**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

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

public class E {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);
	static int mod = 1_000_000_007;

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
		}

		private void solve(int t) {
			long n = sc.nextLong();
			int k = sc.nextInt();

			out.println(solve(n, k));
		}

		private long solve(long n, int k) {
			if (n == 0) {
				return 0;
			}else if (k == 0) {
				return 1;
			}

			long end = 1L << (k + 1);
			end -= 2;

			end = Math.min(n - 1, end);

			// System.out.println(end);

			long val1 = end + 1;
			long val2 = val1 + 1;

			val1 %= mod;
			val2 %= mod;

			long result = val1;
			result *= val2;
			result %= mod;

			result *= multiplicativeInverse(2, mod);
			result %= mod;

			long base = 1L << (k + 1);
			long nextN;

			while (base <= n) {
				nextN = n - base;
				nextN = Math.min(nextN, base);

				result += solve(nextN, k - 1);
				result %= mod;
				base = base << 1;
			}

			return result;
		}

		public static int[] egcd(int a, int b) {
	        if (b == 0)
	            return new int[] { a, 1, 0 };
	        else {
	            int[] arr = egcd(b, a % b);

	            int gcd = arr[0];
	            int X = arr[2];
	            int Y = arr[1] - (a / b) * arr[2];

	            return new int[] { gcd, X, Y };
	        }
	    }

	    public static int multiplicativeInverse(int a, int modulo) {

	        int[] egcdValues = egcd(a, modulo);

	        if (egcdValues[0] != 1)
	            return 0;
	        if (egcdValues[1] > 0)
	            return egcdValues[1];
	        else
	            return egcdValues[1] + modulo;
	    }

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

      public IOHandler() {
         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());
      }

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

      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;
      }

   }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases for \( n \) and \( k \)**:
   - \( n = 1 \): The smallest possible array.
   - \( n = 2 \): A small array to check the transition from single-element subarrays to multiple-element subarrays.
   - \( k = 1 \): To check the behavior when only numbers with a single '1' in their binary representation are considered.
   - \( k = 60 \): The maximum value for \( k \) to check the behavior when almost all numbers are considered.

2. **Critical Values for \( n \) and \( k \)**:
   - \( n = 7 \): To check the transition from \( 2^3 - 1 \) (which has 3 ones in binary) to the next number.
   - \( n = 15 \): To check the transition from \( 2^4 - 1 \) (which has 4 ones in binary) to the next number.
   - \( n = 31 \): To check the transition from \( 2^5 - 1 \) (which has 5 ones in binary) to the next number.

3. **Mixed Cases**:
   - \( n = 100 \): A larger value to check the behavior over a wider range.
   - \( k = 5 \): A moderate value for \( k \) to check the behavior when a reasonable number of elements are considered.

Based on these considerations, a small-scale test input that includes these edge cases and critical values is:

```plaintext
5
1 1
7 1
15 2
31 3
100 5
```

This input includes:
- The smallest possible array with \( n = 1 \) and \( k = 1 \).
- A small array with \( n = 7 \) and \( k = 1 \) to check the transition from single '1' to multiple '1' in binary representation.
- A larger array with \( n = 15 \) and \( k = 2 \) to check the transition from \( 2^4 - 1 \).
- A larger array with \( n = 31 \) and \( k = 3 \) to check the transition from \( 2^5 - 1 \).
- A larger array with \( n = 100 \) and \( k = 5 \) to check the behavior over a wider range.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

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

#define dbg(x) //x
#define prt(x) dbg(cerr << x)
#define pv(x) dbg(cerr << #x << " = " << x << '\n')
#define pv2(x) dbg(cerr << #x << " = " << x.first << ',' << x.second << '\n')
#define parr(x) dbg(prt(#x << " = { "); for (auto y : x) prt(y << ' '); prt("}\n");)
#define parr2(x) dbg(prt(#x << " = { "); for (auto [y, z] : x) prt(y << ',' << z << "  "); prt("}\n");)
#define parr2d(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr(arr);} prt('\n'));
#define parr2d2(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr2(arr);} prt('\n'));

/*
everything up to 2^k - 1 good
goal is to find o(k^2) values
ways(2^a - 1, b)
can we find using dp?
if b >= a
ans = (len+1)(len+2)/2
if b == a - 1
ans = (len)(len+1)/2
if b == a - 2
ans = ways(2^(a-1) - 1, b) + ways(2^(a-1), b - 1)
*/

const long long mod = 1e9 + 7;

long long expo(long long base, long long pwr) {
  base %= mod;
  if (pwr == 0) return 1ll;
  if (pwr % 2 == 1) return base * expo(base, pwr - 1) % mod;
  long long semi = expo(base, pwr / 2);
  return semi * semi % mod;
}

long long madd(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}

long long msub(long long a, long long b) {
  return ((a % mod) + mod - (b % mod)) % mod;
}

long long mmul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}

long long mdiv(long long a, long long b) {
  return ((a % mod) * expo(b, mod - 2)) % mod;
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    long long n, k;
    cin >> n >> k;
    if (n == 0) {
      cout << 1 << '\n';
      continue;
    }
    long long mxbit = 0, tmp = 1;
    while ((tmp << 1) <= n) {
      tmp <<= 1;
      mxbit++;
    }
    if (n < (1ll << (k + 1))) {
      cout << mdiv(mmul(n, n + 1), 2ll) << '\n';
      continue;
    }
    assert(k <= mxbit);
    vector<vector<long long>> dp(mxbit + 1, vector<long long>(mxbit + 1, 0));
    for (int i = 0; i <= mxbit; i++) {
      long long len = (1ll << i) - 1;
      dp[i][0] = 1;
      for (int j = 1; j <= mxbit; j++) {
        if (j >= i) {
          dp[i][j] = mdiv(mmul(len + 1, len + 2), 2);
        } else if (j == i - 1) {
          dp[i][j] = mdiv(mmul(len, len + 1), 2);
        } else {
          dp[i][j] = madd(dp[i - 1][j], dp[i - 1][j - 1]);
        }
      }
    }
    parr2d(dp);
    pv(mxbit); pv(k);
    long long ans = dp[mxbit][k], x = (1ll << mxbit) - 1, bcnt = 1;
    long long prv = (k == mxbit ? (1ll << mxbit) : 0ll);
    for (int rbit = mxbit - 1; rbit >= 0; rbit--) {
      if ((n & ((1ll << rbit)))) {
        if (bcnt <= k) {
          pv(ans); pv(bcnt); pv(rbit); pv(k - bcnt); pv(prv); pv(min(k - bcnt, (long long) rbit));
          pv(mmul(prv, (1ll << (min(k - bcnt, (long long) rbit)))));
          ans = madd(ans, dp[rbit][k - bcnt]);
          ans = madd(ans, mmul(prv, (1ll << (min(k - bcnt, (long long) rbit)))));
          if (bcnt + rbit <= k) {
            prv = madd(prv, (1ll << rbit));
          } else {
            prv = 0;
          }
        }
        bcnt++;
      }
    }
    pv(bcnt);
    cout << ans << '\n';
  }
}

