**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

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

public class Main {
    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;
        }
    }
    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    static long p(long a, long b, long m) {
        a %= m;
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1) ans = (ans * a) % m;
            a = (a * a) % m;
            b >>= 1;
        }
        return ans;
    }
    static long gcd(long a, long b)  {
        return b == 0? a: gcd(b, a % b);
    }
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }
    static long modinv(long a) {
        return a <= 1 ? a : mod - (mod / a) * modinv(mod % a) % mod;
    }
    static boolean isPrime(int n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i * i <= n; i = i + 6) {
            if (n % i == 0 || n % (i + 2) == 0) return false;
        }
        return true;
    }
    static FastReader sc = new FastReader();
    static int mod = 1000000007;
    static long inf = (long)1e17;
//----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
    public static void main(String[] args) {
        int T = sc.nextInt();
        while(T-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();

            int r1 = -1;

            int l2 = m;
            int r2 = m;

            int l3 = -1;

            while(q-- > 0) {
                int a = sc.nextInt();

                int n_r1 = r1;
                int n_l2 = l2;
                int n_r2 = r2;
                int n_l3 = l3;

                if(l2 != -1) {
                    if(a < l2) n_l2 = Math.max(1, l2-1);
                    else if(a > r2) n_r2 = Math.min(r2+1, n);
                    else {
                        if(l2 == r2) {
                            n_l2 = -1;
                            n_r2 = -1;
                        }
                        if(r1 == -1) n_r1 = 1;
                        if(l3 == -1) n_l3 = n;
                    }
                }

                if(r1 != -1) {
                    if(l3 == -1) n_l3 = n;
                    //else n_l3 = Math.max(1, l3-1);
                    if(a > r1) {
                        r1 = Math.min(n, r1+1);
                    }
                }

                if(l3 != -1) {
                    if(r1 == -1) n_r1 = 1;
                    //else n_r1 = Math.min(n, r1+1);
                    if(a < l3) {
                        n_l3 = Math.max(1, l3-1);
                    }
                }
                r1 = n_r1;
                l2 = n_l2;
                r2 = n_r2;
                l3 = n_l3;

                int ans = 0;
                if(r1 != -1) ans += r1;
                if(l2 != -1) {
                    ans += (r2-l2+1);
                    ans = Math.min(ans, r2);
                }
                if(l3 != -1) {
                    ans += (n-l3+1);
                    ans = Math.min(ans, n);
                }

                //System.out.println(r1+" "+l2+" "+r2+" "+l3);
                System.out.print(ans+" ");

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

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

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n;
    TreeSet<Integer> start  = new TreeSet<>(Collections.reverseOrder()), end = new TreeSet<>();
    public void solve() throws IOException {
        t = sc.nextInt();
        while (t-- > 0) {
            int n = i(), m =i(), q = i();
            TreeSet<Integer> toBeAdded = new TreeSet<>(), toBeRemoved = new TreeSet<>();
            boolean[] c = new boolean[n+4];
            start.clear();
            end.clear();
            start.add(m);
            end.add(m);
            c[m] = true;
            int color = 1;
            for(int i = 1;i<=q;i++){
                int a = i();
                if(color==n){
                    out.print(n+" ");
                    continue;
                }
                boolean prev = c[a];
                if(a!=1&&a!=n&&c[a]&&!c[a+1]&&!c[a-1]){
                    c[a] = false;
                    start.remove(a);
                    end.remove(a);
                    color--;
                }
                toBeRemoved.clear();
                toBeAdded.clear();
                for(int e : end){
                    if(e>=a||e+1>n) break;
                    c[e+1] = true;
                    toBeRemoved.add(e);
                    if(e+2<=n&&c[e+2]){
                        start.remove(e+2);
                    } else{
                        toBeAdded.add(e+1);
                    }
                    color++;
                }
                for(int e: toBeRemoved) end.remove(e);
                for(int e: toBeAdded) end.add(e);
                for(int e : start){
                    if(e<=a||e-1<1) break;
                    c[e-1] = true;
                    toBeRemoved.add(e);
                    if(e-2>=1&&c[e-2]){
                        end.remove(e-2);
                    } else{
                        toBeAdded.add(e-1);
                    }
                    color++;
                }
                for(int e: toBeRemoved) start.remove(e);
                for(int e: toBeAdded) start.add(e);
                if(prev&&!c[1]){
                    c[1] = true;
                    start.remove(2);
                    end.remove(2);
                    start.add(1);
                    if(!c[2]) end.add(1);
                    color++;
                }
                if(prev&&!c[n]){
                    c[n] = true;
                    start.remove(n-1);
                    end.remove(n-1);
                    end.add(n);
                    if(!c[n-1]) start.add(n);
                    color++;
                }
                out.print(color);
                out.print(" ");
            }
            out.print("\n");
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

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

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

    int t;
    cin >> t;

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

        vector<int> a(q);
        for (int i = 0; i < q; ++i) {
            cin >> a[i];
            a[i]--;
        }

        vector<int> b(n, 0);
        b[m - 1] = 1;

        int cnt = 1;
        deque<int> l, r;
        vector<int> new_values;

        if (m != 1) l.push_back(m - 1);
        if (m != n) r.push_back(m - 1);

        for (int j = 0; j < q; ++j) {
            int z = b[a[j]];

            if (z && cnt == 1) {
                cnt--;
                b[a[j]] = 0;
                l.clear();
                r.clear();
            }

            while (!l.empty() && l.back() > a[j]) {
                int y = l.back();
                l.pop_back();
                if (b[y - 1] == 0) {
                    cnt++;
                    b[y - 1] = 1;
                }
                if (y - 1 > 0 && b[y - 2] == 0) {
                    new_values.push_back(y - 1);
                }
            }

            while (!new_values.empty()) {
                l.push_back(new_values.back());
                new_values.pop_back();
            }

            while (!r.empty() && r.back() < a[j]) {
                int y = r.back();
                r.pop_back();
                if (b[y + 1] == 0) {
                    cnt++;
                    b[y + 1] = 1;
                }
                if (y + 1 < n - 1 && b[y + 2] == 0) {
                    new_values.push_back(y + 1);
                }
            }

            while (!new_values.empty()) {
                r.push_back(new_values.back());
                new_values.pop_back();
            }

            if (z) {
                if (b[0] == 0) {
                    cnt++;
                    b[0] = 1;
                    if (b[1] == 0) {
                        r.push_back(0);
                    }
                }
                if (b[n - 1] == 0) {
                    cnt++;
                    b[n - 1] = 1;
                    if (b[n - 2] == 0) {
                        l.push_back(n - 1);
                    }
                }
            }

            cout << cnt << " ";
        }
        cout << "\n";
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define f(i, n) for(int i = 0; i < (n); i++)
#define sd(i) scanf("%d", &i)
#define tcase int t; sd(t); for(int casee = 1; casee <= t; casee++)
#define en printf("\n")
#define mod 1000000007
using namespace std;
int bs(vector<ll>& arr, ll tar, int n) {
  int l =0,r=n-1, ans = n;
  while(l <= r) {
     int m=(l+r)/2;
     if(arr[m] >= tar) {
      ans=m;
      r = m-1;
     } else{
       l = m+1;
     }
  }
  return ans;
}
int main() {
    tcase {
      int n,m,k;
      sd(n);sd(m);sd(k);
      vector<pair<int,int>> vec;
      vector<int> vis(n+1,0);
      vec.push_back({m,m});
      int f = 0;
      int ans = 1;
      vis[m] = 1;
      for(int i = 0; i<k;i++) {
         int mv; sd(mv);

         for(int kk = 0; kk < vec.size(); kk++) {
          // printf("%d %d %d \n",vec[kk].first,vec[kk].second,mv);
           if( mv < vec[kk].first ) {
              if(vis[vec[kk].first-1]==0) {
                vis[vec[kk].first-1] =1;
                vec[kk].first = vec[kk].first -1;
                ans++;
              }
              } else if( mv > vec[kk].second )  {
                if(vis[vec[kk].second+1]==0) {
                vis[vec[kk].second+1] = 1;
                vec[kk].second = vec[kk].second+1;
                ans++;
              }
              }
              else {
                if(mv == vec[kk].first && mv==vec[kk].second && mv  != 1 && mv != n) {ans--;vec.pop_back(); vis[mv]=0;}
                int xyz = 0;
                 if(vis[1] == 0) {
                    vis[1] = 1;
                    vec.push_back({1,1});
                    ans++;
                    xyz=1;
                 }
                 if(vis[n] == 0) {
                  // printf("hi\n");
                    vis[n] = 1;
                    vec.push_back({n,n});
                    ans++;
                   xyz=1;
                 }
                if(xyz) break;
              }
           }
            printf("%d ",ans);
         }
        en;

