**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
// Source: https://usaco.guide/general/io
#include <bits/stdc++.h>
using namespace std;
int main() {
	int t;
	cin >> t;
	for (int i = 0; i < t; i++){
		int n, m, q;
		cin >> n >> m >> q;
		int a, b;
		cin >> a >> b;
		int david;
		cin >> david;
		if (david > a && david < b){
			cout << max(david - a, b - (david + 1)) << endl;
		}
		else if (david < a && david < b){
			cout << a - david << endl;
		}
		else if (david > a && david > b){
			cout << david - b << endl;
		}
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
#include<bits/stdc++.h>
#include <algorithm>
#include <numeric>
#include <random>
#include <ranges>
#include <chrono>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>

#define ll long long
#define ld long  double
#define vi vector<int>
#define vc vector<char>
#define vd vector<double>
#define vvi vector<vector<int>>
#define vvd vector<vector<double>>
#define vvc vector<vector<char>>
#define vvp vector<vector<pair<int,int>>>
#define vvpl vector<vector<pair<ll,ll>>>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vpl vector<pair<ll,ll>>
#define vpi vector<pair<int,int>>
#define vppi vector<pair<int,pair<int,int>>>
#define sz(v) sizeof(v)
#define scn(v) for (auto& i : v)cin >> i;
#define scnp(v) for (auto& i : v)cin >> i.first>>i.second;
#define srt(v) sort(v.begin(),v.end())
#define vb vector<bool>
#define vvb vector<vector<bool>>
//#define f first
//#define s second
const ll INF = 1e10+5;
#define LINF  2e6+9

#define en cout<<'\n'
#define lp(i,n) for(int i=0;i<n;i++)
#define rlp(i,n) for(int i=n-1;i>=0;i--)
#define MAXN 200005
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()

ll MOD = 998244353;
ll MOD2=  998244353;
#define pll pair<ll,ll>
#define X first
#define Y second
#define pb push_
#define sINF 9e10;

#define mpi map<int,int>
#define mpl map<ll,ll>
#define YA cout<<"YES"; return;
#define NA cout<<"NO"; return;

ll setBit(ll num, ll idx, ll val = 1) {
    return (val) ? (num | (1 << idx)) : (num & ~(1 << idx));
}
int getBit(ll num, int idx) { return ((num >> idx) & 1) == 1; }

int cntBits(ll num) { int ret = 0; while (num) { if (num % 2)ret++; num /= 2; } return ret; }

ll fastPow(ll x, ll n, ll m) {
    if (n == 0) {
        return 1;
    }
    ll po = fastPow(x, n / 2, m);
    po = (po * po) % m;
    if (n & 1) {
        po *= x;
    }
    return po % m;
}

/*

const int N = 2e5 + 10;
ll Fact[N], FactInv[N];

ll modInv(ll x, ll m) {
    return fastPow(x, m - 2, m) % m;
}
void preFact() {
    Fact[0] = 1;
    for (int i = 1; i < N; i++) {
        Fact[i] = (i * Fact[i - 1]) % MOD;
    }
    FactInv[N - 1] = fastPow(Fact[N - 1], MOD - 2, MOD) % MOD;

    for (int i = N - 2; i >= 0; i--) {
        FactInv[i] = (i + 1) * FactInv[i + 1] % MOD;
    }
}
ll nCr(int n, int r) {

    return Fact[n] * ((FactInv[n - r]) % MOD * FactInv[r] % MOD) % MOD;
}
*/

ll mul(ll a, ll b) {
    return (((a % MOD + MOD) * (b % MOD + MOD)) % MOD);
}
ll add(ll a, ll b) {
    return ((a + b) % MOD);
}
ll sub(ll a,ll b){

    return ((a%MOD-b%MOD)%MOD+MOD)%MOD;
}

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

ll sumDig(int n){
    ll sum=0;
    while(n>0){
        sum+=n%10;
        n/=10;
    }
    return sum;
}