/*
any observations help

check every line
IF YOUR LINES AREN'T WRONG
CHECK IF YOUR LINES ARE IN THE RIGHT ORDER

NEVER GIVE UP
*/

**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
31 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define pb push_back
#define pii pair<int,int>
#define endl '\n'
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int N=2e5+5;
const int p=1e9+7;
int n,m,k;
int qpow(int a,int b)
{
    a%=p;
    int res=1;
    while(b)
    {
        if(b&1) res=res*a%p;
        b>>=1;
        a=a*a%p;
    }
    return res%p;
}
int inv(int x)
{
    return qpow(x,p-2)%p;
}
map<int,int>dp[70];
int F(int n)
{
    n%=p;
    return (n+1)%p*n%p*inv(2)%p;
}
int f(int n,int k)
{
    //if(n==0) return 0;
    if(k==0||n==1) return 1;
    if(dp[k].count(n)) return dp[k][n];
    int t=(1LL<<(k+1));
    if(n<t)
    {
        int res=F(n);
        return dp[k][n]=res;
    }
    else if(n==t)
    {
        t%=p;
        return dp[k][n]=(t-1)%p*t%p*inv(2)%p;
    }
    while(t*2<n) t*=2;
    int res=f(t,k)+f(n-t,k-1);
    //cout<<n<<" "<<k<<" "<<res<<endl;
    return dp[k][n]=res;
}
void solve()
{
    cin>>n>>k;
    int t=(1LL<<(k+1));
    if(n<t) n%=p,cout<<((n+1)%p*n%p*inv(2)%p)%p<<endl;
    else if(n==t) n%=p,cout<<((n-1)%p*n%p*inv(2)%p)%p<<endl;
    else
    {
        cout<<f(n,k)%p<<endl;
    }
}
signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    int _=1;
    cin>>_;
    while(_--)
    {
        solve();
    }
    return 0;
}
/*   /\_/\
*   (= ._.)
*   / >  \>
*/
/*
1
16 2
*/

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

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

Think step by step.

Response:
```plaintext
1
31 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
import sys
from functools import cache
input = sys.stdin.readline
to_print = []
modul = 1000 * 1000 * 1000 + 7

t = int(input())

for _ in range(t):
    n, k = list(map(int, input().split()))
    k += 1
    @cache
    def go(n, k):
        if not n:
            return 0
        if not k:
            return 0
        if n < (1 << k):
            return (n * (n - 1) // 2 + n) % modul
        if n == (1 << k):
            return ((n - 1) * (n - 2) // 2 + (n - 1)) % modul
        pow = 1
        while 2 * pow < n:
            pow <<= 1
        return (go(pow, k) + go(n - pow, k - 1)) % modul
    to_print.append(go(n, k))

print('\n'.join(map(str, to_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:
```plaintext
3
8 2
15 3
31 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
import sys
from functools import lru_cache

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

# 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))
MOD = 10 ** 9 + 7

def cal(x):
    return (x * (x + 1) // 2) % MOD

def cf1982E():
    n, k = mint()

    @lru_cache(None)
    def dfs(x: int, y: int) -> tuple:
        m = x.bit_length()
        if m <= y:
            return x + 1, x + 1, cal(x + 1)
        if x == 0:
            return 1, 1, cal(1)
        if y == 0:
            return 1, 0, cal(1)

        m -= 1
        l1, r1, ans1 = dfs((1 << m) - 1, y)
        l2, r2, ans2 = dfs(x - (1 << m), y - 1)
        ans1 -= cal(r1)
        ans2 -= cal(l2)
        res = ans1 + ans2 + cal(r1 + l2)
        res = res % MOD
        if l1 == 1 << m and l2 == x - (1 << m) + 1:
            return l1 + l2, l1 + l2, res
        elif l1 == 1 << m:
            return l1 + l2, r2, res
        elif l2 == x - (1 << m) + 1:
            return l1, r1 + l2, res

        else:
            return l1, r2, res

    ans = dfs(n - 1, k)[-1]
    print(ans)
    return

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        cf1982E()

**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
1 1
2 1
3 1
4 1
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

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

public class Main {

    static int MOD = 1000000007, INV2 = 500000004;

    static void solve() {
        long n = sc.nextLong();
        int k = sc.nextInt();
        memo = new HashMap<>();
        out.println(process(n, k)[2]);
    }

    static HashMap<Long, HashMap<Integer, long[]>> memo;

    static long[] process(long n, int k) {
        long[] ans;
        if (k < 0) {
            ans = new long[3];
        } else if (n == 1) {
            ans = new long[]{1, 1, 1};
        } else {
            long m = high(n) >> 1;
            if (m << 1 == n) {
                if (memo.containsKey(n) && memo.get(n).containsKey(k)) {
                    return memo.get(n).get(k);
                }
            }
            long[] l = process(m, k);
            long[] r = process(n - m, k - 1);
            ans = new long[3];
            ans[0] = l[0] == m ? l[0] + r[0] : l[0];
            ans[1] = r[1] == n - m ? r[1] + l[1] : r[1];
            ans[2] = l[2] + r[2];
            ans[2] -= l[1] % MOD * ((l[1] + 1) % MOD) % MOD * INV2 % MOD;
            ans[2] -= r[0] % MOD * ((r[0] + 1) % MOD) % MOD * INV2 % MOD;
            ans[2] += (l[1] + r[0]) % MOD * ((l[1] + r[0] + 1) % MOD) % MOD * INV2 % MOD;
            ans[2] = (ans[2] + MOD + MOD) % MOD;
            if (m << 1 == n) {
                memo.computeIfAbsent(n, v -> new HashMap<>()).put(k, ans);
            }
        }
        return ans;
    }

    static long high(long n) {
        n--;
        n |= n >>> 1;
        n |= n >>> 2;
        n |= n >>> 4;
        n |= n >>> 8;
        n |= n >>> 16;
        n |= n >>> 32;
        return n + 1;
    }

    static boolean retest = true;
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    public static void main(String[] args) {
        if (retest) {int T = sc.nextInt(); while (T-- > 0) solve();} else solve(); out.flush(); out.close();
    }
    static class FastReader {
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        String next() {
            try {while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(r.readLine()); return st.nextToken();}
            catch (Exception e) {return null;}
        }
        int nextInt() {return Integer.parseInt(next());}
        long nextLong() {return Long.parseLong(next());}
        double nextDouble() {return Double.parseDouble(next());}
    }
}

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

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

Think step by step.

