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

T = int(input())

for __ in range(T):
    # print(__)
    # n=int(input())

    n,b,c=map(int,input().split())
    if b==0:
        if c>(n-1):
            print(n)
        elif c==n-1 or c==n-2:
            print(n-1)
        else:
            print(-1)
        continue
    if n==1:
        if b==0 and c==0:
            print(0)
        else:
            print(1)
        continue
    print(n-max(((n-1-c)//b)+1,0))

**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
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

Buggy Code:
import java.util.Scanner;

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

        int t = scanner.nextInt();
        scanner.nextLine();

        while (t-- > 0) {
            long n = scanner.nextLong();
            long b = scanner.nextLong();
            long c = scanner.nextLong();

            if (c > n){
                System.out.println(n);
                continue;
            }

            if (b==0){
                if (c<n-2){
                    System.out.println(-1);
                    continue;
                }
                if (c<n){
                    System.out.println(n-1);
                    continue;
                }
            }

            System.out.println(n - ((n-c+b-1)/b));

        }
    }
}

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

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

Think step by step.

Response:
```plaintext
5
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

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

public class B_Alice_s_Adventures_in_Permuting {
    static FastReader fr;

    public static void main(String[] args) throws IOException {
        fr = new FastReader();
        int T = fr.nextInt();
        while (T-- > 0) {
            solve();
        }
    }

    private static void solve() {
        long n = fr.nextLong();
        long b = fr.nextLong();
        long c = fr.nextLong();

        if (b == 0) {
            long maxFormed = c + 1;
            if (maxFormed == n - 1) {
                System.out.println(n - 1);
            } else {
                System.out.println(-1);
            }
            return;
        }

        if (c == 0 && b == 1) {
            System.out.println(0);
            return;
        }

        if (c >= n) {
            System.out.println(n);
            return;
        }

        long elements = n /b  - c / b;

        System.out.println(n - elements);
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            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:
```plaintext
3
10 1 0
100 2 1
3 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main() {
	long long t,b,c,l,n,r,mid;
	scanf("%lld",&t);
	while(t--) {
		scanf("%lld%lld%lld",&n,&b,&c);
		if(b==0) {
			if(c>n-3) printf("%d\n",n-(c<n));
			else printf("-1\n");
			continue;
		}
		if(n<=c) {
			printf("%lld\n",n);
			continue;
		}
		printf("%lld\n",n-1-(n-1-c)/b);
	}
}

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

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

Think step by step.

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

1. **Case where `b == 0`**:
   - The code has a special condition for `b == 0`, but it might not handle all edge cases correctly.
   - For example, if `c` is exactly `n-1`, the array will already be a permutation, so no operations are needed.

2. **Case where `b != 0`**:
   - The formula used to calculate the number of operations might be incorrect or not handle all edge cases.
   - For example, if `n` is very large and `b` is also large, the number of operations might be incorrect.

3. **Special cases for `c`**:
   - If `c` is very close to `n-1`, the array might already be a permutation or very close to being a permutation.

4. **General edge cases**:
   - Small values of `n`, `b`, and `c` to ensure the code handles basic cases correctly.
   - Large values of `n`, `b`, and `c` to ensure the code handles large inputs correctly.

Based on these considerations, here is a small-scale test input that includes various edge cases:

```plaintext
6
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
10 0 9
```

Explanation of the test cases:
1. `10 1 0`: The array is already a permutation, so no operations are needed.
2. `1 2 3`: The array is `[3]`, and the MEX is `0`, so one operation is needed.
3. `100 2 1`: The array starts with `[1, 3, 5, ..., 199]`, and it will take 50 operations to become a permutation.
4. `3 0 1`: The array is `[1, 1, 1]`, and it will take 2 operations to become a permutation.
5. `3 0 0`: The array is `[0, 0, 0]`, and it will never become a permutation, so the answer is `-1`.
6. `10 0 9`: The array is `[9, 9, 9, ..., 9]`, and it will never become a permutation, so the answer is `-1`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Permuting
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice mixed up the words transmutation and permutation! She has an array $a$ specified via three integers $n$, $b$, $c$: the array $a$ has length $n$ and is given via $a_i = b\cdot (i - 1) + c$ for $1\le i\le n$. For example, if $n=3$, $b=2$, and $c=1$, then $a=[2 \cdot 0 + 1, 2 \cdot 1 + 1, 2 \cdot 2 + 1] = [1, 3, 5]$.

Now, Alice really enjoys permutations of $[0, \ldots, n-1]$$^{\text{∗}}$ and would like to transform $a$ into a permutation. In one operation, Alice replaces the maximum element of $a$ with the $\operatorname{MEX}$$^{\text{†}}$ of $a$. If there are multiple maximum elements in $a$, Alice chooses the leftmost one to replace.

Can you help Alice figure out how many operations she has to do for $a$ to become a permutation for the first time? If it is impossible, you should report it.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $0$ to $n-1$ in arbitrary order. Please note, this is slightly different from the usual definition of a permutation. For example, $[1,2,0,4,3]$ is a permutation, but $[0,1,1]$ is not a permutation ($1$ appears twice in the array), and $[0,2,3]$ is also not a permutation ($n=3$ but there is $3$ in the array).