vb isprime(1005+10,true);
vl primsss;
const int N = 1005;
void sieve(){

    isprime[0]=0;
    isprime[1]=0;

    for(int i=2;i<=N;i++){
        if(!isprime[i])continue;
        primsss.push_back(i);
        for(int j=2*i;j<=N;j+=i){
            isprime[j]=0;
        }
    }
}

ll sum(ll n){
    return (n*(n+1))/2;
}

void burnx(){
   ll n,m,q;
   cin>>n>>m>>q;
   ll x,y;
   cin>>x>>y;
   ll p;
   cin>>p;
   if(x>y){
       swap(x,y);
   }
   if(y<p){
       cout<<n-y;
       return;
   }
   if(p<x){
       cout<<x-1;
       return;
   }
   ll d1=p-x,d2=y-p;
   ll dd=y-x;
   cout<<max(min(y-dd,dd-x),min(y-(dd+1),dd+1-x));

}

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

    int t = 1;
    cin >> t;

    int tx=1;
    while (t--) {
       burnx();
       en;

    }

    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 where the logic of the code might fail. The code provided has a complex logic for calculating the number of moves, and it might not handle certain edge cases correctly. Let's analyze the potential issues and generate a test case accordingly.

### Potential Issues in the Buggy Code
1. **Edge Cases for Initial Positions**: The code might not handle the cases where the initial positions of the teachers and David are at the boundaries of the classroom.
2. **Optimal Moves Calculation**: The code might not correctly calculate the optimal moves for both David and the teachers in certain configurations.

### Challenging Test Case
We will create a test case where David is initially positioned such that the teachers need to coordinate their moves carefully to catch him. This will test the logic of the code in handling optimal moves.

#### Test Case
- **Classroom Size (n)**: 10
- **Number of Teachers (m)**: 2
- **Number of Queries (q)**: 1
- **Initial Positions of Teachers**: 1 and 9
- **Initial Position of David**: 5

### Test Input
```plaintext
1
10 2 1
1 9
5
```

### Explanation
- **Classroom Size (n = 10)**: The classroom has 10 cells.
- **Number of Teachers (m = 2)**: There are 2 teachers.
- **Number of Queries (q = 1)**: There is 1 query.
- **Initial Positions of Teachers**: The teachers are initially at positions 1 and 9.
- **Initial Position of David**: David is initially at position 5.

### Expected Output
The expected output for this test case is 2, as it takes 2 moves for the teachers to catch David if they act optimally. Here's the reasoning:
- In the first move, David can move to position 6.
- In the second move, David can move to position 7.
- In the third move, one of the teachers (initially at position 1) can move to position 7, catching David.

This test case will challenge the code to correctly calculate the optimal moves for both David and the teachers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
t=int(input())
for _ in range(t):
      n,m,q=map(int, input().split())
      a,b=map(int, input().split())
      s=int(input())
      if b>a:
           if a< s <b:
               print((b-a)//2)
           elif s<a:
               print(a-1)
           else:
               print(n-b)
      else:
           if b< s <a:
               print((a-b)//2)
           elif s<b:
               print(b-1)
           else:
               print(n-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:
```plaintext
1
5 2 1
1 3
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
#input the test cases

tests = int(input())
for test in range(tests):
    n,m,q=map(int,input().split())
    teach1,teach2 = map(int,input().split())
    david = int(input())

    #alist contains the queries of davids cell number
    #blist contains the positionjs of the teachers
    if teach1 == david or teach2 == david:
        print(0)
        continue

    #three cases can be there
    #first case , both teachers on the right.
    if teach1>david and teach2>david:
        #find the lower one
        lower = min(teach1,teach2)
        print(lower-1)
        continue

    #second both teachers on the left
    elif teach1<david and teach2<david:
        higher = max(teach1,teach2)
        print(n-higher)
        continue

    #last case one on right one on left

    else:
        right = max(teach1,teach2)
        left = min(teach1,teach2)
        diff = right-left+1

        if (david - left) == (right - david):
            print(int((diff/2)))
            continue
        else:
            print(int(diff/2)-1)
            continue

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

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

Think step by step.

Response:
```plaintext
3
10 2 1
1 10
5
10 2 1
2 9
5
10 2 1
3 8
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
def davidcatcher(n, a, b1, b2):
    b1=min(b1,b2)
    b2=max(b1,b2)
    if a > b2 :