Response:
```plaintext
1
8 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
import java.io.*;
import java.util.*;
public class NumberOfKGoodSubarrays {
    static final long MOD = (long) 1e9 + 7;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(br.readLine());
        while(T --> 0) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            long n = Long.parseLong(st.nextToken());
            int k = Integer.parseInt(st.nextToken());
            int pow = Long.toBinaryString(n).length()-1;
            long[][][] dp = new long[pow+1][k+1][3];
            for(int i = 0; i <= k; i++) {
                dp[0][i][0] = dp[0][i][1] = dp[0][i][2] = 1;
            }
            for(int i = 1; i <= pow; i++) {
                dp[i][0][0] = dp[i][0][2] = 1;
            }
            for(int i = 1; i <= pow; i++) {
                for(int j = 1; j <= k; j++) {
                    dp[i][j] = add(dp[i-1][j], dp[i-1][j-1]);
                }
            }
//            System.out.println(Arrays.deepToString(dp));
            long[] arr = dp[pow][k--];
            n -= (long) Math.pow(2, pow);
            while(n > 0) {
                pow = Long.toBinaryString(n).length()-1;
                if(k >= 0)
                    arr = add(arr, dp[pow][k--]);
                n -= (long) Math.pow(2, pow);
            }
            System.out.println(arr[2]);
        }
    }
    static long[] add(long[] a, long[] b) {
        long l = a[0], r = b[1];
        if(a[0] == a[1]) {
            l += b[0];
            l %= MOD;
        }
        if(b[0] == b[1]) {
            r += a[1];
            r %= MOD;
        }
        long ans = a[2] + b[2];
        ans %= MOD;
        ans -= a[1] * (a[1] + 1) / 2 - MOD;
        ans %= MOD;
        ans -= b[0] * (b[0] + 1) / 2 - MOD;
        ans %= MOD;
        ans += (a[1] + b[0]) * (a[1] + b[0] + 1) / 2;
        ans %= MOD;
        return new long[]{l, r, ans};
    }
}

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

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

Think step by step.

Response:
```plaintext
3
15 1
7 2
1 1
```


**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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <utility>

using namespace std;

typedef long long ll;

void solve();

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		solve();
	}
}

void solve()
{
	ll n,m,q;
	cin>>n>>m>>q;
	m--;
	vector<ll> ans;
	ll p1=-1,p2=m,p3 = m, p4 = n;
	while(q--)
	{
		ll x;
		cin>>x;
		x--;
		if(x<= p1)
		{
			p2--;
			p4--;
		}
		else if(p1 < x && x < p2 && p2>= 0)
		{
			if(p1>=0)	p1++;
			p2--;
			if(p4<n)	p4--;
		}
		else if(p2<=x && x<=p3 && p2 >= 0)
		{
			p1++;
			p4--;

			if(p2 == p3)
			{
				p2 = -1;
				p3 = n + 1;
			}
		}
		else if(p3 < x && x < p4 && p3 <= n)
		{
			if(p1 >= 0)	p1++;
			p3++;
			if(p4 < n)	p4--;
		}
		else if(x >= p4)
		{
			p1++;
			p3++;
		}

		// if(p2 <= p1 && p2 >= 0)
		// 	p2 = p1+1;
		// if(p3 >= p4 && p3 < n)
		// 	p4 = p3 + 1;
		// if(p1 >= p4)
		// {
		// 	p1 = n-1;
		// 	p4 = n;
		// 	p2 = -1;
		// 	p3 = n+1;
		// }
		if(p1 >= p2 && p2 >= 0)
		{
			p1 = p3;
			p2 = -1;
			p3 = n+1;
		}
		if(p3 >= p4 && p3 < n)
		{
			p4 = p2;
			p2 = -1;
			p3 = n+1;
		}
		if(p1 >= p4)
		{
			p1 = p4 - 1;
		}
		// cout<<x<<" "<<p1<< " "<<p2<< " "<<p3<<  " "<<p4<<endl;
		ll sum = (p1 + 1) + ((p2 >= 0)*(p3 - p2 + 1)) +  (n - p4);
		ans.push_back(sum);

	}
	for(auto i:ans)
	{
		cout<<i<<" ";
	}
	cout<<"\n";
}

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

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

Think step by step.

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


**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;
// #pragma GCC optimize("O3,unroll-loops")
// #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

typedef long long int ll;
const int zz = 1'000'000'007;
const int mz = 998244353;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define loop(i, n) for (ll i = 0; i < (n); ++i)
#define pool(i, n) for (ll i = ((n) - 1); i >= 0; --i)
#define rep(i, a, b) for (ll i = (a); i <= (b); ++i)
#define repi(i, a, b) for (int i = (a); i <= (b); ++i)
#define per(i, a, b) for (ll i = (a); i >= (b); --i)
#define in(v) \
    for (auto& i : v) cin >> i
#define out(v)            \
    for (auto& i : v) {   \
        cout << i << " "; \
    }                     \
    cout << "\n"
#define outp(v)                                      \
    for (auto& i : v) {                              \
        cout << "(" << i.fi << "," << i.se << "), "; \
    }                                                \
    cout << "\n"
#define foa(v) for (auto& i : v)
#define vi vector<int>
#define vll vector<ll>
#define vbb vector<bool>
#define vvi vector<vi>
#define vvll vector<vll>
#define vvvll vector<vvll>
#define vvbb vector<vbb>
#define vpi vector<pair<int, int>>
#define vpll vector<pair<ll, ll>>
#define all(v) (v).begin(), (v).end()
#define LLMAX 9223372036854775807
#define gcd __gcd
#define lcm(a, b) ((a / gcd(a, b)) * (b))
#define myceil(a, b) ((a + b - 1) / b)

/* Debugging Stuff */

#ifdef LOCAL
#include "debug.hpp"
#define dbg(...) cerr << "Ln " << __LINE__ << ": [", dbg_out(#__VA_ARGS__, __VA_ARGS__);
#define dbgarr(...) cerr << "Ln " << __LINE__ << ": [", dbgarr_out(#__VA_ARGS__, __VA_ARGS__);
#else
#define dbg(...)
#define dbgarr(...)
#endif

/* PBDS

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

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class T>
using ordered_set_dec = tree<T, null_type, greater<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

*/

// /* Atcoder Lib

// #include <atcoder/all>
// #include <atcoder/scc>
// #include <atcoder/twosat>
// using namespace atcoder;

// */

// bs : lo < hi, hi = mid, lo = mid + 1

/* MOD INT from https://usaco.guide/gold/modular */

const int MOD = 998244353;
struct mi {
    int v;
    explicit operator int() const { return v; }
    mi() { v = 0; }
    mi(long long _v) : v(_v % MOD) { v += (v < 0) * MOD; }
};
mi& operator+=(mi& a, mi b) {
    if ((a.v += b.v) >= MOD) a.v -= MOD;
    return a;
}
mi& operator-=(mi& a, mi b) {
    if ((a.v -= b.v) < 0) a.v += MOD;
    return a;
}
mi operator+(mi a, mi b) { return a += b; }
mi operator-(mi a, mi b) { return a -= b; }
mi operator*(mi a, mi b) { return mi((long long)a.v * b.v); }
mi& operator*=(mi& a, mi b) { return a = a * b; }
mi pow(mi a, long long p) {
    assert(p >= 0);
    return p == 0 ? 1 : pow(a * a, p / 2) * (p & 1 ? a : 1);
}
mi inv(mi a) {
    assert(a.v != 0);
    return pow(a, MOD - 2);
}
mi operator/(mi a, mi b) { return a * inv(b); }

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

    // freopen("feast.in", "r", stdin);
    // freopen("feast.out", "w", stdout);

    // int tc = 1;
    int tt = 1;
    cin >> tt;
    while (tt--) {
        ll n, m, q;
        cin >> n >> m >> q;

        vll a(q);
        in(a);

        ll l = m, r = m;
        ll act = 0;

        ll left = 0, right = n + 1;
        ll nomid = 0;

        if (a[0] == m) {
            nomid = 1;
            l = n + 1;
            r = 0;
            act = 1;
        }

        loop(i, q) {
            ll curr = a[i];

            auto expand = [&](auto& left, auto& right) {
                if (nomid) {
                    left = min({n, l - 1, left + 1});
                    right = max({left + 1, right - 1, 1ll, r + 1});
                } else {
