**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
import java.util.*;

public class codebymeet {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            int d = sc.nextInt();
            int[] arr=new int[101];
            for (int i = a; i <=b; i++) {
                arr[i]++;
            }
            for (int i = c; i <=d; i++) {
                arr[i]++;
            }
            if(b==c-1){
                System.out.println(1);
                continue;
            }
            int ans=0;
            for (int i=0;i<101;i++){
                if (arr[i]>1){
                    ans++;
                }
                else if (arr[i]==1 && i<100 && arr[i+1]>1){
                    ans++;
                } else if (arr[i]==1 && i>0 && arr[i-1]>1 ) {
                    ans++;
                }
            }
            System.out.println(ans-1);
        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

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

using namespace std;

using ll = long long;
using vl = vector<ll>;

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

    while(t--){
        int l, r, L, R;

        cin >> l >> r >> L >> R;
        if (r <= L || R <= l){
            cout << 1 << endl;
            continue;
        }
        if (l > L and r < R){
            cout << (r - l + 2) << endl;
            continue;
        }
        if (L > l and R < r){
            cout << (R - L + 2) << endl;
            continue;
        }

        if ((l > L and r == R)or (l == L and r < R)){
            cout << (r - l + 1) << endl;
            continue;
        }
        if ((L > l and R == r)or ( L == l and R < r)){
            cout << (R - L + 1) << endl;
            continue;
        }

        int inter=0;
        for (int i=0; i<105;i++){
            if (i >=l and i <=r and i >= L and i <= R){
               inter++;
            }
        }
        cout << inter -1 << endl;

    }

}

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

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

Think step by step.

Response:
```plaintext
4
1 3
4 6
2 4
3 5
1 5
2 6
1 50
51 100
```

### Explanation:
1. **Case 1: Non-overlapping segments (1 3, 4 6)**
   - Alice is in rooms 1 to 3, and Bob is in rooms 4 to 6.
   - The segments do not overlap, so locking one door (between rooms 3 and 4) is sufficient to prevent Alice and Bob from meeting.
   - Expected output: 1

2. **Case 2: Partially overlapping segments (2 4, 3 5)**
   - Alice is in rooms 2 to 4, and Bob is in rooms 3 to 5.
   - The segments overlap, and the overlap is 2 rooms (3 and 4).
   - Locking the doors between rooms 2 and 3, and between rooms 4 and 5 is necessary to prevent Alice and Bob from meeting.
   - Expected output: 2

3. **Case 3: Larger partially overlapping segments (1 5, 2 6)**
   - Alice is in rooms 1 to 5, and Bob is in rooms 2 to 6.
   - The segments overlap, and the overlap is 4 rooms (2, 3, 4, and 5).
   - Locking the doors between rooms 1 and 2, between rooms 5 and 6, and one more door within the overlap (e.g., between rooms 3 and 4) is necessary to prevent Alice and Bob from meeting.
   - Expected output: 3

4. **Case 4: Completely disjoint segments (1 50, 51 100)**
   - Alice is in rooms 1 to 50, and Bob is in rooms 51 to 100.
   - The segments do not overlap, so locking one door (between rooms 50 and 51) is sufficient to prevent Alice and Bob from meeting.
   - Expected output: 1

These test cases cover a range of scenarios, including non-overlapping, partially overlapping, and completely disjoint segments, which should help in exposing any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

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

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define int long long
#define pb push_back
#define pii pair<int, int>
#define vi vector<int>
#define all(x) x.begin(), x.end()
#define fi first
#define se second

const int MOD = 1e9 + 7;

void solve() {
    int l,r;
    cin >> l >> r ;
    int L,R;
    cin >> L >> R ;

    if(r<=L || R<=l) cout << "1\n" ;
    else{
        int x = max(l,L) ;
        int y = min(r,R) ;
        if(l==L && r==R) cout << r-l << "\n" ;
        else if(l==L || r==R) cout << y-x+1 << "\n" ;
        else cout << y-x+2 << "\n" ;
    }

}

int32_t main() {
    fastio;

    int t;
    cin >> t;

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

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the logic might fail. The code seems to handle most cases correctly, but there are a few scenarios where it might not work as expected. Let's consider the following edge cases:

1. **Overlapping Segments with a Single Room in Common:**
   - If the segments overlap by a single room, the code might incorrectly count the number of doors to lock.

2. **Adjacent Segments:**
   - If the segments are adjacent, the code might not correctly handle the locking of doors.

3. **Identical Segments:**
   - If the segments are identical, the code might not correctly count the number of doors to lock.