        return n - b2
    if a<b1:

        return b1- 1
    else:
        d=(b1+b2)//2
        return min(d-b1,b2-d)
x=int(input())
l=[]
for i in range(x):
    v=input()
    n=int(v.split()[0])
    b1b2=input()
    b1b2=b1b2.split()
    b1=int(b1b2[0])
    b2=int(b1b2[1])
    a=int(input())
    l.append(davidcatcher(n,a,b1,b2))
for i in l:
    print(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(); // number of test cases
        for(int j = 0; j < t; j++) {
            int n = sc.nextInt(); // number of cells
            int m = sc.nextInt(); // number of teachers (always 2)
            int q = sc.nextInt(); // number of queries (always 1)

            int b1 = sc.nextInt(); // position of first teacher
            int b2 = sc.nextInt(); // position of second teacher
            int q1 = sc.nextInt(); // David's position

            // Ensure b1 is always the smaller value
            if (b1 > b2) {
                int temp = b1;
                b1 = b2;
                b2 = temp;
            }

            // Case 1: David is between the two teachers
            if (q1 > b1 && q1 < b2) {
                System.out.println(1); // Caught in 1 move
            }
            // Case 2: David is to the left of both teachers
            else if (q1 < b1) {
                System.out.println(b1 - q1); // Distance to the nearest teacher
            }
            // Case 3: David is to the right of both teachers
            else if (q1 > b2) {
                System.out.println(q1 - b2); // Distance to the nearest teacher
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
import java.util.*;
 public class Main{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int i=0;i<t;i++){
            int n=sc.nextInt();
            int m=sc.nextInt();
            int q=sc.nextInt();
            int t1=sc.nextInt();
            int t2=sc.nextInt();
            int david=sc.nextInt();
            int ans=Math.abs(david-t1);
            int ans2=Math.abs(david-t2);
            System.out.println(Math.min(ans,ans2));
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

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

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int t; cin >> t;
    for(int z = 0; z < t; z++){
        int n, m, q; cin >> n >> m >> q;
        int A[m], B[q];
        for (int i = 0; i < m; i++){
            cin >> A[i];
        }
        for (int i = 0; i < q; i++){
            cin >> B[i];
        }
        if (B[0] < A[0]){
            cout << (B[0] - 1) + (A[0] - B[0]) << endl;
        }
        else if (B[0] < A[1] && B[0] > A[0]){
            cout << abs(B[0] - (A[1] + A[0])/2) + min(A[1] - B[0], B[0] - A[0]) << endl;
        }
        else if(B[0] > A[1]){
            cout << (n - B[0]) + (B[0] - A[1]) << endl;
        }
        else{
            cout << 0;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the faulty logic in the code can lead to incorrect results. Let's analyze the code and identify potential issues:

1. **Initial Positioning of Teachers and David**:
   - The code assumes that the teachers are always positioned in a way that they can catch David in a straightforward manner. However, if the positions are such that David can strategically move to avoid being caught quickly, the code might fail.

2. **Edge Cases**:
   - When David is positioned at the boundaries (i.e., cell 1 or cell n).
   - When David is positioned between the two teachers, but the teachers are not symmetrically placed.

3. **Symmetry and Asymmetry**:
   - The code might not handle cases where the teachers are not symmetrically placed around David correctly.

Let's construct a test case that challenges these assumptions:

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

### Explanation:
- **n = 10**: The classroom has 10 cells.
- **m = 2**: There are 2 teachers.
- **q = 1**: There is 1 query for David's position.
- **Teachers' positions**: Teacher 1 is at cell 3, and Teacher 2 is at cell 7.
- **David's position**: David is at cell 5.

