**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();
            long[] arr = new long[m];
            for (int i = 0; i < m; i++) {
                arr[i]=sc.nextLong();
            }
            long[] answ = new long[q];
            for (int i = 0; i < q; i++) {
                int pos = sc.nextInt();
                int ll =0;
                int ul = arr.length-1;
                int ans =-1;
                while(ll<=ul){
                    int mid = ll+(ul-ll)/2;
                    if(arr[mid]<=pos){
                        ans=mid;
                        ll=mid+1;
                    }
                    else ul = mid-1;
                }
                if(ans==-1){
                    answ[i] = arr[0]-1;
                }
                else if(ans==arr.length-1){
                    answ[i]= n-arr[arr.length-1];
                }
                else{
                    long len = arr[ans+1]-arr[ans]+1;
                    if(len%2==0) answ[i]=(len/2)-1;
                    else answ[i]=len/2;
                }
            }
            for (int i = 0; i < q; i++) {
                System.out.print(answ[i]+" ");
            }
            System.out.println();
        }

    }
    static class pair{
        long a;
        long b;
        public pair(long a, long b){
            this.a=a;
            this.b=b;
        }
    }
    public static int[] mex(int[] array) {
        // Use a Set to store the elements for O(1) average-time complexity checks
        Set<Integer> set = new HashSet<>();

        // Populate the Set with elements from the array
        for (int num : array) {
            if (num >= 0) {
                set.add(num);
            }
        }

        // Array to store the result
        int[] result = new int[2];
        int count = 0;
        int current = 0;

        // Find the smallest two non-negative numbers not in the set
        while (count < 2) {
            if (!set.contains(current)) {
                result[count] = current;
                count++;
            }
            current++;
        }

        return result;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code uses a binary search to find the closest teacher positions and then calculates the number of moves based on the distances. Here are some key points to consider:

1. **Edge Cases for Teacher Positions**:
   - Teachers are at the very start (1) and very end (n) of the line.
   - Teachers are closely packed together.
   - Teachers are spread out across the line.

2. **Edge Cases for David's Positions**:
   - David is at the very start (1) or very end (n) of the line.
   - David is between two teachers.
   - David is far from all teachers.

3. **Large and Small Values**:
   - Use large values for \( n \) and \( m \) to test the performance and correctness of the code.
   - Use small values for \( n \) and \( m \) to test the edge cases more thoroughly.

4. **Multiple Queries**:
   - Test with multiple queries to ensure the code handles multiple test cases correctly.

Given these considerations, let's create a test input that includes these scenarios:

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

### Explanation:
1. **Test Case 1**:
   - \( n = 10 \)
   - \( m = 2 \)
   - Teachers at positions: 1, 10
   - Queries: 2, 5, 8
   - **Expected Output**:
     - For \( a = 2 \): The closest teacher is at 1, so it takes 1 move.
     - For \( a = 5 \): The closest teachers are at 1 and 10, so it takes 4 moves.
     - For \( a = 8 \): The closest teacher is at 10, so it takes 2 moves.

2. **Test Case 2**:
   - \( n = 10 \)
   - \( m = 3 \)
   - Teachers at positions: 1, 5, 10
   - Queries: 2, 3, 8
   - **Expected Output**:
     - For \( a = 2 \): The closest teacher is at 1, so it takes 1 move.
     - For \( a = 3 \): The closest teacher is at 5, so it takes 2 moves.
     - For \( a = 8 \): The closest teacher is at 10, so it takes 2 moves.

3. **Test Case 3**:
   - \( n = 10 \)
   - \( m = 5 \)
   - Teachers at positions: 1, 2, 3, 9, 10
   - Queries: 1, 5, 8
   - **Expected Output**:
     - For \( a = 1 \): The closest teacher is at 1, so it takes 0 moves.
     - For \( a = 5 \): The closest teachers are at 3 and 9, so it takes 2 moves.
     - For \( a = 8 \): The closest teacher is at 9, so it takes 1 move.

This test input covers a range of scenarios, including edge cases and multiple queries, which should help expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
#include<bits/stdc++.h>
#define vectorll vector<long long>
#define vectorcc vector<char>
#define forll for(long long i=0;i<n;i++)
#define mapll map<long long,long long> mp
#define maplc map<char,long long> mpc
#define setll set<long long> sti
#define setcc set<char> stc
#define str string st
#define ll long long
//Speed
#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define Asquare cout.tie(NULL);

//Constants
const double pi= 3.141592653589793238;
const ll INF= LONG_LONG_MAX;
const ll mod=1e9+7;
using namespace std;

//TypeDEf
#define pll  pair<ll, ll>
#define vpll vector<pll>
#define  vs vector<string>
 #define umll unordered_map<ll,ll>
 #define vvll vector<vector<ll>>

#define py cout<<'YES'<<endl;
#define pm cout<<'-1'<<endl;
#define pn cout<<'NO'<<endl;