                    left = min({n, l - 1, left + 1});
                    right = max({right - 1, 1ll, r + 1});
                }
            };

            if (!nomid) {
                if (curr <= left) {
                    expand(left, right);
                } else if (curr < l) {
                    l = max(1ll, l - 1);
                    if (act)
                        expand(left, right);
                } else if (l <= curr && curr <= r) {
                    if (act == 0) {
                        act = 1;
                        left = 1;
                        right = n;
                    } else {
                        expand(left, right);
                    }
                } else if (curr >= right) {
                    expand(left, right);
                } else if (curr > r) {
                    r = min(r + 1, n);
                    if (act)
                        expand(left, right);
                }
            } else {
                if (left < curr && curr < right) {
                    expand(left, right);
                } else if (curr <= left) {
                    left = min({n, left + 1});
                    right = max({right, left + 1, 1ll});
                } else if (curr >= right) {
                    right = max({right - 1, 1ll});
                    left = min({left, n, right - 1});
                }
            }

            ll pos;
            if (act) {
                ll l_ = min(l - 1, left);
                ll r_ = max(r + 1, right);

                if (nomid) {
                    pos = (l_ > 0) * l_ + (r_ <= n) * (n - r_ + 1);
                } else {
                    pos = (r - l + 1) + (l_ > 0) * l_ + (r_ <= n) * (n - r_ + 1);
                }
            } else {
                pos = (r - l + 1);
            }

            // dbg(curr, l, r, left, right, n, act, pos, nomid);

            cout << pos << " ";
        }

        cout << "\n";
    }

    // cerr << "[TC : " << tc++ << "]" << endl;
    cerr << "\nTime:" << setprecision(15) << 1000 * ((long double)clock()) / (long double)CLOCKS_PER_SEC << "ms\n";
    cout << flush;

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
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
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `6 5 3` with `a = [1, 2, 3]`
   - **Description:** This test case checks the basic functionality of the algorithm. The joker starts at position 5, and the operations move cards from positions 1, 2, and 3. The expected output is `2 3 5`.

2. **Test Case 2:**
   - **Input:** `2 1 4` with `a = [2, 1, 1, 2]`
   - **Description:** This test case checks the behavior when the deck size is very small (2 cards). The joker starts at position 1, and the operations move cards from positions 2, 1, 1, and 2. The expected output is `2 2 2 2`.

3. **Test Case 3:**
   - **Input:** `5 3 1` with `a = [3]`
   - **Description:** This test case checks the behavior when the joker is the card being moved. The joker starts at position 3, and the operation moves the card from position 3. The expected output is `2`.

4. **Test Case 4:**
   - **Input:** `3 2 4` with `a = [2, 1, 1, 1]`
   - **Description:** This test case checks the behavior when the joker is moved multiple times. The joker starts at position 2, and the operations move cards from positions 2, 1, 1, and 1. The expected output is `2 3 3 3`.

5. **Test Case 5:**
   - **Input:** `18 15 4` with `a = [13, 15, 1, 16]`
   - **Description:** This test case checks the behavior with a larger deck and a joker starting in the middle. The joker starts at position 15, and the operations move cards from positions 13, 15, 1, and 16. The expected output is `2 4 6 8`.

These test cases cover a range of scenarios, including edge cases and typical operations, to ensure the algorithm is thoroughly tested.


**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 T,n,m,q,a[300000],t,l,r,b;
int main(){
	scanf("%d",&T);
	while(T --){
		scanf("%d%d%d",&n,&m,&q);
		l = r = m,t = (m == 1),b = (m == n);
		for(int i = 1;i <= q;i ++) scanf("%d",&a[i]);
		for(int i = 1;i <= q;i ++){
			if(l > a[i]) l --;
			if(r < a[i]) r ++;
			if(t && a[i] != t) t ++;
			if(b && a[i] != n - b + 1) b ++;
			if(l <= a[i] && a[i] <= r){
				t = max(t,1),b = max(b,1);
				if(l == r) l = r = 1;
			}
			if(l == 1) t = max(t,r);
			if(r == n) b = max(b,n - l + 1);
			t = min(t,n),b = min(b,n);
			printf("%d ",r - l + 1 + min(l - 1,t) + min(n - r,b));
		}
		printf("\n");
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to think through this:

1. **Edge Cases for `m` and `a_i`**:
   - `m` is at the beginning (1) or the end (n).
   - `a_i` is at the beginning (1) or the end (n).

2. **Special Cases for `q`**:
   - `q` is very small (e.g., 1 or 2) to ensure the code handles single operations correctly.
   - `q` is large enough to observe the behavior over multiple operations.

3. **Special Cases for `n`**:
   - `n` is small (e.g., 2 or 3) to ensure the code handles small decks correctly.
   - `n` is large (e.g., 10^9) to ensure the code handles large decks correctly.

4. **Mixed Operations**:
   - Operations that move cards to both the beginning and the end of the deck.

Based on these considerations, here is a small-scale test input that includes edge cases and mixed operations:

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

This test input includes:
1. A test case with `n = 2`, `m = 1`, and `q = 2` to check the behavior when the joker is at the beginning and operations move it around.
2. A test case with `n = 3`, `m = 2`, and `q = 3` to check the behavior when the joker is in the middle and operations move it around.
3. A test case with `n = 10`, `m = 5`, and `q = 4` to check the behavior with a larger deck and multiple operations.

This should help expose any issues in the buggy 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:
t = int(input())

def merge_positions(positions):
    positions = sorted(positions)
    new_positions = [positions[0]]

    for i in range(1, len(positions)):
        if positions[i][0] <= new_positions[-1][1]:
            new_positions[-1] = (new_positions[-1][0], max(new_positions[-1][1], positions[i][1]))
        else:
            new_positions.append(positions[i])

    return new_positions

for _ in range(t):
    n, m, q = map(int, input().split())

    a = list(map(int, input().split()))

    positions = []

    print(2, end=" ")
    if a[0] == m:
        positions.append((1, 2))
        positions.append((n, n + 1))
    else:
        if a[0] < m:
            positions.append((m - 1, m + 1))
        else:
            positions.append((m, m + 2))

    for ai in a[1:]:
        for i in range(len(positions)):
            if ai < positions[i][0]:
                positions[i] = (positions[i][0] - 1, positions[i][1])

            elif ai > positions[i][1]:
                positions[i] = (positions[i][0], positions[i][1] + 1)

            else:
                positions.append((1, 2))
                positions.append((n, n + 1))

        positions = merge_positions(positions)

        print(sum([r - l for l, r in positions]), end=" ")
        # print(positions)
    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:
```plaintext
1
5 3 3
1 3 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:
t=int(input())
for _ in range(t):
    n,m,q = map(int, input().split())
    a = list(map(int, input().split()))