### Why This Test Case is Challenging:
1. **David is between the two teachers**: This scenario tests the code's ability to handle a situation where David is not at the boundary but is in the middle of the two teachers.
2. **Asymmetric positions**: The teachers are not symmetrically placed around David (i.e., the distance from David to Teacher 1 is 2, and the distance from David to Teacher 2 is 2). This tests the code's ability to handle non-symmetric positions.

### Expected Output:
- **Output**: 2

### Reasoning:
- **Move 1**: David can move to cell 4 or cell 6. Let's say David moves to cell 4.
- **Move 2**: Teacher 1 moves to cell 4, and Teacher 2 moves to cell 6. David is caught.

This test case should expose the bug in the code if it incorrectly calculates the number of moves required for the teachers to catch David.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

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

const int mod = 1e9+7;
#define ll long long
#define maxel *max_element(a, a+n)
#define minel *min_element(a, a+n)
#define llmax LLONG_MAX
#define insort(v) sort(v.begin(), v.end())
#define desort(v) sort(v.rbegin(), v.rend())
#define cint int t; cin >> t

int main() {
   cint;
   while(t--)
   {

    int n,m,q; cin>>n>>m>>q;
    int b[m];
   for(int i=0;i<m;i++) cin>>b[i];
   int pos; cin>>pos;
      sort(b,b+m);
   // for(int i=0;i<m;i++) cout<<b[i]<<" ";
   //    cout<<endl;
      int mx=-1;
      if(pos<b[0])
      {
         cout<<b[0]-pos<<endl;
      }
      else if(pos>b[m-1]) cout<<pos-b[m-1]<<endl;
      else
      {
         ll x = abs(pos - b[0]);
         ll y = abs(pos - b[1]);
         cout<<abs(x-y)<<endl;

      }
   }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider scenarios where the code might fail. The key points to consider are:

1. **Edge Cases**: Where David is at the boundary of the classroom.
2. **Middle Cases**: Where David is between the two teachers.
3. **Symmetric Cases**: Where the positions of the teachers and David are symmetrically placed.

The buggy code has a specific issue in the `else` block where it calculates the number of moves based on the absolute difference between David's position and the positions of the two teachers. This can lead to incorrect results in certain scenarios.

Let's create a test input that exposes this issue:

1. **David at the boundary**: This should be straightforward and should work correctly.
2. **David between the two teachers**: This is where the buggy code might fail because it calculates the difference incorrectly.
3. **Symmetric positions**: This can also reveal the issue in the `else` block.

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

```plaintext
3
10 2 1
1 4
2
10 2 1
3 6
5
10 2 1
1 9
5
```

Explanation:
1. **First Test Case**:
   - Classroom size: 10
   - Teachers: 1, 4
   - David: 2
   - Expected output: 1 (David is caught in one move by the teacher at position 1)

2. **Second Test Case**:
   - Classroom size: 10
   - Teachers: 3, 6
   - David: 5
   - Expected output: 2 (David is caught in two moves by the teacher at position 6)

3. **Third Test Case**:
   - Classroom size: 10
   - Teachers: 1, 9
   - David: 5
   - Expected output: 2 (David is caught in two moves by the teacher at position 1)

This test input should help expose the bug in the provided code, particularly in the `else` block where the number of moves is calculated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
t = int(input())
for it in range(t):
    n, m, q = map(int, input().split())
    b1, b2 = map(int, input().split())
    a = int(input())
    b1, b2 = sorted((b1, b2))
    if b1 < a < b2:
        d = b2 - b1
        if d % 2 != 0:
            print((d-1)//2)
        else:
            print((d-1)//2+1)
    else:
        if a < b1:
            d = b1 - a
        else:
            d = a - b2
        print(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
3
10 2 1
1 10
2
8 2 1
3 6
1
8 2 1
3 6
8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
for _ in range(int(input())):
    n,m,q=[int(i) for i in input().split()]
    b1,b2=[int(i) for i in input().split()]
    a=int(input())
    ans=0
    if a<min(b1,b2):
        ans=a-1+(abs(a-min(b1,b2)))
    elif a>max(b1,b2):
        ans=(n-a)+(abs(a-max(b1,b2)))