Let's generate a test input that covers these edge cases:

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

Explanation:
1. **Test Case 1:**
   - Alice's segment: [1, 2]
   - Bob's segment: [2, 3]
   - Expected output: 1 (lock the door between rooms 2 and 3)

2. **Test Case 2:**
   - Alice's segment: [1, 2]
   - Bob's segment: [3, 4]
   - Expected output: 1 (lock the door between rooms 2 and 3)

3. **Test Case 3:**
   - Alice's segment: [1, 2]
   - Bob's segment: [1, 2]
   - Expected output: 1 (lock the door between rooms 1 and 2)

4. **Test Case 4:**
   - Alice's segment: [1, 2]
   - Bob's segment: [3, 3]
   - Expected output: 1 (lock the door between rooms 2 and 3)

This test input should help expose any issues in the logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
import sys
input=lambda: sys.stdin.readline().strip()
from math import gcd,ceil,factorial
from collections import deque
from types import GeneratorType
from itertools import combinations
from heapq import heapify,heappop,heappush
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        to = f(*args, **kwargs)
        if stack:
            return to
        else:
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        return to
                    to = stack[-1].send(to)
    return wrappedfunc
def lcm(n,m):
    return (n*m)//gcd(n,m)
def inp():
     return input()
def ii():
     return int(input())
def lii():
     return list(map(int,input().split()))
def li():
     return input().split()
def mii():
     return map(int,input().split())
def dtb(n): return bin(n).replace("0b","")
def isprime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            return False
    return True
def npr(n, r):
    return factorial(n) // factorial(n - r) if n >= r else 0
def ncr(n, r):
    return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
def l_b(li, num):
    a=-1
    s=0
    e=len(li)-1
    while (s<=e):
        middle = (e+s) // 2
        if li[middle] <= num:
            a = middle
            e = middle - 1
        else:
            s= middle + 1
    return a
def u_b(li, num):
    a=-1
    s=0
    e=len(li) - 1
    while (s<=e):
        middle = (e+s) // 2
        if li[middle] >= num:
            a = middle
            s=middle + 1
        else:
            e=middle - 1
    return a
def binary_search(li, val, lb, ub):
    # print(lb, ub, li)
    ans = -1
    while (lb <= ub):
        mid = (lb + ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid - 1
        elif val > li[mid]:
            lb = mid + 1
        else:
            ans = mid  # return index
            break
    return ans
def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far
def primefactors(n):
    factors = []
    while (n % 2 == 0):
        factors.append(2)
        n //= 2
    for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:  # incase of prime
        factors.append(n)
    return factors
def abs(x):
    return x if x >= 0 else -x
def band():
     global bl
     k=len(l)
     bl=[]
     for i in range(0,k+1):
         bl.append([0]*32)
     p=2
     for j in range(0,32):
          for i in range(0,k):
               if l[i]%p==0:
                    bl[i+1][j]=bl[i][j]+1
               else:
                    bl[i+1][j]=bl[i][j]
                    l[i]=l[i]-(l[i]%p)
          p=p*2
def bor():
    global bl
    k=len(l)
    bl=[]
    for i in range(0,k+1):
        bl.append([0]*32)
    p=2
    for j in range(0,32):
        for i in range(0,k):
            if l[i]%p==0:
                bl[i+1][j]=bl[i][j]
            else:
                bl[i+1][j]=bl[i][j]+1
                l[i]=l[i]-(l[i]%p)
        p=p*2
def make(n):
    parent[n]=n
    size[n]=1
@bootstrap
def find(n):
    if n==parent[n]:
        yield n
    parent[n]=yield find(parent[n])
    yield parent[n]
def uni(a,b):
    a=find(a)
    b=find(b)
    if a!=b:
        if size[a]<size[b]:
            a,b=b,a
        parent[b]=a
        size[a]+=size[b]
for _ in range(ii()):
    a,b=mii()
    c,d=mii()
    if b<c or a>d:
        print(1)
    elif a>c and b<d:
        print(b-a+2)
    elif c>a and d>b:
        print(d-b+2)
    elif a<c and b<d:
        print(b-c+2)
    elif c<a and d<b:
        print(d-a+2)
    elif a==c and b<d:
        print(b-a+1)
    elif a==c and d>b:
        print(d-a+1)
    elif b==d and a<c:
        print(d-c+1)
    elif b==d and c>a:
        print(d-a+1)
    elif a==c and b==d:
        print(d-a)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause the code to fail. The code appears to have issues with handling overlapping segments and calculating the correct number of doors to lock.

### Step-by-Step Analysis

1. **Non-overlapping Segments:**
   - The code correctly handles non-overlapping segments (e.g., Alice in [1, 2] and Bob in [3, 4]) by printing 1. This case is straightforward and likely to be correct.

2. **Partially Overlapping Segments:**
   - The code has multiple conditions to handle partially overlapping segments, but it might not handle them correctly. For example, if Alice is in [2, 5] and Bob is in [3, 7], the code should print 2 (locking doors between rooms 2-3 and 5-6).

3. **Completely Overlapping Segments:**
   - If the segments are completely overlapping (e.g., Alice in [2, 5] and Bob in [2, 5]), the code should print 3 (locking doors between rooms 2-3, 3-4, and 4-5).

4. **Edge Cases:**
   - Edge cases where the segments are adjacent but not overlapping (e.g., Alice in [1, 2] and Bob in [3, 4]) should also be considered. The code should print 1.

