**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, 378QAQ gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha thinks that $a$ is beautiful if there exist two numbers $i$ and $j$ ($1\leq i,j\leq n$, $i\neq j$) such that for all $k$ ($1 \leq k \leq n$), $a_k$ is divisible$^\dagger$ by either $a_i$ or $a_j$.

Determine whether $a$ is beautiful.

$^\dagger$ $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y \cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output "Yes" if array $a$ is beautiful, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
4
3
7 3 8
5
7 1 9 3 5
5
4 12 2 6 3
5
7 49 9 3 1000000000
output:
No
Yes
Yes
No

Note: In the first test case, any two numbers in the array are coprime, so the answer is "No".

In the second test case, we can pick $i=2$ and $j=1$. Since every number in the array is divisible by $a_i = 1$, the answer is "Yes".

In the third test case, we can pick $i=3$ and $j=5$. $2$ and $4$ is divisible by $a_i = 2$ while $3$, $6$ and $12$ is divisible by $a_j = 3$, so the answer is "Yes".

Buggy Code:
for _ in range(int(input().strip())):
    l = int(input().strip())
    arr = [int(x) for x in input().split()]
    arr1 = set(arr)
    arr1 = list(arr1)
    first = arr1[0]
    second = 0
    ans = "YES"
    for i in range(1, len(arr1)):
        if arr1[i] % first != 0:
            second =  arr1[i]
            break
    for i in range(l):
        if (arr[i] % first != 0) and (arr[i] % second != 0):
            ans = "NO"
            break
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, 378QAQ gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha thinks that $a$ is beautiful if there exist two numbers $i$ and $j$ ($1\leq i,j\leq n$, $i\neq j$) such that for all $k$ ($1 \leq k \leq n$), $a_k$ is divisible$^\dagger$ by either $a_i$ or $a_j$.

Determine whether $a$ is beautiful.

$^\dagger$ $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y \cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output "Yes" if array $a$ is beautiful, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
4
3
7 3 8
5
7 1 9 3 5
5
4 12 2 6 3
5
7 49 9 3 1000000000
output:
No
Yes
Yes
No

Note: In the first test case, any two numbers in the array are coprime, so the answer is "No".

In the second test case, we can pick $i=2$ and $j=1$. Since every number in the array is divisible by $a_i = 1$, the answer is "Yes".

In the third test case, we can pick $i=3$ and $j=5$. $2$ and $4$ is divisible by $a_i = 2$ while $3$, $6$ and $12$ is divisible by $a_j = 3$, so the answer is "Yes".

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2024/6/8 14:37
 * @description https://codeforces.com/problemset/problem/1975/B
 */
public class Main {
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int N = 100010, INF = 0x3f3f3f3f;
    static int[] a = new int[N];
    static boolean[] st = new boolean[N];
    static int n;

    public static void main(String[] args) {
        int T = sc.nextInt();
        while (T-- > 0) {
            n = sc.nextInt();
            Arrays.fill(st, 1, n + 1, false);
            int minV = INF;
            for (int i = 1; i <= n; i++) {
                a[i] = sc.nextInt();
                minV = Math.min(a[i], minV);
            }
//            Arrays.sort(a, 1, n + 1);

            if (minV == 1) out.println("Yes");
            else {
                int minV2 = INF;
                for (int i = 1; i <= n; i++) {
                    if (a[i] % minV > 0) {
                        st[i] = true;
                        minV2 = Math.min(a[i], minV2);
                    }
                }
                boolean isFill = false;
                for (int i = 2; i <= n; i++) {
                    if (st[i] && a[i] % minV2 > 0) {
                        isFill = true;
                        break;
                    }
                }
                if (isFill) out.println("No");
                else out.println("Yes");
            }
            out.flush();
        }
    }
}

class FastReader {
    StringTokenizer st;
    BufferedReader br;

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

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

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

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

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

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

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The buggy code has a few potential issues:

1. **Handling of the minimum value**: The code assumes that if the minimum value in the array is 1, the array is automatically beautiful. This is correct, but it doesn't handle other edge cases well.
2. **Handling of the second minimum value**: The code tries to find a second minimum value that is not divisible by the first minimum value. This logic can fail in certain cases where the array has multiple elements that are not divisible by the first minimum value but still form a valid beautiful array.

Let's construct a test case that exposes these issues:

### Test Case:
We will create an array where the minimum value is not 1, and the second minimum value is not easily identifiable. This will test the logic for finding the second minimum value and checking divisibility.

```plaintext
1
5
2 4 6 8 10
```

### Explanation:
- **Array**: [2, 4, 6, 8, 10]
- **Minimum value**: 2
- **Second minimum value**: 4

In this array:
- The minimum value is 2.
- Every element in the array is divisible by 2.
- There is no need for a second minimum value to check divisibility because all elements are already divisible by 2.

However, the buggy code will try to find a second minimum value and check if any element is not divisible by both 2 and 4. Since all elements are divisible by 2, the code should correctly output "Yes". But if the code has a bug in handling this scenario, it might incorrectly output "No".

### Expected Output:
```plaintext
Yes
```

This test case should help expose any issues in the buggy code's handling of arrays where the minimum value is not 1 but all elements are divisible by it.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, 378QAQ gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha thinks that $a$ is beautiful if there exist two numbers $i$ and $j$ ($1\leq i,j\leq n$, $i\neq j$) such that for all $k$ ($1 \leq k \leq n$), $a_k$ is divisible$^\dagger$ by either $a_i$ or $a_j$.

