**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
import java.util.*;

public class GameSimulation {

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

        // Reading the number of test cases
        int t = sc.nextInt();

        // Looping through each test case
        for (int i = 0; i < t; i++) {
            // Reading the length of the sequence
            int n = sc.nextInt();

            // Reading the sequence elements
            int[] a = new int[n];
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt();
            }

            // Variable to alternate turns (0 -> Turtle's turn, 1 -> Piggy's turn)
            int turn = 0;

            // Game simulation
            while (a.length > 1) {
                if (turn == 0) { // Turtle's turn (maximize a1)
                    // Turtle selects the maximum of a[0] and a[1]
                    int[] newArr = new int[a.length - 1];
                    newArr[0] = Math.max(a[0], a[1]);
                    for (int j = 2; j < a.length; j++) {
                        newArr[j - 1] = a[j];
                    }
                    a = newArr;
                } else { // Piggy's turn (minimize a1)
                    // Piggy selects the minimum of a[0] and a[1]
                    int[] newArr = new int[a.length - 1];
                    newArr[0] = Math.min(a[0], a[1]);
                    for (int j = 2; j < a.length; j++) {
                        newArr[j - 1] = a[j];
                    }
                    a = newArr;
                }

                // Alternate turns
                turn = 1 - turn;
            }

            // The game ends with the value of a[0]
            System.out.println(a[0]);
        }

        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class Sol2 {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            ArrayList<Integer> l=new ArrayList<>();
            for(int i=0;i<n;i++){
                int val=sc.nextInt();
                l.add(val);
            }
            boolean tur=true;
            int len=n;
            while(len>1){
                if(tur){
                    int i=0;
                    while(i<len-1 && l.get(i)>=l.get(i+1)){
                        i++;
                    }
                    if(i==len-1){
                        l.remove(len-1);
                    }
                    else{
                        l.add(i,Math.max(l.get(i),l.get(i+1)));
                        l.remove(i+1);
                        l.remove(i+1);
                    }

                    tur=false;
                }
                else{
                    int i=0;
                    while(i<len-1 && l.get(i)<=l.get(i+1)){
                        i++;
                    }
                    if(i==len-1){
                        l.remove(len-1);
                    }
                    else{
                        l.add(i,Math.min(l.get(i),l.get(i+1)));
                        l.remove(i+1);
                        l.remove(i+1);
                    }

                    tur=true;
                }
                len--;
            }
            System.out.println(l.get(0));
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
def final_value(a):
    turn = 0  # 0 for Turtle's turn, 1 for Piggy's turn
    while len(a) > 1:
        if turn % 2 == 0:  # Turtle's turn
            for i in range(len(a) - 1):
                if a[i] < a[i + 1]:
                    a[i] = a[i + 1]
                a.pop(i + 1)
                break
        else:  # Piggy's turn
            for i in range(len(a) - 1):
                if a[i] > a[i + 1]:
                    a[i] = a[i + 1]
                a.pop(i + 1)
                break
        turn += 1
    return a[0]

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    print(final_value(a))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
import sys
# import math
# import heapq as hq
# from collections import defaultdict
# from collections import deque
def main():
    n = ii()
    a = vi()
    if n == 2:
        print(max(a))
        return

    if not n &1 :
        print(a[n//2 - 1])
    else:
        print(a[n//2])
def input(): return sys.stdin.readline().rstrip('\r\n')
def ii():return int(input())
def vi():return list(map(int,input().split()))
# def vvi(rows):return list(vi() for _ in range(rows))
# def vs():return list(i for i in input())
# def vvs(rows):return list(vs() for _ in range(rows))
for _ in range(ii()):
    main()

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

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

Think step by step.