// Mathematical functions
ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);}
ll lcm(ll a, ll b){return (a/gcd(a,b)*b);}
ll moduloMultiplication(ll a,ll b,ll mod){ll res = 0;a %= mod;while (b){if (b & 1)res = (res + a) % mod;b >>= 1;}return res;}
ll powermod(ll x, ll y, ll p){ll res = 1;x = x % p;if (x == 0) return 0;while (y > 0){if (y & 1)res = (res*x) % p;y = y>>1;x = (x*x) % p;}return res;}

bool sorta(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b){return (a.second > b.second);}

bool isPrime(ll n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
bool isPowerOfTwo(ll n){if(n==0)return false;return (ceil(log2(n)) == floor(log2(n)));}
bool isPerfectSquare(ll x){if (x >= 0) {ll sr = sqrt(x);return (sr * sr == x);}return false;}

ll power(ll base, ll exponent) {if (exponent == 0)return 1;long long result = 1;while (exponent > 0) {if (exponent % 2 == 1)result =(result%mod* base%mod)%mod;base =(base%mod* base%mod)%mod;exponent /= 2;}return result%mod;}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll t;
cin>>t;
while(t--){
  ll n,m,q;
  cin>>n>>m>>q;
 vectorll b(m);
 for(int i=0;i<m;i++){
  cin>>b[i];
 }
 sort(b.begin(),b.end());
 vectorll ans;
 while(q--){
  ll q1;
  cin>>q1;
ll a=b[0],b1=b[m-1];
 if(q1<a)ans.push_back(a-1ll);
  else if(q1>b1)ans.push_back(n-b1);
   else {
    ll in2=*upper_bound(b.begin(),b.end(),q1);
    ll in1=in2-1;
    ll mini=b[in1];
    ll maxi=b[in2];
    ll v1=maxi-mini-1;
    if(v1<=2)ans.push_back(1);
    else if(v1>2&&v1%2!=0)ans.push_back(v1/2+1);
    else ans.push_back(v1/2);
   }
 }
for(int i=0;i<ans.size();i++){
  cout<<ans[i]<<endl;
}

}
  return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
using ll = long long;

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

    while(t--){
        int n, m, q;
        cin >> n >> m >> q;

        vector<int> mrr(m);
        vector<int> qrr(q);
        vector<pair<int, int>> prr(q);
        int minm = 1e9;
        int maxm = 0;
        for(int i = 0; i < m; i++) {
            cin >> mrr[i];
            maxm = max(maxm, mrr[i]);
            minm = min(minm, mrr[i]);
        }

        int ind = 0;
        for(int j = 0; j < q; j++) {
            int x;
            cin >> x;
            qrr[j] = x;

            int low = 1;
            int high = n;

            int i = ind;
            while(i < m){
                if(mrr[i] > x){
                    if(mrr[i] <= high){
                        high = mrr[i];
                        break;
                    }
                }
                i++;
            }
            if(i == m) i--;

            while(i >= 0){
                if(mrr[i] < x){
                    if(low >= mrr[i]){
                        low = mrr[i];
                        break;
                    }
                }

                i--;
            }

            ind = i == -1 ? 0 : i;

            prr[j] = {low, high};
        }

        // sort(mrr.begin(), mrr.end());
        // sort(qrr.begin(), qrr.end());

        int ans = 0;

        for(int i = 0; i < q; i++){

            if(qrr[i] > maxm) {
                ans = n - maxm;
            }
            else if(qrr[i] < minm) {
                ans = minm - 1;
            }
            else {
                // int d1 = abs(qrr[i] - prr[i].first);
                // int d2 = abs(qrr[i] - prr[i].second);
                ans = abs(prr[i].first - prr[i].second) / 2;

                // auto it = lower_bound(mrr.begin(), mrr.end(), qrr[i]) - mrr.begin();

                // int d = mrr[it] - mrr[it-1];
                // ans = d/2;

            }

            cout << ans;
            cout << "\n";
        }

        // cout << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 3 3
1 5 9
2 4 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 (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
import bisect

t = int(input())
for _ in range(t):
    a = list(map(int, input().split()))
    n = a[0]
    m = a[1]
    q = a[2]
    z = list(map(int, input().split()))
    y = list(map(int, input().split()))

    # Sort z only once
    z.sort()
    mini = z[0]
    maxi = z[-1]

    for i in range(q):
        if y[i] < mini:
            ans = mini - 1
        elif y[i] > maxi:
            ans = n - maxi
        else:
            # Find the closest element in z using binary search
            idx = bisect.bisect_left(z, y[i])

            # Compute the minimum distance based on the closest elements
            if idx == 0:
                ans = (z[idx] - 1)
            elif idx == m:
                ans = (n - z[idx-1])
            else:
                ans = min((z[idx] - y[i]), (y[i] - z[idx-1]))

        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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has issues with how it handles the edge cases where David's position is outside the range of the teachers' positions or within the range but not directly adjacent to any teacher.

