**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
/******************************************************************************

                            Online Java Compiler.
                Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.

*******************************************************************************/
import java.util.*;
public class Main
{
	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
	    int t = sc.nextInt();
	    while(t--  > 0){
	        int n = sc.nextInt() , k = sc.nextInt();
	       // List<Integer> li = new ArrayList<>()
	       Map<Integer,Integer> ma = new HashMap<>();
	        for(int i = 0 ; i < n ; i++){
	            ma.put(sc.nextInt(), 1);
	        }
	        List<Integer> l = new ArrayList<>();
	        boolean b = false;
	        for(int i = 1 ; i < 200001 ; i++){
	            if(ma.containsKey(i)){
    	            int m = ma.get(i);
    	            while(m > 0 ){
    	                l.add(0);
    	                m--;
    	            }
	            }
	            while(ma.containsKey(i-k)){
	                if(ma.get(i-k) == 1){
	                    ma.remove(i-k);
	                }
	                else{
	                    ma.put(i-k,ma.get(i-k)-1);
	                }
	                ma.put(i+k,ma.getOrDefault(i+k,0)+1);
	                l.remove(0);
	            }
	            if(l.size() == n)
	            {
	                System.out.println(i);
	                b = true;
	                break;
	            }
	        }
	        if(!b)
	            System.out.println(-1);
	    }

	}

}

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

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

Think step by step.

Response:
```plaintext
2
4 2
1 2 3 4
5 3
1 4 7 10 13
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `4 2` with `a = [1, 2, 3, 4]`
   - **Expected Output:** `-1`
   - **Reasoning:** With `k = 2`, the lights will toggle every 2 minutes. The lights installed at times 1, 2, 3, and 4 will never all be on at the same time because their on-off cycles will never align. For example, the light installed at time 1 will be on at times 1, 3, 5, etc., while the light installed at time 2 will be on at times 2, 4, 6, etc. These cycles do not overlap, so the answer should be `-1`.

2. **Test Case 2:**
   - **Input:** `5 3` with `a = [1, 4, 7, 10, 13]`
   - **Expected Output:** `-1`
   - **Reasoning:** With `k = 3`, the lights will toggle every 3 minutes. The lights installed at times 1, 4, 7, 10, and 13 will never all be on at the same time because their on-off cycles will never align. For example, the light installed at time 1 will be on at times 1, 4, 7, 10, etc., while the light installed at time 4 will be on at times 4, 7, 10, etc. These cycles do not overlap, so the answer should be `-1`.

These test cases are designed to expose the bug in the provided code by ensuring that the light cycles never align, which should result in `-1` as the output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

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

public class ApartmentLights {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long t = sc.nextLong();
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            long n = sc.nextLong();
            long k = sc.nextLong();
            long[] a = new long[(int) n];

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextLong();
            }

            // Sort the array
            Arrays.sort(a);

            // Find the maximum installation time
            long max_a = a[(int) (n - 1)];

            // Calculate the earliest moment all lights are on
            long earliestTime = -1;
            long gg = max_a;
            boolean valid = true;

            for (int i = 0; i < n - 1; i++) {
                long diff = max_a - a[i];

                // If the difference is not a multiple of k, adjust k
                long x = diff / k;
                if (x % 2 == 1) {
                    max_a = a[i] + (x + 1) * k;
                    if (max_a >= gg + k) {
                        valid = false;
                    }
                }
            }

            // Validate that this earliestTime is indeed correct
            result.append(valid ? max_a : -1).append("\n");
        }

        System.out.print(result.toString());
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>

using namespace std;
typedef long long ll;

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

    vector <long long> val(n);
    for (long long i = 0; i < n; ++i){
        cin >> val[i];
    }
    long long l = 0, r = 2 * k;
    for (long long i = 0; i < n; ++i){
        long long newl = val[i] % (2 * k);
        long long newr = (val[i] + k) % (2 * k);
        if (newl > newr){
            if (r <= newl){
                newl -= 2 * k;
            } else {
                newr += 2 * k;
            }
        }
        l = max(l, newl);
        r = min(r, newr);
    }
    if (l >= r){
        cout << - 1 << endl;
    } else {
        long long m = *max_element(val.begin(), val.end());
        m += ((l - m % (2 * k)) % (2 * k) + 2 * k) % (2 * k);
        cout << m << endl;
    }
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    long long t = 1;
    cin >> t;
    for (long long i = 0; i < t; ++i){
        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
2
4 4
1 5 9 13
3 2
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;
#define N 200007
#define int long long

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

int num[N];

void solve() {
    int n = in(), k = in();
    int maxn = 0;
    int mod = 2 * k;
    for (int i = 1; i <= n; i++) {
        num[i] = in();
        maxn = max(maxn, num[i]);
        num[i] %= mod;
    }

    int L = num[1], dis = k - 1;

    for (int i = 2; i <= n; i++) {
        int d = (num[i] - L + mod) % mod;
        if (d > dis) {
            int far = (num[i] + k - 1) % mod;
            dis = (far - L + mod) % mod;
            if (dis > k - 1) {
                printf("-1\n");
                return;
            }
        } else {
            L = num[i];
            dis -= d;
        }
    }

    for (int i = 0; i <= mod; i++) {
        int now = maxn + i;
        int t = now % mod;
        int d = t - L + mod;
        d %= mod;
        if (d <=dis) {
            printf("%lld\n", now);
            return;
        }
    }

}

signed main() {
  //  freopen("D:\\Development_Software\\CLion\\CLionProjects\\Test\\in", "r", stdin);
    int T = in();
    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
3 2
1 3 5
4 3
2 4 6 8
2 1
1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
t = int(input())
for _ in range(t):
    # print("_________")
    done = False
    n, k = map(int, input().split())
    arr = list(map(int, input().split()))
    min_ = arr[0]%(2*k)
    max_ = min_ + k - 1
    start = arr[0]
    for i in arr[1:]:
        temp_min = i%(2*k)
        temp_max = temp_min + k - 1
        start = max(start, i)
        if temp_max < 2*k:
            if temp_min > max_ or temp_max<min_:
                print(-1)
                done = True
                break
            min_ = max(min_, temp_min)
            max_ = min(max_, temp_max)
        else:
            if temp_min > max_ and temp_max%(2*k) < min_:
                done = True
                print(-1)
                break
            elif temp_min <= max_:
                min_ = temp_min
            elif temp_max%(2*k) >= min_:
                max_ = temp_max
    if not done:
        temp = start%(2*k)
        start = start - temp + min_
        if temp > min_:
            start+=(2*k)
        print(start)

**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 4
2 3 4 5
4 3
2 3 4 5
7 5
14 34 6 25 46 7 17
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
rr2 = []
t = int(input())
for i in range(t):
    n, k = map(int, input().split(' '))
    A = list(map(int, input().split(' ')))
    mxa = max(A)
    lb = mxa
    ub = mxa+k
    flag = False
    for num in A:
        lam = (lb - num)//k
        lb1 = -1
        ub1= -1
        if lam%2==0:
            lb1 = num + lam * k
            ub1 = lb1 + k
        else:
            lb1 = num + lam * (k)+k
            ub1 = lb1 + k
        # print(lb1,ub1)
        if ub<lb1 or ub1<lb:
            flag = True
            break
        else:
            lb = max(lb,lb1)
            ub = min(ub,ub1)
        # print(lb,ub)
    if flag:
        print(-1)
        # rr2.append(-1)
    else:
        print(lb)
        # rr2.append(lb)
# print(rr2)

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

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

Think step by step.

Response:
```plaintext
1
4 4
1 5 9 13
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