Response:
```plaintext
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
import java.util.Scanner;

public class Problem_B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = Integer.parseInt(in.nextLine());
        while (t-- > 0)
            validString(in);
    }

    static void validString(Scanner in) {
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        if (n %2 !=0)
            System.out.println(a[n/2]);
        else if (n == 2)
            System.out.println(a[1]);
        else System.out.println(a[(n/2) - 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
5
2
1 2
3
1 1 2
4
3 1 2 3
6
3 1 2 2 3 4
8
10 2 5 2 7 9 2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
import java.util.Objects;
import java.util.Set;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

@SuppressWarnings("DuplicatedCode")
public class Solution {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

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

        List<Integer> nextIntList(int n) {
            List<Integer> list = new ArrayList<>();
            for (int i=0; i<n; i++) {
                list.add(nextInt());
            }
            return list;
        }
        List<Long> nextLongList(int n) {
            List<Long> list = new ArrayList<>();
            for (int i=0; i<n; i++) {
                list.add(nextLong());
            }
            return list;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    static final int MOD = (int) 998244353;

    private static long[] sort(long[] arr, boolean ascending) {
        List<Long> list = new ArrayList<>();
        for (long num : arr) {
            list.add(num);
        }

        if (ascending) {
            Collections.sort(list);
        } else {
            list.sort(Comparator.reverseOrder());
        }

        arr = list.stream().mapToLong(Long::longValue).toArray();
        return arr;
    }

//    private static final int N = (int) 1e6+5;
//    private static List<Boolean> sieveList = new ArrayList<>(Collections.nCopies(N, true));
//
//    private static List<Integer> primes = new ArrayList<>();
//    private static List<Integer> spf = new ArrayList<>(Collections.nCopies(N, -1));
//
//
//    private static void sieve() {
//        sieveList.set(0, false);
//        sieveList.set(1, false);
//
//        for (int i = 2; (long) i * i < N; i++) {
//            if (sieveList.get(i)) {
//                for (int j = i*i; j<N; j+=i) {
//                    sieveList.set(j, false);
//                    if (spf.get(j).equals(-1)) {
//                        spf.set(j, i);
//                    }
//                }
//            }
//        }
////        for (int i=2; i<Math.min((int) 3e5+1, N); i++) {
////            if (sieveList.get(i)) {
////                primes.add(i);
////            }
////        }
//    }

    private static class Pair {
        int first;
        int second;

        public Pair(int a, int b) {
            this.first = a;
            this.second = b;
        }
    }

    public static void main(String[] args) {
        //         new Thread(null, new Solution(), "whatever", (long) 1 << 31).
        //                     start();

        try {
            FastReader cin = new FastReader();
            FastWriter cout = new FastWriter();

            int t = cin.nextInt();

            while (t-- > 0) {
                int n = cin.nextInt();
                List<Integer> list = new ArrayList<>();

                for (int i=0; i<n; i++) {
                    list.add(cin.nextInt());
                }

                int max = 0;
                int min = (int) 1e6;

                for (int i=0; i<n; i++) {
                    min = Math.min(min, list.get(i));
                }
                for (int i=0; i<n; i++) {
                    max = Math.max(max, list.get(i));
                }

                if ((n & 1) == 1) {
                    int tempMin = (int) 1e6;
                    for (int i = 1; i < n; i++) {
                        if (list.get(i).equals(max) && i - 1 >= 0) {
                            tempMin = Math.min(tempMin, list.get(i - 1));
                        }
                    }
                    cout.println(tempMin);
                } else {
                    if (n == 2) {
                        cout.println(max);
                    } else {
                        int tempMax = 0;
                        for (int i = 0; i < n; i++) {
                            if (list.get(i).equals(min) && i + 1 < n) {
                                tempMax = Math.max(tempMax, list.get(i + 1));
                            }
                        }
                        cout.println(tempMax);
                    }
                }
            }

            cout.close();
        } catch (Exception e) {
            e.printStackTrace();
            return;
        }
    }

//    @Override
//    public void run() {
//        try {
//            FastReader cin = new FastReader();
//            FastWriter cout = new FastWriter();
//
//            cout.close();
//        } catch (Exception e) {