$^{\text{†}}$The $\operatorname{MEX}$ of an array is the smallest non- negative integer that does not belong to the array. For example, the $\operatorname{MEX}$ of $[0, 3, 1, 3]$ is $2$ and the $\operatorname{MEX}$ of $[5]$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $b$, $c$ ($1\le n\le 10^{18}$; $0\le b$, $c\le 10^{18}$) — the parameters of the array.

Output: For each test case, if the array can never become a permutation, output $-1$. Otherwise, output the minimum number of operations for the array to become a permutation.

Examples:
input:
7
10 1 0
1 2 3
100 2 1
3 0 1
3 0 0
1000000000000000000 0 0
1000000000000000000 1000000000000000000 1000000000000000000
output:
0
1
50
2
-1
-1
1000000000000000000

Note: In the first test case, the array is already $[0, 1, \ldots, 9]$, so no operations are required.

In the third test case, the starting array is $[1, 3, 5, \ldots, 199]$. After the first operation, the $199$ gets transformed into a $0$. In the second operation, the $197$ gets transformed into a $2$. If we continue this, it will take exactly $50$ operations to get the array $[0, 1, 2, 3, \ldots, 99]$.

In the fourth test case, two operations are needed: $[1,1,1] \to [0,1,1] \to [0,2,1]$.

In the fifth test case, the process is $[0,0,0] \to [1,0,0] \to [2,0,0] \to [1,0,0] \to [2,0,0]$. This process repeats forever, so the array is never a permutation and the answer is $-1$.

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

int main() {
    int t;
    cin >> t;
    while (t--) {
        long long n, c, b;
        cin >> n >> b >> c;
        if(c>=n){
            cout<<n<<endl;
            continue;
        }
        if(b!=0){
            cout<<n-(ceil(((n-1)-(c-1))/static_cast<double>(b)))<<endl;
        }
       else{
           if(c==n){
               cout<<n<<endl;
           }
           else if(c==n-1 || c==n-2){
               cout<<n-1<<endl;
           }
           else{
               cout<<-1<<endl;
           }
       }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
10 1 0
100 2 1
3 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=2e5+100;
int tree[N];
int T,n,idx;
struct edge
{
    int l,r,pos,dir;
}e[2*N],tp[N];
int ans[N];
int p[N*2],ix,id;
bool cmp(edge x,edge y)
{
    if(x.l==y.l) {
        if(x.pos==y.pos){
            return x.r>y.r;
        }
        return x.pos<y.pos;

    }
    return x.l<y.l;
}
bool cmp2(edge x,edge y)
{
    if(x.l==y.l) {
        if(x.pos==y.pos) return x.r>y.r;
        return x.pos<y.pos;
    }
    return x.l<y.l;
}
int lowbit(int x)
{
    return x&-x;
}
void add(int pos,int x)
{
    for(int i=pos;i<=ix;i+=lowbit(i))
    {
        tree[i]+=x;
    }
}
int cnt(int r)
{
    int res=0;
    for(int i=r;i;i-=lowbit(i))
    {
        res+=tree[i];
    }
    return res;
}
int query(int l,int r)
{
    return cnt(r)-cnt(l-1);
}
int find(int j)
{
    return lower_bound(p+1,p+ix+1,j)-p;
}
int get1(int l,int r)
{
    while(l<r)
    {
        int mid=l+r+1>>1;
        if(query(mid,r)) l=mid;
        else r=mid-1;
    }
    return l;
}
int get2(int l,int r)
{
    while(l<r)
    {
        int mid=l+r>>1;
        if(query(l,mid)) r=mid;
        else l=mid+1;
    }
    return l;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>T;
    while(T--)
    {
        idx=0,ix=0,id=0;
        cin>>n;
        int l,r;
        for(int i=1;i<=n;i++)
        {
            cin>>l>>r;
            p[++ix]=l,p[++ix]=r;
            if(l==r){
                tp[++id]={l,r,1,i};
                continue;
            }
            e[++idx]={l,r,1,i},e[++idx]={r,l,-1,i};