Let's break down the scenarios:

1. **David's position is outside the range of the teachers' positions:**
   - David's position is less than the smallest teacher's position.
   - David's position is greater than the largest teacher's position.

2. **David's position is within the range of the teachers' positions:**
   - David's position is exactly between two teachers.
   - David's position is adjacent to a teacher.

We will create a test input that includes these scenarios.

### Test Input
```plaintext
3
8 1 1
6
3
10 3 3
1 4 8
2 3 10
15 2 2
1 14
7 8
```

### Explanation
1. **First Test Case:**
   - **n = 8, m = 1, q = 1**
   - **Teachers' positions: [6]**
   - **David's position: [3]**
   - **Expected Output: 5**
   - **Reason:** David can run to cell 1, and it will take the teacher 5 moves to reach from cell 6 to cell 1.

2. **Second Test Case:**
   - **n = 10, m = 3, q = 3**
   - **Teachers' positions: [1, 4, 8]**
   - **David's positions: [2, 3, 10]**
   - **Expected Output:**
     - **1** (David at cell 2, closest teacher at cell 1)
     - **1** (David at cell 3, closest teacher at cell 4)
     - **2** (David at cell 10, closest teacher at cell 8)

3. **Third Test Case:**
   - **n = 15, m = 2, q = 2**
   - **Teachers' positions: [1, 14]**
   - **David's positions: [7, 8]**
   - **Expected Output:**
     - **6** (David at cell 7, closest teacher at cell 8)
     - **6** (David at cell 8, closest teacher at cell 7)

This test input covers a range of scenarios including edge cases and positions within the range of the teachers. It should help expose any issues in the buggy code's handling of these 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 (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
t = int(input())  # number of test cases

for _ in range(t):
    n, m, q = map(int, input().split())
    b = list(map(int, input().split()))  # positions of teachers
    a = list(map(int, input().split()))  # positions of David

    if m == 1:
        # Only one teacher: distance to the nearest end
        print(max(n - b[0], b[0] - 1))
    else:
        # More than one teacher
        # Sort teacher positions for easier range calculations
        b.sort()

        for A in a:
            # Initialize closest teachers
            min1 = None  # Closest teacher on the left
            min2 = None  # Closest teacher on the right

            # Find closest teacher to the left of A
            for k in b:
                if k < A:
                    min1 = k
                else:
                    break

            # Find closest teacher to the right of A
            for k in reversed(b):
                if k > A:
                    min2 = k
                else:
                    break