//            e.printStackTrace();
//            return;
//        }
//    }
}

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

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

Think step by step.

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

1. **Odd and Even Length Handling**: The code handles odd and even lengths differently. It's crucial to test both cases.
2. **Max and Min Values**: The code uses `max` and `min` values to determine the final output, but it might not correctly handle sequences where the max or min values are not in specific positions.
3. **Special Cases**: Sequences with all elements being the same, sequences with alternating max and min values, and sequences with specific patterns that might confuse the logic.

### Step-by-Step Analysis

1. **Odd Length Test Case**:
   - The code checks for the minimum value adjacent to the maximum value when the length is odd.
   - A sequence where the maximum value is in the middle and has different values around it can be challenging.

2. **Even Length Test Case**:
   - The code checks for the maximum value adjacent to the minimum value when the length is even.
   - A sequence where the minimum value is in the middle and has different values around it can be challenging.

3. **Special Cases**:
   - All elements being the same.
   - Alternating max and min values.
   - Sequences with a specific pattern that might confuse the logic.

### Challenging Test Input

Let's create a test input that includes these edge cases:

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

### Explanation

1. **Test Case 1**: `3 1 2 3`
   - Length is odd.
   - Maximum value is 3, and the minimum adjacent value is 2.

2. **Test Case 2**: `4 3 2 1 4`
   - Length is even.
   - Minimum value is 1, and the maximum adjacent value is 4.

3. **Test Case 3**: `5 1 1 2 2 3`
   - Length is odd.
   - Maximum value is 3, and the minimum adjacent value is 2.

4. **Test Case 4**: `6 1 2 3 2 1 4`
   - Length is even.
   - Minimum value is 1, and the maximum adjacent value is 4.

5. **Test Case 5**: `7 1 1 1 1 1 1 1`
   - All elements are the same.
   - Length is odd.
   - Maximum and minimum values are the same.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

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

using namespace std;

#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif

template <typename T, typename SizeT>
vector<T> readVector(SizeT n) {
  vector<T> res(n);
  for (SizeT i = 0; i < n; ++i) {
    cin >> res[i];
  }
  return res;
}

template <typename T, typename SizeT>
vector<T> prefixSum(vector<T> v) {
  vector<T> res(v.size() + 1);
  for (SizeT i = 0; i < v.size(); ++i) {
      res[i + 1] = res[i] + v[i];
  }
  return res;
}

template <typename T>
T readNum() {
  T n;
  cin >> n;
  return n;
}

void SolutionA() {
  auto n = readNum<int64_t>();
  auto s = readNum<std::string>();
  char c = s[s.length() - 1];
  bool isGood = false;

  if (s[0] != c) {
    std::cout << "yes" << std::endl;
  } else {
    std::cout << "no" << std::endl;
  }

  // auto p = prefixSum(x);
  // debug(x);
}

std::vector<int64_t> x(10);

void SolutionB() {
  auto n = readNum<int64_t>();
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
  }

  int l = 0;
  int r = n;
  int findEl = n / 2;
  debug(findEl);

  while (l < r - 1) {
      int el = x[l + std::rand() % (r - l)];
      debug(el);
      auto p = std::partition (x.begin() + l, x.begin() + r, [el](int i) { return i <= el; });
      debug(l);
      debug(r);
      debug(x);
      bool allEq = true;
      for (int i = l; i < r; ++i) {
          if (x[i] != el) {
              allEq = false;
              break;
          }
      }
      if (allEq) {
          break;
      }
      auto newP = p - x.begin();
      debug(newP);
      if (newP <= findEl) {
            l = newP;
      } else {
            r = newP;
      }
  }

  std::cout << x[findEl] << std::endl;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);

  int tt;
  cin >> tt;
  while (tt--) {
    // SolutionA();
    debug(tt);
    SolutionB();
    // SolutionC();
    // SolutionD();
    // SolutionE();
    // SolutionF();
}
}

**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
2
1 2
3
2 1 1
4
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game 2
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a game on a sequence. They are given a sequence $a_1, a_2, \ldots, a_n$, and Turtle goes first. Turtle and Piggy alternate in turns (so, Turtle does the first turn, Piggy does the second, Turtle does the third, etc.).

The game goes as follows:

  * Let the current length of the sequence be $m$. If $m = 1$, the game ends.    * If the game does not end and it's Turtle's turn, then Turtle must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\max(a_i, a_{i + 1})$, and remove $a_{i + 1}$.    * If the game does not end and it's Piggy's turn, then Piggy must choose an integer $i$ such that $1 \le i \le m - 1$, set $a_i$ to $\min(a_i, a_{i + 1})$, and remove $a_{i + 1}$.

Turtle wants to maximize the value of $a_1$ in the end, while Piggy wants to minimize the value of $a_1$ in the end. Find the value of $a_1$ in the end if both players play optimally.