      }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

Buggy Code:
import bisect

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    for _ in range(t):
        n = int(data[idx])
        m = int(data[idx+1])
        q = int(data[idx+2])
        idx += 3
        a = list(map(int, data[idx:idx+q]))
        idx += q
        # Initialize the possible positions with m
        positions = [m]
        answers = []
        for operation in a:
            new_positions = []
            for p in positions:
                # Move card at position 'operation' to the beginning
                if p < operation:
                    new_p = p + 1
                elif p == operation:
                    new_p = 1
                else:
                    new_p = p
                bisect.insort(new_positions, new_p)
                # Move card at position 'operation' to the end
                if p < operation:
                    new_p = p
                elif p == operation:
                    new_p = n
                else:
                    new_p = p - 1
                bisect.insort(new_positions, new_p)
            # Remove duplicates
            positions = [new_positions[0]]
            for p in new_positions[1:]:
                if p != positions[-1]:
                    positions.append(p)
            answers.append(str(len(positions)))
        print(' '.join(answers))

if __name__ == "__main__":
    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:
To generate a challenging test input that exposes the bug in the provided code, we need to consider scenarios that can cause the code to fail. The code uses a list to keep track of possible positions of the joker and updates this list for each operation. The main issues that can arise are:

1. **Duplicates in the positions list**: If the code does not handle duplicates correctly, it might lead to incorrect results.
2. **Edge cases**: Operations that move the joker to the beginning or end of the deck can be tricky, especially if the joker is already at the beginning or end.
3. **Large number of operations**: The code might have performance issues with a large number of operations, but since we need a small-scale test, we will focus on the logic.

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

1. **Initial position of the joker at the beginning or end of the deck**.
2. **Operations that move the joker to the beginning or end of the deck**.
3. **Operations that move other cards, which can affect the joker's position**.

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

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

### Explanation:
1. **Test Case 1:**
   - **n = 5, m = 1, q = 3**
   - **a = [1, 2, 3]**
   - The joker starts at the beginning. Moving the card at position 1 to the beginning or end should not change the joker's position. The subsequent operations will move other cards, potentially affecting the joker's position.

2. **Test Case 2:**
   - **n = 5, m = 5, q = 3**
   - **a = [1, 2, 3]**
   - The joker starts at the end. Moving the card at position 1 to the beginning or end will not change the joker's position. The subsequent operations will move other cards, potentially affecting the joker's position.

3. **Test Case 3:**
   - **n = 5, m = 3, q = 3**
   - **a = [1, 2, 3]**
   - The joker starts in the middle. Moving the card at position 1, 2, and 3 will affect the joker's position in various ways, testing the logic of moving the joker to different positions.

This test input should help identify any issues in the code's handling of duplicates and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

Buggy Code:
t = int(input())
for i in range(t):
    n,m,k = map(int, input().split())
    q = list(map(int, input().split()))
    lister = [i for i in range(n+1)]
    max_start = 0
    max_last = n+1
    max_left = m
    max_right = m
    set1 = [m]
    alpha =0
    for i in range(k):
        if max_left<=q[i]<=max_right:
            alpha =1
        if max_left==max_right==q[i]:
            set1.pop(0)
            max_left = 0
            max_right =n+1

        if max_left > q[i]:
            max_left -=1
            set1.append(max_left)

        if max_right<q[i]:
            max_right+=1
            # print('rr')
            set1.append(max_right)

        if alpha and max_start<q[i]:
            # print('pp')
            max_start +=1
            set1.append(max_start)

        if alpha and max_last>q[i]:
            # print('qq')
            max_last -=1
            set1.append(max_last)

        print(len(set(set1)),end=' ')
        # print(alpha,set1)
    # print(set(set1))
    print('')

**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 consider edge cases and scenarios that can maximize the failure probability. Here are the steps to create such a test input:

1. **Edge Cases for \( n \), \( m \), and \( q \)**:
   - \( n \) should be large enough to stress the algorithm but small enough to fit within the problem constraints.
   - \( m \) should be chosen to be at the boundary of the deck (e.g., 1 or \( n \)).
   - \( q \) should be large enough to ensure multiple operations are performed.

2. **Operations**:
   - Include operations that move the joker to the beginning and end of the deck.
   - Include operations that move cards around the joker to see how the joker's position changes.

3. **Multiple Test Cases**:
   - Include multiple test cases to ensure the solution handles different scenarios.

Given these considerations, here is a small-scale test input that can expose potential bugs in the code:

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

Explanation:
- **Test Case 1**: \( n = 5 \), \( m = 1 \), \( q = 3 \)
  - Operations: \( 2, 3, 4 \)
  - This case tests the joker starting at the beginning of the deck and moving other cards around it.

- **Test Case 2**: \( n = 5 \), \( m = 5 \), \( q = 3 \)
  - Operations: \( 2, 3, 4 \)
  - This case tests the joker starting at the end of the deck and moving other cards around it.

- **Test Case 3**: \( n = 10 \), \( m = 1 \), \( q = 5 \)
  - Operations: \( 2, 3, 4, 5, 6 \)
  - This case tests a larger deck and more operations, ensuring the joker's position changes multiple times.

This input should help in identifying issues in the buggy code, especially with how it handles the joker's position and the set of possible positions after each operation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

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

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n;
    TreeSet<Integer> start  = new TreeSet<>(Collections.reverseOrder()), end = new TreeSet<>();
    public void solve() throws IOException {
        t = sc.nextInt();
        while (t-- > 0) {
            int n = i(), m =i(), q = i();
            boolean[] c = new boolean[n+4];
            start.clear();
            end.clear();
            start.add(m);
            end.add(m);
            c[m] = true;
            int color = 1;
            for(int i = 1;i<=q;i++){
                int a = i();
                if(color==n){
                    out.print(n+" ");
                    continue;
                }
                boolean prev = c[a];
                if(a!=1&&a!=n&&c[a]&&!c[a+1]&&!c[a-1]){
                    c[a] = false;
                    start.remove(a);
                    end.remove(a);
                    color--;
                }
                TreeSet<Integer> toBeAdded = new TreeSet<>(), toBeRemoved = new TreeSet<>();
                for(int e : end){
                    if(e>=a||e+1>n) break;
                    c[e+1] = true;
                    toBeRemoved.add(e);
                    if(e+2<=n&&c[e+2]){
                        start.remove(e+2);
                    } else{
                        toBeAdded.add(e+1);
                    }
                    color++;
                }
                for(int e: toBeRemoved) end.remove(e);
                for(int e: toBeAdded) end.add(e);
                toBeRemoved.clear();
                toBeAdded.clear();
                for(int e : start){
                    if(e<=a||e-1<1) break;
                    c[e-1] = true;
                    toBeRemoved.add(e);
                    if(e-2>=1&&c[e-2]){
                        end.remove(e-2);
                    } else{
                        toBeAdded.add(e-1);
                    }
                    color++;
                }
                for(int e: toBeRemoved) start.remove(e);
                for(int e: toBeAdded) start.add(e);
                if(prev&&!c[1]){
                    c[1] = true;
                    start.remove(2);
                    end.remove(2);
                    start.add(1);
                    if(!c[2]) end.add(1);
                    color++;
                }
                if(prev&&!c[n]){
                    c[n] = true;
                    start.remove(n-1);
                    end.remove(n-1);
                    end.add(n);
                    if(!c[n-1]) start.add(n);
                    color++;
                }
                out.print(color);
                out.print(" ");
            }
            out.print("\n");
        }
    }
}

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

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

Think step by step.