        }
        sort(tp+1,tp+id+1,cmp);
        tp[id+1]={0,0,0,0};
        for(int i=1;i<=id;i++)
        {
            if(tp[i].l==tp[i-1].l||tp[i].l==tp[i+1].l)
            {
                ans[tp[i].dir]=0;
            }
            else{
                e[++idx]={tp[i].l,tp[i].r,-1,tp[i].dir};
            }
        }
        sort(e+1,e+idx+1,cmp);
        e[idx+1]={0,0,0,0};
        sort(p+1,p+ix+1);
        ix=unique(p+1,p+ix+1)-p-1;
        for(int i=1;i<=idx;i++)
        {
            edge tmp=e[i];
            if(tmp.pos==1){
                int l=find(tmp.l);
                add(l,1);
            }
            else{
                int r=find(tmp.r);
                if(tmp.l!=tmp.r) add(r,-1);
                if(query(1,r)==0) ans[tmp.dir]=0;
                else{
                    if(tmp.r==e[i-1].r&&tmp.l==e[i-1].l&&tmp.pos==e[i-1].pos) {ans[tmp.dir]=0;continue;}
                    ans[tmp.dir]=tmp.r-p[get1(1,r)];
                }
            }
        }
        for(int i=1;i<=idx;i++) {
            if(e[i].l==e[i].r) e[i].pos=1;
        }
        sort(e+1,e+idx+1,cmp2);
        for(int i=idx;i;i--)
        {
            edge tmp=e[i];
            if(tmp.pos==-1)
            {
                int l=find(tmp.l);
                add(l,1);
            }
            else{
                int r=find(tmp.r);
                if(e[i].l!=e[i].r)add(r,-1);
                if(query(r,ix)==0) ans[tmp.dir]=0;
                else{
                    if(tmp.r==e[i+1].r&&tmp.l==e[i+1].l&&tmp.pos==e[i+1].pos) {ans[tmp.dir]=0;continue;}
                    ans[tmp.dir]+=p[get2(r,ix)]-tmp.r;
                }
            }
        }
        for(int i=1;i<=n;i++) cout<<ans[i]<<'\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
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<random>
#include<vector>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#include<ext/pb_ds/trie_policy.hpp>
#include<ext/pb_ds/priority_queue.hpp>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
#define lowbit(x) (x&-x)
#define fi first
#define se second

const int N = 2e5 + 10;
int n, l[N], r[N];
struct node
{
    int l, r, id;
};
bool cmp1(const node &x, const node &y)
{
    if(x.l != y.l) return x.l < y.l;
    return x.r > y.r;
}
bool cmp2(const node &x, const node &y)
{
    if(x.r != y.r) return x.r > y.r;
    return x.l < y.l;
}

    // for(auto i : p) cout << i.l << ' ' << i.r << '\n';
    // cout << "=============\n";
void solve()
{
    cin >> n;
    vector<node> p;
    for(int i = 1; i <= n; i ++)
        cin >> l[i] >> r[i], p.push_back({l[i], r[i], i - 1});
    set<int> st;

    sort(p.begin(), p.end(), cmp1);
    vector<LL> ans(n, 0);
    for(auto [l, r, id] : p){
        auto it = st.lower_bound(r);
        if(it != st.end()) ans[id] += *it - r;
        st.insert(r);
        // cout << ans[i] << '\n';
    }

    st.clear();
    for(int i = 0; i < n; i ++){
        p[i].l = 1e9 - p[i].l;
        p[i].r = 1e9 - p[i].r;
        swap(p[i].l, p[i].r);
    }
    sort(p.begin(), p.end(), cmp1);

    for(auto [l, r, id] : p){
        auto it = st.lower_bound(r);
        if(it != st.end()) ans[id] += *it - r;
        st.insert(r);
        // cout << ans[i] << '\n';
    }

    // for(auto i : p) cout << i.l << ' ' << i.r << '\n';
    // cout << "=============\n";
    for(int i = 0; i < n; i ++){
        if (p[i].l == p[i + 1].l and p[i].r == p[i + 1].r)
            ans[p[i].id] = ans[p[i + 1].id] = 0;
    }

    for(int i = 0; i < n; i ++) cout << ans[i] << '\n';
}
int main()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    //freopen("", "r", stdin); freopen("", "w", stdout);
	int Case = 1;
	cin >> Case;
	while(Case --) 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 code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are some key points to consider:

1. **Overlapping Segments**: Ensure that some users have overlapping segments but are not exact duplicates.
2. **Predictor Relationships**: Create scenarios where users have multiple predictors, and the strongly recommended tracks are not straightforward.
3. **Edge Cases**: Include users with segments that are at the extreme ends of the range (1 and 10^9).
4. **Single User Cases**: Test with a single user to ensure the code handles this scenario correctly.
5. **Multiple Test Cases**: Include multiple test cases to ensure the code handles different scenarios.

Given these considerations, here is a small-scale test input that should expose the bug:

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

Explanation:
1. **Test Case 1**:
   - User 1: [1, 10] (Predictors: none)
   - User 2: [3, 5] (Predictors: User 1)
   - User 3: [3, 7] (Predictors: User 1, User 2)
   - User 4: [5, 7] (Predictors: User 1, User 2, User 3)
   - User 5: [4, 4] (Predictors: User 1, User 2, User 3, User 4)
   - Expected Output: 0, 2, 3, 2, 4

2. **Test Case 2**:
   - User 1: [1, 10] (Predictors: none)
   - User 2: [2, 2] (Predictors: User 1)
   - Expected Output: 0, 9

3. **Test Case 3**:
   - User 1: [1, 10] (Predictors: none)
   - User 2: [2, 5] (Predictors: User 1)
   - User 3: [6, 10] (Predictors: User 1)
   - Expected Output: 0, 4, 4

This input should help in identifying issues with the code's logic for handling predictor relationships and strongly recommended tracks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