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

// import java.math.BigInteger;
public class Main {
  public static FastReader in;
  public static PrintWriter out;
  public static long mod = (long) (1e9 + 7);

  public static void solve(int nTestCase) {
    int N = in.nextInt(), K = in.nextInt();
    int[] a = in.readArray(N);
    long[] pref = new long[2 * K + 2];
    int KTwice = 2 * K;
    for (int i : a) {
      // int div = (i / KTwice) % 2;
      int mod = (i % KTwice);
      int start = mod, end = ((mod + K - 1) % KTwice);
      pref[start]++;
      // if(end==2*K) continue;
      if (end < start) {
        pref[0]++;
      }
      pref[end + 1]--;
      // out.println("start: " + start + " end: " + end);
    }
    int indx = -1;
    for (int i = 0; i < KTwice; i++) {
      if (i - 1 >= 0)
        pref[i] += pref[i - 1];
      if (pref[i] >= N) {
        indx = i;
        break;
      }
    }
    if (indx == -1) {
      out.println(indx);
      return;
    }
    int max = a[0];
    for (int i = 0; i < N; i++)
      max = Math.max(max, a[i]);
    int finalAns = (max / KTwice) * KTwice + indx;
    // out.println(finalAns + " " + max + " " + indx);
    if (finalAns < max)
      finalAns += KTwice;
    out.println(finalAns);
  }

  public static int[] reverse(int[] a) {
    int[] b = new int[a.length];
    for (int i = 0; i < a.length; i++)
      b[i] = a[a.length - 1 - i];
    return b;
  }

  public static int gcd(int a, int b) {
    if (a == 0)
      return b;
    else
      return gcd(b % a, a);
  }

  public static void fastSort(int[] a) {
    fastSort(a, false);
  }

  public static void fastSort(int[] a, boolean isReverse) {
    Random random = new Random();
    for (int i = 0; i < a.length; i++) {
      int next = random.nextInt(a.length);
      int curr = a[next];
      a[next] = a[i];
      a[i] = curr;
    }
    Arrays.sort(a);
    if (isReverse) {
      int n = a.length;
      for (int i = 0; i < a.length / 2; i++) {
        int temp = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = temp;
      }
    }
    // return a;
  }

  public static void main(String[] args) throws IOException {
    boolean isInputOutputFiles = args != null && args.length > 0 && args[0].equals("LOCAL_RUN");
    // isInputOutputFiles = false;
    if (isInputOutputFiles) {
      in = new FastReader(new BufferedReader(new FileReader("./input/input.txt")));
      out = new PrintWriter(new BufferedWriter(new FileWriter("./output/output.txt")));
    } else {
      in = new FastReader(new InputStreamReader(System.in));
      out = new PrintWriter(System.out);
    }
    long startT = 0;
    if (isInputOutputFiles)
      startT = System.currentTimeMillis();
    int T = 1;
    T = in.nextInt();
    for (int tt = 1; tt <= T; tt++) {
      solve(tt);
    }
    if (isInputOutputFiles)
      out.println("Total time: " + (System.currentTimeMillis() - startT) + "ms");
    out.close();
  }

  static class Pair implements Comparable<Pair> {
    int a;
    int b;

    // int c;
    public Pair(int a, int b) {
      this.a = a;
      this.b = b;
      // this.c = c;
    }

    public int compareTo(Pair o) {
      if (this.a == o.a)
        return (int) (o.b - this.b);
      if ((this.a - o.a) > 0)
        return 1;
      else if ((this.a - o.a) < 0)
        return -1;
      else
        return (int) (this.a - o.a);
    }
  }

  static class SegTree {
    int leftmost, rightmost;
    long minAns;
    SegTree left, right;

    public SegTree(int leftmost, int rightmost, int[] a) {
      this.leftmost = leftmost;
      this.rightmost = rightmost;
      if (leftmost == rightmost) {
        this.minAns = a[leftmost];
        return;
      }
      int mid = (leftmost + rightmost) / 2;
      left = new SegTree(leftmost, mid, a);
      right = new SegTree(mid + 1, rightmost, a);
      recalc();
    }

    public void recalc() {
      if (leftmost == rightmost)
        return;
      minAns = Math.min(left.minAns, right.minAns);
      // System.out.println("xorAns: "+xorAns);
      return;
    }

    public void pointUpdate(int index, long newVal) {
      if (leftmost == rightmost) {
        this.minAns = newVal;
        return;
      }
      if (index <= left.rightmost)
        left.pointUpdate(index, newVal);
      else
        right.pointUpdate(index, newVal);
      recalc();
    }

    public long rangeQuery(int l, int r) {
      if (l > rightmost || r < leftmost)
        return Long.MAX_VALUE;
      if (l <= leftmost && r >= rightmost)
        return minAns;
      return Math.min(left.rangeQuery(l, r), right.rangeQuery(l, r));
    }

  }

  static class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader(Reader rd) {
      br = new BufferedReader(rd);
    }

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

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

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

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

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

    public int[] readArray(int n) {
      int[] a = new int[n];
      for (int i = 0; i < n; i++)
        a[i] = nextInt();
      return a;
    }
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can highlight potential issues in the algorithm. Here are the steps to think through this:

1. **Identify Potential Bugs:**
   - The code uses a prefix sum array to track the light status changes. Ensure that the prefix sum array is correctly handling the wrap-around of the 2K period.
   - Check how the code handles the case where the lights are never all on at the same time.
   - Verify the logic for calculating the final answer, especially the handling of the `finalAns` calculation.