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

### Explanation:
1. **Initial Setup**:
   - Deck: `[1, 2, 3, 4, 5]`
   - Joker at position: `3`

2. **Operations**:
   - **First Operation (a_1 = 1)**:
     - Move card at position 1 to the beginning or end.
     - Possible decks: `[1, 2, 3, 4, 5]` (no change), `[2, 3, 4, 5, 1]`
     - Joker positions: `3`
     - **Expected output**: `1`
   - **Second Operation (a_2 = 2)**:
     - Move card at position 2 to the beginning or end.
     - Possible decks: `[1, 2, 3, 4, 5]` (no change), `[1, 3, 4, 5, 2]`, `[2, 1, 3, 4, 5]`, `[3, 4, 5, 2, 1]`
     - Joker positions: `3`, `4`
     - **Expected output**: `2`
   - **Third Operation (a_3 = 3)**:
     - Move card at position 3 to the beginning or end.
     - Possible decks: `[1, 2, 3, 4, 5]` (no change), `[1, 2, 4, 5, 3]`, `[3, 1, 2, 4, 5]`, `[4, 5, 3, 1, 2]`
     - Joker positions: `3`, `4`, `5`
     - **Expected output**: `3`
   - **Fourth Operation (a_4 = 4)**:
     - Move card at position 4 to the beginning or end.
     - Possible decks: `[1, 2, 3, 4, 5]` (no change), `[1, 2, 3, 5, 4]`, `[4, 1, 2, 3, 5]`, `[5, 4, 1, 2, 3]`
     - Joker positions: `3`, `4`, `5`
     - **Expected output**: `3`

This test case is designed to expose potential issues with the algorithm's handling of the joker's position and the set of distinct positions it can occupy after each operation. The operations are chosen to maximize the number of distinct positions the joker can be in, which should help in identifying any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Joker
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a deck of $$$n$$$ cards. The positions in the deck are numbered from $$$1$$$ to $$$n$$$ from top to bottom. A joker is located at position $$$m$$$.

$$$q$$$ operations are applied sequentially to the deck. During the $$$i$$$-th operation, you need to take the card at position $$$a_i$$$ and move it either to the beginning or to the end of the deck. For example, if the deck is $$$[2, 1, 3, 5, 4]$$$, and $$$a_i=2$$$, then after the operation the deck will be either $$$[1, 2, 3, 5, 4]$$$ (the card from the second position moved to the beginning) or $$$[2, 3, 5, 4, 1]$$$ (the card from the second position moved to the end).

Your task is to calculate the number of distinct positions where the joker can be after each operation.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$2 \le n \le 10^9$$$; $$$1 \le m \le n$$$; $$$1 \le q \le 2 \cdot 10^5$$$).

The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le n$$$).

Additional constraint on the input: the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print $$$q$$$ integers — the number of distinct positions where the joker can be after each operation.

Examples:
input:
5
6 5 3
1 2 3
2 1 4
2 1 1 2
5 3 1
3
3 2 4
2 1 1 1
18 15 4
13 15 1 16
output:
2 3 5
2 2 2 2
2
2 3 3 3
2 4 6 8

Note:

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

public class Main {

    public static void solve() throws IOException {
        int t = readInt();
        for(int _t = 0; _t < t; _t++) {
            int[] opt = readIntArray();
            int n = opt[0];
            int m = opt[1];
            int q = opt[2];

            boolean[] cur = new boolean[n + 1];
            boolean[] newPos = new boolean[n + 1];
            cur[m] = true;

            StringBuilder sb = new StringBuilder();
            int[] arr = readIntArray();
            int maxC = -1;
            for(int j = 0; j < q; j++) {
                if (maxC ==n){
                    sb.append(n).append(" ");
                    continue;
                }
                int move = arr[j];

                Arrays.fill(newPos, false);

                for(int pos = 1; pos <= n; pos++) {
                    if(cur[pos]) {
                        if(move == 1) {
                            if(pos == 1) {
                                newPos[n] = true;
                                newPos[pos] = true;
                            } else {
                                newPos[pos - 1] = true;
                                newPos[pos] = true;
                            }
                        } else if(move == n) {
                            if(pos == n) {
                                newPos[1] = true;
                                newPos[pos] = true;
                            } else {
                                newPos[pos + 1] = true;
                                newPos[pos] = true;
                            }
                        } else {
                            if(pos == move) {
                                newPos[1] = true;
                                newPos[n] = true;
                            } else if(move < pos) {
                                newPos[pos - 1] = true;
                                newPos[pos] = true;
                            } else {
                                newPos[pos + 1] = true;
                                newPos[pos] = true;
                            }
                        }
                    }
                }
                boolean[] temp = cur;
                cur = newPos;
                newPos = temp;
                int count = 0;
                for(int pos = 1; pos <= n; pos++) {
                    if(cur[pos]) count++;
                }
                maxC = Math.max(maxC,count);
                sb.append(count).append(" ");
            }

            sb.setLength(sb.length() - 1);
            print(sb.toString());
        }
    }

    private static class Pair<A, B> {
        A a;
        B b;

        private Pair(A a, B b) {
            this.a = a;
            this.b = b;
        }
    }

    /* 控制流 */
    private static BufferedReader br;
    private static BufferedWriter bw;
    public static void print(long s) throws IOException {
        bw.write(s+"");
        bw.newLine();
    }
    public static void print(String s, Object... obj) throws IOException {
        bw.write(obj.length == 0 ? s : String.format(s, obj));
        bw.newLine();
    }
    public static long[] readLongArray() throws IOException {
        String[] parts = br.readLine().split(" ");
        long[] arr = new long[parts.length];
        for (int i = 0; i < parts.length; i++) arr[i] = Long.parseLong(parts[i]);
        return arr;
    }
    public static List<Integer> readIntList() throws IOException {
        String[] parts = br.readLine().split(" ");
        List<Integer> arr = new ArrayList<>();
        for (int i = 0; i < parts.length; i++) arr.add(Integer.parseInt(parts[i]));
        return arr;
    }
    public static int[] readIntArray() throws IOException {
        String[] parts = br.readLine().split(" ");
        int[] arr = new int[parts.length];
        for (int i = 0; i < parts.length; i++) arr[i] = Integer.parseInt(parts[i]);
        return arr;
    }
    public static String read() throws IOException {
        return br.readLine();
    }
    public static int readInt() throws IOException {
        return Integer.parseInt(br.readLine());
    }
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
        solve();
        bw.flush();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided Java code, we need to focus on edge cases and scenarios that can highlight issues in the algorithm. The main problem with the code is how it handles the movement of the joker and the update of the possible positions. The code has a flaw in how it updates the positions when the joker is moved to the beginning or the end of the deck.

### Key Points to Consider:
1. **Initial Position of the Joker**: The joker starts at position `m`.
2. **Operations**: The operations move the card at position `a_i` to either the beginning or the end of the deck.
3. **Possible Positions**: The code should correctly track the possible positions of the joker after each operation.