Determine whether $a$ is beautiful.

$^\dagger$ $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y \cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output "Yes" if array $a$ is beautiful, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
4
3
7 3 8
5
7 1 9 3 5
5
4 12 2 6 3
5
7 49 9 3 1000000000
output:
No
Yes
Yes
No

Note: In the first test case, any two numbers in the array are coprime, so the answer is "No".

In the second test case, we can pick $i=2$ and $j=1$. Since every number in the array is divisible by $a_i = 1$, the answer is "Yes".

In the third test case, we can pick $i=3$ and $j=5$. $2$ and $4$ is divisible by $a_i = 2$ while $3$, $6$ and $12$ is divisible by $a_j = 3$, so the answer is "Yes".

Buggy Code:
import java.util.*;
public class frough {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0) {
            int n =sc.nextInt();
            int arr[]=new int[n];
            int min=Integer.MAX_VALUE;
            int secMin=Integer.MAX_VALUE;
            for(int i =0;i<n;i++){
                arr[i]=sc.nextInt();
                if(arr[i]<min){
                    secMin=min;
                    min=arr[i];
                }else if(secMin>arr[i] && min!=arr[i]){
                    secMin=arr[i];
                }
            }
            if(secMin==Integer.MAX_VALUE){
                secMin=min;
            }
            int flag=0;
            for(int i =0;i<n;i++){
                if(arr[i]%min!=0 && arr[i]%secMin!=0){
                    flag=1;
                    break;
                }
            }
            if(flag==0){
                System.out.println("YES");
            }else{
                System.out.println("No");
            }
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to craft a test case that highlights the issue in the logic. The buggy code assumes that the smallest and second smallest elements are the only candidates for the beauty condition. However, this is not always correct, as the beauty condition requires that every element in the array is divisible by either of two specific elements, not just the smallest and second smallest.

### Steps to Craft the Test Input:
1. **Identify the Bug**: The code fails when the smallest and second smallest elements are not sufficient to make the array beautiful. Specifically, it fails when there are other elements that are not divisible by these two but can be made beautiful with other pairs.
2. **Create a Test Case**:
   - Use a small array where the smallest and second smallest elements do not satisfy the beauty condition.
   - Ensure that there are other pairs of elements that do satisfy the condition.

### Example Test Case:
Consider an array where the smallest and second smallest elements are not sufficient to make the array beautiful, but another pair of elements can make it beautiful.

```plaintext
1
5
2 4 6 8 12
```

### Explanation:
- **Array**: [2, 4, 6, 8, 12]
- **Smallest Element**: 2
- **Second Smallest Element**: 4
- **Check**:
  - 2 is divisible by 2.
  - 4 is divisible by 2.
  - 6 is divisible by 2.
  - 8 is divisible by 2.
  - 12 is divisible by 2.
- **Conclusion**: All elements are divisible by 2, so the array is beautiful. However, the code will incorrectly check if all elements are divisible by 2 and 4, which is not necessary.

This test case should expose the bug in the code, as the code will incorrectly output "No" instead of "Yes".

### Output:
```plaintext
1
5
2 4 6 8 12
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
// Problem: D. Med-imize
// Contest: Codeforces - Codeforces Round 963 (Div. 2)
// URL: https://codeforces.com/contest/1993/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

ll lower(ll l, ll r, ll target, function<ll(ll)> f) {
	if(f(r) < target) return r + 1;
	while(l < r) {
		ll mid = (l + r) / 2;
		if(f(mid) < target) {
			l = mid + 1;
		} else {
			r = mid;
		}
	}
	return l;
}
void solve() {
	int n, k; cin >> n >> k;
	int a[n + 1];
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}

	cout << lower(1, 1e9, 1, [&](int limit){

		vector<int> dp(n + 1, -1e9);
		vector<int> maxv(k + 1, -1e9);

		int r = n % k;
		if(r == 0) r = k;

		maxv[0] = 0;
		// 控制 "0 -> 1"
		// for(int i = 1; i <= n; i++) {
//
		// }
		int maxt = -1e9;
		for(int i = 1; i <= n; i++) {
			if(i % k > r) continue;
			if(n % k == 0 && i % k == 0) {
				maxt = max(maxt, maxv[(i - 1) % k] + (a[i] >= limit ? 1 : -1));
			} else {
				dp[i] = maxv[(i - 1) % k] + (a[i] >= limit ? 1 : -1);
				maxv[i % k] = max(maxv[i % k], dp[i]);
			}
		}
		// cout << limit << " " << maxt << '\n';
		if(n % k == 0) return maxt <= 0;
		return maxv[n % k] <= 0;
	}) - 1 << '\n';
}

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

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

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

Think step by step.

Response:
```plaintext
1
5 3
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: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

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

bool pr(int c, int k, vector<long long> &a) {
    int n = a.size();
    long long di = 0;
    vector<long long> dp(n+1, 0);
    for (int i = 0; i < n; i++) {
        if (a[i] >= c) di++;
        else di--;
    }
    long long curr = 0; // maggiori di c
    for (int i = 0; i < k; i++) {
        if (a[i] >= c) curr++;
    }
    for (int i = k; i <= n; i++) {
        dp[i] = max(dp[i-1], k-2*curr+dp[i-k]);
        if (a[i-k] >= c) curr--;
        if (i < n && a[i] >= c) curr++;
    }
    if (di+dp[n] > 0) return 1;
    return 0;
}

void solve() {
    long long n, k; cin >> n >> k;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    long long l = 0, r = 1e9+7;
    while (l < r-1) {
        long long m = (l+r)/2;
        if (pr(m, k, a)) l = m;
        else r = m;
    }
    cout << l << "\n";
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) solve();
}

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

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

Think step by step.