2. **Edge Cases:**
   - **All lights are installed at the same time:** This should be easy to handle, but it's a good starting point.
   - **Lights are installed at times that are multiples of K:** This can cause issues with the periodic on/off cycles.
   - **Lights are installed at times that are close to each other but not aligned:** This can cause the prefix sum array to incorrectly calculate the light status.
   - **Lights are installed at times that are far apart:** This can test the long-term periodic behavior.

3. **Generate a Small-Scale Test Input:**
   - We need a test case that is small enough to be easily verified but challenging enough to expose the bug.

Let's consider a test case where the lights are installed at times that are close to each other but not aligned, which can cause the prefix sum array to incorrectly calculate the light status.

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

**Explanation:**
- **Test Case:**
  - `t = 1`: One test case.
  - `n = 4`, `k = 3`: Four rooms with a period of 3 minutes.
  - `a = [1, 2, 3, 4]`: The chips are installed at times 1, 2, 3, and 4.

- **Expected Output:**
  - The expected output should be `-1` because the lights will never all be on at the same time.
  - The 1st light will be on at minutes `1, 2, 3, 7, 8, 9, ...`
  - The 2nd light will be on at minutes `2, 3, 4, 8, 9, 10, ...`
  - The 3rd light will be on at minutes `3, 4, 5, 9, 10, 11, ...`
  - The 4th light will be on at minutes `4, 5, 6, 10, 11, 12, ...`

This test case should highlight the issue with the prefix sum array and the logic for calculating the final answer. If the code fails to correctly handle this case, it will likely produce an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.HashSet;
import java.util.Scanner;
import java.util.TreeSet;

public class C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int cases = in.nextInt();
        for (int i = 0; i < cases; i++) {
            int n = in.nextInt();
            int k = in.nextInt();
            int out = 0;
            ArrayList<Integer> arr = new ArrayList<>();
            ArrayList<Integer> arr2 = new ArrayList<>();
            HashSet<Integer> set = new HashSet<>();
            for (int j = 0; j < n; j++) {
                arr.add(in.nextInt());
                arr2.add(arr.get(j));
            }
            Collections.sort(arr);
            int m = arr.get(n - 1);

            for (int j = 0; j < n; j++) {
                arr.set(j, arr.get(j) % (k * 2));

            }

            for (int j = 0; j < n; j++) {
                set.add(arr.get(j));
            }
            TreeSet<Integer> s2 = new TreeSet<>();

            ArrayList<ArrayList<Integer>> matrix = new ArrayList<>();
            int count = 0;
            for (int j : set) {
                matrix.add(new ArrayList<Integer>());
                for (int l = 0; l < k; l++) {
                    matrix.get(count).add((l + j) % (2 * k));
                }
                count++;
            }
            for (int d = 0; d < 4 * k; d++) {
                s2.add(d);
            }
            for (ArrayList<Integer> a : matrix) {
                if (s2.size() == 0) {
                    break;
                }

                for (int d = 0; d < 4 * k; d++) {
                    if (s2.contains(d) && !a.contains(d)) {
                        s2.remove(d);
                    }
                }

            }

            if (s2.size() == 0) {
                System.out.println(-1);
            } else {
                if (i == 2003) {

                    System.out.print(k);
                }
                int add = 0;
                if (s2.contains(m % (2 * k))) {
                    add = 0;
                } else {
                    int tar = s2.first();
                    while ((m % (2 * k)) != tar) {
                        m++;

                    }
                }

                System.out.println(m + add);
            }
        }
    }
}

**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: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

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

int main() {
	// your code goes here
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        vector<int>arr(n);
        int maxi=0;
        for(int i=0;i<n;i++)
        {
            cin>>arr[i];

        }
        sort(arr.begin(),arr.end());
        maxi=arr[n-1];
        // for(int i=0;i<n;i++)
        // {
        //     int x=arr[i];
        //     int inc=(maxi-x)/(2*k);

        //     arr[i]+=inc*(2*k);
        // }
        int ans=-1;
       for(int i=maxi;i<maxi+k;i++)
       {
           int cnt=0;
           for(int j=0;j<n;j++)
           {
               int x=(i-arr[j])/(k);
               if(x%2==0)cnt++;
           }
           if(cnt==n)
           {
               ans=i;
               break;
           }
       }
       cout<<ans<<endl;

    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
// Road to Expert: 67 pts left
// Code by grinding_codexp

void precp() {

}

void solve() {
    int n, k; cin >> n >> k;
    vector<int> a(n);
    int mx = 0;
    for (int i = 0; i < n; ++i) cin >> a[i], mx = max(a[i], mx);
    for (int i = 0; i < k; ++i) {
        int t = mx + i;
        bool on = 1;
        for (int i = 0; i < n; ++i) {
            int cl = (t - a[i]) / k;
            if (cl % 2 == 1) {
                on = 0;
            }
        }
        if (on) {
            cout << t << endl;
            return;
        }
    }

    cout << -1 << endl;
}

signed main() {
    // freopen("", "r", stdin);
    // freopen("", "w", stdout);

    std::ios_base::sync_with_stdio(0);
    std::cin.tie(0); std::cout.tie(0);

    int T_ = 1;
    cin >> T_;
    precp();
    while (T_--) solve();

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The main issue in the code is the way it checks if all lights are on at a given time. It iterates over a range of possible times and checks if all lights are on at those times, but it doesn't handle the periodic nature of the light toggles correctly.

### Key Points to Consider:
1. **Initial Installation Times**: Ensure that the installation times are spread out in a way that could cause issues with the periodic toggling.
2. **Periodic Toggling**: The code should correctly handle the periodic toggling of lights, especially when the period \( k \) is small or when the installation times are close to each other.
3. **Edge Cases**: Consider cases where \( k = 1 \) or \( k \) is a large number, and the installation times are close to each other or far apart.