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

using namespace std;

inline void eraseOne(multiset<int> &s, int value) {
    auto itr = s.find(value);
    if (itr != s.end()){
        s.erase(itr);
    }
}

void testCase() {
    int N;
    cin >> N;
    vector<pair<int, int>> pairs(N);
    for (int i = 0; i < N; i++) {
        int l, r;
        cin >> l >> r;
        pairs[i] = { l, r };
    }
    struct Event {
        int p, l, r, add, id;
        bool operator<(const Event& other) const {
            if (p == other.p) {
                return r > other.r;
            }
            return p < other.p;
        }
    };
    map<int, vector<Event>> eventsAdd, eventsRemove;
    for (int i = 0; i < N; i++) {
        eventsAdd[pairs[i].first].push_back({ pairs[i].first, pairs[i].first, pairs[i].second, 1, i });
        eventsRemove[pairs[i].second].push_back({ pairs[i].second, pairs[i].first, pairs[i].second, 0, i });
    }
    vector<int> ans(N);
    multiset<int> left, right;
    auto pa = eventsAdd.begin();
    auto pr = eventsRemove.begin();
    auto add_event = [&](Event e) {
        auto [p, l, r, add, u] = e;
        // cout << l << " " << r << " " << add << " : " << endl;
        // for (auto el: left) cout << el << " ";
        // cout << endl;
        if (add) {
            left.insert(l);
            right.insert(r);
        } else {
            eraseOne(left, l);
            eraseOne(right, r);
        }
    };
    for (auto& [k, v]: eventsAdd) {
        sort(v.begin(), v.end(), [&](auto& a, auto& b) {
            if (a.r == b.r) return (a.id < b.id);
            return (a.r > b.r);
        });
    }
    vector<int> same;
    while (true) {
        if (pr != eventsRemove.end() && pr->second[0].p <= pa->second[0].p) {
            for (auto el : pr->second) {
                if (el.l == el.r) continue;
                add_event(el);
            }
            pr++;
        } else if (pa != eventsAdd.end()) {
            for (int i = 0; i < pa->second.size() - 1; i++) {
                if (pa->second[i].r == pa->second[i+1].r) {
                    same.push_back(pa->second[i].id);
                    same.push_back(pa->second[i+1].id);
                }
            }
            for (auto el : pa->second) {
                auto [p, l, r, add, u] = el;
                auto it = right.lower_bound(r);
                if (it != right.end() && left.size() > 0) {
                    ans[u] = *it - *left.rbegin() + 1 - (r - l + 1);
                }
                if (el.l != el.r) add_event(el);
            }

            pa++;
        } else {
            break;
        }
    }
    for (auto el: same) ans[el] = 0;
    for (int i = 0; i < N; i++) {
        cout << ans[i] << endl;
    }
}

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

    while (t--) {
        testCase();
    }

    cout << flush;
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
3
1 5
2 6
3 7
2
1 1
1 1000000000
3
1 1
1 1000000000
1 1
6
1 10
2 10
3 10
4 10
5 10
6 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
#from heapq import heapify,heappush,heappop

def cmax(a,b):
    if b is None:
        return a
    else:
        return max(a,b)

class Tree:
    def __init__(self, val,a):
        self.val=val
        self.a = a
        self.left=None
        self.right=None

    def add(self, x, a):
        if x<=self.val:
            a=max(a, self.a)
            if self.left is None:
                self.left=Tree(x, a)
            else:
                self.left.add(x,a)
        else:
            if self.a<a:
                self.a=a
            if self.right is None:
                self.right=Tree(x,a)
            else:
                self.right.add(x,a)

    def find(self,x, curr_best=None, curr_best_a=None):
        if x==self.val:
            return (0, self.a)
        elif x<self.val:
            if self.left is None:
                return (self.val-x, cmax(self.a, curr_best_a))
            else:
                return self.left.find(x, curr_best=self.val, curr_best_a=cmax(self.a, curr_best_a))
        else:
            if self.right is None:
                return (curr_best-x,self.a) if curr_best is not None else (None,None)
            else:
                return self.right.find(x, curr_best=curr_best, curr_best_a=cmax(self.a, curr_best_a))

def sol():
    n=int(input())
    intv=[]

    for i in range(n):
        line=input()
        a,b=line.split()
        a,b=int(a),int(b)
        intv.append((a,b,i))

    intv.sort(key=lambda x:(x[0],-x[1]))
    #print(intv)

    sol={}
    sol[intv[0][2]]=0
    rights=Tree(intv[0][1], intv[0][0])
    for a,b,idx in intv[1:]:
        best_right, best_a=rights.find(b)
        if best_right is None:
            sol[idx]=0
        else:
            sol[idx]=best_right+a-best_a
            #print((a,b),(best_right+a-best_a), best_right, best_a)
        #lasta=a
        rights.add(b,a)

    ints=[(a,b) for a,b,i in intv]
    reps={}
    for a,b in ints:
        if (a,b) in reps.keys():
            reps[(a,b)]+=1
        else:
            reps[(a,b)]=1

    for a,b,idx in intv:
        if reps[(a,b)]>1:
            sol[idx]=0

    for i in range(n):
        print(sol[i])

    return

t=int(input())

while t>0:
    sol()
    t-=1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
print(0)
print(0)
print(1)
print(999999999)
print(0)
print(0)
print(0)
print(0)
print(0)
print(2)
print(3)
print(2)
print(4)
print(8)