    visited = set()
    ans = [0]*q
    ans[0] = 2
    for i in range(1,q):
        if i == 1 and a[i] == m and a[i] not in visited:
            ans[i] = ans[i-1]+2
        elif i == 1 and a[i] not in visited:
            ans[i] = ans[i-1]+1
        elif i> 1 and a[i] not in visited:
           ans[i] = ans[i-1]+2
        else :
            ans[i] = ans[i-1]
        ans[i] = min(ans[i], n)
        visited.add(a[i])

    print(" ".join(map(str, ans)))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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:
# cook your dish here
t=int(input())
for _ in range(t):
    n,m,q=map(int,input().split())
    L=list(map(int,input().split()))
    vis=[False for i in range(n+1)]
    e1,e2=0,n+1
    i=1
    ans=[]
    if L[0]==m:
        e1=1
        e2=n
        ans.append(2)
        vis[e1]=True
        vis[e2]=True
        while i<q and ans[-1]!=n:
            if L[i]<=e1:
                z=0
                e2-=1
                e2=max(e2,1)
                if vis[e2]==False:
                    z+=1
                vis[e2]=True
                ans.append(ans[-1]+z)
            elif L[i]>=e2:
                z=0
                e1+=1
                e1=min(e1,n)
                if vis[e1]==False:
                    z+=1
                vis[e1]=True
                ans.append(ans[-1]+z)
            else:
                z=0
                e1+=1
                e2-=1
                e1=min(e1,n)
                e2=max(e2,1)
                if vis[e1]==False:
                    z=1
                    vis[e1]=True
                if vis[e2]==False:
                    vis[e2]=True
                    z+=1
                ans.append(ans[-1]+z)
            i+=1
        while i<q:
            ans.append(ans[-1])
            i+=1
        print(*ans)

    else:
        i=0
        vis[m]=True
        v1,v2=m,m
        ans=[1]
        while i<q and ans[-1]!=n:
            if vis[L[i]]==False:
                z=0
                if L[i]>v2:
                    v2+=1
                    v2=min(v2,n)
                    if vis[v2]==False:
                        z+=1
                    vis[v2]=True
                else:
                    v1-=1
                    v1=max(v1,1)
                    if vis[v1]==False:
                        z+=1
                    vis[v1]=True
                if e2<n+1:
                    e2-=1
                    e2=max(e2,1)
                    if vis[e2]==False:
                        z+=1
                    vis[e2]=True
                if e1>0:
                    e1+=1
                    e1=min(e1,n)
                    if vis[e1]==False:
                        z+=1
                    vis[e1]=True
                ans.append(ans[-1]+z)
            elif L[i]<=e1:
                z=0
                e2-=1
                e2=max(e2,1)
                if vis[e2]==False:
                    z+=1
                v1-=1
                v1=max(v1,1)
                if vis[v1]==False:
                    z+=1
                vis[e2]=True
                vis[v1]=True
                ans.append(ans[-1]+z)
            elif L[i]>=e2:
                z=0
                e1+=1
                e1=min(e1,n)
                if vis[e1]==False:
                    z+=1
                v2+=1
                v2=min(v2,n)
                if vis[v2]==False:
                    z+=1
                vis[e1]=True
                vis[v2]=True
                ans.append(ans[-1]+z)
            else:
                z=0
                e1+=1
                e1=min(e1,n)
                if vis[e1]==False:
                    z+=1
                vis[e1]=True
                e2-=1
                e2=max(e2,1)
                if vis[e2]==False:
                    z+=1
                vis[e2]=True
                ans.append(ans[-1]+z)
            i+=1
        while i<q:
            ans.append(ans[-1])
            i+=1
        for i in range(1,len(ans)):
            print(ans[i],end=' ')
        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:
```plaintext
5
6 1 3
1 2 3
2 1 4
1 1 1 2
5 1 1
1
3 1 4
1 1 1 1
18 1 4
1 2 1 18
```