### Test Input:
Let's create a test input that exposes the bug by having a small \( k \) and installation times that are close to each other but not perfectly aligned with the period.

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

### Explanation:
- **Test Case**: 1
- **Number of Rooms (n)**: 5
- **Period (k)**: 2
- **Installation Times**: 1, 2, 3, 4, 5

### Why This Test Input?
- **Period (k = 2)**: The lights will toggle every 2 minutes.
- **Installation Times**: The installation times are consecutive, which means the lights will be toggled at different moments:
  - Room 1: On at 1, 3, 5, ...
  - Room 2: On at 2, 4, 6, ...
  - Room 3: On at 3, 5, 7, ...
  - Room 4: On at 4, 6, 8, ...
  - Room 5: On at 5, 7, 9, ...

- **Expected Output**: -1
- **Reason**: There is no time when all lights are on simultaneously. For example:
  - At time 1: Only room 1 is on.
  - At time 2: Only room 2 is on.
  - At time 3: Only room 3 is on.
  - At time 4: Only room 4 is on.
  - At time 5: Only room 5 is on.
  - And so on...

This test input should expose the bug in the code by showing that the algorithm incorrectly determines that there is no time when all lights are on, when in fact, the periodic toggling makes it impossible for all lights to be on at the same time.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
def solve():
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    maxi = max(a)
    sets = []
    if n == 1 and k == 1:
        return 1
    for i in range(n):
        temp = (maxi - a[i]) // k
        temp_set = set()
        if temp % 2 == 1:
            for j in range(1, k+1):
                temp_set.add((a[i] + (temp * k)) + (k-1) + j)
        else:
            for j in range(1, k+1):
                temp_set.add((a[i] + (temp * k)) + -1 + j)
        sets.append(temp_set)
    # print(sets)
    if len(sets) < 2:
        return -1
    else:
        ans = sorted(list(set(sets[0]).intersection(*sets[1:])))
        if ans:
            return min(ans)

for _ in range(int(input())):
    a = solve()
    if a is None:
        print(-1)
    else:
        print(a)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
t = int(input())
for i in range(t):
    a = list(map(int, input().split()))
    n = a[0]  # kol-vo komnat
    k = a[1]  # period raboty
    b = list(map(int, input().split()))  # spisok tochki ystanovki po vremeny

    v = {}
    for x in b:
        l = x % (2 * k)
        if l in v:
            v[l] += 1
        else:
            v[l] = 1
        for j in range(k - 1):
            l += 1
            c = l % (2 * k)
            if c in v:
                v[c] += 1
            else:
                v[c] = 1

    m = []
    for x in v:
        if v[x] == n:
            m.append(x)
    if len(m) == 0:
        d = -1
    else:
        ma = max(b)
        p =[]
        for x in m:
            ma = max(b)
            while True:
                if (ma % (2 * k)) == x:
                    p.append(ma)
                    break
                else:
                    ma += 1
        d = min(p)
    print(d)

**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
3 2
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

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

public class C_Light_Switches {
    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {}
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }

        char[] nextCharArray() {
            try {
                String line = br.readLine();
                return line.toCharArray();
            } catch (IOException e) {
                e.printStackTrace();
                return new char[0];
            }
        }
    }

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        FastScanner s = new FastScanner();

        int t = s.nextInt();

        while(t-- != 0) {
            int n = s.nextInt();
            int k = s.nextInt();

            Long[] arr = new Long[n];
            for(int i = 0; i < n; i++) {
                arr[i] = s.nextLong();
            }

            Arrays.sort(arr);

            long max = arr[n-1];
            for(int i = 0; i < n-1; i++) {
                arr[i] += (k << 1) * ((max - arr[i]) / (k << 1));
            }

            int[] slot = new int[k];
            for(long i : arr) {
                for(int o = 0; o < 2; o++) {
                    for(int j = 0; j < k; j++) {
                        long idx = i + j;
                        if(idx >= max && (idx - max) < k) {
                            slot[(int)(idx - max)]++;
                        }
                    }
                    i += (k << 1);
                }
            }

            max = 0;
            int max_idx = -1;
            for(int i = 0; i < k; i++) {
                if(slot[i] > max) {
                    max = slot[i];
                    max_idx = i;
                }
            }

            if(max >= n) {
                out.println(arr[n-1] + max_idx);
            } else {
                out.println(-1);
            }
        }
        out.close();
    }
}

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

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

Think step by step.

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

1. **Edge Cases for `k` and `n`**:
   - `k = 1` or `k = n`: These cases can be tricky because the light status changes very frequently.
   - `n = 1`: This is a simple case but can help verify the basic logic.
   - `k` and `n` are large: This can test the performance and correctness of the algorithm.

2. **Edge Cases for `a_i`**:
   - All `a_i` are very close to each other: This can test the logic for handling nearly simultaneous installations.
   - `a_i` values are very large: This can test the bounds and performance.
   - `a_i` values are spread out: This can test the logic for handling widely different installation times.

3. **Specific Scenarios**:
   - All `a_i` are the same: This can test the logic for handling identical installation times.
   - `a_i` values are such that the lights will never be on simultaneously: This can test the logic for detecting impossible scenarios.

Given these considerations, let's generate a small-scale test input that covers some of these edge cases:

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

### Explanation:
1. **Test Case 1:**
   - `n = 3`, `k = 1`
   - `a = [1, 2, 3]`
   - This case tests the scenario where `k` is 1, and the lights will be on and off very frequently. The lights will be on at minute 1, 2, and 3, and then off at minute 2, 3, and 4, respectively. The earliest moment when all lights are on is minute 3.

2. **Test Case 2:**
   - `n = 3`, `k = 3`
   - `a = [1, 3, 5]`
   - This case tests the scenario where `k` is 3, and the lights are installed at different times. The light in room 1 will be on at minutes 1, 4, 7, ..., the light in room 3 will be on at minutes 3, 6, 9, ..., and the light in room 5 will be on at minutes 5, 8, 11, ... These sequences do not overlap, so the answer should be -1.

3. **Test Case 3:**
   - `n = 3`, `k = 2`
   - `a = [1, 2, 3]`
   - This case tests the scenario where `k` is 2, and the lights are installed at different times. The light in room 1 will be on at minutes 1, 3, 5, ..., the light in room 2 will be on at minutes 2, 4, 6, ..., and the light in room 3 will be on at minutes 3, 5, 7, ... The earliest moment when all lights are on is minute 3.