You can refer to notes for further clarification.

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 a single integer $n$ ($2 \le n \le 10^5$) — the length of the sequence.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the elements of the sequence $a$.

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

Output: For each test case, output a single integer — the value of $a_1$ in the end if both players play optimally.

Examples:
input:
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
output:
2
1
2
2
7

Note: In the first test case, initially $a = [1, 2]$. Turtle can only choose $i = 1$. Then he will set $a_1$ to $\max(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ becomes $[2]$. Then the length of the sequence becomes $1$, and the game will end. The value of $a_1$ is $2$, so you should output $2$.

In the second test case, one of the possible game processes is as follows:

  * Initially $a = [1, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[1, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 1$ and remove $a_2$. The sequence $a$ will become $[1]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $1$ in the end.

In the fourth test case, one of the possible game processes is as follows:

  * Initially $a = [3, 1, 2, 2, 3]$.    * Turtle can choose $i = 4$. Then he will set $a_4$ to $\max(a_4, a_5) = 3$ and remove $a_5$. The sequence $a$ will become $[3, 1, 2, 3]$.    * Piggy can choose $i = 3$. Then he will set $a_3$ to $\min(a_3, a_4) = 2$ and remove $a_4$. The sequence $a$ will become $[3, 1, 2]$.    * Turtle can choose $i = 2$. Then he will set $a_2$ to $\max(a_2, a_3) = 2$ and remove $a_3$. The sequence $a$ will become $[3, 2]$.    * Piggy can choose $i = 1$. Then he will set $a_1$ to $\min(a_1, a_2) = 2$ and remove $a_2$. The sequence $a$ will become $[2]$.    * The length of the sequence becomes $1$, so the game will end. The value of $a_1$ will be $2$ in the end.

Buggy Code:
#include <iostream>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <math.h>
#include <limits.h>
#include <string.h>
#include <algorithm>
#include <queue>

using namespace std;

int findLow(int i,vector<int>arr,vector<bool>used){
  while(i>=0){
    if(!used[i]){return i;}
    i--;
  }
  return -1;
}

int findHigh(int i,vector<int>arr,vector<bool>used){
  while(i<arr.size()){
    if(!used[i]){return i;}
    i++;
  }
  return -1;
}
int ans;
void poss(priority_queue<int>q,priority_queue<int,vector<int>,greater<int>>q1,unordered_map<int,vector<int>>rec,vector<bool>used,bool chance,int a,vector<int>arr){
  if(a==1){
    for(int i=0;i<used.size();i++){
      if(!used[i]){ans=i;}
    }
    return;
  }
  if(!chance){
    int minEle = INT_MAX,minInd=-1;
    while(!q.empty()){
      auto k = rec.find(q.top());
      if(k!=rec.end() && k->second.size()!=0){
        for(int i:k->second){
          if(!used[i]){
            int low = findLow(i-1,arr,used);
            int high = findHigh(i+1,arr,used);
            if(low!=-1){
              if(minEle>arr[low]){
                minInd = i;
                minEle = arr[low];
              }
            }
            if(high!=-1){
              if(minEle>arr[high]){
                minInd = i;
                minEle = arr[high];
              }
            }
          }
        }
        if(minInd !=-1){
          used[minInd] = true;
          q.pop();
          poss(q,q1,rec,used,!chance,a-1,arr);
          break;
        }
        else{
          q.pop();
        }
      }
      else{q.pop();}
    }
  }
  else{
    int minEle = INT_MIN,minInd=-1;
    while(!q1.empty()){
      auto k = rec.find(q1.top());
      if(k!=rec.end() && k->second.size()!=0){
        for(int i:k->second){
          if(!used[i]){
            int low = findLow(i-1,arr,used);
            int high = findHigh(i+1,arr,used);
            if(low!=-1){
              if(minEle<arr[low]){
                minInd = i;
                minEle = arr[low];
              }
            }
            if(high!=-1){
              if(minEle<arr[high]){
                minInd = i;
                minEle = arr[high];
              }
            }
          }
          }

        if(minInd !=-1){

          used[minInd] = true;
          q1.pop();
          poss(q,q1,rec,used,!chance,a-1,arr);
          break;
        }
        else{
          q1.pop();
        }
      }
      else{q.pop();}
    }
  }
  return;
}

int main() {

    int n;
    cin>>n;
    while(n>0){
      int a;
      cin>>a;
      vector<int>arr(a);
      for(int i=0;i<a;i++){
        cin>>arr[i];
      }
      unordered_map<int,vector<int>>rec;
      priority_queue<int>q;
      priority_queue<int,vector<int>,greater<int>>q1;
      for(int i=0;i<arr.size();i++){
        q.push(arr[i]);
        q1.push(arr[i]);
        rec[arr[i]].push_back(i);
      }
      vector<bool>used(a,false);
      poss(q,q1,rec,used,true,a,arr);
      cout<<arr[ans]<<endl;
      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
5
2
1 2
3
1 1 2
3
1 2 3
5
3 1 2 2 3
10
10 2 5 2 7 9 2 5 10 7
```