Response:
```plaintext
3
5 3
1 2 3 4 5
6 2
1 2 3 4 5 6
8 4
1 2 3 4 5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
#include <bits/stdc++.h>
const int maxn = 5e5 + 5;
int a[maxn], p[maxn], b[maxn];
void solve()
{
	int n, k;
	std::cin >> n >> k;
	for (int i = 1; i <= n; i++) std::cin >> a[i];
	if (n < k)
	{
		std::sort(a + 1, a + n + 1);
		std::cout << a[n / 2 + n % 2] << "\n";
		return;
	}
	for (int i = 1; i <= n; i++) p[i] = i;
	std::sort(p + 1, p + n + 1, [&](int x, int y)
	{
		return a[p[x]] < a[p[y]];
	});
	std::map<int, int> mp;
	int m = (n % k > 0 ? n % k : k);
	int L = m / 2 + m % 2 - 1;
	int R = m - L - 1;
	int ans = 0;
	std::map<int, int> check;
	for (int i = 1; i <= n; i++)
	{
		if ((p[i] - 1) % k + (n - p[i]) % k != L + R) continue;
		int tt = p[i] % k;
		if (mp[(tt - 1 + k) % k] >= L)
		{
			check[p[i]] += 1;
		}
		mp[tt]++;
	}
	mp.clear();
	for (int i = n; i >= 1; i--)
	{
		if ((p[i] - 1) % k + (n - p[i]) % k != L + R) continue;
		int tt = p[i] % k;
		if (mp[(tt - 1 + k) % k] >= R)
		{
			check[p[i]] += 1;
		}
		mp[tt]++;
	}
	for (auto iter : check)
	{
		if (iter.second == 2)
		{
			ans = std::max(ans, a[iter.first]);
		}
	}
	std::cout << ans << "\n";
}
int main ()
{
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	int _;
	std::cin >> _;
	while (_--)
	{
		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
5
4 2
1 5 2 4
5 2
1 2 3 4 5
6 3
1 2 3 4 5 6
7 2
1 2 3 4 5 6 7
8 3
1 2 3 4 5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
ans = []
for _ in range(int(input())):
	n,k = [int(y) for y in input().split()]
	a = [int(x) for x in input().split()]

	end = 1+((n-1)%k)
	L = 1
	R = 10**9 + 9

	while L < R:
		X = (L+R)//2

		v = [0]*n
		for res in range(end-1,-1,-1):
			st = (n//k + 1)*k + res
			while st >= n:
				st -= k
			f = 0
			for i in range(st,-1,-k):
				if a[i] >= X:
					f = 1
				v[i] = f
				if i%k < end-1:
					v[i] += v[i+1]

		if v[0] >= end//2 + 1:
			L = X+1
		else:
			R = X
	ans.append(L-1)

for z in ans:
	print(z)

**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
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
import random
from heapq import heapify, heappop, heappush

# xor = random.randint(10**7, 10**8)
xor = 0

def median_achievable(target, a, k):
    n = len(a)
    a = a.copy()
    larger_count = 0
    smaller_count = 0
    count_removed = [0 for i in range(n)]
    count_removed_big = [0 for i in range(n)]
    above_target = 0
    below_target = 0
    for i in range(n):
        #
        if a[i] >= target:
            above_target += 1
            larger_count += 1
        else:
            below_target += 1
            smaller_count += 1
        #
        if larger_count + smaller_count > k:
            removed_ele_idx = i - k
            removed_ele = a[removed_ele_idx]
            if removed_ele >= target:
                larger_count -= 1
            else:
                smaller_count -= 1
        #
        if larger_count + smaller_count == k:
            if smaller_count > larger_count:
                # count_removed[i] = (count_removed[i - k] if i >= k else 0) + smaller_count
                choice1 = (count_removed[i - k] if i >= k else 0) + smaller_count
                choice2 = count_removed[i-1] if i >= 1 else 0
                if choice1 > choice2:
                    count_removed[i] = choice1
                    count_removed_big[i] = larger_count + (count_removed_big[i-k] if i >= k else 0)
                else:
                    count_removed[i] = choice2
                    count_removed_big[i] = count_removed_big[i-1] if i >= 1 else 0
            else:
                count_removed[i] = count_removed[i-1] if i >= 1 else 0
                count_removed_big[i] = count_removed_big[i-1] if i >= 1 else 0

    # print(count_removed, count_removed_big, a, k, target)
    if above_target - count_removed_big[-1] > below_target - count_removed[-1]:
        return True
    else:
        return False

def solve():
    n, k = list(map(int, input().split()))
    a = list(map(int, input().split()))
    head = min(a) # true
    tail = max(a)+1 # false
    while head+1 != tail:
        mid = (head + tail) // 2
        if median_achievable(mid, a, k):
            head = mid
        else:
            tail = mid
    print(head, end=' ')

for case in range(int(input())):
    if case == 267:
        print()
        n, k = list(map(int, input().split()))
        a = list(map(int, input().split()))
        print(n, k, a)
        exit()
    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
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
from re import I
import sys
import os.path
from collections import *
from math import *

if os.path.exists("input.txt"):
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("output.txt", "w")

input = sys.stdin.readline

##########################################################

mod = 10 ** 9 + 7

def modularExponentiation(a,b):
    if b == 0:
        return 1
    if b % 2 == 0:
        return modularExponentiation((a * a) % mod, b // 2)
    return (a * modularExponentiation((a * a) % mod,(b - 1) // 2)) % mod

def solve():
    n, k = [int(x) for x in input().split()]
    arr = [int(x) for x in input().split()]
    val = 0
    if n <= k:
        arr.sort()
        print(arr[(n - 1)// 2])
        return
    if n % k == 0:
        val = k
    else:
        val = n % k
    l = (val - 1) // 2
    r = val // 2
    val1 = 0
    val2 = 0
    for i in range(n):
        if i % k == l:
            val1 = max(val1, arr[i])
        if i % k == r:
            val2 = max(val2, arr[i])
    print(min(val1, val2))

##########################################################

t = 1
t = int(input())
while t:
    t -= 1
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
//  package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF25 {

    public static long findElementOrNext(long[] arr, long target) {
        int low = 0;
        int high = arr.length - 1;
        long result = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;

            if (arr[mid] == target) {
                return arr[mid];
            } else if (arr[mid] < target) {
                low = mid + 1;
            } else {
                result = arr[mid];
                high = mid - 1;
            }
        }

        return result;
    }

    public static boolean isValid(int[] arr, int idx, int K){

        int[] dp = new int[arr.length];

        for(int i = 0;i<K;i++) dp[i] = arr[i] >= arr[idx] ? 1 : -1;

        for(int i = K;i<arr.length;i++) {

            if( i % K == 0) {
                dp[i] = Math.max(arr[i] >= arr[idx] ? 1 : -1 , dp[i-K]);
            } else {
                int temp = arr[i] >= arr[idx] ? 1 : -1;
                int temp2 = dp[i-1];
                int temp3 = temp + temp2;
                int temp4 = dp[i-K];
                dp[i] = Math.max(temp3 , temp4);
            }
        }

        return dp[arr.length-1] > 0;
    }

    public static void task() throws IOException{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int t = Integer.parseInt(st.nextToken());

        while(t-- > 0){
            st = new StringTokenizer(br.readLine());
            int N = Integer.parseInt(st.nextToken());
            int K = Integer.parseInt(st.nextToken());
            int max = Integer.MIN_VALUE;

            int[] arr = new int[N];
            PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
                @Override
                public int compare(int[] o1, int[] o2) {