    elif abs(b1-a)<abs(b2-a):
        ans+=(abs(b2-a)-abs(b1-a))//2
        #print(ans)
        ans+=abs(b1-a)

    else:
        if abs(b2-a)<abs(b1-a):
            ans+=(abs(b1-a)-abs(b2-a))//2
        #print(ans)
            ans+=abs(b2-a)
    print(ans)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases where the logic for calculating the number of moves might fail. The code seems to handle the cases where David is to the left or right of both teachers, but it might have issues when David is between the two teachers.

Here's a step-by-step approach to generate a challenging test input:

1. **David is between the two teachers**: This is a critical case where the teachers need to coordinate to minimize the number of moves. The code might not correctly calculate the optimal moves in this scenario.
2. **David is close to one of the teachers**: This can help verify if the code correctly handles the scenario where David is already close to one of the teachers.
3. **David is at the edge of the classroom**: This can help verify if the code correctly handles the edge cases where David is at the start or end of the classroom.

Let's create a test input that covers these scenarios:

```plaintext
3
10 2 1
1 8
5
10 2 1
1 8
2
10 2 1
1 8
10
```

Explanation:
1. **First Test Case**: David is at cell 5, between the two teachers at cells 1 and 8. This tests the scenario where David is between the two teachers.
2. **Second Test Case**: David is at cell 2, close to one of the teachers at cell 1. This tests the scenario where David is close to one of the teachers.
3. **Third Test Case**: David is at cell 10, at the edge of the classroom, far from the teachers at cells 1 and 8. This tests the scenario where David is at the edge of the classroom.

This test input should help expose any issues in the buggy code's logic for handling these edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

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

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int test = sc.nextInt();
        while (test > 0) {
            long length = sc.nextLong();
            int teachers = sc.nextInt();
            int q = sc.nextInt();
            int t1 = sc.nextInt();
            int t2 = sc.nextInt();
            int posi = sc.nextInt();
            int x;
            int y;
            if (t1 > t2) {
                x = t1;
                y = t2;
            } else {
                y = t1;
                x = t2;
            }
            int output = Math.min(Math.abs(t1 - posi), Math.abs(t2 - posi));

            if (posi > x) {
                output += length - posi;
                // System.out.println(output + length - posi);
            } else if (posi < y) {
                output += posi - 1;
                // System.out.println(output + posi - 1);
            } else {
                if ((x + y) % 2 == 1) {
                    output += Math.min(Math.abs(((x + y + 1) / 2) - posi), Math.abs(((x + y - 1) / 2) - posi));
                } else {
                    output += ((x + y) / 2) - posi;
                }
                // System.out.println(output);
            }
            System.out.println(output);
            test--;
        }
        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
5 2 1
1 4
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
import java.util.Scanner;

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

        int t = sc.nextInt();

        while (t-- > 0) {
            int n = sc.nextInt(); // number of cells
            int m = sc.nextInt(); // number of teachers
            int q = sc.nextInt(); // number of queries

            int[] qD = new int[m];
            for (int i = 0; i < m; i++) {
                qD[i] = sc.nextInt();
            }

            int[] qT = new int[q];
            for (int i = 0; i < q; i++) {
                qT[i] = sc.nextInt();
            }

            for (int i = 0; i < q; i++) {
                if (qD[0] < qT[0] && qT[0] < qD[1]) {
                    System.out.println(Math.min(qT[0]-qD[0], qD[1]-qT[0]));
                } else {
                    if (qT[0] < qD[0]) {
                        System.out.println(qD[0]-1);
                    } else if (qT[0] > qD[1]) {
                        System.out.println(n-qD[1]);
                    }
                }
            }
        }

        sc.close();
        return;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

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

int main() {
    int t;
    cin >> t;
    while(t--) {
        int n,m,qe;
        cin >> n >> m >> qe;
        int t1,t2;
        cin >> t1 >> t2;
        int q; cin >> q;
        int ans = 0;
        if(q<t1 && q<t2) {
            ans = min(t1,t2) - 1;
        }
        else if(q>t1 && q>t2) {
            ans = n - max(t1,t2);
        }
        else {
            ans = abs(t2-t1) - 2;
        }
        cout << ans << endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the logic of the code and identify potential edge cases or scenarios where the code might fail.