This test input should help expose the bug in the provided buggy code by covering a range of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Light Switches
time_limit_ms: 2000
memory_limit_mb: 256
Description: There is an apartment consisting of $n$ rooms, each with its light initially turned off.

To control the lights in these rooms, the owner of the apartment decided to install chips in the rooms so that each room has exactly one chip, and the chips are installed at different times. Specifically, these times are represented by the array $a_1, a_2, \ldots, a_n$, where $a_i$ is the time (in minutes) at which a chip is installed in the $i$-th room.

As soon as a chip is installed, it changes the room's light status every $k$ minutes — it turns on the light for $k$ minutes, then turns it off for the next $k$ minutes, then turns it back on for the next $k$ minutes, and so on. In other words, the light status is changed by the chip at minute $a_i$, $a_i + k$, $a_i + 2k$, $a_i + 3k$, $\ldots$ for the $i$-th room.

What is the earliest moment when all rooms in the apartment have their lights turned on?

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 $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of rooms in the apartment and the period of the chips.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the moments when the chips are installed.

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

Output: For each test case, print a single integer — the answer to the question (in minutes). If there is no such moment that the lights are turned on in all the rooms, print $-1$ instead.

Examples:
input:
9
4 4
2 3 4 5
4 3
2 3 4 5
4 3
3 4 8 9
3 3
6 2 1
1 1
1
7 5
14 34 6 25 46 7 17
6 5
40 80 99 60 90 50
6 5
64 40 50 68 70 10
2 1
1 1000000000
output:
5
-1
10
8
1
47
100
-1
-1

Note: In the first test case, all lights will be on by the minute $5$ without any of them being turned off by the chips. The answer is $5$.

In the second test case, due to $k=3$, the $1$-st light will be on at minutes $2, 3, 4, 8, 9, 10, 14, \ldots$; meanwhile, the $4$-th light will be on at minutes $5, 6, 7, 11, 12, 13, 17, \ldots$. These two sequences don't have any number in common, so they will never be on at the same time.

In the third test case, it can be seen that the $1$-st and $2$-nd lights will be turned off at minutes $6$ and $7$, but the chips will turn them back on at minutes $9$ and $10$. The $3$-rd and $4$-th lights will also be on at minute $10$, so the answer is $10$.

Buggy Code:
import java.util.*;

public class C{
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt(),k=sc.nextInt(),l=-1;
            Integer[]arr=new Integer[n];
            for(int i=0;i<n;i++)arr[i]=sc.nextInt();
            Arrays.sort(arr);
            int[]num=new int[k];
            for(int i=0;i<n-1;i++){
                if((arr[n-1]-arr[i])/k%2!=0){
                    for(int j=arr[n-1]+k-1;j>arr[n-1]+k-1-(arr[n-1]-arr[i])%k;j--){
                        num[j-arr[n-1]]++;
                    }
                }else{
                    for(int j=arr[n-1];j<arr[n-1]+k-(arr[n-1]-arr[i])%k;j++){
                        num[j-arr[n-1]]++;
                    }
                }
            }

            for(int i=0;i<k;i++){
                if(num[i]==n-1){
                    l=arr[n-1]+i;
                    break;
                }
            }
            System.out.println(l);
        }
    }
}

**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
3 2
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Counting Pairs
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence $$$a$$$, consisting of $$$n$$$ integers, where the $$$i$$$-th element of the sequence is equal to $$$a_i$$$. You are also given two integers $$$x$$$ and $$$y$$$ ($$$x \le y$$$).

A pair of integers $$$(i, j)$$$ is considered interesting if the following conditions are met:

- $$$1 \le i < j \le n$$$;
- if you simultaneously remove the elements at positions $$$i$$$ and $$$j$$$ from the sequence $$$a$$$, the sum of the remaining elements is at least $$$x$$$ and at most $$$y$$$.

Your task is to determine the number of interesting pairs of integers for the given sequence $$$a$$$.

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

Each test case consists of two lines:

- The first line contains three integers $$$n, x, y$$$ ($$$3 \le n \le 2 \cdot 10^5$$$, $$$1 \le x \le y \le 2 \cdot 10^{14}$$$);
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$).

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

Output: For each test case, output one integer — the number of interesting pairs of integers for the given sequence $$$a$$$.

Examples:
input:
7
4 8 10
4 6 3 6
6 22 27
4 9 6 3 4 5
3 8 10
3 2 1
3 1 1
2 3 4
3 3 6
3 2 1
4 4 12
3 3 2 1
6 8 8
1 1 2 2 2 3
output:
4
7
0
0
1
5
6

Note: In the first example, there are $$$4$$$ interesting pairs of integers:

1. $$$(1, 2)$$$;
2. $$$(1, 4)$$$;
3. $$$(2, 3)$$$;
4. $$$(3, 4)$$$.

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

int main()
{
     int t;
     cin>>t;
      for(int i=0; i<t; i++){
            long long  int n,x,y;
              cin>>n>>x>>y;
              long long  int arr[n];
              for(long long int i=0; i<n; i++){
                   cin>>arr[i];
              }
              long long int sum=0;
               for(long long int i=0; i<n; i++){
                   sum= sum+ arr[i];
               }
               sort(arr, arr+n);
                 long long int st=0;
  long long int ans=0;
           for(long long int i=0; i<n-1; i++){
               long long int str;
                int chk=0;
                if(i+1>=st){
                   str= i+1;
                }
                else{
                   str= st;
                }

               for(long long int j= str ; j<n; j++ ){
                      if((sum-arr[i]-arr[j]<=y) && (chk==0)){
                            st= j;
                            chk= 1;
                      }
                    if((sum-arr[i]-arr[j]<=y) && (sum-arr[i]-arr[j]>= x)){
                          ans++;
                    }
                  if(sum-arr[i]-arr[j]<x){
                     break;
                  }

               }
           }
           cout<<ans<<endl;

      }
  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that are likely to cause issues. The code sorts the array and then iterates through it to find pairs that meet the conditions. Here are some potential edge cases to consider:

1. **All elements are the same**: This can cause issues if the code does not handle duplicate elements correctly.
2. **Very large or very small values**: This can cause overflow or underflow issues.
3. **Edge cases for the sum conditions**: Pairs where the sum of the remaining elements is exactly equal to `x` or `y`.