                    return Integer.compare(o1[0], o2[0]);
                }
            });
            st = new StringTokenizer(br.readLine());

            for(int i = 0;i<N;i++){
                int ele = Integer.parseInt(st.nextToken());
                pq.add(new int[]{ele, i});
                arr[i] = ele;
                max = Math.max(max, ele);
            }

            int max_median = -1;
            int low = 0, high = N-1;
            ArrayList<int[]> lst = new ArrayList<>();

            while(!pq.isEmpty()){
                lst.add(pq.poll());
            }

            if(N <= K){
                if(N % 2 == 1){
                    System.out.println(lst.get(N/2)[0]);
                }else{
                    N = N/2;
                    System.out.println(lst.get(N-1)[0]);
                }
                continue;
            }else if(K == 1){
                System.out.println(max);
                continue;
            }

            while(low <= high){
                int mid = (high + low) >> 1;

                if(isValid(arr, lst.get(mid)[1], K)){
                    max_median = lst.get(mid)[0];
                    low = mid+1;
                }else{
                    high = mid-1;
                }
            }

            System.out.println(max_median);
        }

    }

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

}

**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
4 2
1 3 2 4
6 3
1 2 3 4 5 6
5 2
10 20 30 40 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
// package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF25 {

    public static long findElementOrNext(long[] arr, long target) {
        int low = 0;
        int high = arr.length - 1;
        long result = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;

            if (arr[mid] == target) {
                return arr[mid];
            } else if (arr[mid] < target) {
                low = mid + 1;
            } else {
                result = arr[mid];
                high = mid - 1;
            }
        }

        return result;
    }

    public static boolean isValid(int[] arr, int idx, int K){

        int[] dp = new int[arr.length];

        for(int i = 0;i<arr.length;i++){

            if(i-K < 0){
                if(arr[i] >= arr[idx]){
                    dp[i] = 1;
                }else{
                    dp[i] = -1;
                }
            }else if( (i+1) % K == 1){

                if(arr[i] >= arr[idx]){
                    dp[i] = Math.max(1, dp[i-K]);
                }else{
                    dp[i] = Math.max(-1, dp[i-K]);
                }

            }else{
                if(arr[i] >= arr[idx]){
                    dp[i] = Math.max(dp[i-1] + 1, dp[i-K]);
                }else{
                    dp[i] = Math.max(dp[i-1] - 1, dp[i-K]);
                }
            }
        }

        int count = 0;
        for(int i  : dp) count += i;

        return dp[arr.length-1] > 0 && count > 0;

    }

    public static void task() throws IOException{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int t = Integer.parseInt(st.nextToken());

        while(t-- > 0){
            st = new StringTokenizer(br.readLine());
            int N = Integer.parseInt(st.nextToken());
            int K = Integer.parseInt(st.nextToken());
            int max = Integer.MIN_VALUE;

            int[] arr = new int[N];
            PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
                @Override
                public int compare(int[] o1, int[] o2) {
                    return Integer.compare(o1[0], o2[0]);
                }
            });
            st = new StringTokenizer(br.readLine());

            for(int i = 0;i<N;i++){
                int ele = Integer.parseInt(st.nextToken());
                pq.add(new int[]{ele, i});
                arr[i] = ele;
                max = Math.max(max, ele);
            }

            int max_median = -1;
            int low = 0, high = N-1;
            ArrayList<int[]> lst = new ArrayList<>();

            while(!pq.isEmpty()){
                lst.add(pq.poll());
            }

            if(N <= K){
                if(N % 2 == 1){
                    System.out.println(lst.get(N/2)[0]);
                }else{
                    N = N/2;
                    System.out.println(lst.get(N-1)[0]);
                }
                continue;
            }else if(K == 1){
                System.out.println(max);
                continue;
            }

            while(low <= high){
                int mid = (high + low) >> 1;

                if(isValid(arr, lst.get(mid)[1], K)){
                    max_median = lst.get(mid)[0];
                    low = mid+1;
                }else{
                    high = mid-1;
                }
            }

            System.out.println(max_median);
            }

        }

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

}

**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
4 2
1 3 2 4
5 2
1 2 3 4 5
6 3
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
// "We create our own demons"
#include <bits/stdc++.h>

using namespace std;

#ifdef ONPC
    #include "debug.h"
#else
    #define dbg(...)
#endif

#define int long long
#define ll long long
#define ld long double
#define pi pair<int, int>
// vector
#define sz(a) (int)((a).size())
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define maxel(x) (*max_element(all(x)))
#define minel(x) (*min_element(all(x)))
#define maxpos(x) (max_element(all(x)) - x.begin())
#define minpos(x) (min_element(all(x)) - x.begin())
#define rev(x) (reverse(all(x)))
// bits
#define popcount(n) __builtin_popcountll(n)
#define clz(n) __builtin_clzll(n)
// math
#define sqr(n) ((n) * (n))
#define divup(a, b) (((a) + (b)-1) / (b))
// ostream
#define Fixed(a) cout << fixed << setprecision(12) << a;

template <class T> bool chmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template <class T> bool chmax(T& a, const T& b) { return b > a ? a = b, 1 : 0; }

template <class T> using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T> using max_queue = priority_queue<T, vector<T>, less<T>>;

template <class T> using V = vector<T>;
using vi = V<int>;
using vd = V<ld>;
using vb = V<bool>;
using vpi = V<pi>;
using vvi = V<vi>;
using vvb = V<vb>;

const int mod = 1e9 + 7; // 998244353 1e9 + 7
const ll inf = (int)(1e18) + 7;
const int inf_s = 1e9 + 7;
const ld eps = 1e-9;

const int B = 32;
const int N = 1000 + 3;
const int logn = 20;
const int maxn = 2e5 + 7;
/////////////////////////solve/////////////////////////

void solve(int t) {
    int n, k;
    cin >> n >> k;

    vpi a(n);
    if (t == 176) cout << n << "//";
    for (int i = 0; i < n; ++i) {
        cin >> a[i].first;
        a[i].second = i % k;
        if (t == 176) cout << "/";
    }
    if (t == 176) cout << "//";

    sort(rall(a));
    int need = n % k;
    if (need == 0) need = k;
    --need;

    set <int> s;
    for (int i = 0; i <= need; ++i) {
        s.insert(i);
    }

    vi ans;
    for (int i = 0; i < n; ++i) {
        if (s.count(a[i].second)) {
            ans.push_back(a[i].first);
            s.erase(a[i].second);
        }
    }

    sort(all(ans));
    int pos = (sz(ans) + 1) / 2 - 1;
    cout << ans[pos] << "\n";
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

#ifdef ONPC
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif

    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; ++i) {