### Challenging Test Input

Based on the analysis, a challenging test input that covers these scenarios is:

```plaintext
4
1 2
3 4
2 5
3 7
2 5
2 5
1 100
50 51
```

### Explanation

1. **Test Case 1:**
   - Alice in [1, 2], Bob in [3, 4]
   - Expected output: 1 (lock door between 2 and 3)

2. **Test Case 2:**
   - Alice in [2, 5], Bob in [3, 7]
   - Expected output: 2 (lock doors between 2-3 and 5-6)

3. **Test Case 3:**
   - Alice in [2, 5], Bob in [2, 5]
   - Expected output: 3 (lock doors between 2-3, 3-4, and 4-5)

4. **Test Case 4:**
   - Alice in [1, 100], Bob in [50, 51]
   - Expected output: 49 (lock doors between 1-2, 2-3, ..., 49-50, 50-51)

This input should help expose any issues in the buggy code, especially in handling overlapping and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
from collections import Counter
def solve():
    # x=int(input(""))
        # n,k= list(map(int, input("").split()))
    a,b= list(map(int, input("").split()))
    c,d= list(map(int, input("").split()))
    if a==c:
        if b<d:
            print(b-a+1)
        elif b==d:
            print(d-c)
        else:
            print(d-c+1)
    elif b==d:
        d1,d2 = b-a,d-c
        print(min(d1,d2)+1)
    elif c<a and d>b:
        print(b-a+2)
    elif a<c and b>d:
        print(d-c+2)
    else:
        if c>b:
            print(1)
        else:
            print(abs(c-b))
n=int(input(""))
for i in range(n):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
1 50
51 100
1 49
51 100
1 50
50 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
//        int t=1;
        while(t-->0){
            int l=sc.nextInt();
            int r=sc.nextInt();
            int L=sc.nextInt();
            int R=sc.nextInt();
            int arr1[]=new int[101];
            int arr2[]=new int[101];
            for(int i=l;i<=r;i++){
                arr1[i]++;
            }
            for(int i=L;i<=R;i++){
                arr2[i]++;
            }
            int cnt1=0;
            for(int i=0;i<100;i++){
                if(arr1[i]!=0 && arr2[i+1]!=0){
                    cnt1++;
                }
            }
            int cnt2=0;
            for(int i=0;i<100;i++){
                if(arr2[i]!=0 && arr1[i+1]!=0){
                    cnt2++;
                }
            }
            if(l!=0 && arr1[l]!=0 && arr2[l-1]!=0){
                cnt1++;
            }
            if(L!=0 && arr2[L]!=0 && arr1[L-1]!=0){
                cnt2++;
            }
            System.out.println(Math.max(cnt1,cnt2));
        }
    }
}

**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, let's consider the following points:

1. **Edge Cases**: The code might fail when the segments of Alice and Bob are right next to each other or overlap in a way that requires careful handling.
2. **Maximizing Failure Probability**: We need to create a scenario where the code's logic for counting the doors to lock is likely to be incorrect.

### Challenging Test Input
We will create a test case where the segments of Alice and Bob are adjacent but not overlapping. This will test the logic for counting the doors that need to be locked.

### Test Case
```plaintext
1
1 2
3 4
```

### Explanation
- **Alice's Segment**: [1, 2]
- **Bob's Segment**: [3, 4]

In this case, the only way to prevent Alice and Bob from meeting is to lock the door between rooms 2 and 3. The correct output for this test case should be 1.

### Why This Test Case is Challenging
- The segments are adjacent, which means there is exactly one door that needs to be locked to separate them.
- The code provided has a logic for counting the doors, but it might not handle this specific edge case correctly, especially the part where it checks the boundaries and the overlap.

This test input should help expose any issues in the code's logic for counting the doors that need to be locked.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
import java.util.Scanner;

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

        Scanner sc = new Scanner(System.in);
        int time = sc.nextInt();
        while (time-- > 0) {
            int bobleft = sc.nextInt();
            int bobright = sc.nextInt();
            int aliceleft = sc.nextInt();
            int aliceright = sc.nextInt();
            if (bobright+1<aliceleft||aliceright+1<bobleft) {
                System.out.println("1");
                break;
            }
            int littlehead =bobright - aliceright<=0?bobright:aliceright;
            int littletail =bobleft - aliceleft>=0?bobleft:aliceleft;
            int ans = littlehead-littletail+2;
            ans +=bobright-aliceright==0?-1:0;
            ans +=bobleft-aliceleft==0?-1:0;
            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:
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 maximize the failure probability. The buggy code has issues in handling certain conditions, particularly when the segments of Alice and Bob overlap or are adjacent in a way that the code does not correctly handle.

Here's a step-by-step approach to generating the test input:

1. **Identify the Bug**: The main issue in the buggy code is the incorrect handling of the conditions where the segments of Alice and Bob overlap or are adjacent. The code uses a complex and incorrect logic to determine the number of doors to lock.

2. **Edge Cases**:
   - **Non-overlapping segments with a single room gap**: This should return 1.
   - **Overlapping segments with a single room overlap**: This should return the number of rooms in the overlap.
   - **Adjacent segments with no overlap**: This should return 1.
   - **Segments with no gap and no overlap**: This should return the number of rooms in the intersection.