Let's create a test case that combines these elements:

```plaintext
1
5 10 15
5 5 5 5 5
```

### Explanation:
- **Test Case**: 1
- **n**: 5
- **x**: 10
- **y**: 15
- **Array**: [5, 5, 5, 5, 5]

### Expected Output:
- **Number of interesting pairs**: 10

### Reasoning:
- The sum of the entire array is `25`.
- Removing any two elements (each of which is 5) results in a sum of `20` for the remaining elements.
- Since `10 <= 20 <= 15` is true, every pair of elements is an interesting pair.
- There are `10` pairs of elements in a 5-element array: (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5).

This test case is designed to be small and manageable, but it should effectively expose any issues in the code related to handling duplicate elements and the sum conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Counting Pairs
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence $$$a$$$, consisting of $$$n$$$ integers, where the $$$i$$$-th element of the sequence is equal to $$$a_i$$$. You are also given two integers $$$x$$$ and $$$y$$$ ($$$x \le y$$$).

A pair of integers $$$(i, j)$$$ is considered interesting if the following conditions are met:

- $$$1 \le i < j \le n$$$;
- if you simultaneously remove the elements at positions $$$i$$$ and $$$j$$$ from the sequence $$$a$$$, the sum of the remaining elements is at least $$$x$$$ and at most $$$y$$$.

Your task is to determine the number of interesting pairs of integers for the given sequence $$$a$$$.

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

Each test case consists of two lines:

- The first line contains three integers $$$n, x, y$$$ ($$$3 \le n \le 2 \cdot 10^5$$$, $$$1 \le x \le y \le 2 \cdot 10^{14}$$$);
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$).

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

Output: For each test case, output one integer — the number of interesting pairs of integers for the given sequence $$$a$$$.

Examples:
input:
7
4 8 10
4 6 3 6
6 22 27
4 9 6 3 4 5
3 8 10
3 2 1
3 1 1
2 3 4
3 3 6
3 2 1
4 4 12
3 3 2 1
6 8 8
1 1 2 2 2 3
output:
4
7
0
0
1
5
6

Note: In the first example, there are $$$4$$$ interesting pairs of integers:

1. $$$(1, 2)$$$;
2. $$$(1, 4)$$$;
3. $$$(2, 3)$$$;
4. $$$(3, 4)$$$.

Buggy Code:
#include <iostream>
#include <cmath>
#include <algorithm>
#include <ios>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdint>

using ll = int64_t;
using ull = uint64_t;
using namespace std;

void Solve() {
    ll n, x, y;
    cin >> n >> x >> y;

    vector<ll> nms(n);

    ll sum = 0;

    for (ll i = 0; i < n; ++i) {
        cin >> nms[i];
        sum += nms[i];
    }

    ll mn = sum - y;
    if (mn < 0) mn = 0;
    ll mx = sum - x;
    if (mx < 1) {
        cout << 0 << '\n';
        return;
    }

    sort(nms.begin(), nms.end());

    //cout << '\n';

    ll it = 0, jt = 1;

    ll res = 0;
    ll count = 0;

    ll s = nms[it] + nms[jt], s0 = nms[jt] + nms[jt - 1];
    while (jt < n && s < mn) {
        ++jt;
        if (jt == n) {
            --jt;
            break;
        }
        s = nms[it] + nms[jt]; s0 = nms[jt] + nms[jt - 1];
    }

    s = nms[it] + nms[jt]; s0 = nms[jt] + nms[jt - 1];
    while (jt < n && s >= mn && s <= mx) {
        ++res;
        if (s0 >= mn && s0 <= mx && jt - it != 1) {
            ++count;
            res += count;
            //cout << count+1;
        }
        //cout << '\t';
        ++jt;
        if (jt == n) {
            --jt;
            break;
        }
        s = nms[it] + nms[jt]; s0 = nms[jt] + nms[jt - 1];
    }

    s = nms[it] + nms[jt]; s0 = nms[jt] + nms[jt - 1];
    if (!(s >= mn && s <= mx)) {
        --jt;
    }

    ++it;
    ll b = nms[jt];
    ll bt = jt - 1;
    ll add = 1;
    while (bt > -1 && nms[bt] == nms[jt]) {
        ++add;
        --bt;
    }
    if (nms[bt + 1] == nms[jt]) --add;
    //cout << '\n' << jt << '\n';
    while (it != jt) {
        s = nms[it] + nms[jt]; s0 = nms[jt] + nms[jt - 1];
        if (s >= mn && s <= mx) {
            res+=add;
            //cout << "ADDED +";
            //cout << '\t';
        }
        ++it;
    }

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

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.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
3 6 6
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Counting Pairs
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence $$$a$$$, consisting of $$$n$$$ integers, where the $$$i$$$-th element of the sequence is equal to $$$a_i$$$. You are also given two integers $$$x$$$ and $$$y$$$ ($$$x \le y$$$).

A pair of integers $$$(i, j)$$$ is considered interesting if the following conditions are met:

- $$$1 \le i < j \le n$$$;
- if you simultaneously remove the elements at positions $$$i$$$ and $$$j$$$ from the sequence $$$a$$$, the sum of the remaining elements is at least $$$x$$$ and at most $$$y$$$.

Your task is to determine the number of interesting pairs of integers for the given sequence $$$a$$$.

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

Each test case consists of two lines:

- The first line contains three integers $$$n, x, y$$$ ($$$3 \le n \le 2 \cdot 10^5$$$, $$$1 \le x \le y \le 2 \cdot 10^{14}$$$);
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$).

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

Output: For each test case, output one integer — the number of interesting pairs of integers for the given sequence $$$a$$$.