#ifdef ONPC
        cerr << "===========" << i << "===========" << '\n';
#endif
        solve(t);
    }

#ifdef ONPC
    cerr << endl << "Time " << clock() * 1.0 / CLOCKS_PER_SEC << " sec" << endl;
#endif
}

**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 2
1 5 4 3 2 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
#include <bits/stdc++.h>
#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define Noway cout << -1 << endl
#define yuyuko                    \
	std::ios::sync_with_stdio(0); \
	std::cin.tie(0);              \
	std::cout.tie(0)
#define int int64_t
using namespace std;
int n, k;
vector<int> v[500010];
int a[500010];
signed main() {
	yuyuko;
	int t;
	cin >> t;
	while (t--) {
		cin >> n >> k;
		for (int i = 1; i <= n; i++) {
			v[i].clear();
		}
		int cnt = 1;
		for (int i = 1; i <= n; i++) {
			cin >> a[i];
			v[cnt].push_back(a[i]);
			if (i % k == 0) {
				cnt++;
			}
		}
		int d = n % k == 0 ? k : n % k;
        int anss = 0;
		for (int begin = 0; begin <= d; begin++) {
			vector<int> ans;
            for (int i = 0;i < begin;i++) {
                ans.push_back(v[1][i]);
            }
			for (int i = begin; i < d; i++) {
				int maxn = 0,last = 1;
				for (int j = last; j <= (n + k - 1) / k; j++) {
					if (v[j][i] > maxn) {
                        maxn = v[j][i];
                        last = j;
                    }
				}
				ans.push_back(maxn);
			}
			ranges::sort(ans);
            anss = max(anss,ans[(d+1)/2-1]);
		}
        cout << anss << endl;
	}
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - **Small n and k:** When `n` is small, the code should handle it correctly.
   - **n = k:** When `n` is equal to `k`, the entire array is removed in one operation.
   - **n < k:** This is not a valid case according to the problem constraints, so we won't consider it.

2. **Large n and k:**
   - **n >> k:** When `n` is much larger than `k`, the code should handle multiple operations correctly.
   - **k = 1:** When `k` is 1, the code should handle the removal of single elements.

3. **Special Cases:**
   - **All elements are the same:** This can test how the code handles uniform arrays.
   - **Array with a single element left:** This can test the final step of the algorithm.

4. **Critical Cases:**
   - **Median Calculation:** The code should correctly calculate the median of the remaining elements after operations.

Based on these considerations, let's generate a test input that maximizes the failure probability in the buggy code:

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

### Explanation:
1. **Test Case 1:**
   - **n = 3, k = 2**
   - **Array: [1, 2, 3]**
   - **Operations:**
     - Remove subarray [1, 2] → Remaining: [3]
   - **Expected Output: 3**

2. **Test Case 2:**
   - **n = 5, k = 2**
   - **Array: [5, 4, 3, 2, 1]**
   - **Operations:**
     - Remove subarray [5, 4] → Remaining: [3, 2, 1]
     - Remove subarray [3, 2] → Remaining: [1]
   - **Expected Output: 1**

3. **Test Case 3:**
   - **n = 7, k = 3**
   - **Array: [1, 3, 5, 7, 9, 11, 13]**
   - **Operations:**
     - Remove subarray [1, 3, 5] → Remaining: [7, 9, 11, 13]
     - Remove subarray [7, 9, 11] → Remaining: [13]
   - **Expected Output: 13**

4. **Test Case 4:**
   - **n = 8, k = 4**
   - **Array: [4, 3, 2, 1, 8, 7, 6, 5]**
   - **Operations:**
     - Remove subarray [4, 3, 2, 1] → Remaining: [8, 7, 6, 5]
     - Remove subarray [8, 7, 6, 5] → Remaining: []
   - **Expected Output: -1** (or an appropriate message indicating no elements left)

This test input should help expose any issues in the buggy code, particularly in handling different array lengths and element values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
import random
from heapq import heapify, heappop, heappush

# xor = random.randint(10**7, 10**8)
xor = 0

def median_achievable(target, a, k):
    n = len(a)
    a = a.copy()
    larger_count = 0
    smaller_count = 0
    count_removed = [0 for i in range(n)]
    count_removed_big = [0 for i in range(n)]
    above_target = 0
    below_target = 0
    for i in range(n):
        #
        if a[i] >= target:
            above_target += 1
            larger_count += 1
        else:
            below_target += 1
            smaller_count += 1
        #
        if larger_count + smaller_count > k:
            removed_ele_idx = i - k
            removed_ele = a[removed_ele_idx]
            if removed_ele >= target:
                larger_count -= 1
            else:
                smaller_count -= 1
        #
        if larger_count + smaller_count == k:
            if smaller_count > larger_count:
                # count_removed[i] = (count_removed[i - k] if i >= k else 0) + smaller_count
                choice1 = (count_removed[i - k] if i >= k else 0) + smaller_count
                choice2 = count_removed[i-1] if i >= 1 else 0
                if choice1 > choice2:
                    count_removed[i] = choice1
                    count_removed_big[i] = larger_count + (count_removed_big[i-k] if i >= k else 0)
                else:
                    count_removed[i] = choice2
                    count_removed_big[i] = count_removed_big[i-1] if i >= 1 else 0
            else:
                count_removed[i] = count_removed[i-1] if i >= 1 else 0
                count_removed_big[i] = count_removed_big[i-1] if i >= 1 else 0

    # print(count_removed, count_removed_big, a, k, target)
    if above_target - count_removed_big[-1] > below_target - count_removed[-1]:
        return True
    else:
        return False

def solve():
    n, k = list(map(int, input().split()))
    a = list(map(int, input().split()))
    head = min(a) # true
    tail = max(a)+1 # false
    while head+1 != tail:
        mid = (head + tail) // 2
        if median_achievable(mid, a, k):
            head = mid
        else:
            tail = mid
    print(head, end=' ')

for case in range(int(input())):
    if case == 267:
        print()
        print()
        n, k = list(map(int, input().split()))
        a = list(map(int, input().split()))
        print(n, k, a)
        # exit()
    else:
        # pass
        if case < 5:
            solve()

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

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

Think step by step.

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

1. **Edge Cases for `n` and `k`**:
   - **Small `n` and `k`**: Test with the smallest possible values.
   - **Large `n` and `k`**: Test with values close to the upper limit.
   - **`n` equal to `k`**: This should result in an empty array after one operation.
   - **`n` slightly greater than `k`**: This should leave a small number of elements.

2. **Edge Cases for Array `a`**:
   - **All elements equal**: This can test the logic for counting and removing elements.
   - **Sorted array**: This can test the logic for finding the median.
   - **Reverse sorted array**: This can test the logic for finding the median in a different order.
   - **Array with a single large element**: This can test the logic for handling large values.

3. **Specific Scenarios**:
   - **Array with a mix of large and small elements**: This can test the logic for balancing the counts of larger and smaller elements.
   - **Array with a large number of elements just above or below the target**: This can test the logic for handling close values.

Based on these considerations, let's generate a small-scale test input that is complete and valid:

```plaintext
3
3 3
1 2 3
4 3
3 2 5 6
7 1
5 9 2 6 5 4 6
```

This test input includes:
1. A case where `n` equals `k` (3, 3, [1, 2, 3]).
2. A case where `n` is slightly greater than `k` (4, 3, [3, 2, 5, 6]).
3. A case with a mix of large and small elements (7, 1, [5, 9, 2, 6, 5, 4, 6]).

These cases should help expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
"""
https://codeforces.com/contest/1993/problem/D
Simplified problem: select 1 number from each group to maximize median

Binary search the maximum median
Notes:
- after removing length K subarrays, we are left with 1 element from each i (mod `rem`)
- selecting largest value != greatest median
"""

def solve():
    def works(med):
        # dp = [0] * N  # dp[i]: maximum possible number of elements >=med up to `i`
        # for i in range(N):
        #     if i - K - 1 < 0:  # can't fit another consecutive segment
        #         dp[i] = dp[i - 1] + (arr[i] >= med)
        #     else:  # transition from previous consecutive segment
        #         dp[i] = max(dp[i - 1], dp[i - K - 1]) + (arr[i] >= med)
        return sum(max(arr[j] for j in range(i, N, K)) >= med for i in range(rem)) >= rem // 2 + 1

    N, K = map(int, input().split())
    arr = list(map(int, input().split()))
    if K >= N:
        arr.sort()
        print(arr[(N-1)//2])
        return
    moves = (N - 1) // K  # number of removals
    rem = N - moves * K  # elements left after removing

    res = [max(arr[j] for j in range(i, N, K)) for i in range(rem)]
    res.sort()
    print(res[(rem - 1) // 2])

    # low = min(arr)
    # high = max(arr)
    # while low <= high:
    #     mid = (low + high) // 2
    #     if works(mid):
    #         low = mid + 1
    #     else:
    #         high = mid - 1
    # print(low - 1)

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

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

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

Think step by step.

Response:
```plaintext
1
5 2
1 1000000000 2 999999999 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
//  package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF25 {