            if min1 is None:
                # David is to the right of all teachers
                print(min2 - A)
            elif min2 is None:
                # David is to the left of all teachers
                print(A - min1)
            else:
                # David is between two closest teachers
                print((min2 - min1) // 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
1
10 2 3
3 7
2 5 8
```

### Explanation:
- **Test Case**: 1
- **Number of Cells (n)**: 10
- **Number of Teachers (m)**: 2
- **Number of Queries (q)**: 3
- **Teacher Positions (b)**: 3, 7
- **David's Positions (a)**: 2, 5, 8

### Scenarios:
1. **David at Position 2**:
   - The closest teacher to the left is at position 3.
   - The distance is \(3 - 2 = 1\).
   - Expected Output: 1

2. **David at Position 5**:
   - David is between the teachers at positions 3 and 7.
   - The distance to the closest teacher on either side is \((7 - 3) // 2 = 2\).
   - Expected Output: 2

3. **David at Position 8**:
   - The closest teacher to the right is at position 7.
   - The distance is \(8 - 7 = 1\).
   - Expected Output: 1

### Why This Test Case?
- **Edge Cases**:
  - David is at the leftmost position relative to the teachers (position 2).
  - David is between two teachers (position 5).
  - David is at the rightmost position relative to the teachers (position 8).
- **Small Scale**: The test case is small enough to be easily understood and verified, but it covers a variety of scenarios to test the correctness of the algorithm.
- **Challenges**:
  - The buggy code may fail to correctly handle the scenario where David is between two teachers, as it calculates the distance to the nearest end instead of the closest teacher.
  - The test case also checks the boundary conditions where David is at the edge of the classroom.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

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

public class B_2_The_Strict_Teacher_Hard_Version {
    public static void main(String[] args) throws IOException {
        // Set up BufferedReader for input and PrintWriter for output
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out), true);

        // Read the number of test cases
        long t = Long.parseLong(br.readLine().trim());

        // Process each test case
        for (int i = 0; i < t; i++) {
            String[] nm = br.readLine().trim().split(" ");
            long n = Long.parseLong(nm[0]);
            int m = Integer.parseInt(nm[1]);
            int q = Integer.parseInt(nm[2]);

            // Read the array elements
            int[] teachers = new int[(int)m];
            String[] arrStr = br.readLine().trim().split(" ");
            for (int j = 0; j < m; j++) {
                teachers[j] = Integer.parseInt(arrStr[j]);
            }
            int[] answers = new int[(int)q];
            arrStr = br.readLine().trim().split(" ");
            for (int j = 0; j < q; j++) {
                answers[j]=Integer.parseInt(arrStr[j]);
            }

            for(int j=0;j<q;j++) pw.println(helper(n, m, teachers, answers[j]));
        }

        // Close the BufferedReader and PrintWriter
        br.close();
        pw.close();
    }

    // Define the helper function
    public static long helper(long n, int m, int[] teachers, int davidCell){
        if(davidCell<teachers[0]) return teachers[0]-1;
        else if(davidCell>teachers[m-1]) return n - teachers[m-1];
        else{
            int iterator = 0;
            for(int i=0;i<m;i++){
                if(teachers[i]>davidCell){
                    iterator=i;
                    break;
                }
            }
            int midVal = (teachers[iterator] + teachers[iterator-1]) / 2;
            return Math.min(Math.abs(midVal - teachers[iterator]), Math.abs(midVal - teachers[iterator-1]));
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
import java.util.*;

public class Demo{
    public static void main(String args[]){
        try (Scanner sc = new Scanner(System.in)) {
            Sorter obj=new Sorter();
            int t,n,m,q,a,b;
            t=sc.nextInt();
            for(int i=0;i<t;i++){
                n=sc.nextInt();
                m=sc.nextInt();
                q=sc.nextInt();
                ArrayList<Integer> arr=new ArrayList<>();
                ArrayList<Integer> brr=new ArrayList<>();
                ArrayList<Integer> qrr=new ArrayList<>();
                for(int j=0;j<m;j++){
                    b=sc.nextInt();
                    brr.add(b);
                    arr.add(b);
                }
                for(int j=0;j<q;j++){
                    a=sc.nextInt();
                    qrr.add(a);
                    arr.add(a);
                }
                obj.sort(arr, 0, arr.size()-1);
                for(int j=0;j<q;j++){
                    int c=obj.binSearch(arr, qrr.get(j));
                    if(c==0){
                        int k=1;
                        while(!brr.contains(arr.get(k))) k++;
                        System.out.println(arr.get(k)-1);
                    }
                    else if(c==arr.size()-1){
                        int k=arr.size()-2;
                        while(!brr.contains(arr.get(k))) k--;
                        System.out.println(n-arr.get(k));
                    }
                    else{
                        int p=c-1;
                        int r=c+1;
                        while(!brr.contains(arr.get(p))){
                            p--;
                            if(p==0) break;
                        }
                        while(!brr.contains(arr.get(r))){
                            r++;
                            if(r==arr.size()-1) break;
                        }
                        int x=(arr.get(p)+arr.get(r))/2;
                        if(x-arr.get(p)>arr.get(r)-x) System.out.println(arr.get(r)-x);
                        else System.out.println(x-arr.get(p));
                    }
                }

            }
        }
    }
}

class Sorter{
    void merge(ArrayList<Integer> arr, int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        // Create temp arrays
        ArrayList<Integer> L=new ArrayList<>();
        ArrayList<Integer> R=new ArrayList<>();

        // Copy data to temp arrays
        for (int i = 0; i < n1; ++i)
            L.add(arr.get(l+i));
        for (int j = 0; j < n2; ++j)
            R.add(arr.get(m+1+j));

        // Merge the temp arrays

        // Initial indices of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarray array
        int k = l;
        while (i < n1 && j < n2) {
            if (L.get(i) <= R.get(j)) {
                arr.set(k, L.get(i));
                i++;
            }
            else {
                arr.set(k, R.get(j));
                j++;
            }
            k++;
        }

        // Copy remaining elements of L[] if any
        while (i < n1) {
            arr.set(k, L.get(i));
            i++;
            k++;
        }

        // Copy remaining elements of R[] if any
        while (j < n2) {
            arr.set(k, R.get(j));
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(ArrayList<Integer> arr, int l, int r)
    {
        if (l < r) {

            // Find the middle point
            int m = l + (r - l) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    int binSearch(ArrayList<Integer> a, int key){
        int low=0, high=a.size() -1;
        while(low<=high){
            int mid=(low+high)/2;
            if(a.get(mid)==key) return mid;
            else if(key>a.get(mid)) low=mid+1;
            else high=mid-1;
        }
        return -1;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code has issues with handling the binary search and the logic for determining the minimum number of moves. Here are the steps to generate a challenging test input:

1. **Edge Case 1: David is at the boundary and teachers are far away**:
   - This can test how the code handles the scenario where David is at the extreme ends of the classroom and the teachers are far away.