3. **Generate Test Cases**:
   - **Test Case 1**: Alice in [1, 2], Bob in [4, 5] (Non-overlapping with a single room gap)
   - **Test Case 2**: Alice in [2, 4], Bob in [3, 5] (Overlapping with a single room overlap)
   - **Test Case 3**: Alice in [3, 4], Bob in [5, 6] (Adjacent with no overlap)
   - **Test Case 4**: Alice in [4, 6], Bob in [4, 6] (Segments with no gap and no overlap)

Here is the test input:

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

This test input will help expose the bugs in the provided buggy code by covering various edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7;

int solve(){
    int l,r,L,R;
    cin >> l>>r>>L>>R;
    if(r<L){
        cout<<1<<endl;
        return 0;
    }
    int mxr = max(r,R);
    int mxl  = min(l,L);
    int a = max(l,L);
    int b = min(r,R);
    int cnt = b-a+2;
    if(b==mxr) cnt--;
    if(mxl==a) cnt--;
    cout<<cnt<<endl;

}

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 generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The code has a logic that calculates the number of doors to lock based on the segments where Alice and Bob are located. The main issue is in how it calculates the number of doors to lock, especially when the segments overlap or are adjacent.

Let's consider the following scenarios:
1. Segments are completely separate.
2. Segments are adjacent.
3. Segments overlap partially.
4. One segment is completely within the other.

We will create a test case for each of these scenarios, but we will focus on a scenario that is likely to expose the bug in the code.

### Test Case 1: Segments are completely separate
This is a straightforward case where the segments are completely separate, and the code should handle it correctly.

### Test Case 2: Segments are adjacent
This is a more challenging case where the segments are adjacent. The code might not handle this correctly.

### Test Case 3: Segments overlap partially
This is a critical case where the segments overlap partially. The code might not handle this correctly.

### Test Case 4: One segment is completely within the other
This is another critical case where one segment is completely within the other. The code might not handle this correctly.

Given the constraints and the need for a small-scale test input, we will focus on a test case that is likely to expose the bug in the code. Let's choose a test case where the segments overlap partially.

### Chosen Test Case
```plaintext
1
1 5
4 8
```

### Explanation
- Alice is in the segment [1, 5].
- Bob is in the segment [4, 8].
- The segments overlap at the room 4 and 5.
- To ensure Alice and Bob cannot meet, we need to lock the doors between rooms 4 and 5, and between rooms 5 and 6.
- The correct output should be 2, but the buggy code might not handle this correctly.

This test case should expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'

int main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int t,l,r,L,R,left=0,right=0,ans=0;
	string hole;
	cin>>t;
	while(t--){
		cin>>l>>r>>L>>R;
		if(r<L)
		{
			cout<<"1"<<endl;
		}
		else
		{
			left=max(l,L);
			right=min(r,R);
			ans=right-left+2;
			if(left==min(l,L))
				ans--;
			if(right==max(r,R))
				ans--;
			cout<<ans<<endl;
		}

	}
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the logic might fail. The code calculates the number of doors to lock based on the overlap between Alice's and Bob's segments. The main issue is in the calculation of `ans` when there is an overlap. Specifically, the code might incorrectly handle the boundaries of the overlapping segment.

Let's consider the following edge cases:

1. **No Overlap**: This is already handled correctly in the code.
2. **Partial Overlap**: The code might incorrectly count the number of doors to lock.
3. **Complete Overlap**: The code might not correctly handle the boundaries of the overlapping segment.

A particularly challenging test case would be one where the segments of Alice and Bob overlap completely or partially, and the code's logic for calculating the number of doors to lock is incorrect.

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

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

### Explanation:
1. **Test Case 1:**
   - Alice is in rooms [1, 3]
   - Bob is in rooms [2, 4]
   - Expected output: 1 (lock the door between rooms 3 and 4)

2. **Test Case 2:**
   - Alice is in rooms [2, 5]
   - Bob is in rooms [3, 6]
   - Expected output: 2 (lock the doors between rooms 2 and 3, and 5 and 6)

3. **Test Case 3:**
   - Alice is in rooms [4, 6]
   - Bob is in rooms [5, 7]
   - Expected output: 1 (lock the door between rooms 6 and 7)

These test cases cover different scenarios of partial and complete overlaps, and they should help expose the bug in the faulty logic of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
def function(la, ra, lb, rb):
    closed = set()
    if ra < lb:
        closed.add((ra, ra+1))
        closed.add((lb-1, lb))
    elif rb < la:
        closed.add((rb, rb+1))
        closed.add((la-1, la))
    elif la==lb and ra==rb:
        for i in range(la, ra):
            closed.add((i, i+1))
    else:
        lmax = max(la, lb)
        rmin = min(ra, rb)

        lmin = min(la, lb)
        rmax = max(ra, rb)

        for i in range(lmax-1, rmin+1):
            if i+1 <= rmax and i>= lmin:
                closed.add((i, i + 1))
    return closed

for _ in range(int(input())):
    la, ra = map(int, input().split())
    lb, rb = map(int, input().split())
    closed = function(la, ra, lb, rb)
    print(len(closed))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
for _ in range(int(input())):
    a,b=map(int,input().split())
    c,d=map(int,input().split())
    if a>c or (a==c and d>b):
        a,b,c,d=c,d,a,b
    if b<c:
        print(1)
        continue
    if b<d:
        print(c-b+2)
        continue
    print(d-c+int(c>a)+int(b>d))

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

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

Think step by step.