### Edge Cases:
1. **Joker at the Beginning or End**: If the joker is at the beginning or end of the deck, moving a card to the beginning or end should not change its position.
2. **Large Number of Operations**: The code should handle a large number of operations efficiently.
3. **Consecutive Moves**: Consecutive moves of the same card can lead to complex updates.

### Test Input:
To expose the bug, we can use a test case where the joker is initially at the beginning or end of the deck and perform operations that move cards around it. This will test how the code handles boundary conditions.

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

### Explanation:
- **Initial Deck**: `[1, 2, 3, 4, 5]` with the joker at position 1.
- **Operations**:
  - Move card at position 2 to the beginning or end.
  - Move card at position 3 to the beginning or end.
  - Move card at position 4 to the beginning or end.

### Expected Output:
- After the first operation, the joker can be at positions 1 or 2.
- After the second operation, the joker can be at positions 1, 2, or 3.
- After the third operation, the joker can be at positions 1, 2, 3, or 4.

This test input should expose the bug in the code by forcing it to handle boundary conditions and consecutive moves correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rendez-vous de Marian et Robin
time_limit_ms: 5000
memory_limit_mb: 256
Description: In the humble act of meeting, joy doth unfold like a flower in bloom.

Absence makes the heart grow fonder. Marian sold her last ware at the Market at the same time Robin finished training at the Major Oak. They couldn't wait to meet, so they both start without delay.

The travel network is represented as $n$ vertices numbered from $1$ to $n$ and $m$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$, and takes $w_i$ seconds to travel (all $w_i$ are even). Marian starts at vertex $1$ (Market) and Robin starts at vertex $n$ (Major Oak).

In addition, $h$ of the $n$ vertices each has a single horse available. Both Marian and Robin are capable riders, and could mount horses in no time (i.e. in $0$ seconds). Travel times are halved when riding. Once mounted, a horse lasts the remainder of the travel. Meeting must take place on a vertex (i.e. not on an edge). Either could choose to wait on any vertex.

Output the earliest time Robin and Marian can meet. If vertices $1$ and $n$ are disconnected, output $-1$ as the meeting is cancelled.

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

The first line of each test case consists of three integers $n$, $m$, $h$ ($2 \le n \le 2 \cdot 10^5, \;1 \le m \le 2 \cdot 10^5, \; 1 \le h \le n$) — the number of vertices, the number of edges and the number of vertices that have a single horse.

The second line of each test case contains $h$ distinct integers $a_1, a_2, \ldots, a_h$ ($1 \le a_i \le n$) — the vertices that have a single horse available.

Then follow $m$ lines of each test case, each with three integers $u_i$, $v_i$, $w_i$ ($1\le u_i,v_i \le n, \; 2\le w_i \le 10^6$) — meaning that there is an edge between vertices $u_i$ and $v_i$ with travel cost $w_i$ seconds without a horse.

There are no self loops or multiple edges. By good fortune, all $w_i$ are even integers.

It is guaranteed that the sums of both $n$ and $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer, the earliest time Robin and Marian can meet. If it is impossible for them to meet, output $-1$.

Examples:
input:
6
2 1 1
1
1 2 10
3 1 2
2 3
1 2 10
3 3 1
2
1 2 4
1 3 10
2 3 6
4 3 2
2 3
1 2 10
2 3 18
3 4 16
3 2 1
2
1 2 4
1 3 16
7 7 1
3
1 5 2
2 6 12
1 2 12
6 4 8
7 3 4
6 3 4
7 6 4
output:
5
-1
6
19
14
12

Note: In the first test case, Marian rides from vertex $1$ to vertex $2$, Robin waits.

In the second test case, vertices $1$ and $3$ are not connected.

In the third test case, both Marian and Robin travel to vertex $2$ to meet.

In the fourth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides to vertex $3$ to meet Robin.

In the fifth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides back to vertex $1$ and then to vertex $3$. Robin waits.

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

#define ll long long
#define ldb long double
#define endl '\n'
#define For(i,l,r) for(int i=l;i<=r;i++)
#define ForD(i,r,l) for(int i=r;i>=l;i--)
#define REP(i,l,r) For(i,l,r-1)
#define PER(i,r,l) ForD(i,r-1,l)
#define ff first
#define ss second
#define pb push_back
#define all(x) x.begin(),x.end()
#define All(x,n) x+1,x+1+n
#define Alll(x,n) x,x+n
#define sz(x) (signed)x.size()
#define unq(x) x.resize(unique(all(x))-x.begin())
#define mpa make_pair

#ifdef NCGM
#include"debug.h"
#else
#define debug(...) "fr";
#endif

using namespace std;

const int N=2e5+3;

int n,m,h;
bool ho[N];
ll d[N][2],d1[N][2];
priority_queue<pair<ll,pair<int,bool>>,vector<pair<ll,pair<int,bool>>>,greater<pair<ll,pair<int,bool>>>> pq;
vector<pair<int,int>> g[N];
void dijkstra(int s,bool c) {
    while(sz(pq)) pq.pop();
    if (c) d1[s][ho[s]]=0;
    else d[s][ho[s]]=0;
    pq.push({0,{s,ho[s]}});
    while(sz(pq)) {
        int u=pq.top().ss.ff;
        bool hor=pq.top().ss.ss;
        ll ccc=d[u][hor];
        if (c) ccc=d1[u][hor];
        ll w=pq.top().ff;
        if (ccc!=w) continue;
        pq.pop();
        for(auto ed: g[u]) {
            int v=ed.ff,ww=(hor?ed.ss/2:ed.ss);
            ll cmp=d[v][hor|ho[v]];
            if (c) cmp=d1[v][hor|ho[v]];
            if (cmp>w+ww) {
                if (c) d1[v][hor|ho[v]]=w+ww;
                else d[v][hor|ho[v]]=w+ww;
                pq.push({w+ww,{v,hor|ho[v]}});
            }

        }
    }
}

void solve() {
    cin >> n >> m >> h;
    For(i,1,n) d[i][0]=d[i][1]=d1[i][0]=d1[i][1]=1e18;
    fill(ho,ho+n+1,0);
    For(i,1,h) {
        int x;
        cin >> x;
        ho[x]=1;
    }
    For(i,1,m) {
        int u,v,w;
        cin >> u >> v >> w;
        g[u].pb({v,w});
        g[v].pb({u,w});
    }
    dijkstra(1,0);
    dijkstra(n,1);
    For(i,1,n) g[i].clear();
    if (d[n][0]>=1e18 && d[n][1]>=1e18) return void(cout << -1 << endl);
    ll ans=1e18;
    For(i,1,n)
        For(j,0,1)
            For(k,0,1) ans=min(ans,max(d[i][j],d1[i][k]));
    cout << ans << endl;
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int t=1;
    cin >> t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4 5 2
2 3
1 2 4
1 3 4
2 4 10
3 4 10
2 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rendez-vous de Marian et Robin
time_limit_ms: 5000
memory_limit_mb: 256
Description: In the humble act of meeting, joy doth unfold like a flower in bloom.

Absence makes the heart grow fonder. Marian sold her last ware at the Market at the same time Robin finished training at the Major Oak. They couldn't wait to meet, so they both start without delay.

The travel network is represented as $n$ vertices numbered from $1$ to $n$ and $m$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$, and takes $w_i$ seconds to travel (all $w_i$ are even). Marian starts at vertex $1$ (Market) and Robin starts at vertex $n$ (Major Oak).

In addition, $h$ of the $n$ vertices each has a single horse available. Both Marian and Robin are capable riders, and could mount horses in no time (i.e. in $0$ seconds). Travel times are halved when riding. Once mounted, a horse lasts the remainder of the travel. Meeting must take place on a vertex (i.e. not on an edge). Either could choose to wait on any vertex.