**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:
//and in that light, I find deliverance
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

//defines
#define int long long
#define debug(x) cerr << "(" << #x << "=" << x << "," << __LINE__ << ")\n";
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);

//constants
const int dx[4]{1, 0, -1, 0}, dy[4]{0, 1, 0, -1};
const char dir[4]{'D','R','U','L'};
const int MOD=1e9+7;
const int maxn=2e5+5;
const double eps=1e-9;

//typedefs
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<string> vs;

//Template
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

//Mods
int add(int a, int b)
{
    if((a+=b)>=MOD)
    {
        a-=MOD;
    }
    return a;
}
int sub(int a, int b)
{
  if((a-=b)<0)
  {
        a+=MOD;
  }
  return a;
}
int mul(int a,int b)
{
    int ret=(a%MOD*b%MOD)%MOD;
    if(ret<0)
    {
        ret+=MOD;
    }
    return ret;
}

void solve()
{
    int n,m,q;
    cin >> n >> m >> q;
    vector<int> b(m);
    for(int i=0;i<m;i++)
    {
        cin >> b[i];
    }
    sort(all(b));
    while(q--)
    {
        int pos;
        cin >> pos;
        int lb=lower_bound(all(b),pos)-b.begin();
        if(lb==m)
        {
            cout << n-b[lb-1];
        }
        else if(lb==0)
        {
            cout << b[lb]-1;
        }
        else
        {
            cout << min(abs(pos-b[lb]),abs(pos-b[lb-1]));
        }
        cout << '\n';
    }
}
signed main()
{
    fastio();
    int t;
    cin >> t;
    while(t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code provided seems to handle the basic cases correctly, but there are a few potential issues that can arise in more complex or edge scenarios. Here are the steps to generate a challenging test input:

1. **Edge Cases:**
   - When David is at the boundary of the classroom (i.e., cell 1 or cell n).
   - When the teachers are at the boundary of the classroom.
   - When David is in the middle of the classroom, but the teachers are spread out in a way that makes it difficult for them to catch him quickly.

2. **Complex Scenarios:**
   - When there are multiple teachers and David is in a position where the optimal strategy for the teachers is not straightforward.
   - When the number of teachers is large, and the classroom size is relatively small, making it difficult for David to avoid being caught.

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

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

### Explanation
1. **First Test Case:**
   - **n = 8, m = 1, q = 1**
   - **b = [6]**
   - **a = [3]**
   - **Expected Output: 5**
   - **Explanation:** David is at cell 3, and the teacher is at cell 6. The teacher needs 5 moves to reach cell 3, so the answer is 5.

2. **Second Test Case:**
   - **n = 10, m = 3, q = 3**
   - **b = [1, 4, 8]**
   - **a = [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)

### Challenging Scenarios
- **Edge Case 1:** David at the boundary (cell 1) and a teacher at the other boundary (cell 8). This tests the code's ability to handle boundary conditions.
- **Edge Case 2:** Multiple teachers and David in the middle, but not directly next to any teacher. This tests the code's ability to handle multiple teachers and find the minimum distance correctly.
- **Edge Case 3:** David at the far end (cell 10) with a teacher close to the other end (cell 8). This tests the code's ability to handle large distances and ensure the correct minimum distance is calculated.

This test input should help expose any bugs in the code by covering a variety of scenarios, including edge cases and complex teacher positions.


**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 <iostream>
using namespace std;
int index(int arr[], int left, int right, int a){
    if(right - left <= 1){
        return left;
    }
    int mid = (left + right) / 2;
    if(a > arr[mid]){
        return index(arr, mid, right, a);
    }
    else return index(arr, left, mid, a);
}
int main(){
     int t; cin >> t;
     while(t){
         t--;
         int n,m,q;
         cin >> n >> m >> q;
         int b[m], a[q], tarberutyun[m - 1];
         cin >> b[0];
         for(int i = 1; i < m; i++){
             cin >> b[i];
             tarberutyun[i - 1] = b[i] - b[i - 1];
         }
         for(int i = 0; i < q; i++){
             cin >> a[i];
             if(a[i] < b[0]){
                 cout << b[0] - 1 << endl;
             }
             else if(a[i] > b[m - 1]){
                 cout << n - b[m - 1] << endl;
             }
             else{
               int ind = index(b,0,m -1,a[i]);
               cout << (b[ind + 1] - b[ind]) / 2 << endl;
             }
         }

     }
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - When David is very close to the teachers.
   - When David is at the boundary of the classroom (cell 1 or cell n).
   - When there are multiple teachers with small gaps between them.
   - When the teachers are spread out uniformly.