Response:
```plaintext
4
1 50
51 100
1 50
49 100
1 30
31 60
1 30
29 60
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

Buggy Code:
import java.util.*;
public class first {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int l=sc.nextInt();
            int r = sc.nextInt();
            int L = sc.nextInt();
            int R = sc.nextInt();
            if(r<L || R<l){
                System.out.println(1);
            }
            else if(l==L && r==R){
                System.out.println(r-l);
            }
            else if(l>L && r<R){
                System.out.println(r-l+2);
            }
            else if(L>l && R<r){
                System.out.println(R-L+2);
            }
            else if((l>L && r==R) || (l==L && r<R)){
                System.out.println(r-l+1);
            }
            else if((L>l && r==R) || (l==L && r>R)){
                System.out.println(R-L+1);
            }
        }

    }
}

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

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

Think step by step.

Response:
```plaintext
4
1 50
51 100
1 50
49 100
1 50
50 100
1 50
48 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game with Doors
time_limit_ms: 2000
memory_limit_mb: 256
Description: There are $100$ rooms arranged in a row and $99$ doors between them; the $i$-th door connects rooms $i$ and $i+1$. Each door can be either locked or unlocked. Initially, all doors are unlocked.

We say that room $x$ is reachable from room $y$ if all doors between them are unlocked.

You know that:

  * Alice is in some room from the segment $[l, r]$;    * Bob is in some room from the segment $[L, R]$;    * Alice and Bob are in different rooms.

However, you don't know the exact rooms they are in.

You don't want Alice and Bob to be able to reach each other, so you are going to lock some doors to prevent that. What's the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments?

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

The first line of each test case contains two integers $l$ and $r$ ($1 \le l < r \le 100$) — the bounds of the segment of rooms where Alice is located.

The second line of each test case contains two integers $L$ and $R$ ($1 \le L < R \le 100$) — the bounds of the segment of rooms where Bob is located.

Output: For each test case, print a single integer — the smallest number of doors you have to lock so that Alice and Bob cannot meet, regardless of their starting positions inside the given segments.

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

Note: In the first test case, it is sufficient to lock the door between rooms $2$ and $3$.

In the second test case, the following doors have to be locked: $(2,3)$, $(3,4)$, $(4,5)$.

In the third test case, the following doors have to be locked: $(5, 6)$ and $(6,7)$.

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

public class div2_edu_rd1969_b {

	static BufferedReader br;
	static StringTokenizer st;
	static StringBuilder sb = new StringBuilder("");

	public static void main (String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		st = new StringTokenizer(br.readLine());

		int t = Integer.parseInt(st.nextToken());

		for (int i = 0; i < t; i++) {
			st = new StringTokenizer(br.readLine());
			solve();
		}

		br.close();
		System.out.print(sb.toString());
	}

	static void solve() throws IOException {
		int l = Integer.parseInt(st.nextToken());
		int r = Integer.parseInt(st.nextToken());

		st = new StringTokenizer(br.readLine());
		int L = Integer.parseInt(st.nextToken());
		int R = Integer.parseInt(st.nextToken());

		if (r < L) {
			sb.append("1\n"); return;
		}

		if (R < l) {
			sb.append("1\n");
		}

		int overlap = Math.min(R,  r) - Math.max(L, l);

		if (L != l) {
			overlap++;
		}

		if (R != r) {
			overlap++;
		}

		sb.append(overlap + "\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
1 50
51 100
1 49
51 100
1 50
50 100
1 49
50 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define MOD 1000000007
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/gcd(a,b)
const int MAX_N = 1e5 + 5;

using namespace std;
void solve()
{
    ll n;
    cin>>n;
    vector<int>a(n);
    vector<int>b;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sort(a.begin(),a.end());
    b.push_back(a[0]);
    ll check=a[0];
    for(int i=1;i<n;i++)
    {
        if(a[i]!=check)
        {
            check=a[i];
            b.push_back(check);
        }
    }
    if(b[0]>1)
    {
        cout<<"Alice"<<endl;
    }
    else
    {

        bool bobfacing=false;
        for(int i=1;i<b.size()-1;i++)
        {
            if(b[i]==b[i-1]+1)
            {
                bobfacing^=true;
            }
        }
        if(bobfacing)
        {
            cout<<"Alice"<<endl;
        }
        else
        {
            cout<<"Bob"<<endl;
        }
    }

}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    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, let's analyze the code and identify potential issues:

1. **Sorting and Removing Duplicates**: The code sorts the array and then removes duplicates. This step is generally correct, but it might not handle cases where the smallest pile is 1 correctly.

2. **Initial Check for Alice**: The code checks if the smallest pile is greater than 1. If it is, Alice wins. This is correct because Alice can always choose `k` such that all piles are emptied in one move.

3. **Bob's Turn Calculation**: The code uses a XOR operation to determine if Bob is facing a situation where he can win. This logic is flawed because it doesn't correctly handle all possible scenarios where Bob can win or lose.

To expose the bug, we need to create a test case where the smallest pile is 1, and the remaining piles have a specific pattern that should cause Bob to win, but the code incorrectly predicts Alice wins.