Output the earliest time Robin and Marian can meet. If vertices $1$ and $n$ are disconnected, output $-1$ as the meeting is cancelled.

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

The first line of each test case consists of three integers $n$, $m$, $h$ ($2 \le n \le 2 \cdot 10^5, \;1 \le m \le 2 \cdot 10^5, \; 1 \le h \le n$) — the number of vertices, the number of edges and the number of vertices that have a single horse.

The second line of each test case contains $h$ distinct integers $a_1, a_2, \ldots, a_h$ ($1 \le a_i \le n$) — the vertices that have a single horse available.

Then follow $m$ lines of each test case, each with three integers $u_i$, $v_i$, $w_i$ ($1\le u_i,v_i \le n, \; 2\le w_i \le 10^6$) — meaning that there is an edge between vertices $u_i$ and $v_i$ with travel cost $w_i$ seconds without a horse.

There are no self loops or multiple edges. By good fortune, all $w_i$ are even integers.

It is guaranteed that the sums of both $n$ and $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer, the earliest time Robin and Marian can meet. If it is impossible for them to meet, output $-1$.

Examples:
input:
6
2 1 1
1
1 2 10
3 1 2
2 3
1 2 10
3 3 1
2
1 2 4
1 3 10
2 3 6
4 3 2
2 3
1 2 10
2 3 18
3 4 16
3 2 1
2
1 2 4
1 3 16
7 7 1
3
1 5 2
2 6 12
1 2 12
6 4 8
7 3 4
6 3 4
7 6 4
output:
5
-1
6
19
14
12

Note: In the first test case, Marian rides from vertex $1$ to vertex $2$, Robin waits.

In the second test case, vertices $1$ and $3$ are not connected.

In the third test case, both Marian and Robin travel to vertex $2$ to meet.

In the fourth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides to vertex $3$ to meet Robin.

In the fifth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides back to vertex $1$ and then to vertex $3$. Robin waits.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 1e9+7;
const int N = 1e7 + 10;
vector<bool> isprime(1005,true);
void sieve(){
	isprime[0] = isprime[1] = true;
	for(int i=2;i<1005;i++){
		if(isprime[i]){
			for(int j=i*i;j<1005;j+=i){
				isprime[j] = false;
			}
		}
	}
}
vector<ll>mar;
vector<ll>rob;
vector<ll>mhorse;
vector<ll>rhorse;
set<ll> st;
void dij(int src,int dest,vector<vector<pair<ll,ll>>>&graph,vector<ll>&dist,vector<ll>&horse){
	bool flag = false;
	priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> pq;
	pq.push({0,src});
	dist[src] = 0;
	// horse[src] = 0;
	while(!pq.empty()){
		ll ct = pq.top().first;
		ll node = pq.top().second;
		pq.pop();
		if(st.count(node)) flag = true;
		for(auto it : graph[node]){
			ll wt = it.second;
			ll child = it.first;
			if(flag){
				if(wt/2+ct<horse[child]){
					pq.push({wt/2+ct,child});
					horse[child]=wt/2+ct;
				}
				if(wt+ct<horse[child]){
					pq.push({wt+ct,child});
					horse[child] = wt+ct;
				}
			}
			if(wt+ct<dist[child]){
				pq.push({wt+ct,child});
				dist[child] = wt+ct;
			}
		}
	}

}
void solve(){
	ll n,m,h;
	cin>>n>>m>>h;
	st.clear();
	mar.clear();
	rob.clear();
	mhorse.clear();
	rhorse.clear();
	for(int i=1;i<=h;i++){
		ll x;
		cin>>x;
		st.insert(x);
	}
	mar.resize(n+1,INT_MAX);
	rob.resize(n+1,INT_MAX);
	mhorse.resize(n+1,INT_MAX);
	rhorse.resize(n+1,INT_MAX);
	vector<vector<pair<ll,ll>>> graph(n+1);
	for(int i=1;i<=m;i++){
		ll x,y,z;
		cin>>x>>y>>z;
		graph[x].push_back({y,z});
		graph[y].push_back({x,z});
	}
	// for(int i=1;i<=m;i++){
	// 	cout<<i<<endl;
	// 	for(auto it : graph[i]){
	// 		cout<<it.first<<" "<<it.second;
	// 		cout<<endl;
	// 	}
	// 	cout<<endl;
	// }
	dij(1,n,graph,mar,mhorse);
	dij(n,1,graph,rob,rhorse);
	ll ans = INT_MAX;
	for(int i=1;i<=n;i++){
		ll m = min(mar[i],mhorse[i]);
		ll r = min(rob[i],rhorse[i]);
		ans = min(ans,max(m,r));
	}
	if(ans==INT_MAX){
		cout<<-1<<endl;
		return;
	}
	cout<<ans<<endl;
}
int main(){
	int t = 1;
	cin>>t;
	while(t--){
		solve();
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Disconnected Graph**: Ensure that vertices 1 and n are not connected, which should result in an output of -1.
2. **Multiple Paths with Horses**: Create a scenario where there are multiple paths with different travel times, and some of these paths have horses available.
3. **Optimal Path with Horses**: Ensure that the optimal path involves using horses to highlight the issue in the distance calculation.
4. **Edge Cases**: Include cases where the optimal meeting point is not immediately obvious, such as when the optimal path involves backtracking or using horses at non-trivial points.

Let's construct a small-scale test input that meets these criteria:

```plaintext
1
4 5 2
2 3
1 2 10
2 3 20
3 4 30
1 4 100
2 4 50
```

### Explanation:

1. **Graph Structure**:
   - **Vertices**: 4 (1, 2, 3, 4)
   - **Edges**:
     - 1-2 with weight 10
     - 2-3 with weight 20
     - 3-4 with weight 30
     - 1-4 with weight 100
     - 2-4 with weight 50
   - **Horses**: Available at vertices 2 and 3

2. **Expected Behavior**:
   - **Path without Horses**:
     - Marian (1) to 2: 10 seconds
     - Marian (2) to 3: 20 seconds
     - Marian (3) to 4: 30 seconds (Total: 60 seconds)
   - **Path with Horses**:
     - Marian (1) to 2: 10 seconds
     - Marian (2) to 4: 50/2 = 25 seconds (Total: 35 seconds)
     - Robin (4) to 2: 50/2 = 25 seconds (Total: 25 seconds)

3. **Optimal Meeting Point**:
   - Marian and Robin can meet at vertex 2 in 25 seconds each, making the optimal meeting time 25 seconds.

4. **Buggy Code Issue**:
   - The buggy code might not correctly handle the horse usage or might not find the optimal path, leading to an incorrect output.

This test input should effectively expose the bug in the provided code by requiring the algorithm to correctly handle horse usage and find the optimal meeting point.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rendez-vous de Marian et Robin
time_limit_ms: 5000
memory_limit_mb: 256
Description: In the humble act of meeting, joy doth unfold like a flower in bloom.

Absence makes the heart grow fonder. Marian sold her last ware at the Market at the same time Robin finished training at the Major Oak. They couldn't wait to meet, so they both start without delay.

The travel network is represented as $n$ vertices numbered from $1$ to $n$ and $m$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$, and takes $w_i$ seconds to travel (all $w_i$ are even). Marian starts at vertex $1$ (Market) and Robin starts at vertex $n$ (Major Oak).

In addition, $h$ of the $n$ vertices each has a single horse available. Both Marian and Robin are capable riders, and could mount horses in no time (i.e. in $0$ seconds). Travel times are halved when riding. Once mounted, a horse lasts the remainder of the travel. Meeting must take place on a vertex (i.e. not on an edge). Either could choose to wait on any vertex.

Output the earliest time Robin and Marian can meet. If vertices $1$ and $n$ are disconnected, output $-1$ as the meeting is cancelled.