Examples:
input:
7
4 8 10
4 6 3 6
6 22 27
4 9 6 3 4 5
3 8 10
3 2 1
3 1 1
2 3 4
3 3 6
3 2 1
4 4 12
3 3 2 1
6 8 8
1 1 2 2 2 3
output:
4
7
0
0
1
5
6

Note: In the first example, there are $$$4$$$ interesting pairs of integers:

1. $$$(1, 2)$$$;
2. $$$(1, 4)$$$;
3. $$$(2, 3)$$$;
4. $$$(3, 4)$$$.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define irange(n) for (ll i = 0; i < n; i++)
#define mod 1000000007
#define pb push_back
#define ff first
#define ss second
#define pii pair<ll,ll>
#define vi vector<ll>
#define vvi vector<vector<ll>>
#define ll long long
#define inf 2000000000
#define pyes cout<<"YES"<<endl
#define pno cout<<"NO"<<endl

int main(){
    fastio;
    ll t;
    cin>>t;
    while(t--){
        ll n,x,y;
        cin>>n>>x>>y;
        vi a(n,0);
        ll sum = 0;
        irange(n){
            cin>>a[i];
            sum+= a[i];
        }
        sort(a.begin(),a.end());

        ll low = sum-y;
        ll high = sum-x;
        if(low<2){
            low = 2;
        }
        if(high<2){
            cout<<0<<endl;
            continue;
        }
        ll count=0;
        ll lowi ;
        ll highi;
        ll lowp=1;
        ll highp=n-1;

        irange(n){
            if(i==n-1) break;
            lowi = low - a[i];
            highi = high- a[i];

                while(a[lowp]<lowi && lowp<n-1){
                    lowp++;
                }
                while(a[lowp]>lowi && lowp>i+1){
                    lowp--;
                }

                while(a[highp]>highi && highp>lowp){
                    highp--;
                }

                if(highp<max(lowp,i+1)) continue;
                if(lowp==highp && (a[lowp]<lowi || a[lowp]>highi)){
                    continue;
                }
                count += highp-max(lowp,i+1)+1;

        }
        cout<<count<<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:
```plaintext
1
3 3 6
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Counting Pairs
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence $$$a$$$, consisting of $$$n$$$ integers, where the $$$i$$$-th element of the sequence is equal to $$$a_i$$$. You are also given two integers $$$x$$$ and $$$y$$$ ($$$x \le y$$$).

A pair of integers $$$(i, j)$$$ is considered interesting if the following conditions are met:

- $$$1 \le i < j \le n$$$;
- if you simultaneously remove the elements at positions $$$i$$$ and $$$j$$$ from the sequence $$$a$$$, the sum of the remaining elements is at least $$$x$$$ and at most $$$y$$$.

Your task is to determine the number of interesting pairs of integers for the given sequence $$$a$$$.

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

Each test case consists of two lines:

- The first line contains three integers $$$n, x, y$$$ ($$$3 \le n \le 2 \cdot 10^5$$$, $$$1 \le x \le y \le 2 \cdot 10^{14}$$$);
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$).

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

Output: For each test case, output one integer — the number of interesting pairs of integers for the given sequence $$$a$$$.

Examples:
input:
7
4 8 10
4 6 3 6
6 22 27
4 9 6 3 4 5
3 8 10
3 2 1
3 1 1
2 3 4
3 3 6
3 2 1
4 4 12
3 3 2 1
6 8 8
1 1 2 2 2 3
output:
4
7
0
0
1
5
6

Note: In the first example, there are $$$4$$$ interesting pairs of integers:

1. $$$(1, 2)$$$;
2. $$$(1, 4)$$$;
3. $$$(2, 3)$$$;
4. $$$(3, 4)$$$.

Buggy Code:
import sys
input = sys.stdin.readline

class BIT:
    def __init__(self, size):
        self.size = size
        self.tree = [0] * (size + 1)

    def add(self, idx, val):
        while idx <= self.size:
            self.tree[idx] += val
            idx += idx & (-idx)

    def sum(self, idx):
        total = 0
        while idx > 0:
            total += self.tree[idx]
            idx -= idx & (-idx)
        return total

    def sum_range(self, left, right):
        if right < left:
            return 0
        return self.sum(right) - self.sum(left - 1)

def solve_test_case(n: int, x: int, y: int, nums: list) -> int:
    # 计算原始序列的总和
    total = sum(nums)
    count = 0
    max_val = 100000

    # 因为 1 ≤ ai ≤ 10^5，直接用值作为索引
    bit = BIT(max_val + 1)

    # 从前往后遍历j
    for j in range(n):
        # 对于每个位置j，计算移除j后的和
        sum_without_j = total - nums[j]

        # 计算需要查找的范围
        # sum_without_j - nums[i] ∈ [x,y]
        # nums[i] ∈ [sum_without_j-y, sum_without_j-x]
        left_val = sum_without_j - y
        right_val = sum_without_j - x

        # 如果范围完全超出可能的值域，跳过
        if left_val > max_val or right_val < 1:
            continue

        # 限制在有效范围内
        left_val = max(1, left_val)
        right_val = min(max_val, right_val)

        # 使用树状数组计算范围内的数的个数
        count += bit.sum_range(left_val, right_val)

        # 将当前数加入树状数组
        bit.add(nums[j], 1)

    return count

def main():
    input_lines = sys.stdin.read().strip().split('\n')
    pos = 0

    t = int(input_lines[pos])
    pos += 1

    for _ in range(t):
        n, x, y = map(int, input_lines[pos].split())
        pos += 1
        nums = list(map(int, input_lines[pos].split()))
        pos += 1

        result = solve_test_case(n, x, y, nums)
        sys.stdout.write(str(result) + '\n')

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The primary issue in the code is that it assumes all values in the array `nums` are within the range `[1, 100000]`. This assumption can lead to incorrect results if the values in `nums` exceed this range.

Here are the steps to create a test input that can expose this bug:

1. **Test with Values Outside the Assumed Range**: Use values in `nums` that are greater than 100000.
2. **Test with Small `n`**: Ensure the test case is small enough to be manageable and valid.
3. **Test with Specific `x` and `y` Values**: Choose `x` and `y` such that the conditions for interesting pairs are met or not met based on the values in `nums`.

Let's create a test input that includes a value greater than 100000 and check if the code fails to handle it correctly.