**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
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
import bisect
import sys

input = sys.stdin.readline

t = int(input())

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

    arr = []
    for i in range(n):
        l, r = map(int, input().split())
        arr.append((l, r, i))

    arr.sort(key=lambda x: (x[0], -x[1]))

    ans = [0] * n
    r_list = [arr[0][1]]

    for a, b, i in arr[1:]:
        if b > r_list[-1]:
            r_list.append(b)
            continue

        idx = bisect.bisect_right(r_list, b)
        if idx and r_list[idx - 1] == b:
            continue
        else:
            ans[i] += r_list[idx] - b
            r_list.insert(idx, b)

    arr.sort(key=lambda x: (-x[1], x[0]))
    l_list = [arr[0][0]]

    for a, b, i in arr[1:]:
        if a > l_list[-1]:
            ans[i] += a - l_list[-1]
            l_list.append(a)
            continue

        idx = bisect.bisect_left(l_list, a)
        if l_list[idx] == a:
            continue
        elif idx == 0:
            l_list.insert(idx, a)
        else:
            ans[i] += a - l_list[idx - 1]

    for i in range(n - 1):
        a1, b1, i1 = arr[i + 1]
        a2, b2, i2 = arr[i]

        if a1 == a2 and b1 == b2:
            ans[i1] = ans[i2] = 0

    for row in ans:
        print(row)

**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
3
1 10
3 10
3 7
2
1 1
1 1000000000
3
1 1
1 1000000000
1 1
6
1 10
3 10
3 7
5 7
4 4
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

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

public class Main{

    static String[] temp;
    static int N = (int) 2e5+5;
    static int[] Tree=new int[N];
    static  int length;

    public static void update(int index,int val){
        while (index<=length){
            Tree[index]=Math.max(val,Tree[index]);
            index+=lowBit(index);
        }
    }

    public static int lowBit(int x){
        return x&-x;
    }

    public static int  query(int index){
        int ans=0;
        while (index>0){
            ans=Math.max(ans,Tree[index]);
            index-=lowBit(index);
        }
        return ans;
    }

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
            int T= Integer.parseInt(br.readLine());
            while (T>0){
                int n= Integer.parseInt(br.readLine());
                length=n;
                for (int i = 0; i <=length ; i++) Tree[i]=0;
                int[][]queries=new int[n][2];
                for (int i = 0; i <n; i++) {
                    temp=br.readLine().split(" ");
                    queries[i][0]= Integer.parseInt(temp[0]);
                    queries[i][1]= Integer.parseInt(temp[1]);
                }
                Map<Integer,Integer> map=new HashMap<>();
                TreeSet<Integer> set=new TreeSet<>();
                Integer[]index=new Integer[n];
                for (int i=0;i<n;i++)index[i]=i;
                Arrays.sort(index, Comparator.comparingInt(a -> queries[a][1]));
                int id=1;
                for (int i = 0; i <n; i++) {
                    int start=i;
                    while (start<n&&queries[index[start]][1]==queries[index[i]][1]){
                        start++;
                    }
                    map.put(queries[index[i]][1],id++);
                    i=start-1;
                }
                for (int i = 0; i <n; i++) index[i]=i;
                Arrays.sort(index,(a,b)->queries[a][0]==queries[b][0]?queries[a][1]-queries[b][1]:queries[a][0]-queries[b][0]);
                int[]ans=new int[n];
                int inf= (int) 1e9+1;
                for (int i = 0; i <n ; i++) {
                    int idx=index[i],L=queries[idx][0],R=queries[idx][1],limitL=0,limitR=inf;
                    if (i+1<n&&queries[index[i+1]][0]==L){
                        limitR=queries[index[i+1]][1];
                        ans[idx]=limitR-R;
                    }else {
                        if (set.ceiling(R)!=null) limitR=Math.min(limitR,set.ceiling(R));
                        limitL=query(id-map.get(R));
                        if (limitR==inf)limitR=R;
                        if (limitL==0)limitL=L;
                        ans[idx]=Math.max(0,limitR-limitL-R+L);
                    }
                    update(id-map.get(R),L);
                    set.add(R);
                }
                for (int i = 0; i <n; i++) {
                    pw.println(ans[i]);
                }
                T--;
            }
        pw.close();
        br.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