2. **Buggy Code Analysis:**
   - The function `index` is used to find the position of David relative to the teachers. It might fail if the binary search logic is incorrect.
   - The logic for calculating the number of moves when David is between two teachers is using `(b[ind + 1] - b[ind]) / 2`, which might not always be correct.

3. **Challenging Test Case:**
   - **Test Case 1:**
     - David is at the boundary (cell 1 or cell n).
     - David is very close to one of the teachers.
     - David is between two teachers with a small gap.

Let's create a test case that includes these scenarios:

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

**Explanation:**
- **Test Case 1:**
  - Classroom size: 10
  - Number of teachers: 3
  - Teachers' positions: 1, 4, 8
  - Queries for David's positions: 2, 3, 10

**Expected Output:**
- For David at position 2: The closest teacher is at position 1, so it will take 1 move for the teacher to catch David.
- For David at position 3: The closest teacher is at position 4, so it will take 1 move for the teacher to catch David.
- For David at position 10: The closest teacher is at position 8, so it will take 2 moves for the teacher to catch David.

This test case should expose the bug in the binary search logic and the calculation of moves when David is between two teachers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (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;
int n,b[100005];
void process()
{
 int n,q,m;cin>>n>>m>>q;
 for(int i=1;i<=m;i++) cin>>b[i];
 sort(b+1,b+m+1);
 for(int i=1;i<=q;i++)
 {
     int x;cin>>x;
     for(int j=1;j<=m;j++)
    if(x==b[j])
     {
         cout<<"0\n";continue;
     }
     if(x>b[m])
     {
         cout<<n-b[q]<<"\n";continue;
     }
     if(x<b[1])
     {
         cout<<b[1]-1<<"\n";continue;
     }
     int vt=upper_bound(b+1,b+m+1,x)-b;
     int u=b[vt-1];
     int v=b[vt];
     cout<<(v-u)/2<<"\n";
 }
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int t;cin>>t;
    while(t--)
    {
        process();
    }
}

**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
```

**Explanation:**
- This test input is designed to expose the bug in the given buggy code.
- The test case is small and valid, ensuring it fits within the constraints.
- The positions of the teachers and David are chosen to highlight the incorrect logic in the code:
  - For David at position 2, the closest teacher is at position 1. The correct answer should be 1 (since the teacher can move to position 2 in one move).
  - For David at position 3, the closest teacher is at position 4. The correct answer should be 1 (since the teacher can move to position 3 in one move).
  - For David at position 10, the closest teacher is at position 8. The correct answer should be 2 (since the teacher can move to position 10 in two moves).

The buggy code has issues with handling edge cases and calculating the correct number of moves, which this test input is designed to expose.


**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:
def nearest_ints(arr,val):
    l=0
    r=len(arr)-1
    lmax=None
    rmin=None
    while l<=r:
        mid=(l+r)//2
        if val==arr[mid]:
            return arr[mid],arr[mid]
        if val<arr[mid]:
            rmin=arr[mid]
            r=mid-1
        else:
            lmax=arr[mid]
            l=mid+1
    return lmax,rmin
t=int(input())
for i in range(t):
    n,m,q=map(int,input().split())
    teacher=list(map(int,input().split()))
    david=list(map(int,input().split()))
    for j in range(q):
        d=david[j]
        lmax,rmin=nearest_ints(teacher,d)
        if lmax==None:
            print(rmin-1)
        elif rmin==None:
            print(n-lmax)
        else:
            dl=d-lmax
            dr=rmin-d
            print((max(dl,dr)+min(dl,dr))//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:
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 might cause the code to fail. The code has a function `nearest_ints` that finds the closest values in the array of teachers' positions to a given value. The main issues to consider are:

1. **Edge Cases:**
   - When the student is at the very start or end of the line.
   - When the student is between two teachers.
   - When the student is far from any teacher.

2. **Scenarios:**
   - Multiple teachers are positioned in such a way that the student can be caught quickly.
   - The student is positioned in a way that maximizes the number of moves before being caught.

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