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

The first line of each test case consists of three integers $n$, $m$, $h$ ($2 \le n \le 2 \cdot 10^5, \;1 \le m \le 2 \cdot 10^5, \; 1 \le h \le n$) — the number of vertices, the number of edges and the number of vertices that have a single horse.

The second line of each test case contains $h$ distinct integers $a_1, a_2, \ldots, a_h$ ($1 \le a_i \le n$) — the vertices that have a single horse available.

Then follow $m$ lines of each test case, each with three integers $u_i$, $v_i$, $w_i$ ($1\le u_i,v_i \le n, \; 2\le w_i \le 10^6$) — meaning that there is an edge between vertices $u_i$ and $v_i$ with travel cost $w_i$ seconds without a horse.

There are no self loops or multiple edges. By good fortune, all $w_i$ are even integers.

It is guaranteed that the sums of both $n$ and $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer, the earliest time Robin and Marian can meet. If it is impossible for them to meet, output $-1$.

Examples:
input:
6
2 1 1
1
1 2 10
3 1 2
2 3
1 2 10
3 3 1
2
1 2 4
1 3 10
2 3 6
4 3 2
2 3
1 2 10
2 3 18
3 4 16
3 2 1
2
1 2 4
1 3 16
7 7 1
3
1 5 2
2 6 12
1 2 12
6 4 8
7 3 4
6 3 4
7 6 4
output:
5
-1
6
19
14
12

Note: In the first test case, Marian rides from vertex $1$ to vertex $2$, Robin waits.

In the second test case, vertices $1$ and $3$ are not connected.

In the third test case, both Marian and Robin travel to vertex $2$ to meet.

In the fourth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides to vertex $3$ to meet Robin.

In the fifth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides back to vertex $1$ and then to vertex $3$. Robin waits.

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

int n;
vector<vector<pii>> adj;

void pr(vector<int> v){
    for(int i=0; i<v.size(); i++){
        cout<<v[i]<<" ";
    }
    cout<<endl;
}

void dij(int s, vector<int> &dis1){
    priority_queue<pair<int,int>, vector<pair<int,int>> , greater<pair<int,int>>> pq;
    for(int i=0; i<n; i++)dis1[i]=1e12;
    vector<bool> vis(n,false);
    dis1[s]=0;
    pq.push({dis1[s],s});
    while(!pq.empty()){
        int u=pq.top().second;
        pq.pop();
        if(vis[u])continue;
        vis[u]=1;
        for(auto p: adj[u]){
            int w=p.second/2;
            int v=p.first;
            //cout<<v<<" "<<w<<" "<<u<<endl;
            if(!vis[v] && dis1[u]+w<dis1[v]){
                dis1[v]=dis1[u]+w;
                pq.push({dis1[v],v});
            }
        }
    }
}

void dijkstra(int s, vector<int> &dis1, map<int,int> mp){
    priority_queue<pair<int,int>, vector<pair<int,int>> , greater<pair<int,int>>> pq;
    for(int i=0; i<n; i++)dis1[i]=1e12;
    vector<bool> vis(n,false);
    dis1[s]=0;
    int yo=0,xx=-1;
    pq.push({dis1[s],s});
    while(!pq.empty()){
        int u=pq.top().second;
        pq.pop();
        if(vis[u])continue;
        if(mp[u]==1 && yo==0){
            yo=1;
            xx=u;
        }
        vis[u]=1;
        for(auto p: adj[u]){
            int w=p.second;
            int v=p.first;
            //cout<<v<<" "<<w<<" "<<u<<endl;
            if(!vis[v] && dis1[u]+w<dis1[v]){
                dis1[v]=dis1[u]+w;
                pq.push({dis1[v],v});
            }
        }
    }
    //pr(dis1);
    vector<int> dis(n);
    if(xx!=-1)dij(xx,dis);
    for(int i=0; i<n; i++){
        dis1[i]=min(dis1[i],dis[i]+dis1[xx]);
    }
    //pr(dis1);
}

void solve(){
    map<int,int> mp;
    adj.clear();
    int m,h;
    cin>>n>>m>>h;
    adj.resize(n);
    for(int i=0; i<h; i++){
        int x;
        cin>>x;
        x--;
        mp[x]=1;
    }
    while(m--){
        int u,v,w;
        cin>>u>>v>>w;
        u--; v--;
        adj[u].push_back({v,w});
        adj[v].push_back({u,w});
    }
    vector<int> dis1(n),dis2(n);
    dijkstra(0,dis1,mp);
    dijkstra(n-1,dis2,mp);
    //pr(dis1);
    //pr(dis2);
    int ans=1e18;
    for(int i=0; i<n; i++){
        ans=min(ans,max(dis1[i],dis2[i]));
    }
    if(ans>=1e12 || ans<=0)ans=-1;
    cout<<ans<<endl;
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    //cout<<fixed<<setprecision(10);
    while(t--)
        solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rendez-vous de Marian et Robin
time_limit_ms: 5000
memory_limit_mb: 256
Description: In the humble act of meeting, joy doth unfold like a flower in bloom.

Absence makes the heart grow fonder. Marian sold her last ware at the Market at the same time Robin finished training at the Major Oak. They couldn't wait to meet, so they both start without delay.

The travel network is represented as $n$ vertices numbered from $1$ to $n$ and $m$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$, and takes $w_i$ seconds to travel (all $w_i$ are even). Marian starts at vertex $1$ (Market) and Robin starts at vertex $n$ (Major Oak).

In addition, $h$ of the $n$ vertices each has a single horse available. Both Marian and Robin are capable riders, and could mount horses in no time (i.e. in $0$ seconds). Travel times are halved when riding. Once mounted, a horse lasts the remainder of the travel. Meeting must take place on a vertex (i.e. not on an edge). Either could choose to wait on any vertex.

Output the earliest time Robin and Marian can meet. If vertices $1$ and $n$ are disconnected, output $-1$ as the meeting is cancelled.

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

The first line of each test case consists of three integers $n$, $m$, $h$ ($2 \le n \le 2 \cdot 10^5, \;1 \le m \le 2 \cdot 10^5, \; 1 \le h \le n$) — the number of vertices, the number of edges and the number of vertices that have a single horse.

The second line of each test case contains $h$ distinct integers $a_1, a_2, \ldots, a_h$ ($1 \le a_i \le n$) — the vertices that have a single horse available.

Then follow $m$ lines of each test case, each with three integers $u_i$, $v_i$, $w_i$ ($1\le u_i,v_i \le n, \; 2\le w_i \le 10^6$) — meaning that there is an edge between vertices $u_i$ and $v_i$ with travel cost $w_i$ seconds without a horse.

There are no self loops or multiple edges. By good fortune, all $w_i$ are even integers.

It is guaranteed that the sums of both $n$ and $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer, the earliest time Robin and Marian can meet. If it is impossible for them to meet, output $-1$.

Examples:
input:
6
2 1 1
1
1 2 10
3 1 2
2 3
1 2 10
3 3 1
2
1 2 4
1 3 10
2 3 6
4 3 2
2 3
1 2 10
2 3 18
3 4 16
3 2 1
2
1 2 4
1 3 16
7 7 1
3
1 5 2
2 6 12
1 2 12
6 4 8
7 3 4
6 3 4
7 6 4
output:
5
-1
6
19
14
12

Note: In the first test case, Marian rides from vertex $1$ to vertex $2$, Robin waits.

In the second test case, vertices $1$ and $3$ are not connected.

In the third test case, both Marian and Robin travel to vertex $2$ to meet.

In the fourth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides to vertex $3$ to meet Robin.

In the fifth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides back to vertex $1$ and then to vertex $3$. Robin waits.