**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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.SortedSet;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.LinkedBlockingQueue;

public class Joker {
    private boolean debug = false;
    private PrintWriter outWriter;

    public static void main(String[] args) {
        new Joker().solve();
    }

    public void solve() {
        MyScanner sc = new MyScanner();
        outWriter = new PrintWriter(new BufferedOutputStream(System.out));
        int testCases = sc.nextInt();
        for (int t = 0; t < testCases; t++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();

            int[] a = new int[q];
            LinkedBlockingQueue<PairIntegers> queue = new LinkedBlockingQueue<PairIntegers>();
            queue.add(new PairIntegers(m, m));
            for (int i = 0; i < q; i++) {
                a[i] = sc.nextInt();
                PriorityQueue<PairIntegers> pq = new PriorityQueue<>();
                while (queue.size() > 0) {
                    PairIntegers pair = queue.poll();
                    if (a[i] < pair.left) {
                        pq.add(new PairIntegers(Math.max(1, pair.left - 1), pair.right));
                    } else if (a[i] > pair.right) {
                        pq.add(new PairIntegers(pair.left, Math.min(pair.right + 1, n)));
                    } else {
                        pq.add(new PairIntegers(1, 1));
                        pq.add(new PairIntegers(n, n));
                        if (a[i] == pair.left && a[i] == pair.right) {
                            continue;
                        }
                        if (a[i] == pair.right || a[i] == pair.left) {
                            pq.add(new PairIntegers(pair.left, pair.right));
                        } else {
                            pq.add(new PairIntegers(Math.max(1, pair.left - 1), Math.min(n ,pair.right + 1)));
                        }
                    }
                }
                PairIntegers currentPair = pq.poll();
                int answer = 0;
                while (pq.size() > 0) {
                    PairIntegers nextPair = pq.poll();
                    if (nextPair.left <= currentPair.right + 1) {
                        currentPair.right = Math.max(currentPair.right, nextPair.right);
                    } else {
                        queue.add(currentPair);
                        answer += (currentPair.right - currentPair.left + 1);
                        currentPair = nextPair;
                    }
                }
                queue.add(currentPair);
                answer += (currentPair.right - currentPair.left + 1);
                outWriter.print(answer + " ");
            }
            outWriter.println();
        }
        outWriter.flush();
    }

    class PairIntegers implements Comparable<PairIntegers> {
        int left;
        int right;

        public PairIntegers(int x, int y) {
            this.left = x;
            this.right = y;
        }

        @Override
        public int compareTo(PairIntegers o) {
            int result = Integer.compare(left, o.left);
            if (result == 0) {
                result = Integer.compare(right, o.right);
            }
            return result;
        }
    }

    public void debugPrintln(String str) {
        debugPrint(str.trim() + "\n");
    }

    public void debugPrint(String str) {
        if (debug) {
            outWriter.print(str);
        }
    }

    static final HashMap<Integer, Long> savedFactorials = new HashMap<>();

    public long ncr(int n, int r) {
        long mod = 998244353;
        savedFactorials.put(1, 1L);
        // f[n, r] = n! / {(n-r)! * r!}
        long res = 1;
        for (long x = n; x > n - r; x--) {
            res = (res * x) % mod;
        }
        long rFactorial = rFactorial(r);
        // debugPrintln(String.format("calculating (%d)C(%d): numerator: %d,
        // denominator: %d, result: %d", n, r, res, rFactorial, modDivision(res,
        // rFactorial, mod)));
        return modDivision(res, rFactorial, mod);
    }

    long rFactorial(int r) {
        if (savedFactorials.containsKey(r)) {
            return savedFactorials.get(r);
        }
        return r * rFactorial(r - 1);
    }

    long modDivision(long a, long b, long primeMod) {
        return a * modInverse(b, primeMod) % primeMod;
    }

    long modInverse(long b, long primeMod) {
        // b^(m-2) mod primeMod
        // using modular exponentation (divide & conquer)
        long result = 1;
        long base = b % primeMod; // Reduce base modulo m initially
        long e = primeMod - 2;
        while (e > 0) {
            if (e % 2 == 1) // If e is odd
                result = (result * base) % primeMod;
            base = (base * base) % primeMod; // Square the base
            e /= 2; // Divide exponent by 2
        }
        // debugPrintln(String.format("Modular inverse of %d is %d", b, result));
        return result;
    }

    class SieveOfErastothenes {
        List<Integer> primes;
        int limit;
        int[] arr;
        Map<Integer, List<Integer>> factors;

        public SieveOfErastothenes(int limit) {
            this.arr = new int[limit + 1];
            this.limit = limit;
            this.factors = new HashMap<>();
        }

        public Map<Integer, List<Integer>> getFactors() {
            for (int i = 2; i <= limit; i++) {
                for (int j = 1; i * j <= limit; j++) {
                    factors.putIfAbsent(i * j, new ArrayList<>());
                    factors.get(i * j).add(i);
                }
            }
            return factors;
        }

        public List<Integer> getPrimes() {
            List<Integer> result = new ArrayList<>();
            int i;
            for (i = 2; i <= Math.sqrt(limit); i++) {
                if (arr[i] == 0) {
                    result.add(i);
                    for (int j = 1; j * i <= limit; j++) {
                        arr[j * i] = -1; // j * i == xyz
                    }
                }
            }
            while (i <= limit) {
                if (arr[i] == 0)
                    result.add(i);
                i++;
            }
            return result;
        }
    }

    class DisJointSetUnion {
        int[] arr;

        public DisJointSetUnion(int size) {
            size++;
            this.arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = i;
            }
        }

        public Map<Integer, Set<Integer>> expandSets() {
            Map<Integer, Set<Integer>> result = new HashMap<>();
            for (int i = 1; i < arr.length; i++) {
                int parent = find(i);
                if (!result.containsKey(parent)) {
                    result.put(parent, new HashSet<>());
                }
                result.get(parent).add(i);
                result.put(i, result.get(parent));
            }
            return result;
        }

        public int find(int i) {
            if (arr[i] == i) {
                return i;
            }
            arr[i] = find(arr[i]);
            return arr[i];
        }

        public int union(int i, int j) {
            int iKaBaap = find(i);
            int jKaBaap = find(j);
            if (iKaBaap > jKaBaap) {
                arr[iKaBaap] = jKaBaap;
            } else {
                arr[jKaBaap] = iKaBaap;
            }
            return jKaBaap;
        }
    }