    public static long findElementOrNext(long[] arr, long target) {
        int low = 0;
        int high = arr.length - 1;
        long result = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;

            if (arr[mid] == target) {
                return arr[mid];
            } else if (arr[mid] < target) {
                low = mid + 1;
            } else {
                result = arr[mid];
                high = mid - 1;
            }
        }

        return result;
    }

    public static boolean isValid(int[] arr, int idx, int K){

        int[] dp = new int[arr.length];

        dp[0] = arr[0] >= arr[idx] ? 1 : -1;

        for(int i = 1;i<arr.length;i++) {

            if (i - (K-1) <= 0) {
                if (arr[i] >= arr[idx]) {
                    dp[i] = 1;
                } else {
                    dp[i] = -1;
                }
            }else if( i % K == 0) {

                if (arr[i] >= arr[idx]) {
                    dp[i] = Math.max(1, dp[i - K]);
                } else {
                    dp[i] = Math.max(-1, dp[i - K]);
                }

            } else {
                if (arr[i] >= arr[idx]) {
                    dp[i] = Math.max(dp[i - 1] + 1, dp[i - K]);
                } else {
                    dp[i] = Math.max(dp[i - 1] - 1, dp[i - K]);
                }
            }
        }

        return dp[arr.length-1] > 0;
    }

    public static void task() throws IOException{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int t = Integer.parseInt(st.nextToken());

        while(t-- > 0){
            st = new StringTokenizer(br.readLine());
            int N = Integer.parseInt(st.nextToken());
            int K = Integer.parseInt(st.nextToken());
            int max = Integer.MIN_VALUE;

            int[] arr = new int[N];
            PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
                @Override
                public int compare(int[] o1, int[] o2) {
                    return Integer.compare(o1[0], o2[0]);
                }
            });
            st = new StringTokenizer(br.readLine());

            for(int i = 0;i<N;i++){
                int ele = Integer.parseInt(st.nextToken());
                pq.add(new int[]{ele, i});
                arr[i] = ele;
                max = Math.max(max, ele);
            }

            int max_median = -1;
            int low = 0, high = N-1;
            ArrayList<int[]> lst = new ArrayList<>();

            while(!pq.isEmpty()){
                lst.add(pq.poll());
            }

            if(N <= K){
                if(N % 2 == 1){
                    System.out.println(lst.get(N/2)[0]);
                }else{
                    N = N/2;
                    System.out.println(lst.get(N-1)[0]);
                }
                continue;
            }else if(K == 1){
                System.out.println(max);
                continue;
            }

            while(low <= high){
                int mid = (high + low) >> 1;

                if(isValid(arr, lst.get(mid)[1], K)){
                    max_median = lst.get(mid)[0];
                    low = mid+1;
                }else{
                    high = mid-1;
                }
            }

            System.out.println(max_median);
        }

    }

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

}

**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
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: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
// package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF25 {

    public static long findElementOrNext(long[] arr, long target) {
        int low = 0;
        int high = arr.length - 1;
        long result = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;

            if (arr[mid] == target) {
                return arr[mid];
            } else if (arr[mid] < target) {
                low = mid + 1;
            } else {
                result = arr[mid];
                high = mid - 1;
            }
        }

        return result;
    }

    public static boolean isValid(int[] arr, int idx, int K){

        int left_start = 0, left_end = idx-1, right_start = idx+1, right_end = arr.length-1;

        int right_swapping = right_end - (right_start) + 1;
        int left_swapping = left_end - (left_start) + 1;

        if(right_swapping % K == 0 && left_swapping % K == 0) return true;

        if(right_swapping % K == 0) {
            for (int i = idx - 1; i >= 0; i--) {
                if (arr[i] >= arr[idx]) {
                    int element_bw_ls_le = (idx - 1) - (i + 1) + 1;
                    int element_bw_le_idx = (i - 1) - left_start + 1;

                    if(element_bw_le_idx % K == 0 && element_bw_ls_le % K == 0) return true;