### Analysis of the Buggy Code
1. **Condition 1: `if(q < t1 && q < t2)`**
   - The code calculates the minimum distance from `q` to the closest teacher and subtracts 1.
   - This logic is incorrect because it doesn't account for the optimal moves of David and the teachers.

2. **Condition 2: `else if(q > t1 && q > t2)`**
   - Similar to the first condition, it calculates the distance from `q` to the farthest teacher and subtracts the farthest teacher's position from `n`.
   - This logic is also incorrect for the same reason as the first condition.

3. **Condition 3: `else`**
   - The code calculates the absolute difference between the positions of the two teachers and subtracts 2.
   - This logic is also flawed because it doesn't consider the optimal moves of David and the teachers.

### Challenging Test Input
To expose the bug, we need a test case where the positions of David and the teachers are such that the optimal moves are not correctly handled by the code. Let's consider a scenario where David is between the two teachers but not equidistant from them.

### Test Input
```plaintext
3
10 2 1
1 7
4
10 2 1
2 8
5
10 2 1
3 9
6
```

### Explanation
1. **Test Case 1:**
   - `n = 10`, `m = 2`, `q = 1`
   - Teachers are at positions `1` and `7`
   - David is at position `4`
   - The correct answer should be `2` because the teachers can catch David in 2 moves. The code might incorrectly calculate it as `1`.

2. **Test Case 2:**
   - `n = 10`, `m = 2`, `q = 1`
   - Teachers are at positions `2` and `8`
   - David is at position `5`
   - The correct answer should be `2` because the teachers can catch David in 2 moves. The code might incorrectly calculate it as `1`.

3. **Test Case 3:**
   - `n = 10`, `m = 2`, `q = 1`
   - Teachers are at positions `3` and `9`
   - David is at position `6`
   - The correct answer should be `2` because the teachers can catch David in 2 moves. The code might incorrectly calculate it as `1`.

These test cases are designed to challenge the buggy code's logic and expose its incorrect handling of the optimal moves of David and the teachers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define endl "\n"
#define fo(i, n) for (i = 0; i < n; i++)
#define Fo(i, k, n) for (i = k; k < n; k++)
#define pb push_back
#define ll long long
#define mp make_pair
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, arr) for (auto it = arr.begin(); it != arr.end(); it++)
#define PI 3.1415926535897932384626
#define suprit ios_base::sync_with_stdio(0); cout.tie(0); cin.tie(0);
#define line cout << endl;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef map<ll, ll> ml;
typedef map<string, ll> msl;
typedef map<ll, string> mls;
typedef unordered_map<ll, ll> uml;
typedef unordered_map<string, ll> umsl;
typedef unordered_map<ll, string> umls;
typedef set<ll> sl;
typedef set<pair<ll, ll>> spl;
typedef ordered_set<ll> osl;
typedef ordered_set<pair<ll, ll>> ospl;

const ll mod = 1e9 + 7;

bool comp2(pair<ll, ll> &arr, pair<ll, ll> &b)
{
    if (arr.first == b.first)
        return arr.second < b.second;
    return arr.first < b.first;
};

template <typename T> void read(T i, T n, vector<T> &arr) { for(T j = i; j < n; j++) cin >> arr[j]; }
template <typename T> void read(vector<T> &arr) { for(auto &j : arr) cin>>j; }