    // -----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    // -----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }

    }
    // --------------------------------------------------------
}

**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 3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Temp {

    private void solve() throws IOException {
        int t = nextInt();
        while (t-- > 0) {
            long n = nextInt();
            long m = nextInt();
            long q = nextInt();
            long[] a = new long[300005];
            for (int i = 0; i < q; i++) {
                a[i] = nextInt();
            }
            // n为1到n张牌， m为小丑牌的位置，q代表q次操作
            // right表示当前节点最右端的点，left表示最左边的点
            // 每次只遍历端点

            Set<Long> root = new HashSet<>();
            long count = 0;
            root.add(m);
            boolean flag = true;
            long head = -1;
            long midl = m;
            long midr = m;
            long tail = -2;
            long heart = -3;
            List<Long> ans = new ArrayList<>();
            for (int i = 0; i < q; i++) {
                if (root.contains(a[i])) {
                    if (head == -1) {
                        head = 0;
//                        root.add(head);
                    }
                    if (tail == -2) {
                        tail = n+1;
//                        root.add(tail);
                    }

                }
                if (head != -1 && head< a[i]) {
                    head = Math.min(n, head + 1);
                    root.add(head);
                }
                if (tail != -2 && tail> a[i]) {
                    tail = Math.max(1, tail - 1);
                    root.add(tail);
                }
                if (i == 0 && root.contains(a[i])) {
                    if (a[i] != 1 && a[i] != n) {
                        root.remove(a[i]);
                    }

                    flag = false;
                    ans.add((long) root.size());
                    continue;
                }
                if (flag) {
                    if (a[i] < midl) {
                        midl = Math.max(1, midl - 1);
                        root.add(midl);
                    }
                    if (a[i] > midr) {
                        midr = Math.min(n, midr + 1);
                        root.add(midr);
                    }
                }
                ans.add((long) root.size());
            }
            for (int i = 0; i < ans.size(); i++) {
                System.out.print(ans.get(i)+" ");
            }
            System.out.println();

        }

    }

    public static void main(String[] args) throws IOException {
        new Temp().run();
        // 使用BufferedReader和BufferedWriter提高输入输出效率

    }

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            writer = new PrintWriter(System.out);
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