### Challenging Test Input

Let's consider the following test case:
- One pile with 1 stone.
- Several piles with consecutive values starting from 2.

This should cause Bob to win because Alice is forced to remove 1 stone from all piles, leaving Bob with a configuration where he can always force Alice to lose.

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

### Explanation
- **Input**: One test case with 4 piles: 1, 2, 3, 4.
- **Expected Output**: Bob wins.
- **Reason**: Alice must choose `k=1` on her first turn, leaving the piles as 0, 1, 2, 3. Bob can then choose `k=1` again, leaving the piles as 0, 0, 1, 2. Alice is forced to choose `k=1` again, leaving the piles as 0, 0, 0, 1. Bob can then choose `k=1` to win the game.

This test input should expose the bug in the code where the XOR logic incorrectly predicts Alice wins.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

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

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int tc;
    cin >> tc;
    for (int ti = 0; ti < tc; ++ti) {
        int n;
        cin >> n;
        vector<int> arr(n);
        for (auto &el : arr) {
            cin >> el;
        }
        sort(arr.begin(), arr.end());
        arr.erase(unique(arr.begin(), arr.end()), arr.end());
        reverse(arr.begin(), arr.end());
        bool ok = true;
        for (int i = 1; i < n; ++i) {
            if (i & 1) {
                if (arr[i] == 1) {
                    ok = false;
                }
            } else if (arr[i] == arr[i + 1] - 1) {
                ok = false;
            }
        }
        cout << (ok ? "Alice" : "Bob") << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code's logic might fail. The code sorts the array, removes duplicates, and then checks certain conditions to determine the winner. The key points to consider are:

1. **Sorting and Removing Duplicates**: The code removes duplicates and sorts the array, which can change the original structure of the piles.
2. **Loop Conditions**: The loop conditions and checks are based on the sorted and unique array, which might not accurately reflect the game state.

A good test case to expose this bug would be one where the original piles have a structure that, when sorted and duplicates are removed, changes the game's outcome. Specifically, we can create a test case where the piles have a specific pattern that the code's logic fails to handle correctly.

Let's create a test case with a small number of piles that highlights this issue:

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

### Explanation:
- **Input**:
  - `t = 1` (one test case)
  - `n = 5` (five piles)
  - `a = [1, 2, 3, 4, 5]` (initial number of stones in each pile)

### Expected Output:
- **Output**: "Alice"

### Why This Test Case:
- **Original Piles**: The piles are `[1, 2, 3, 4, 5]`.
- **After Sorting and Removing Duplicates**: The array remains `[1, 2, 3, 4, 5]` (no duplicates to remove).
- **Game Analysis**:
  - Alice can win by choosing `k = 1` on her first turn, which leaves the piles as `[0, 1, 2, 3, 4]`.
  - Bob then has to choose `k = 1`, leaving the piles as `[0, 0, 1, 2, 3]`.
  - Alice can then choose `k = 1` again, leaving the piles as `[0, 0, 0, 1, 2]`.
  - Bob has to choose `k = 1`, leaving the piles as `[0, 0, 0, 0, 1]`.
  - Alice can then choose `k = 1`, leaving the piles as `[0, 0, 0, 0, 0]`.
  - Bob cannot make a move and loses.

### Bug Exposure:
- The code's logic checks for specific conditions in the sorted and unique array, which might not correctly handle the original structure of the piles. In this test case, the code might incorrectly determine the winner due to its loop conditions.

This test case should help expose 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: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

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

using namespace std;

int main() {
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<int> arr(n);
		for (int i = 0; i < n; i++) {
			cin >> arr[i];
		}
		auto First = [&](vector<int> arr) {
			if (arr.size() == 1) {
				return true;
			}

			for (int i = 0; i < arr.size() - 1; i++) {
				if (arr[i] != arr[i + 1]) {
					return false;
				}
			}
			return true;

		};
		auto Check = [&](vector<int> arr) {
			for (int i = 0; i < arr.size(); i++) {
				if (arr[i] == 1) {
					return true;
				}
			}
			return false;
		};

		auto f = First(arr);
		if (f) {
			cout << "Alice" << '\n';
			continue;
		}

		auto chk = Check(arr);
		if (arr.size() % 2 == 0 && chk) {
			cout << "Bob" << '\n';

		}
		else if (arr.size() % 2 == 0 && !chk) {
			cout << "Alice" << '\n';
		}

		else if (arr.size() % 2 != 0 && chk) {
			cout << "Alice" << '\n';
		}

		else if (arr.size() % 2 != 0 && !chk) {
			cout << "Bob" << '\n';
		}

	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios where the code's logic might fail. The code has several issues, including incorrect handling of the game's rules and some logical errors. Let's break down the potential issues and create a test case that can expose these problems:

1. **Incorrect Handling of Equal Piles**: The code assumes that if all piles are equal, Alice wins. However, this is not always true. For example, if all piles have an odd number of stones, Bob can always win by forcing Alice to make the last move.

2. **Incorrect Handling of Piles with 1 Stone**: The code incorrectly assumes that if there is at least one pile with 1 stone and the number of piles is even, Bob wins. This is not always true. For example, if there are multiple piles with 1 stone and other piles with an even number of stones, Alice can still win.