#ifndef ONLINE_JUDGE

template <typename T, typename V> void print(set<pair<T, V>> &arr) { for(auto &it : arr) cout<<it.first<<" "<<it.second<<endl; line}
template <typename T, typename V> void print(pair<T, V>& pa) { cout<<pa.first<<" "<<pa.second<<endl; }
template <typename T> void print(T i, T last, vector<T> &arr) { for(T j = i; j < last; j++) cout<<arr[j]<<" "; line}
template <typename T> void print(T i, vector<T> &arr) { for(T j = i; j < arr.size(); j++) cout<<arr[j]<<" "; line}
template <typename T> void print(vector<T> &arr) { for(auto &i : arr) cout<<i<<" "; line}
template <typename T, typename V> void print(unordered_map<T, V>& arr) { for(auto &it : arr) cout<<it.first<<" "<<it.second<<endl; line}
template <typename T, typename V> void print(map<T, V>& arr) { for(auto &it : arr) cout<<it.first<<" "<<it.second<<endl;}
template <typename T> void print(unordered_set<T> &arr) { for(auto &it : arr) cout<<it<<" "; line }
template <typename T> void print(ordered_set<T> &arr) { for(auto &it : arr) cout<<it<<" "; line }
template <typename T> void print(set<T> &arr) { for(auto &it : arr) cout<<it<<" "; line }
template <typename T, typename... Args> void print(T t, Args... args) { cout << t << " "; print(args...); }
template <typename T> void print(T t) { cout<<t<<"\n"; }
#define debug(x) cout<<#x<<" "<<x<<endl;
#define debug2(x, y) cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<endl;
#define debug3(x, y, z) cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<" "<<#z<<" "<<z<<endl;
#define debug4(x, y, z, a) cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<" "<<#z<<" "<<z<<" "<<#a<<" "<<a<<endl;

#else

#define print(x)
#define debug(x)
#define debug2(x, y)
#define debug3(x, y, z)
#define debug4(x, y, z, a)

#endif

void solve()
{
    int n, m, k;
    cin >> n >> m >> k;
    int fir, sec, david;
    cin >> fir >> sec >> david;
    if(david < fir) {
        cout << fir - 1 << endl;
        return;
    }
    if(david > sec) {
        cout << n - sec << endl;
        return;
    }
    cout << (sec - fir) / 2 << endl;
}

int main()
{
    suprit;
    clock_t start = clock();

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

    #ifndef ONLINE_JUDGE
    double elapsed = double(end - start) / CLOCKS_PER_SEC;
    cout << setprecision(10) << elapsed << endl;
    #endif
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
for i in range(int(input())):
    n,m,q = map(int,input().split())
    teacher_pos = [int(x) for x in input().split()]
    david_pos = int(input())
    if david_pos < teacher_pos[0]:
        print(teacher_pos[0]-1)
    elif david_pos > teacher_pos[1]:
        print(n-teacher_pos[1])
    elif teacher_pos[0] < david_pos < teacher_pos[1]:
        mid_point = teacher_pos[0] + (teacher_pos[1]-teacher_pos[0]) // 2
        a = abs(mid_point - teacher_pos[0])
        b = abs(mid_point - teacher_pos[1])
        print(min(a,b))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases**:
   - David's position at the boundaries of the classroom (i.e., cell 1 or cell n).
   - David's position exactly in the middle of the two teachers.
   - David's position close to one of the teachers.

2. **Optimal Moves**:
   - The teachers should always move towards David in such a way that they minimize the number of moves to catch him.
   - David should always move in such a way that he maximizes the number of moves before being caught.

3. **Buggy Code Analysis**:
   - The code incorrectly calculates the number of moves in certain scenarios, especially when David is in the middle of the two teachers.

Given these considerations, let's create a test case where David is positioned in the middle of the two teachers. This will expose the bug in the mid-point calculation and the incorrect handling of the number of moves.

Here is the test input:

```plaintext
1
10 2 1
3 8
5
```

**Explanation**:
- **n = 10**: The classroom has 10 cells.
- **m = 2**: There are 2 teachers.
- **q = 1**: There is 1 query.
- **Teacher positions**: The teachers are initially at cells 3 and 8.
- **David's position**: David is initially at cell 5.