```plaintext
1
3 10 20
100001 2 3
```

### Explanation:
- **Test Case Count**: 1
- **Sequence Length `n`**: 3
- **Range for Remaining Sum `x` and `y`**: 10 and 20
- **Sequence `nums`**: [100001, 2, 3]

### Expected Output:
- The total sum of the sequence is `100001 + 2 + 3 = 100006`.
- Removing any pair of elements:
  - Removing (1, 2): Remaining elements are [3], sum = 3 (not in [10, 20])
  - Removing (1, 3): Remaining elements are [2], sum = 2 (not in [10, 20])
  - Removing (2, 3): Remaining elements are [100001], sum = 100001 (not in [10, 20])
- No interesting pairs.

The expected output is `0`.

### Why This Test Input Exposes the Bug:
- The value `100001` in `nums` is outside the range `[1, 100000]` that the code assumes.
- The code will fail to correctly handle this value in the tree状数组 (BIT), leading to incorrect results.

This test input should help identify the issue with the code's assumption about the range of values in `nums`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Counting Pairs
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence $$$a$$$, consisting of $$$n$$$ integers, where the $$$i$$$-th element of the sequence is equal to $$$a_i$$$. You are also given two integers $$$x$$$ and $$$y$$$ ($$$x \le y$$$).

A pair of integers $$$(i, j)$$$ is considered interesting if the following conditions are met:

- $$$1 \le i < j \le n$$$;
- if you simultaneously remove the elements at positions $$$i$$$ and $$$j$$$ from the sequence $$$a$$$, the sum of the remaining elements is at least $$$x$$$ and at most $$$y$$$.

Your task is to determine the number of interesting pairs of integers for the given sequence $$$a$$$.

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

Each test case consists of two lines:

- The first line contains three integers $$$n, x, y$$$ ($$$3 \le n \le 2 \cdot 10^5$$$, $$$1 \le x \le y \le 2 \cdot 10^{14}$$$);
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$).

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

Output: For each test case, output one integer — the number of interesting pairs of integers for the given sequence $$$a$$$.

Examples:
input:
7
4 8 10
4 6 3 6
6 22 27
4 9 6 3 4 5
3 8 10
3 2 1
3 1 1
2 3 4
3 3 6
3 2 1
4 4 12
3 3 2 1
6 8 8
1 1 2 2 2 3
output:
4
7
0
0
1
5
6

Note: In the first example, there are $$$4$$$ interesting pairs of integers:

1. $$$(1, 2)$$$;
2. $$$(1, 4)$$$;
3. $$$(2, 3)$$$;
4. $$$(3, 4)$$$.

Buggy Code:
import sys, math
#import numpy as np

lines = []
for line in sys.stdin.readlines():
  lines.append(line)

"""
dd = open("input-D.txt", "r")
lines = dd.readlines()
dd.close()
"""

def bs(arr, x, y, left, rght):
  if left >= rght:
     return -1

  mid = (rght - left) // 2 + left
  if x > arr[mid]:
     if mid+1 <= len(arr)-1:
        if x > arr[mid+1]:
           left = mid + 1
           return bs (arr, x, y, left, rght)
        else:
           return mid+1
     return -1
  elif x <= arr[mid]:
     if mid-1>=0:
        if x <= arr[mid-1]:
           rght = mid - 1
           return bs (arr, x, y, left, rght)
        else:
           return mid
     else:
        return mid

def bs2(arr, x, y, left, rght):
  if left >= rght:
     return -1

  mid = (rght - left) // 2 + left
  if arr[mid] <= y:
     if mid+1 <= len(arr)-1:
        if arr[mid+1] <= y:
           left = mid + 1
           return bs2 (arr, x, y, left, rght)
        else:
           return mid
     else:
        return mid
  elif y < arr[mid]:
     if mid-1>=0:
        if y < arr[mid-1]:
           rght = mid - 1
           return bs2 (arr, x, y, left, rght)
        else:
           return mid-1
     else:
        return -1

t = int(lines[0][:-1])

for i in range(t):
  ln1 = lines[2*i+1][:-1].split()
  n = int(ln1[0])
  x = int(ln1[1])
  y = int(ln1[2])

  ln2 = lines[2*i+2][:-1].split()
  A = []
  S = 0
  for j in range(n):
    A.append(int(ln2[j]))
    S += int(ln2[j])
  #print(n,x,y)
  #print("A =", A)

  # solution
  N = 0
  C = []
  for ii in range(0,n-1):
    for jj in range(ii+1,n):
       C.append(S - A[ii]-A[jj])

  C.sort()
  #print("C =", C)

  idx1 = bs(C, x, y, 0, len(C))
  idx2 = bs2(C, x, y, 0, len(C))

  if idx2 >= idx1 and idx1 > -1 and idx2 > -1:
     print(idx2 - idx1 + 1)
  else:
     print(0)

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

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

Think step by step.

Response:
```plaintext
3
3 8 10
2 3 4
4 4 12
1 2 3 4
5 1 15
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: Counting Pairs
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence $$$a$$$, consisting of $$$n$$$ integers, where the $$$i$$$-th element of the sequence is equal to $$$a_i$$$. You are also given two integers $$$x$$$ and $$$y$$$ ($$$x \le y$$$).

A pair of integers $$$(i, j)$$$ is considered interesting if the following conditions are met:

- $$$1 \le i < j \le n$$$;
- if you simultaneously remove the elements at positions $$$i$$$ and $$$j$$$ from the sequence $$$a$$$, the sum of the remaining elements is at least $$$x$$$ and at most $$$y$$$.

Your task is to determine the number of interesting pairs of integers for the given sequence $$$a$$$.

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

Each test case consists of two lines:

- The first line contains three integers $$$n, x, y$$$ ($$$3 \le n \le 2 \cdot 10^5$$$, $$$1 \le x \le y \le 2 \cdot 10^{14}$$$);
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$).

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

Output: For each test case, output one integer — the number of interesting pairs of integers for the given sequence $$$a$$$.

Examples:
input:
7
4 8 10
4 6 3 6
6 22 27
4 9 6 3 4 5
3 8 10
3 2 1
3 1 1
2 3 4
3 3 6
3 2 1
4 4 12
3 3 2 1
6 8 8
1 1 2 2 2 3
output:
4
7
0
0
1
5
6

Note: In the first example, there are $$$4$$$ interesting pairs of integers:

1. $$$(1, 2)$$$;
2. $$$(1, 4)$$$;
3. $$$(2, 3)$$$;
4. $$$(3, 4)$$$.

Buggy Code:
for _ in range(int(input())):
    n, x, y = map(int, input().split())
    a = list(map(int, input().split()))
    c, a, l, r, ans = sum(a), sorted(a), 1, n - 1, 0
    for i in range(n):
        if l == i: l += 1
        while l < n and c - (a[l] + a[i]) > y: l += 1
        while r >= 0 and c - (a[r] + a[i]) < x: r -= 1
        if l > r: break
        ans += r - l + 1
    print(ans)

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

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

Think step by step.

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