Buggy Code:
import math
from queue import PriorityQueue
def solve():
    n, m, h = map(int, input().split())
    list_h = list(map(int, input().split()))
    list_edges = {}
    for _ in range(m):
        u, v, w = map(int, input().split())
        if u not in list_edges:
            list_edges[u] = [(v, w)]
        else:
            list_edges[u].append((v, w))
        if v not in list_edges:
            list_edges[v] = [(u, w)]
        else:
            list_edges[v].append((u, w))
    # print("list_edges", list_edges)
    # print("list_h", list_h)
    # a solution is two paths, which are list of edges, one start from 1, another start from n
    def get_score(sol):
        return max(sol[0][1], sol[1][1])

    def is_valid(sol):
        return sol[0][0][-1] == sol[1][0][-1]

    init_sol = ([1], 0, 1 in list_h, None if 1 in list_h else 0), ([n], 0, n in list_h, None if n in list_h else 0)

    # heuristic search
    queue = PriorityQueue()
    queue.put((get_score(init_sol), init_sol))

    while not queue.empty():
        cur_score, cur_sol = queue.get()
        # print("cur_sol", cur_score, cur_sol)
        if is_valid(cur_sol):
            # print("sol", cur_sol)
            print(cur_score)
            return

        for smaller_path_idx in [0, 1]:
            cur_path = cur_sol[smaller_path_idx][0]
            cur_vertex = cur_path[-1]
            if cur_vertex not in list_edges:
                continue
            for next_vertex, w in list_edges[cur_vertex]:
                cur_path, cur_path_score, cur_riding_hourse, cur_horse_idx = cur_sol[smaller_path_idx]
                if cur_horse_idx is not None and next_vertex in cur_path[cur_horse_idx:]:
                    continue
                if cur_riding_hourse:
                    w = w//2
                new_path_score = cur_path_score + w
                new_path = cur_path + [next_vertex]
                if not cur_riding_hourse and next_vertex in list_h:
                    cur_riding_hourse = True
                    cur_horse_idx = len(new_path) - 1
                new_sol = (cur_sol[0], (new_path, new_path_score, cur_riding_hourse, cur_horse_idx)) if smaller_path_idx == 1 else \
                      ((new_path, new_path_score, cur_riding_hourse, cur_horse_idx), cur_sol[1])
                new_score = get_score(new_sol)
                # print("new_sol", new_score, new_sol)
                queue.put((new_score, new_sol))

    print(-1)

n_tests = int(input())
for i in range(n_tests):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rendez-vous de Marian et Robin
time_limit_ms: 5000
memory_limit_mb: 256
Description: In the humble act of meeting, joy doth unfold like a flower in bloom.

Absence makes the heart grow fonder. Marian sold her last ware at the Market at the same time Robin finished training at the Major Oak. They couldn't wait to meet, so they both start without delay.

The travel network is represented as $n$ vertices numbered from $1$ to $n$ and $m$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$, and takes $w_i$ seconds to travel (all $w_i$ are even). Marian starts at vertex $1$ (Market) and Robin starts at vertex $n$ (Major Oak).

In addition, $h$ of the $n$ vertices each has a single horse available. Both Marian and Robin are capable riders, and could mount horses in no time (i.e. in $0$ seconds). Travel times are halved when riding. Once mounted, a horse lasts the remainder of the travel. Meeting must take place on a vertex (i.e. not on an edge). Either could choose to wait on any vertex.

Output the earliest time Robin and Marian can meet. If vertices $1$ and $n$ are disconnected, output $-1$ as the meeting is cancelled.

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

The first line of each test case consists of three integers $n$, $m$, $h$ ($2 \le n \le 2 \cdot 10^5, \;1 \le m \le 2 \cdot 10^5, \; 1 \le h \le n$) — the number of vertices, the number of edges and the number of vertices that have a single horse.

The second line of each test case contains $h$ distinct integers $a_1, a_2, \ldots, a_h$ ($1 \le a_i \le n$) — the vertices that have a single horse available.

Then follow $m$ lines of each test case, each with three integers $u_i$, $v_i$, $w_i$ ($1\le u_i,v_i \le n, \; 2\le w_i \le 10^6$) — meaning that there is an edge between vertices $u_i$ and $v_i$ with travel cost $w_i$ seconds without a horse.

There are no self loops or multiple edges. By good fortune, all $w_i$ are even integers.

It is guaranteed that the sums of both $n$ and $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer, the earliest time Robin and Marian can meet. If it is impossible for them to meet, output $-1$.

Examples:
input:
6
2 1 1
1
1 2 10
3 1 2
2 3
1 2 10
3 3 1
2
1 2 4
1 3 10
2 3 6
4 3 2
2 3
1 2 10
2 3 18
3 4 16
3 2 1
2
1 2 4
1 3 16
7 7 1
3
1 5 2
2 6 12
1 2 12
6 4 8
7 3 4
6 3 4
7 6 4
output:
5
-1
6
19
14
12

Note: In the first test case, Marian rides from vertex $1$ to vertex $2$, Robin waits.

In the second test case, vertices $1$ and $3$ are not connected.

In the third test case, both Marian and Robin travel to vertex $2$ to meet.

In the fourth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides to vertex $3$ to meet Robin.

In the fifth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides back to vertex $1$ and then to vertex $3$. Robin waits.

Buggy Code:
import heapq

def dijkstra(graph, start, half):
    priority_queue = [(0, start)]
    distances = {vertex: float('infinity') for vertex in graph}
    distances[start] = 0

    while priority_queue:
        current_distance, current_vertex = heapq.heappop(priority_queue)

        if current_distance > distances[current_vertex]:
            continue

        for neighbor, weight in graph[current_vertex].items():
            adjusted_weight = weight / half
            distance = current_distance + adjusted_weight

            if distance < distances[neighbor]:
                distances[neighbor] = distance
                heapq.heappush(priority_queue, (distance, neighbor))

    return distances

t = int(input())
for _ in range(t):
    n, m, h = map(int, input().split())
    hourses = map(int, input().split())  # use iterator to save memory
    graph = {i: {} for i in range(1, n + 1)}

    for _ in range(m):
        u, v, w = map(int, input().split())
        graph[u][v] = w
        graph[v][u] = w

    # Avoid copying the graph by directly updating the horse nodes' distances
    for horse in hourses:
        graph[horse] = dijkstra(graph, horse, 2)

    # Calculate distances for Robin and Lover directly on the original graph
    Robin = dijkstra(graph, 1, 1)
    Lover = dijkstra(graph, n, 1)

    result = min(max(Robin[i], Lover[i]) for i in range(1, n + 1))

    if result != float("inf"):
        print(int(result))
    else:
        print(-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
1
4 4 2
2 3
1 2 10
1 3 10
2 4 20
3 4 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rendez-vous de Marian et Robin
time_limit_ms: 5000
memory_limit_mb: 256
Description: In the humble act of meeting, joy doth unfold like a flower in bloom.

Absence makes the heart grow fonder. Marian sold her last ware at the Market at the same time Robin finished training at the Major Oak. They couldn't wait to meet, so they both start without delay.

The travel network is represented as $n$ vertices numbered from $1$ to $n$ and $m$ edges. The $i$-th edge connects vertices $u_i$ and $v_i$, and takes $w_i$ seconds to travel (all $w_i$ are even). Marian starts at vertex $1$ (Market) and Robin starts at vertex $n$ (Major Oak).

In addition, $h$ of the $n$ vertices each has a single horse available. Both Marian and Robin are capable riders, and could mount horses in no time (i.e. in $0$ seconds). Travel times are halved when riding. Once mounted, a horse lasts the remainder of the travel. Meeting must take place on a vertex (i.e. not on an edge). Either could choose to wait on any vertex.