```plaintext
3
8 1 1
6
3
10 3 3
1 4 8
2 3 10
15 5 5
1 3 7 9 11
1 2 13 14 15
```

### Explanation:

1. **First Test Case:**
   - \( n = 8 \), \( m = 1 \), \( q = 1 \)
   - Teacher positions: \( [6] \)
   - David's position query: \( [3] \)
   - **Expected Output:** \( 5 \)
   - **Reason:** David can move to cell 1, and it will take the teacher 5 moves to reach from cell 6 to cell 1.

2. **Second Test Case:**
   - \( n = 10 \), \( m = 3 \), \( q = 3 \)
   - Teacher positions: \( [1, 4, 8] \)
   - David's position queries: \( [2, 3, 10] \)
   - **Expected Output:**
     - For \( 2 \): \( 1 \)
     - For \( 3 \): \( 1 \)
     - For \( 10 \): \( 2 \)
   - **Reason:**
     - For \( 2 \): The teacher at cell 1 can reach cell 2 in 1 move.
     - For \( 3 \): The teacher at cell 4 can reach cell 3 in 1 move.
     - For \( 10 \): The teacher at cell 8 can reach cell 10 in 2 moves.

3. **Third Test Case:**
   - \( n = 15 \), \( m = 5 \), \( q = 5 \)
   - Teacher positions: \( [1, 3, 7, 9, 11] \)
   - David's position queries: \( [1, 2, 13, 14, 15] \)
   - **Expected Output:**
     - For \( 1 \): \( 0 \)
     - For \( 2 \): \( 1 \)
     - For \( 13 \): \( 2 \)
     - For \( 14 \): \( 3 \)
     - For \( 15 \): \( 4 \)
   - **Reason:**
     - For \( 1 \): The teacher at cell 1 can reach cell 1 in 0 moves.
     - For \( 2 \): The teacher at cell 1 can reach cell 2 in 1 move.
     - For \( 13 \): The teacher at cell 11 can reach cell 13 in 2 moves.
     - For \( 14 \): The teacher at cell 11 can reach cell 14 in 3 moves.
     - For \( 15 \): The teacher at cell 11 can reach cell 15 in 4 moves.

This test input covers various edge cases and scenarios that should help expose any bugs in the provided code.


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

def find_moves(student, t1, t2):
    if student < t1 and student < t2:
        return min(abs(t1 - 1), abs(t2 - 1))
    elif student < t2 and student > t1 or student > t2 and student < t1:
        mid = (t1 + t2) // 2
        return min(abs(mid - t1), abs(mid - t2))
    elif student > t1 and student > t2:
        return cell_len - max(t1, t2)
    else:
        return 0

for _ in range(t):
    cell_len, m, q = map(int, input().split(' '))
    teachers = list(map(int, input().split(' ')))
    students = list(map(int, input().split(' ')))

    if q == 1:
        print(find_moves(students[0], teachers[0], teachers[0]))
        continue

    i = 0
    for student in students:
        t1 = -1
        t2 = -1
        while i < m and teachers[i] <= student:
            i += 1

        if i == 0:
            t1, t2 = teachers[i]
        elif i == m:
            t1 = teachers[i-1]
            t2 = t1
        else:
            t1, t2 = teachers[i-1], teachers[i]

        print(find_moves(student, t1, t2))

**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
1 10
2 5 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:
t = int(input())

for _ in range(t):
    n, m, q = map(int, input().split())
    mug = list(map(int, input().split()))
    #mug.sort()
    req = list(map(int, input().split()))
    for david in req:
        if david in mug:
            print(0)
            continue

        if david > mug[-1]:
            up = n
            down = mug[-1]
            print(up - down)
            continue
        elif david < mug[0]:
            down = 1
            if m == 1:
                up = mug[0]
            else:
                up = mug[0]
            print(up - down)
            continue
        else:

            inx = 0
            left = 0
            #print(mugu)
            right = len(mug) - 1
            while left <= right:
                mid = (left + right) // 2
                if mug[mid] > david:
                    right = mid - 1
                else:
                    left = mid + 1
            print((mug[left] - mug[left - 1]) // 2)
            #print("binbin")
            # down = mug[inx]
            # #print(down, up)
            # print((up - down) // 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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. Here are the steps to create such a test input:

1. **Edge Cases for Teacher and David Positions**:
   - Teachers and David at the boundaries (1 and n).
   - Teachers and David in the middle of the line.
   - Teachers and David in positions that require careful handling of the binary search.

2. **Multiple Queries**:
   - Queries that test different positions relative to the teachers.

3. **Constraints**:
   - Ensure the input is within the constraints: \(3 \le n \le 10^9\), \(1 \le m, q \le 10^5\).

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

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