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

public class CodeForcesSolution {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;
    static int mod = (int) 1e9+7;
    public static void main(String[] args) throws IOException {
        int t = readInt();
        while (t-->0) {
            int n = readInt();
            int[] ans = new int[n];
            List<int[]> pr = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                pr.add(new int[] {readInt(), readInt(), i});
            }
            pr.sort((a, b) -> { if (a[0] == b[0]) return b[1] - a[1]; else return a[0] - b[0]; });
            TreeSet<int[]> LSet = new TreeSet<>((a, b) -> { if (a[0] == b[0]) return a[1] - b[1]; else return a[0] - b[0]; });
            TreeSet<int[]> RSet = new TreeSet<>((a, b) -> { if (a[1] == b[1]) return a[0] - b[0]; else return a[1] - b[1]; });
            for (int i = 0; i < n; i++) {
                int[] p = pr.get(i);
                while (!RSet.isEmpty() && RSet.first()[1] < p[0]) {
                    int[] tmp = RSet.first();
                    RSet.remove(tmp);
                    LSet.remove(tmp);
                }
                int j = i, cnt = 0;
                while (j < n && pr.get(j)[0] == p[0] && pr.get(j)[1] == p[1]) {
                    cnt++;
                    j++;
                }
                if (cnt > 1) {
                    for (int k = i; k < j; k++) {
                        ans[pr.get(k)[2]] = 0;
                    }
                    i = j - 1;
                    continue;
                }
                if (!LSet.isEmpty() && !RSet.isEmpty()) {
                    int[] mxLft = LSet.last(), miRht = RSet.ceiling(new int[] {0, p[1]});
                    if (mxLft != null && miRht != null) {
                        ans[p[2]] = (p[0] - mxLft[0]) + (miRht[1] - p[1]);
                    }
                }
                LSet.add(p);
                RSet.add(p);
            }
            for (int i = 0; i < n; i++) {
                System.out.println(ans[i]);
            }
        }
    }

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();

    }
    static long readLong() throws IOException {
        return Long.parseLong(next());
    }
    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter() throws IOException {
        return next().charAt(0);
    }
    static String readLine() throws IOException {
        return br.readLine().trim();
    }
    static int readALotInt() throws IOException{
        int x = 0, c;
        while((c = br.read()) != ' ' && c != '\n')
            x = x * 10 + (c - '0');
        return x;
    }
}

**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 probability of failure. Here are the steps to consider:

1. **Identify the Bug**: The bug in the code seems to be related to how it handles the predictors and the calculation of strongly recommended tracks. Specifically, the code might not correctly handle cases where multiple users have the same segment or where the segments overlap in a way that is not straightforward.

2. **Edge Cases**:
   - **Users with the same segment**: This can cause the code to incorrectly handle the predictors.
   - **Overlapping segments**: This can cause issues in the calculation of strongly recommended tracks.
   - **Single user segments**: This can test the code's ability to handle the simplest case.
   - **Large range segments**: This can test the code's ability to handle large ranges and ensure it doesn't overflow or have performance issues.

3. **Generate Test Input**:
   - Create a test case with multiple users having the same segment.
   - Create a test case with overlapping segments.
   - Create a test case with a single user segment.
   - Create a test case with large range segments.

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

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

This test input includes:
1. A case with users having overlapping segments.
2. A case with users having the same segment.
3. A case with a single user segment.
4. A case with a mix of overlapping and non-overlapping segments.

This should help in exposing the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
#include <iostream>
#include <iomanip>
#include <vector>
#include <string>
#include <cmath>
#include <utility>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <functional>
#include <set>
#include <stack>
#include <map>
#include <random>
#include <chrono>
// #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;

typedef long long ll;
//typedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;
#define fw(i, l, r) for (int i = l; i < r; i += 1)
#define yes(ok) cout << ( ok == 1 ? "YES" : "NO") << endl
#define fi first
#define se second
#define pb push_back
#define mr make_pair
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll rnd(ll x, ll y) {
return uniform_int_distribution<ll>(x, y)(rng);
}
ll lmax(ll x, ll y) {
    return x >= y ? x : y;
}
ll lmin(ll x, ll y) {
    return x <= y ? x : y;
}
void solve() {
    ll n;
    cin >> n;
    vector<ll> l(n), r(n);
    map<pair<ll, ll>, ll> st;
    fw(i, 0, n) {
        cin >> l[i] >> r[i];
        st[mr(l[i], r[i])] += 1;
    }
    vector<ll> p(n);
    fw(i, 0, n) {
        p[i] = i;
    }
    sort(p.begin(), p.end(), [&] (ll i, ll j) {
        if(l[i] == l[j]) {
            return r[i] > r[j];
        }
        return l[i] < l[j];
    });
    // for(ll i = 0; i < n; i += 1) {
    //     cout << l[p[i]] << " " << r[p[i]] << endl;
    // }
    // li >= lj ri <= rj 的最小 rj
    vector<ll> R(n, -1);
    set<ll> s;
    for(ll i = 0; i < n; i += 1) {
        //
        ll ll = l[p[i]], rr = r[p[i]];
        if(st[{ll, rr}] >= 2) {
            continue;
        }
        if(!s.empty()) {
            auto it = s.lower_bound(rr);
            if(it != s.end()) {
                R[p[i]] = *it;
                // cout << R[p[i]] << " " << ll << " " << rr << " 888" << endl;
            }
        }
        s.insert(rr);
        // cout << rr << " " << "fafafa" << endl;
    }
    // ri <= rj lj <= li 的最大 lj
    sort(p.begin(), p.end(), [&] (ll i, ll j) {
        if(r[i] == r[j]) {
            return l[i] < l[j];
        }
        return r[i] > r[j];
    });
    // for(ll i = 0; i < n; i += 1) {
    //     cout << l[p[i]] << " " << r[p[i]] << endl;
    // }
    vector<ll> L(n, -1);
    s.clear();
    for(ll i = 0; i < n; i += 1) {
        ll ll = l[p[i]], rr = r[p[i]];
        if(st[{ll, rr}] >= 2) {
            continue;
        }
        if(!s.empty()) {
            auto it = s.lower_bound(ll + 1);
            if(it != s.begin()) {
                it--;
                L[p[i]] = *it;
                // cout << L[p[i]] << " " << ll << " " << rr << " 999" << endl;
            }
        }
        s.insert(ll);
    }
    for(ll i = 0; i < n; i += 1) {
        ll res = 0;
        // cout << l[i] << " " << r[i] << " " << L[i] << " " << R[i] << endl;
        if(L[i] != -1) {
            res += l[i] - L[i];
        }
        if(R[i] != -1) {
            res += R[i] - r[i];
        }
        cout << res << endl;
    }
}
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To 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 highlight potential issues in the algorithm. Here are the steps to think through this:

1. **Understand the Bug**: The provided code seems to have issues with handling cases where there are multiple users with the same segment of liked tracks. It also might not correctly handle the calculation of strongly recommended tracks for users with overlapping segments.

2. **Identify Edge Cases**:
   - **Single User with a Unique Segment**: This should be straightforward and should not expose the bug.
   - **Multiple Users with the Same Segment**: This can cause issues with the `st` map and the logic that skips segments with a count of 2 or more.
   - **Overlapping Segments**: This can expose issues in the logic that calculates the range of strongly recommended tracks.
   - **Edge Cases in Range Limits**: Segments that are very close to the maximum or minimum range limits.

3. **Construct the Test Input**:
   - **Multiple Users with the Same Segment**: This will test the logic that skips segments with a count of 2 or more.
   - **Overlapping Segments**: This will test the logic that calculates the range of strongly recommended tracks.
   - **Edge Cases in Range Limits**: This will test the boundary conditions.

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

```plaintext
3
3
1 5
1 5
1 5
4
1 3
2 5
3 7
4 6
2
1 1000000000
1 1000000000
```

### Explanation:
1. **First Test Case**:
   - 3 users with the same segment `[1, 5]`.
   - This should expose the bug in the logic that skips segments with a count of 2 or more.

2. **Second Test Case**:
   - 4 users with overlapping segments:
     - User 1: `[1, 3]`
     - User 2: `[2, 5]`
     - User 3: `[3, 7]`
     - User 4: `[4, 6]`
   - This will test the logic that calculates the range of strongly recommended tracks for users with overlapping segments.

3. **Third Test Case**:
   - 2 users with the same segment `[1, 1000000000]`.
   - This will test the boundary conditions and the logic that skips segments with a count of 2 or more.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Recommendations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Suppose you are working in some audio streaming service. The service has $$$n$$$ active users and $$$10^9$$$ tracks users can listen to. Users can like tracks and, based on likes, the service should recommend them new tracks.

Tracks are numbered from $$$1$$$ to $$$10^9$$$. It turned out that tracks the $$$i$$$-th user likes form a segment $$$[l_i, r_i]$$$.

Let's say that the user $$$j$$$ is a predictor for user $$$i$$$ ($$$j \neq i$$$) if user $$$j$$$ likes all tracks the $$$i$$$-th user likes (and, possibly, some other tracks too).

Also, let's say that a track is strongly recommended for user $$$i$$$ if the track is not liked by the $$$i$$$-th user yet, but it is liked by every predictor for the $$$i$$$-th user.

Calculate the number of strongly recommended tracks for each user $$$i$$$. If a user doesn't have any predictors, then print $$$0$$$ for that user.

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

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of users.

The next $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ per line ($$$1 \le l_i \le r_i \le 10^9$$$) — the segment of tracks the $$$i$$$-th user likes.

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$n$$$ integers, where the $$$i$$$-th integer is the number of strongly recommended tracks for the $$$i$$$-th user (or $$$0$$$, if that user doesn't have any predictors).

Examples:
input:
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
6
1 10
3 10
3 7
5 7
4 4
1 2
output:
0
0
1
999999999
0
0
0
0
0
2
3
2
4
8

Note: In the first test case:

- the first user has no predictors;
- the second user has no predictors;
- the third user has two predictors: users $$$1$$$ and $$$2$$$; only track $$$3$$$ is liked by both of them and not liked by the third user.

In the second test case, the second user is a predictor for the first user. Therefore, all tracks, except $$$42$$$, are strongly recommended for the first user.