Output the earliest time Robin and Marian can meet. If vertices $1$ and $n$ are disconnected, output $-1$ as the meeting is cancelled.

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

The first line of each test case consists of three integers $n$, $m$, $h$ ($2 \le n \le 2 \cdot 10^5, \;1 \le m \le 2 \cdot 10^5, \; 1 \le h \le n$) — the number of vertices, the number of edges and the number of vertices that have a single horse.

The second line of each test case contains $h$ distinct integers $a_1, a_2, \ldots, a_h$ ($1 \le a_i \le n$) — the vertices that have a single horse available.

Then follow $m$ lines of each test case, each with three integers $u_i$, $v_i$, $w_i$ ($1\le u_i,v_i \le n, \; 2\le w_i \le 10^6$) — meaning that there is an edge between vertices $u_i$ and $v_i$ with travel cost $w_i$ seconds without a horse.

There are no self loops or multiple edges. By good fortune, all $w_i$ are even integers.

It is guaranteed that the sums of both $n$ and $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer, the earliest time Robin and Marian can meet. If it is impossible for them to meet, output $-1$.

Examples:
input:
6
2 1 1
1
1 2 10
3 1 2
2 3
1 2 10
3 3 1
2
1 2 4
1 3 10
2 3 6
4 3 2
2 3
1 2 10
2 3 18
3 4 16
3 2 1
2
1 2 4
1 3 16
7 7 1
3
1 5 2
2 6 12
1 2 12
6 4 8
7 3 4
6 3 4
7 6 4
output:
5
-1
6
19
14
12

Note: In the first test case, Marian rides from vertex $1$ to vertex $2$, Robin waits.

In the second test case, vertices $1$ and $3$ are not connected.

In the third test case, both Marian and Robin travel to vertex $2$ to meet.

In the fourth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides to vertex $3$ to meet Robin.

In the fifth test case, Marian travels to vertex $2$ without a horse, mounts the horse at vertex $2$ and rides back to vertex $1$ and then to vertex $3$. Robin waits.

Buggy Code:
import sys

# region fastio
from collections import Counter
from functools import reduce
from typing import List
from heapq import heappop, heappush
from math import inf, gcd
from random import getrandbits
from types import GeneratorType

input = lambda: sys.stdin.readline().rstrip()
sint = lambda: int(input())
mint = lambda: map(int, input().split())
ints = lambda: list(map(int, input().split()))

def bootstrap(f, stack=None):
    if stack is None:
        stack = []

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

    return func

MOD = 10 ** 9 + 7

mx = 2 * 10 ** 5

# f = [0] * MX
# f[0] = f[1] = 1
# for i in range(2, MX):
#     f[i] = f[i - 1] + 2 * (i - 1) * f[i - 2]
#     f[i] %= MOD
# 阶乘
# fac = [1] * (mx + 1)
# for i in range(1, mx + 1):
#     fac[i] = fac[i - 1] * i % MOD
# # 逆元
# ifac = [0] * (mx + 1)
# ifac[mx] = pow(fac[mx], MOD - 2, MOD)
# for i in range(mx, 0, -1):
#     ifac[i - 1] = ifac[i] * i % MOD
#
#
# # 组合数
# def comb(n: int, m: int, MOD=MOD) -> int:
#     if m < 0 or m > n:
#         return 0
#     return fac[n] * ifac[m] % MOD * ifac[n - m] % MOD

# endregion fastio

# MOD = 998_244_353
# MOD = 10 ** 9 + 7
# DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1)) #URDL
# DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))
def dijstra(n: int, g: List[List[int]], start: int, vis: List[int]) -> List[int]:
    dis = [inf] * n
    dis[start] = 0
    h = [(0, start)]
    cnt = [0] * n
    while h:
        dx, x = heappop(h)
        if (dx > dis[x] and vis[x] == 0) or (dx > dis[x] and vis[x] == 1 and cnt[x] == 1):  # x 之前出堆过
            continue
        if vis[x] == 1:
            cnt[x] += 1
        for y, wt in g[x]:
            new_dis = dx + wt if vis[x] == 0 else dx + wt // 2
            if vis[x] == 1:
                if vis[y] == 0:
                    vis[y] = 1
                heappush(h, (new_dis, y))
                if new_dis < dis[y]:
                    dis[y] = new_dis  # 更新 x 的邻居的最短路
            else:
                if new_dis < dis[y]:
                    dis[y] = new_dis  # 更新 x 的邻居的最短路
                    heappush(h, (new_dis, y))
    return dis

def cf2002A():
    n, m, h = mint()
    a = ints()
    vis = [0] * n
    for x in a:
        vis[x-1] = 1
    g = [[] for _ in range(n)]
    for _ in range(m):
        u, v, w = mint()
        u -= 1
        v -= 1
        g[u].append((v, w))
        g[v].append((u, w))

    dis_a = dijstra(n, g, 0, vis)
    if dis_a[-1] == inf:
        print(-1)
        return
    vis = [0] * n
    for x in a:
        vis[x-1] = 1
    dis_b = dijstra(n, g, n-1, vis)
    res = inf
    for i in range(n):
        cur = max(dis_a[i], dis_b[i])
        if cur < res:
            res = cur
    print(res)
    return

if __name__ == "__main__":
    t_ = sint()
    for _ in range(t_):
        cf2002A()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code's handling of horses and travel times might fail. The code uses Dijkstra's algorithm to find the shortest path, but it has a specific condition for handling vertices with horses. We need to create a test case that challenges this condition.

### Key Points to Consider:
1. **Vertices with Horses**: Ensure that the vertices with horses are strategically placed to test the code's logic for mounting and dismounting horses.
2. **Travel Times**: Use even travel times, but consider scenarios where the travel times are such that the logic for halving the travel time when riding a horse might lead to incorrect results.
3. **Disconnection**: Test cases where the graph is disconnected to ensure the code correctly handles the `-1` output.
4. **Optimal Meeting Points**: Ensure the test case has multiple potential meeting points to test the logic for finding the earliest meeting time.

### Step-by-Step Test Case Generation:
1. **Vertices and Edges**: Create a small graph with a few vertices and edges.
2. **Horses**: Place horses in a way that challenges the code's logic.
3. **Travel Times**: Use even travel times that might cause the code to incorrectly compute the shortest path.
4. **Disconnection**: Create a disconnected graph to test the `-1` output.

### Example Test Case:
```plaintext
3
4 5 2
2 3
1 2 10
2 3 10
3 4 10
1 4 20
2 4 5
3 2 1
1
1 2 10
2 3 10
5 4 2
1 3
1 2 10
2 3 10
3 4 10
4 5 10
```

### Explanation:
1. **First Test Case**:
   - **Vertices and Edges**: 4 vertices, 5 edges.
   - **Horses**: Horses at vertices 2 and 3.
   - **Travel Times**: Even travel times.
   - **Expected Output**: 15. Marian can ride from 1 to 2, then to 3, and finally to 4. Robin can ride from 4 to 3, then to 2, and finally to 1. They meet at vertex 3 at time 15.

2. **Second Test Case**:
   - **Vertices and Edges**: 3 vertices, 2 edges.
   - **Horses**: Horse at vertex 1.
   - **Travel Times**: Even travel times.
   - **Expected Output**: -1. The graph is disconnected, so Marian and Robin cannot meet.

3. **Third Test Case**:
   - **Vertices and Edges**: 5 vertices, 4 edges.
   - **Horses**: Horses at vertices 1 and 3.
   - **Travel Times**: Even travel times.
   - **Expected Output**: 20. Marian can ride from 1 to 3, then to 4, and finally to 5. Robin can ride from 5 to 4, then to 3, and finally to 1. They meet at vertex 4 at time 20.

This test case is designed to expose potential issues with the code's handling of horses and travel times, as well as to test the handling of disconnected graphs.