In this scenario:
- The teachers are at cells 3 and 8.
- David is at cell 5, which is exactly in the middle of the two teachers.
- The correct number of moves for the teachers to catch David is 2 (both teachers move towards David, and they catch him in 2 moves).

The buggy code incorrectly calculates the number of moves in this case, making it a good test input to expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
testCount = int(input())

for i in range(testCount):
    n, m, q = (int(x) for x in input().strip().split())
    #n = int(input())
    #m = 2
    #q = 1

    teachersPos = [int(x) for x in input().strip().split()]
    #teachersPos = [ int(input()), int(input()) ]

    david = int(input())

    #print('')

    if david < teachersPos[0]:
        #print('david:', david)
        #print('teacher 0:', teachersPos[0])
        print(teachersPos[0] - 1)
    elif david > teachersPos[1]:
        #print('david:', david)
        #print('teacher 1:', teachersPos[1])
        print(n - teachersPos[1])
    else:
        print((teachersPos[1] - teachersPos[0]) // 2)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Easy Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the easy version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m=2$ and $q=1$. You can make hacks only if both versions of the problem are solved.

Narek and Tsovak were busy preparing this round, so they have not managed to do their homework and decided to steal David's homework. Their strict teacher noticed that David has no homework and now wants to punish him. She hires other teachers to help her catch David. And now $m$ teachers together are chasing him. Luckily, the classroom is big, so David has many places to hide.

The classroom can be represented as a one-dimensional line with cells from $1$ to $n$, inclusive.

At the start, all $m$ teachers and David are in distinct cells. Then they make moves. During each move

  * David goes to an adjacent cell or stays at the current one.    * Then, each of the $m$ teachers simultaneously goes to an adjacent cell or stays at the current one.

This continues until David is caught. David is caught if any of the teachers (possibly more than one) is located in the same cell as David. Everyone sees others' moves, so they all act optimally.

Your task is to find how many moves it will take for the teachers to catch David if they all act optimally.

Acting optimally means the student makes his moves in a way that maximizes the number of moves the teachers need to catch him; and the teachers coordinate with each other to make their moves in a way that minimizes the number of moves they need to catch the student.

Also, as Narek and Tsovak think this task is easy, they decided to give you $q$ queries on David's position. Note: this is the easy version, and you are given only one query.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of each test case follows.

In the first line of each test case, you are given three integers $n$, $m$, and $q$ ($3 \le n \le 10^9$, $m=2$, $q=1$) — the number of cells on the line, the number of teachers, and the number of queries.

In the second line of each test case, you are given $m$ distinct integers $b_1, b_2, \ldots, b_m$ ($1 \le b_i \le n$) — the cell numbers of the teachers.

In the third line of each test case, you are given $q$ integers $a_1, a_2, \ldots, a_q$ ($1 \le a_i \le n$) — David's cell number for every query.

It is guaranteed that for any $i$, $j$ such that $1 \le i \le m$ and $1 \le j \le q$, $b_i \neq a_j$.

Output: For each test case, output $q$ lines, the $i$-th of them containing the answer of the $i$-th query.

Examples:
input:
3
10 2 1
1 4
2
8 2 1
3 6
1
8 2 1
3 6
8
output:
1
2
2

Note: In the first example, the student can just stay at cell $2$. The teacher, initially located in cell $1$, can reach cell $2$ in one move. Therefore, the answer is $1$.

In the second example, the student should just stay at cell $1$. The teacher, initially located in cell $3$, can reach cell $1$ in two moves. Therefore, the answer is $2$.

Buggy Code:
import java.util.Scanner;

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

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();

            int b1 = sc.nextInt();
            int b2 = sc.nextInt();
            int a1 = sc.nextInt();

            int dist1 = Math.abs(b1 - a1);
            int dist2 = Math.abs(b2 - a1);

            int result = Math.min(dist1, dist2);
            System.out.println(result);
        }

        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
10 2 1
1 10
5
```