In the third test case, the first user has two predictors: users $$$2$$$ and $$$3$$$, but there is no track that is liked by them and not liked by the first user himself.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
typedef long long ll;
struct node{
	int l,r,id;
};
node per[N];
bool cmp1(node a,node b){
	if (a.l!=b.l) return a.l<b.l;
	return a.r>b.r;
}
bool cmp2(node a,node b){
	if (a.r!=b.r) return a.r>b.r;
	return a.l<b.l;
}
int ans[N],l_per[N],r_per[N];
int num=0;

void solve(){
	int n;
	cin>>n;
	for (int i=1;i<=n;i++){
		cin>>per[i].l>>per[i].r;
		per[i].id=i;
		r_per[i]=0;
		l_per[i]=0;
		ans[i]=0;
	}
	sort(per+1,per+n+1,cmp1);
	r_per[per[1].id]=-1;

	set<int > per_set_l,per_set_r;
	per_set_r.insert(per[1].r);
	set<int >::iterator it_pre,it_last;
	for (int i=2;i<=n;i++){
		it_last=per_set_r.lower_bound(per[i].r);
		if (it_last==per_set_r.end()){
			r_per[per[i].id]=-1;
		}
		else{
//			cout<<(*it_last)<<" "<<(*it_pre)<<"\n";
//			int sum=((*it_last)-(*it_pre)+1)-per[i].r+per[i].l-1;
//			ans[per[i].id]=sum;
			r_per[per[i].id]=(*it_last);
		}
		per_set_r.insert(per[i].r);
	}

	sort(per+1,per+n+1,cmp2);
	l_per[per[1].id]=-1;

	per_set_l.insert(per[1].l);
	for (int i=2;i<=n;i++){
		it_pre=--per_set_l.upper_bound(per[i].l);
		if ((*it_pre)>per[i].l){
			l_per[per[i].id]=-1;
		}
		else{
			l_per[per[i].id]=(*it_pre);
		}
		per_set_l.insert(per[i].l);
	}

	for (int i=1;i<=n;i++){
		if (l_per[per[i].id]==-1 || r_per[per[i].id]==-1) ans[per[i].id]=0;
		else if (i>1 && per[i].l==per[i-1].l && per[i].r==per[i-1].r){
			ans[per[i].id]=0;
			ans[per[i-1].id]=0;
		}
		else{
			ans[per[i].id]=r_per[per[i].id]-l_per[per[i].id]+1-(per[i].r-per[i].l+1);
		}
//		cout<<l_per[i]<<" "<<r_per[i]<<" "<<ans[i]<<"\n";
	}

	for (int i=1;i<=n;i++){
		num++;
		if (num==25961 && n>=20){
			for (int j=1;j<=n;j++) cout<<per[j].l<<","<<per[j].r<<",";
		}
		cout<<ans[i]<<"\n";
	}

}

int main(){
	int t;
	cin>>t;
	while (t--){
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can highlight issues in the logic. Here are the steps to think through the process:

1. **Identify the Bug**: The code seems to have issues with handling overlapping segments and ensuring that the predictors are correctly identified. Specifically, the code might not correctly handle cases where multiple users have the same segment of liked tracks.

2. **Edge Cases**:
   - **Single User**: A test case with only one user should result in 0 recommendations.
   - **Overlapping Segments**: Test cases where multiple users have overlapping segments, especially where some users are predictors of others.
   - **Disjoint Segments**: Test cases where users have completely disjoint segments of liked tracks.
   - **Equal Segments**: Test cases where multiple users have the same segment of liked tracks.

3. **Small-Scale Test Input**:
   - We will create a test case with a small number of users to ensure it is manageable and can be easily validated.
   - We will include a mix of overlapping and equal segments to test the code's ability to handle these scenarios.

Here is a small-scale test input that includes the above considerations:

```plaintext
4
3
3 8
2 5
4 5
2
42 42
1 1000000000
3
42 42
1 1000000000
42 42
4
1 5
2 6
3 7
4 8
```

Explanation of the test input:
1. **Test Case 1**:
   - User 1 likes tracks 3 to 8.
   - User 2 likes tracks 2 to 5.
   - User 3 likes tracks 4 to 5.
   - User 3 has predictors User 1 and User 2. Only track 4 is liked by both predictors and not by User 3, so the recommendation for User 3 is 1.

2. **Test Case 2**:
   - User 1 likes track 42.
   - User 2 likes all tracks from 1 to 1,000,000,000.
   - User 1 has User 2 as a predictor. All tracks except 42 are strongly recommended, so the recommendation for User 1 is 999,999,999.

3. **Test Case 3**:
   - User 1 and User 3 like track 42.
   - User 2 likes all tracks from 1 to 1,000,000,000.
   - User 1 and User 3 have User 2 as a predictor, but there is no track that is liked by User 2 and not by User 1 or User 3, so the recommendations for User 1 and User 3 are 0.

4. **Test Case 4**:
   - User 1 likes tracks 1 to 5.
   - User 2 likes tracks 2 to 6.
   - User 3 likes tracks 3 to 7.
   - User 4 likes tracks 4 to 8.
   - Each user has a predictor, and the number of strongly recommended tracks can be calculated based on the segments.

This test input should expose issues in the code's handling of overlapping and equal segments, as well as the correct identification of predictors and strongly recommended tracks.