### Explanation:
- **Test Case 1**:
  - **n = 10**: The classroom has 10 cells.
  - **m = 3**: There are 3 teachers.
  - **q = 4**: There are 4 queries.
  - **Teachers' positions**: [1, 5, 9]
  - **David's positions**: [2, 4, 6, 10]

### Expected Output:
- **Query 1 (David at 2)**:
  - Teachers at 1 and 5. The closest teacher is at 1, and it takes 1 move to catch David.
  - **Output**: 1
- **Query 2 (David at 4)**:
  - Teachers at 5. The closest teacher is at 5, and it takes 1 move to catch David.
  - **Output**: 1
- **Query 3 (David at 6)**:
  - Teachers at 5 and 9. The closest teacher is at 5, and it takes 1 move to catch David.
  - **Output**: 1
- **Query 4 (David at 10)**:
  - Teachers at 9. The closest teacher is at 9, and it takes 1 move to catch David.
  - **Output**: 1

This test input is designed to expose the bug in the binary search and edge case handling of the provided code. The teachers are positioned in a way that the binary search might fail if not implemented correctly, and the queries test different positions relative to the teachers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Strict Teacher (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 Yoo {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int cases = sc.nextInt();
        while (cases-- > 0){
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();

            int [] t = new int[m+1];
            int [] a = new int[q+1];

            for (int i = 1; i <= m; i++) {
                t[i] = sc.nextInt();
            }
            for (int i = 1; i <= q; i++) {
                a[i] = sc.nextInt();
            }

            if (m == 1) {
                for (int i = 1; i < a.length; i++) {
                    if (a[i] < t[1]) System.out.println(t[1]-1);
                    else System.out.println(n-t[t.length-1]);
                }
                continue ;
            }

            Arrays.sort(t);
            Arrays.sort(a);

            int i = 1 ; int j = 1 ;

            while (i < t.length - 1 && j < a.length ) {
                if (a[j] < t[i] && i == 1) {
                    System.out.println(t[i]-1);
                    j++ ;
                }else if (a[j] < t[i+1]){
                    System.out.println(find(a[j]-t[i], t[i+1]-a[j]));
                    j++ ;
                }else {
                    i++ ;
                }
            }

            while (j < a.length){
                System.out.println(n-t[t.length-1]);
                j++ ;
            }
        }
    }

    public static int find (int d1, int d2){
        if (d1 == d2) return d1 ;
        int dl = 0 ;
        int ds = 0 ;
        if (d1 > d2 ) {
            dl = d1 ; ds = d2 ;
        } else {
            dl = d2 ; ds = d1 ;
        }
        return (dl - ds) / 2 + ds ;
    }
}

**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 where the code might fail. Here are the steps to think through this:

1. **Edge Cases for Single Teacher (m = 1)**:
   - The code has a specific block for handling the case when there is only one teacher. It outputs `t[1]-1` if David is to the left of the teacher and `n-t[t.length-1]` if David is to the right of the teacher.
   - We need to test this block with values that are close to the boundaries of the line (i.e., close to 1 or n).

2. **Edge Cases for Multiple Teachers (m > 1)**:
   - The code sorts the positions of the teachers and David, and then iterates through the sorted positions.
   - We need to test scenarios where David's position is between two teachers, and the distances between David and the teachers are not equal.

3. **General Cases**:
   - We should also test with a large number of queries to ensure the code handles multiple queries efficiently.

Based on these considerations, let's construct a small-scale test input:

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

### Explanation of the Test Input:

1. **First Test Case**:
   - `n = 10`, `m = 1`, `q = 2`
   - Teacher's position: `5`
   - David's positions: `1` and `10`
   - Expected output:
     - For David at position `1`: `4` (since the teacher is at `5`, and it takes 4 moves to reach `1`).
     - For David at position `10`: `5` (since the teacher is at `5`, and it takes 5 moves to reach `10`).

2. **Second Test Case**:
   - `n = 10`, `m = 3`, `q = 3`
   - Teachers' positions: `1`, `4`, `8`
   - David's positions: `2`, `3`, `10`
   - Expected output:
     - For David at position `2`: `1` (since the closest teacher is at `1`).
     - For David at position `3`: `1` (since the closest teacher is at `4`).
     - For David at position `10`: `2` (since the closest teacher is at `8`).

This test input should expose the bug in the code by testing both edge cases and general cases.