2. **Edge Case 2: David is in the middle, and teachers are symmetrically placed**:
   - This can test how the code handles the scenario where David is in the middle, and teachers are symmetrically placed around him.

3. **Edge Case 3: David is in a position that is not the boundary, and teachers are close but not adjacent**:
   - This can test how the code handles the scenario where David is in a position that is not the boundary, and teachers are close but not adjacent.

4. **Edge Case 4: David is in a position where the teachers are already very close**:
   - This can test how the code handles the scenario where David is in a position where the teachers are already very close, and the number of moves should be minimal.

Based on these scenarios, here is a small-scale test input that should expose the bugs:

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

Explanation:
1. **Test Case 1**:
   - Classroom size: 10
   - Number of teachers: 1
   - Number of queries: 1
   - Teacher position: 5
   - David position: 1
   - Expected output: 4 (David can move to cell 5 in 4 moves)

2. **Test Case 2**:
   - Classroom size: 10
   - Number of teachers: 2
   - Number of queries: 2
   - Teacher positions: 3, 8
   - David positions: 2, 7
   - Expected output: 1 (for both positions, David can be caught in 1 move)

3. **Test Case 3**:
   - Classroom size: 10
   - Number of teachers: 3
   - Number of queries: 3
   - Teacher positions: 1, 4, 8
   - David positions: 2, 3, 10
   - Expected output: 1 (for position 2), 1 (for position 3), 2 (for position 10)

This input should help expose any issues in the binary search and the logic for determining the minimum number of moves.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
int main()
{
    int t ; cin>>t; while(t--)
    {
        ll n,m,q; cin>>n>>m>>q;
        vector<ll> a(m),qi(q);
        for(ll i=0;i<m;i++)
        cin>>a[i];
        for(ll i=0;i<q;i++)
        cin>>qi[i];
        sort(a.begin(),a.end());
        for(ll i=0;i<q;i++)
        {
            ll p=qi[i];
            if(p<a[0])
            cout<<a[0]-1<<endl;
            else if(a[m-1]<p)
            cout<<n-a[m-1]<<"\n";
            else{
                ll lo=0,hi=m-1,ans=m;
                while(lo<hi)
                {
                    //cout<<lo<<" "<<hi<<endl;
                    ll mid=lo+(hi-lo)/2;
                    if(a[mid]>p)
                    hi=mid;
                    else
                    lo=mid;
                     if(hi-lo==1) break;
                }
                ll l=p-a[lo],r=a[hi]-p;
                ll maxi=max(l,r),mini=min(l,r);
                 ans=(maxi-mini)/2+mini;
                 if(t==76032-3628)
                 cout<<a[lo]<<endl;
                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 code, we need to focus on edge cases and scenarios that can highlight the faulty logic. The code provided has a binary search to find the closest teachers to David's position. The bug might be in how it calculates the optimal number of moves when David is between two teachers.

Here are the steps to create a challenging test input:

1. **Edge Case with Minimum Classroom Size**: Use the smallest possible classroom size that still fits the constraints.
2. **Multiple Teachers**: Use a small number of teachers to ensure the test case is small but effective.
3. **Queries on Positions**: Place David in positions that are between teachers or at the boundaries to test the edge cases.

Let's create a test input that fits these criteria:

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

### Explanation:
- **Number of Test Cases (t)**: 1
- **Classroom Size (n)**: 3 (smallest possible size that fits the constraints)
- **Number of Teachers (m)**: 2
- **Number of Queries (q)**: 3
- **Teachers' Positions (b)**: 1 and 3
- **Queries (a)**: 2, 1, 3

### Expected Output:
- For the first query (David at position 2):
  - The teachers are at positions 1 and 3.
  - David is in the middle, and the optimal move for the teachers is to move towards David.
  - The answer should be 1, as the teachers can catch David in one move.

- For the second query (David at position 1):
  - One teacher is already at position 1.
  - The answer should be 0, but since the problem states that David and teachers are in distinct cells initially, the answer should be 1.

- For the third query (David at position 3):
  - One teacher is already at position 3.
  - The answer should be 0, but since the problem states that David and teachers are in distinct cells initially, the answer should be 1.

This test input should expose any issues in the binary search logic or the calculation of the optimal number of moves.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

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

int main() {
    int t; cin >> t;
    for (int i=0; i<t; i++) {
        long n;
        int m, q;
        cin >> n >> m >> q;
        int t[m];
        for (int j=0; j<m; j++) {
            cin >> t[j];
        }
        int qr[q];
        for (int j=0; j<q; j++) {
            cin >> qr[j];
        }
        sort(t, t+m);
        int avgT[m+1]; avgT[0] = 0;
        for (int j=1; j<m; j++) {
            avgT[j] = (t[j] - t[j-1])/2;
        }
        avgT[m] = avgT[m-1];
        int grid[n];
        int j = 0, k = 0;
        while (j<n and k<m) {
            if (j+1 == t[k]) {
                grid[j] = avgT[k+1];
                k++;
                j++;
            }
            else {
                grid[j] = avgT[k];
                j++;
            }
        }
        for (int j=0; j<q; j++) {
            if (qr[j] < t[0]) {
                cout << (qr[j] - 1) + (t[0] - qr[j]) << "\n";
            }
            else if (qr[j] > t[m-1]) {
                cout << (n - qr[j]) + (qr[j] - t[m-1]) << "\n";
            }
            else {
                cout << grid[qr[j]-1] << "\n";
            }
        }
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
x = int(input())
for i in range(x):

    y=[int(i) for i in input().split()]
    z=[int(i) for i in input().split()]
    a=[int(i) for i in input().split()]
    for i in a:
        j=i-1
        l=i+1
        neg=0
        pos=0
        while j>0:
            if j in z:
                pos=i-j
                break
            j-=1
        while l<=y[0]:
            if l in z:
                neg=l-i
                break
            l+=1

        if neg==0:
            print(pos+y[0]-i)
        elif pos==0:
            print(neg+i-1)
        else:
            print((pos+neg)//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
1
10 2 2
3 8
1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
from math import ceil

t = int(input())

for i in range(t):
    n, m, q = list(map(int, input().split()))
    classroom = list(range(n+1))
    b = list(map(int, input().split()))
    a = list(map(int, input().split()))

    b.sort()

    for j in a:
        step = 0
        if b[0] < j < b[-1]:
            left, right = 0, len(b) - 1

            while left <= right:
                mid = left + (right - left) // 2

                if b[mid] < j:
                    left = mid + 1
                else:
                    right = mid - 1

            l = b[left - 1]
            r = b[left]
            mid = (l + r) / 2
            step = min(b[1] - ceil(mid), ceil(mid) - b[0])
        elif j < b[0]:
            step = b[0] - 1
        elif j > b[-1]:
            step = n - b[-1]

        print(step)

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

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

Think step by step.

Response:
```plaintext
1
10 3 3
1 4 8
2 3 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();
        while (t-- > 0) {
            int n = scn.nextInt();
            int m = scn.nextInt();
            int q = scn.nextInt();
            TreeSet<Integer> st = new TreeSet<>();
            for (int i = 0; i < m; i++) st.add(scn.nextInt());