3. **Incorrect Handling of Odd and Even Piles**: The code's logic for determining the winner based on the number of piles being odd or even is flawed. It does not consider the parity of the stones in the piles correctly.

Given these issues, a good test case to expose the bugs would be:

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

Explanation:
- **Test Case 1**: `3 1 2 3` - Alice can win by choosing `k=1` on her first turn, leaving the piles as `0 1 2`. Bob will then be forced to choose `k=1` again, leaving `0 0 1`, and Alice can win by choosing `k=1`.
- **Test Case 2**: `2 1 1` - Bob can win because Alice must choose `k=1` on her first turn, and Bob can then win by choosing `k=1` on his turn.
- **Test Case 3**: `4 1 2 3 4` - Alice can win by choosing `k=1` on her first turn, leaving the piles as `0 1 2 3`. Bob will then be forced to choose `k=1` again, leaving `0 0 1 2`, and Alice can win by choosing `k=1`.
- **Test Case 4**: `3 2 2 2` - Alice can win by choosing `k=2` on her first turn, leaving the piles empty.
- **Test Case 5**: `6 1 1 1 1 1 1` - Bob can win because Alice must choose `k=1` on her first turn, and Bob can then win by choosing `k=1` on his turn.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
def winner_of_modified_nim_game(piles):
    """
    Определяет победителя в модифицированной игре Ним, где на каждом ходу игрок удаляет камни из всех непустых куч.

    Args:
        piles: Список целых чисел, представляющий начальные размеры кучек.

    Returns:
        Строка "Alice" или "Bob", обозначающая победителя.
    """

    # Сортируем кучи по возрастанию размера.
    piles.sort()

    # Подсчитываем количество куч с количеством камней больше, чем их индекс.
    greater_count = sum(1 for i, pile_size in enumerate(piles) if pile_size > i)

    # Если количество таких куч четное, то второй игрок (Боб) выигрывает.
    # В противном случае, первый игрок (Алиса) выигрывает.
    return "Bob" if greater_count % 2 == 0 else "Alice"

# Считываем количество тестов.
num_tests = int(input())

# Обрабатываем каждый тест.
for _ in range(num_tests):
    # Считываем количество кучек и их размеры.
    num_piles = int(input())
    piles = list(map(int, input().split()))

    # Определяем и выводим победителя.
    winner = winner_of_modified_nim_game(piles)
    print(winner)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
def calculateResult(array, currTurn):
    minVal = max(min(array),1)
    maxVal = max(array)

    if (minVal == maxVal):
        return currTurn;

    if (minVal < 2):
        arrayCopy = [max(0, x - minVal) for x in array if (x - minVal) > 0]
        currTurn += 1
        return calculateResult(arrayCopy, currTurn)
    else:
        arrayCopy = [max(0, x - minVal) for x in array if (x - minVal) > 0]
        arrayCopy2 = [max(0, x - minVal+1) for x in array if (x - minVal + 1) > 0]
        currTurn+=1
        turnMin = calculateResult(arrayCopy, currTurn)
        turnMinPlusOne = calculateResult(arrayCopy2, currTurn)

        if (turnMin % 2 == 0):
            return turnMin
        if (turnMinPlusOne % 2 == 0):
            return turnMinPlusOne
        return currTurn

t = (int)(input())
for i in range(t):
    n = (int)(input())
    a = list(map(int, input().split(' ')))
    res = calculateResult(a, 0) % 2 == 0
    if (res):
        print("Alice")
    else:
        print("Bob")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
import heapq
import sys

input = sys.stdin.readline

def solve(arr):
    A = False
    arr.sort()
    if arr[0] != 1:
        return 'Alice'
    set_ = list(set(arr))
    not_c = True
    for i in range(1, len(set_)):
        if set_[i] - set_[i - 1] > 1:
            not_c = False
            break
        A = not A
    if not_c:
        A = not A

    return 'Alice' if A else 'Bob'

t = int(input())
c = 1
while t > 0:
    n = list(map(int, input().rstrip().split()))

    arr = list(map(int, input().rstrip().split()))
    r = solve(arr)
    if t <= 7:
        print(r)
    if c == 183:
        print(arr)
    c += 1
    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
4
1 2 3 4
5
1 2 2 3 3
6
1 1 2 2 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
import com.sun.source.tree.Tree;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.*;
import java.io.PrintWriter;

public class Codeforces {
    public static void main (String[] args) throws IOException {
        // your code goes here
        FastScanner fs = new FastScanner();
        //BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=fs.nextInt();
        PrintWriter out=new PrintWriter(System.out);
        while(t-->0) {
            int n=fs.nextInt();
            int[] a=fs.readArray(n);
            Arrays.sort(a);
            if (a[0]>=2) out.println("Alice");
            else{
                int flag=1;
                boolean check=false;
                for(int i=1;i<n-1;i++){
                    if (a[i]-i>=2){
                        check=true;
                        if (flag==0) out.println("Alice");
                        else out.println("Bob");
                        break;
                    }
                    if (a[i]-i<=0){
                        check=true;
                        if (flag==0) out.println("Bob");
                        else out.println("Alice");
                        break;
                    }
                    flag=1-flag;
                }
                if (check) continue;
                if (a[n-1]==n-1){
                    if (flag==0) out.println("Bob");
                    else out.println("Alice");
                }else{
                    if (flag==0) out.println("Alice");
                    else out.println("Bob");
                }
            }
        }
        out.close();