                }
            }
        }

        if(left_swapping % K == 0) {
            for (int i = idx + 1; i < right_end + 1; i++) {
                if (arr[i] >= arr[idx]) {
                    int element_bw_idx_rs = i - idx - 1;
                    int element_bw_rs_re = right_end - (i+1) + 1;
                    if (element_bw_idx_rs % K == 0 && element_bw_rs_re % K == 0) return true;
                }
            }
        }

        return false;
    }

    public static void task() throws IOException{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int t = Integer.parseInt(st.nextToken());

        while(t-- > 0){
            st = new StringTokenizer(br.readLine());
            int N = Integer.parseInt(st.nextToken());
            int K = Integer.parseInt(st.nextToken());

            int[] arr = new int[N];
            PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
                @Override
                public int compare(int[] o1, int[] o2) {
                    return Integer.compare(o1[0], o2[0]);
                }
            });
            st = new StringTokenizer(br.readLine());

            for(int i = 0;i<N;i++){
                int ele = Integer.parseInt(st.nextToken());
                pq.add(new int[]{ele, i});
                arr[i] = ele;
            }

            int max_median = -1;
            int low = 0, high = N-1;
            ArrayList<int[]> lst = new ArrayList<>();

            while(!pq.isEmpty()){
                lst.add(pq.poll());
            }

            if(N <= K){
                if(N % 2 == 1){
                    System.out.println(lst.get(N/2)[0]);
                }else{
                    N = N/2;
                    System.out.println(lst.get(N-1)[0]);
                }
                continue;
            }

            while(low <= high){
                int mid = (high + low) >> 1;

                if(isValid(arr, lst.get(mid)[1], K)){
                    max_median = lst.get(mid)[0];
                    low = mid+1;
                }else{
                    high = mid-1;
                }
            }

            System.out.println(max_median);
            }

        }

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

}

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

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

Think step by step.

Response:
```plaintext
1
5 2
1 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: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
vector<int>a,ast;
int n,k;
bool check(int x){
    x=ast[x];
	vector<int>dp(n+1,0),s0(n+1,0);
    for (int i=1;i<=n;i++){
        dp[i]=dp[i-1]+(a[i]<x);
        if (i>=k){
            if (dp[i]>dp[i-k]) s0[i]=s0[i-k]+1;
            dp[i]=min(dp[i],dp[i-k]);
        }
    }
    if (s0[n]*k==n){
        vector<int>dp2(n+1,0);
        for (int i=1;i<=n;i++){
            dp2[i]=dp2[i-1]+(a[i]<x);
            if (i>=k+1){
                dp2[i]=min(dp2[i],dp2[i-k]);
            }
        }
        dp[n]=dp2[n];
    }
    int md=n%k;
    if (md==0) md=k;
    return (md+1)/2>dp[n];
}
void solve(){
	cin>>n>>k;
	a.push_back(0),ast.push_back(0);
	for (int i=1;i<=n;i++){
		int x;
		cin>>x;
		a.push_back(x);
		ast.push_back(x);
	}
	sort(ast.begin(),ast.end());
	int l=1,r=n;
	while(l<r){
		int mid=(l+r+1)>>1;
		if (check(mid)){
			l=mid;
		}
		else{
			r=mid-1;
		}
	}
	cout<<ast[l]<<endl;
	a.clear();
	ast.clear();
	return;
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int T;
	cin>>T;
	while(T--){
		solve();
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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 two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the array $a$.

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
#include "bits/stdc++.h"

using namespace std;

#define ar array
typedef long long ll;
//~ #define int ll

template<class T> bool umin(T& a, const T b) { if(a > b) { a = b; return true; } return false; }
template<class T> bool umax(T& a, const T b) { if(a < b) { a = b; return true; } return false; }
const int inf = 1e9;

void solve(){
	int n, k; cin >> n >> k;

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

	int sz = n % k;
	if(sz == 0) sz = k;
	if(sz > n) sz = n;

	vector<int> b(n);
	vector<int> dp(n);
	vector<int> mx(sz);

	auto check = [&](int v){
		fill(mx.begin(), mx.end(), -inf);
		//~ fill(dp.begin(), dp.end(), -inf);

		for(int i=0;i<n;i++){
			b[i] = (v <= a[i]);
		}

		for(int i=0;i<n;i++){
			int r = i % k;
			if(r > 0){
				dp[i] = mx[r - 1] + b[i];
			} else {
				dp[i] = b[i];
			}

			umax(mx[r], dp[i]);
		}

		return mx[sz - 1] * 2 > sz;
	};

	int l = 0, r = inf;

	while(l < r){
		int m = (l + r + 1) >> 1;
		if(check(m)) l = m;
		else r = m - 1;
	}

	cout<<l<<"\n";
}

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

	int t = 1;

	cin >> t;

	while(t--){
		solve();
	}
}

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

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

Think step by step.

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