            while (q-- > 0) {
                int x = scn.nextInt();

                if (x < st.first()) {
                    System.out.println(st.first() - 1);
                } else if (x > st.last()) {
                    System.out.println(n - st.last());
                } else {
                    int lb = st.lower(x);
                    int ub = st.higher(x);
                    System.out.println((ub-lb)/2);
                }
            }
        }
        scn.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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has issues with calculating the correct number of moves when David is in a position that is not directly adjacent to a teacher but is within a range that can be caught efficiently.

### Step-by-Step Analysis:
1. **Edge Case: David at the Edge of the Classroom:**
   - If David is at the very first or last cell, the code should correctly calculate the distance to the nearest teacher.
2. **Middle Case: David Between Two Teachers:**
   - The code calculates the number of moves as `(ub - lb) / 2`, which might be incorrect if the teachers are not optimally positioned.
3. **Multiple Queries:**
   - Ensure the test case includes multiple queries to test the code's consistency and correctness across different scenarios.

### Challenging Test Input:
Let's create a test case where David is in a position that is not directly adjacent to a teacher but is within a range that can be caught efficiently. We will also include edge cases for David being at the start or end of the classroom.

```plaintext
1
10 3 3
1 4 8
2 3 10
```

### Explanation:
- **Test Case:**
  - **n = 10**: The classroom has 10 cells.
  - **m = 3**: There are 3 teachers.
  - **q = 3**: There are 3 queries.
  - **Teachers' positions: 1, 4, 8**
  - **David's positions: 2, 3, 10**

### Expected Output:
- **For David at cell 2:**
  - The nearest teacher is at cell 1. David can be caught in 1 move.
- **For David at cell 3:**
  - The nearest teacher is at cell 4. David can be caught in 1 move.
- **For David at cell 10:**
  - The nearest teacher is at cell 8. David can be caught in 2 moves.

### Correct Output:
```
1
1
2
```

This test case should expose the bug in the provided code, particularly in the calculation of the number of moves when David is not directly adjacent to a teacher but is within a range that can be caught efficiently.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only differences between the two versions are the constraints on $m$ and $q$. In this version, $m, q \le 10^5$. 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.

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 $q$ ($3 \le n \le 10^9$, $1 \le m, q \le 10^5$) — 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 $1 \le j \le q$, $b_i \neq a_j$.

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

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

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

Examples:
input:
2
8 1 1
6
3
10 3 3
1 4 8
2 3 10
output:
5
1
1
2

Note: In the only query of the first example, the student can run to cell $1$. It will take the teacher five moves to reach from cell $6$ to cell $1$, so the answer is $5$.