/*
1
6 1 3
1 1 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
6 1 3
1 1 1
```


**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 fastio ios_base::sync_with_stdio(0);    cin.tie(0);
#define ff first
#define ss second
//#define endl "\n"
using namespace std;
const int MAX=1e4+5;
int A[MAX];
int B[MAX];

void go(){
    int n,m,q;  cin>>n>>m>>q;
    set<pair<int,int>>s;
    s.insert({m,m});

    while(q--){
        int a;  cin>>a;
        set<pair<int,int>>w;
        bool one=false;
        bool nn=false;
        for(auto it:s){
            //cout<<it.ff<<" "<<it.ss<<endl;
            if(it.ff==1 and it.ff!=1)    one=true;
            if(it.ss==n and it.ff!=1)    nn=true;
        }
        for(auto it:s){
            if(a<it.ff){
                int x=it.ff;
                int y=it.ss;
                w.insert({x-1,y});
                if(x==2)    one=true;

            }
            else if(a>it.ss){
                int x=it.ff;
                int y=it.ss;
                w.insert({x,y+1});
                if(y==n-1)  nn=true;

            }
            else{
                w.insert(it);
                if(it.ff==1 and it.ss==n)   continue;
                if(it.ff==it.ss){

                    w.erase(it);
                }
                if(!one){
                    w.insert({1,1});
                }
                if(!nn){
                    w.insert({n,n});
                }
            }
        }
        s=w;
        int ans=0;
        int awa=0;
        for(auto it:s){
            awa=max(awa,it.ff-1);
            ans+=max(0,it.ss-awa);
            awa=max(awa,it.ss);
        }
        cout<<ans<<" ";
    }
    cout<<endl;
}

int main(){
    fastio;
    int t;  cin>>t;
    while(t--)  go();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 5 3
1 10 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>
//#include<iostream>
//#include<stdio.h>
//#include<vector>
//#include<algorithm>
//#include<cmath>
//#include<map>
using namespace std;

#define int long long
const int mod = 998244353;
const int N = 200005;
const int P = 31;
typedef pair<int, int> PII;

int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};

int ksm(int a, int b)
{
    int res = 1;
    while(b)
    {
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0' && ch<='9')
        x=x*10+ch-'0',ch=getchar();
    return x*f;
}

inline void writeS(std::string s){
    int len=s.length();
    for(int i=0;i<len;i++)
        putchar(s[i]);
}

const double pi = acos(-1.0);
const double eps = 1e-8;
const double INF = 1e18;

//判断浮点数大小
int sgn(double x) {
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}
// 比较浮点数
int dcmp(double x, double y) {
    if(fabs(x - y) < eps) return 0;
    else return x < y ? -1 : 1;
}

void solve()
{
    int n, m, q;
    cin >> n >> m >> q;
    int ans = 0;
    int x;
    vector<PII> a(3);
    a[0] = {1,-1}, a[1] = {m, -m}, a[2] = {n, -n};
    auto change = [&](PII r)
    {
        PII t;
        if(x == r.first || x == r.second) t = r;
        else if(x < r.first) t = {max(r.first - 1, (int)1), r.second};
        else if(x > r.second) t = {r.first, min(r.second + 1, n)};
        else t = {max(r.first - 1, (int)1), min(r.second + 1, n)};
        return t;
    };
    int f;
    for(int i = 1; i <= q; i ++)
    {
        cin >> x;
        ans = 0;
        if(i == 1)
        {
            if(x == m) f = 0, a[0] = {1, 1}, a[2] = {n, n}, cout << 2 << " ";
            else
            {
                f = 1;
                a[1] = {m, m};
                for(int j = 0; j < 3; j ++)
                {
                    if(a[j].first <= a[j].second) a[j] = change(a[j]), ans += a[j].second - a[j].first + 1;
                }
                cout << ans << " ";
            }
//            cout << endl;
//            for(int j = 0; j < 3; j ++) cout << a[j].first << " " << a[j].second << endl;
            continue;
        }
        if(f)
        {
            for(int j = 0; j < 3; j ++) if(a[j].first <= a[j].second) a[j] = change(a[j]);
            if(x >= a[1].first && x <= a[1].second && a[0].first > a[0].second) a[0] = {1,1}, a[2] = {n,n};
            int ed = 0;
            for(int j = 0; j < 3; j ++)
                if(a[j].first <= a[j].second)
                {
                    ans += a[j].second - a[j].first + 1;
                    if(ed >= a[j].first) ans -= (ed - a[j].first + 1);
                    ed = a[j].second;
                }
            cout << ans << " ";
        }
        else
        {
            for(int j = 0; j < 3; j ++) if(a[j].first <= a[j].second) a[j] = change(a[j]);
            int ed = 0;
            for(int j = 0; j < 3; j ++)
                if(a[j].first <= a[j].second)
                {
                    ans += a[j].second - a[j].first + 1;
                    if(ed >= a[j].first) ans -= (ed - a[j].first + 1);
                    ed = a[j].second;
                }
            cout << ans << " ";
        }
//        cout << endl;
//        for(int j = 0; j < 3; j ++) cout << a[j].first << " " << a[j].second << endl;
    }
    cout << endl;
}

signed main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int T = 1;

    cin >> T;
//    scanf("%d", &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
6 3 3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 sys
input = sys.stdin.readline

def solve():
    n, m, q = map(int, input().split())
    moves = list(map(int, input().split()))

    upper_range = [0, 0]
    mid_range = [m - 1, m]
    lower_range = [n, n]
    result = []
    for i in range(q):
        card = moves[i] - 1
        if mid_range and card in range(*mid_range):
            if mid_range[0] == card and mid_range[1] - 1 == card:
                mid_range = None
            elif mid_range[0] == card or mid_range[1] - 1 == card:
                pass
            else:
                mid_range = [mid_range[0] - 1, mid_range[1] + 1]
            upper_range[1] += 1
            lower_range[0] -= 1
        elif card in range(*upper_range):
            if mid_range:
                mid_range[0] -= 1
            lower_range[0] -= 1
        elif card in range(*lower_range):
            if mid_range:
                mid_range[1] += 1
            upper_range[1] += 1
        else:
            if mid_range and card < mid_range[0]:
                mid_range[0] -= 1
            elif mid_range and mid_range[1] <= card:
                mid_range[1] += 1
            if upper_range[1] - upper_range[0] > 0:
                upper_range[1] += 1
            if lower_range[1] - lower_range[0] > 0:
                lower_range[0] -= 1

        if mid_range and upper_range[1] >= mid_range[0]:
            upper_range[1] = mid_range[1]
            mid_range = None
        if mid_range and lower_range[0] <= mid_range[1]:
            lower_range[0] = mid_range[0]
            mid_range = None

        num = 0
        if upper_range[1] >= lower_range[0]:
            num = n
        else:
            if mid_range:
                num += mid_range[1] - mid_range[0]
            num += lower_range[1] - lower_range[0]
            num += upper_range[1] - upper_range[0]
        result.append(num)

    print(*result, sep=' ')

T = int(input().strip())
for _ in range(T):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
10 1 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:
# cook your dish here
t=int(input())
for _ in range(t):
    n,m,q=map(int,input().split())
    L=list(map(int,input().split()))
    vis=[False for i in range(n+1)]
    e1,e2=0,n+1
    i=1
    ans=[]
    if L[0]==m:
        e1=1
        e2=n
        ans.append(2)
        vis[e1]=True
        vis[e2]=True
        while i<q and e1<e2:
            if L[i]<=e1:
                e2-=1
                if vis[e2]:
                    break
                vis[e2]=True
                ans.append(ans[-1]+1)
            elif L[i]>=e2:
                e1+=1
                if vis[e1]:
                    break
                vis[e1]=True
                ans.append(ans[-1]+1)
            else:
                e1+=1
                e2-=1
                if vis[e1] or vis[e2]:
                    break
                vis[e1]=True
                vis[e2]=True
                ans.append(ans[-1]+2)
            i+=1
        while i<q:
            ans.append(ans[-1])
            i+=1
        print(*ans)

    else:
        i=0
        vis[m]=True
        v1,v2=m,m
        ans=[1]
        while i<q and ans[-1]!=n:
            if vis[L[i]]==False:
                z=0
                if L[i]>v2:
                    v2+=1
                    if vis[v2]==False:
                        z+=1
                    vis[v2]=True
                else:
                    v1-=1
                    if vis[v1]==False:
                        z+=1
                    vis[v1]=True
                if e2<n+1:
                        e2-=1
                        if vis[e2]==False:
                            z+=1
                        vis[e2]=True
                if e1>0:
                    e1+=1
                    if vis[e1]==False:
                        z+=1
                    vis[e1]=True
                ans.append(ans[-1]+z)
            elif L[i]<=e1:
                z=0
                e2-=1
                if vis[e2]==False:
                    z+=1
                v1-=1
                if vis[v1]==False:
                    z+=1
                vis[e2]=True
                vis[v1]=True
                ans.append(ans[-1]+z)
            elif L[i]>=e2:
                z=0
                e1+=1
                if vis[e1]==False:
                    z+=1
                v2+=1
                if vis[v2]==False:
                    z+=1
                vis[e1]=True
                vis[v2]=True
                ans.append(ans[-1]+z)
            else:
                z=0
                e1+=1
                e2-=1
                if vis[e1]==False:
                    z+=1
                if vis[e2]==False:
                    z+=1
                vis[e1]=True
                vis[e2]=True
                ans.append(ans[-1]+z)
            i+=1
        while i<q:
            ans.append(ans[-1])
            i+=1
        for i in range(1,len(ans)):
            print(ans[i],end=' ')
        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:
```plaintext
1
10 5 5
5 1 10 6 4
```