    }

    public static boolean isPrime(int n){
        if (n%2==0) return false;
        for(int i=3;i<=Math.sqrt(n);i+=2){
            if (n%i==0) return false;
        }
        return true;
    }

    public static int sumOfDigits(long n){
        int sum=0;
        while(n!=0){
            sum+=n%10;
            n/=10;
        }
        return sum;
    }

    public static long gcd(long a, long b){
        if (a==0) return b;
        return gcd(b%a,a);
    }
    public static boolean larger(ArrayList<Integer> list1, ArrayList<Integer> list2){
        if (list2.size()==0) return true;
        int x=list1.size(),y=list2.size(),i=0,j=0;
        while(i<x && j<y){
            int first=list1.get(i++),second=list2.get(j++);
            if (first>second) return true;
            else if (first<second) return false;
        }
        return false;
    }
    public static String minString(String s1, String s2){
        if (s1.compareTo(s2)<=0) return s1;
        return s2;
    }
    public static String maxString(String s1, String s2){
        if (s1.compareTo(s2)<0) return s2;
        return s1;
    }
    public static int uppperBound(ArrayList<Integer> a, long x){
        int lo=0,hi=a.size()-1,res=a.size();
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            int curr=a.get(mid);
            if (curr>x){
                res=mid;
                hi=mid-1;
            }else lo=mid+1;
        }
        return res;
    }
    public static int lowerBound(ArrayList<Long> a, long x){
        int res=-1,lo=0,hi=a.size()-1;
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            long curr=a.get(mid);
            if (curr<=x){
                res=mid;
                lo=mid+1;
            }else hi=mid-1;
        }
        return res;
    }
    public static int uppperBound(long[] a, long x){
        int lo=0,hi=a.length-1,res=a.length;
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            long curr=a[mid];
            if (curr>x){
                res=mid;
                hi=mid-1;
            }else lo=mid+1;
        }
        return res;
    }
    public static long GCD(long a, long b){
        if (a==0) return b;
        return GCD(b%a,a);
    }

    public static long lcm_of_array(long[] arr)
    {
        long lcm = arr[0];
        for (int i = 1; i < arr.length; i++) {
            long num1 = lcm;
            long num2 = arr[i];
            long gcd_val = GCD(num1, num2);
            lcm = (lcm * arr[i]) / gcd_val;
        }
        return lcm;
    }
    static long power(long x, long y, long p) {
        long res = 1;
        x = x % p;
        if (x == 0)
            return 0;

        while (y > 0) {
            if ((y & 1) != 0)
                res = (res * x) % p;

            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        long[] readArray1(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
import java.util.*;

public class Codechef2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        while (t-->0) {
            int n = sc.nextInt();
            int []a=new int[n];
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
            }

            HashMap<Integer,Integer> map=new HashMap<>();
            for(int i=0;i<n;i++){
                map.put(a[i], map.getOrDefault(a[i],0)+1);
            }

            Arrays.sort(a);

            if(a[0]==1){
                if(map.size()%2==0){
                    System.out.println("Bob");
                }else{
                    System.out.println("Alice");
                }
            }else{
                System.out.println("Alice");
            }

        }
    }
}
//1 2 3 4 7 9 100
//1 2 3 6 8 99
//1 2 5 7 98
//1 4 6 97
//3 5 96
//2 93
//91
//0

//1 3 5 6 7 9
//1 3 4 5 7
//2 3 4 6
//1 2 4
//1 3
//2
//0

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

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

int main(){
    int q;cin>>q;
    while(q--){
        int n,x;cin>>n;
        set<int>s;
        for(int i = 0;i<n;i++){
            cin>>x;
            s.insert(x);
        }
        if(*(s.begin()) != 1){
            cout<<"Alice\n";
        }
        else{
            auto p = s.end();--p;
            int l  = *p;
            if(s.size()== l){
                if(l&1){
                    cout<<"Alice\n";
                }
                else{
                    cout<<"Bob\n";
                }
            }
            else{
                int st = 1;
                for(auto &x:s){
                      if(s.find(st) == s.end()){
                        break;
                      }
                      st++;
                }
                if(((s.size()-st))== 0){
                    cout<<"Bob\n";
                }
                else{
                    cout<<"Alice\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
3
1
1
2
1 2
4
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

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

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

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

#define int long long

void solve(){
    int n; cin >> n;
    set<int> st;
    for(int i = 0; i < n; i++) {
        int x; cin >> x;
        st.insert(x);
    }
    vector<int> v(st.begin(), st.end());
    int sum = 0;
    int k = 0;
    for(int i = 0; i < st.size()-1; i++) {
        int y = v[i];
        v[i] -= sum;
        if(v[i] > 1) {
            k += 2;
        }
        else {
            k += 1;
        }
        sum += y;
    }
    // cout << k << "\n";
    if(k & 1) cout << "Bob\n";
    else cout << "Alice\n";
}

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

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

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

Think step by step.

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