In the second query of the second example, the student can just stay at cell $3$. The teacher, initially located in cell $4$, can reach cell $3$ in one move. Therefore, the answer is $1$.

Buggy Code:
// "static void main" must be defined in a public class.
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();
        while (t-- > 0) {
            int n = scn.nextInt();
            int m = scn.nextInt();
            int q = scn.nextInt();
            TreeSet<Integer> ts = new TreeSet<>();
            int leftMostT = Integer.MAX_VALUE;
            int rightMostT = Integer.MIN_VALUE;
            for (int i = 0; i < m; i++) {
                int temp = scn.nextInt();
                ts.add(temp);
                leftMostT = Math.min(leftMostT , temp);
                rightMostT = Math.max(rightMostT , temp);
            }
            while (q-- > 0) {
                int davidPos = scn.nextInt();
                Integer teacherCell1 = ts.floor(davidPos);
                Integer teacherCell2 = ts.ceiling(davidPos);

                if(teacherCell1 != null && teacherCell2 != null){
                    int distance = davidPos - teacherCell1;
                    int distance2 = teacherCell2 - davidPos;
                    int ans = (distance + distance2) / 2;
                    System.out.println(ans);
                }
                else if(teacherCell1 == null){
                    System.out.println(leftMostT - 1);
                }
                else{
                    System.out.println(n - rightMostT);
                }

//                 int minMoves = Integer.MAX_VALUE;
//                 if (teacherCell1 != null) {
//                     minMoves = Math.min(minMoves, davidPos - teacherCell1);
//                 }
//                 if (teacherCell2 != null) {
//                     minMoves = Math.min(minMoves, teacherCell2 - davidPos);
//                 }

//                 System.out.println(minMoves);
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code might fail. The main issue in the buggy code is that it incorrectly calculates the number of moves required for the teachers to catch David. Specifically, it calculates the average distance between the closest teachers on either side of David, which is not the correct approach.

Here are the steps to create a challenging test input:

1. **Edge Case with Single Teacher**: Test a scenario where there is only one teacher. The code should correctly handle this by calculating the distance from David to the single teacher.

2. **Edge Case with Multiple Teachers**: Test a scenario where there are multiple teachers, and David is positioned in a way that the teachers are not directly adjacent to him but still within a short distance.

3. **Edge Case with David at the Ends**: Test scenarios where David is at the ends of the classroom (i.e., cell 1 or cell n). The code should handle these cases by correctly calculating the distance from the closest teacher.

4. **Edge Case with Large Classroom**: Test a scenario with a large classroom to ensure the code can handle large values of \( n \) and \( m \).

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

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

Explanation of the test input:
1. **First Test Case**:
   - \( n = 8 \), \( m = 1 \), \( q = 1 \)
   - Teacher position: 6
   - David's position: 3
   - Expected output: 3 (The teacher at position 6 will take 3 moves to reach David at position 3).

2. **Second Test Case**:
   - \( n = 10 \), \( m = 3 \), \( q = 3 \)
   - Teacher positions: 1, 4, 8
   - David's positions: 2, 3, 10
   - Expected outputs:
     - For David at position 2: 1 (The teacher at position 1 will take 1 move to reach David at position 2).
     - For David at position 3: 1 (The teacher at position 4 will take 1 move to reach David at position 3).
     - For David at position 10: 2 (The teacher at position 8 will take 2 moves to reach David at position 10).

This test input should help identify the bug in the provided code, as it covers various edge cases and scenarios where the incorrect distance calculation might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $b \cdot \gcd(a,b)$ is a multiple of $a+b$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
int gcd(int a, int b){
    if(b == 0) return a;
    return gcd(b, a%b);
}
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(18);
    int t;
    cin>>t;
    while(t--){

        int n, m;
        cin>>n>>m;
        int res = 0;

        for(int c = 1; c <= m + 2; c++){
            for(int g = c; g <= m; g += c){
                for(int b = g; b <= m; b += g){

                    int b0 = b / g;
                    int a0 = g / c - b0;
                    if(a0 <= 0)break;

                    //if(gcd(a0, b0) != 1)continue;
                    if(a0 * g <= n and gcd(a0, b0) == 1)res++;

                }
            }
        }
        cout<<res<<'\n';
    }
}

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

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

Think step by step.

Response:
```plaintext
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $b \cdot \gcd(a,b)$ is a multiple of $a+b$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$ satisfy the conditions.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int64_t mx = 1e15;
void test() {
    int n, m;
    cin >> n >> m;
    int64_t res = 0;
    for (int a = 1; a * a <= n; a++) {
        for (int b = 1; b * b <= m; b++) {
            int d = gcd(a, b);
            if (d > 1)
                continue;
            for (int d = a + b; d <= min(n, m); d += (a + b)) {
                if (a * d <= n && b * d <= m)
                    res++;
            }
        }
    }
    cout << res;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        test();
        cout << "\n";
    }
}

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

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

Think step by step.

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

1. **Small Values of \( n \) and \( m \)**: The code might fail for small values of \( n \) and \( m \) where the conditions are more likely to be met.
2. **Equal Values of \( n \) and \( m \)**: The code might have issues when \( n \) and \( m \) are equal.
3. **Special Cases**: Values where \( \gcd(a, b) \) is 1 or where \( a + b \) is a divisor of \( a \cdot b \).

Given these considerations, let's create a small-scale test input:

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

This input includes:
- The simplest case (1, 1) where no pairs satisfy the conditions.
- A case (2, 2) where there is one valid pair.
- A case (3, 3) where there is one valid pair.

This should help in identifying the bug in the code by testing various small values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $b \cdot \gcd(a,b)$ is a multiple of $a+b$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$ satisfy the conditions.

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

typedef long long ll;
typedef vector<vector<int>> Graph;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define FOR(i,l,r) for (int i = l;i < (int)(r); i++)
#define rep(i,n) for (int i = 0;i < (int)(n); i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define my_sort(x) sort(x.begin(), x.end())
#define my_max(x) *max_element(all(x))
#define my_min(x) *min_element(all(x))
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = (1<<30) - 1;
const ll LINF = (1LL<<62) - 1;
const int MOD = 998244353;
const int MOD2 = 1e9+7;
const double PI = acos(-1);
vector<int> di = {1,0,-1,0};
vector<int> dj = {0,1,0,-1};

#ifdef LOCAL
#  include <debug_print.hpp>
#  define debug(...) debug_print::multi_print(#__VA_ARGS__, __VA_ARGS__)
#else
#  define debug(...) (static_cast<void>(0))
#endif

void solve(){
    ll n, m; cin >> n >> m;
    if(n > m) swap(n, m);
    ll ans = 0LL;

    unordered_set<ll> st;
    for(ll g = 1; g <= n; g++){
        for(ll c = 1; c <= (n + m) / g; c++){
            if(g % c) continue;
            if(st.find(g * c) != st.end()) continue;
            st.insert(g * c);

            ll tmp = min(n / g, c - 1) - max(1LL, c - m / g) + 1;
            ans += max(0LL, tmp);
        }
    }
    cout << ans << endl;
}

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios where the code's logic might fail. Let's break down the problem and the code to identify potential issues:

### Problem Analysis
The problem requires calculating the number of ordered pairs \((a, b)\) such that:
1. \(1 \le a \le n\)
2. \(1 \le b \le m\)
3. \(b \cdot \gcd(a, b)\) is a multiple of \(a + b\)

### Code Analysis
The code uses a nested loop to iterate over possible values of \(g\) (the greatest common divisor) and \(c\) (a coefficient). It then calculates the number of valid pairs using the formula:
\[ \text{tmp} = \min\left(\frac{n}{g}, c - 1\right) - \max(1, c - \frac{m}{g}) + 1 \]

### Potential Issues
1. **Edge Cases**: Values of \(n\) and \(m\) that are very small or very large.
2. **Boundary Conditions**: Values of \(a\) and \(b\) that are at the boundaries of the ranges.
3. **Special GCD Values**: Values of \(g\) that might cause unexpected behavior in the loop.

### Test Input
To expose the bug, we can use a small-scale test case that includes values where the code might fail due to boundary conditions or special GCD values. Let's consider the following test case:

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

### Explanation
1. **Test Case 1: \(n = 1, m = 1\)**:
   - No pairs \((a, b)\) satisfy the conditions.
   - Expected output: 0

2. **Test Case 2: \(n = 2, m = 2\)**:
   - The pairs \((2, 2)\) satisfy the conditions.
   - Expected output: 1

3. **Test Case 3: \(n = 3, m = 3\)**:
   - The pairs \((3, 3)\) satisfy the conditions.
   - Expected output: 1

This test input is small and complete, and it includes edge cases that can help identify potential issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $b \cdot \gcd(a,b)$ is a multiple of $a+b$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$ satisfy the conditions.

Buggy Code:
def gcd(a, b):
  while a != 0:
    b %= a
    if b == 0: return a
    a %= b
  return b

import sys
input = lambda: sys.stdin.readline().strip()

def solve():
  n,m = map(int,input().split())
  ans = 0
  for x in range(1,min(n,m)+1):
    if x*x > n + m:
      break
    for p in range(1,min(n,m)+1):
      g = p*x
      if g > min(n,m):
        break
      for a in range(1,n//g + 1):
        b = x - a
        if g*b > m:
          continue
        if b <= 0:
          break
        if gcd(a,b) == 1:
          ans += 1
  print(ans)
  return

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

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

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

Think step by step.

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