**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
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){
        long n = sc.nextLong() ;
        int count = 1 ;
        while(n>=4){
          n = (long)(Math.floor(n/4)) ;
          count = count*2 ;
        }
        System.out.println(count) ;
      }
  }
}

**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
12
13
```

**Explanation:**
- For `n = 4`: The expected output is `2` because you can transform a coin of value `4` into two coins of value `1`.
- For `n = 12`: The expected output is `4` because you can transform a coin of value `12` into two coins of value `3`, and each of those can be transformed into two coins of value `0` (but since `3` is not greater than `3`, no further transformation is possible).
- For `n = 13`: The expected output is `4` because you can transform a coin of value `13` into two coins of value `3`, and each of those can be transformed into two coins of value `0` (but since `3` is not greater than `3`, no further transformation is possible).

These test cases are designed to challenge the buggy code, especially the handling of the transformation when `n` is exactly `4` or just above `4`. The buggy code incorrectly multiplies the count by 2 even when `n` is `3`, which is not allowed according to the problem statement.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

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

public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		long n;
		while(t-- > 0) {
		    n = sc.nextLong();
		    double power = Math.log10(n) / Math.log10(4);
		    long ans = (long) Math.pow(2, (long) power);
		    System.out.println(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
5
4
7
11
15
1024
```

### Explanation:
1. **Edge Case: 4**
   - When `n = 4`, the transformation rule does not apply because `x` must be greater than 3. The expected output is 1, but the buggy code might incorrectly calculate it as 2.

2. **Edge Case: 7**
   - When `n = 7`, the transformation rule applies, and the coin can be transformed into two coins of value 1. The expected output is 2, but the buggy code might incorrectly calculate it as 2 due to the logarithmic approximation.

3. **Edge Case: 11**
   - When `n = 11`, the transformation rule applies, and the coin can be transformed into two coins of value 2. Each of these coins can be further transformed into two coins of value 0. The expected output is 4, but the buggy code might incorrectly calculate it as 4 due to the logarithmic approximation.

4. **Edge Case: 15**
   - When `n = 15`, the transformation rule applies, and the coin can be transformed into two coins of value 3. Each of these coins can be further transformed into two coins of value 0. The expected output is 4, but the buggy code might incorrectly calculate it as 4 due to the logarithmic approximation.

5. **Large Value: 1024**
   - When `n = 1024`, the transformation rule applies multiple times. The expected output is 256, but the buggy code might incorrectly calculate it due to the logarithmic approximation.

These test cases are designed to expose the bug in the logarithmic calculation and the incorrect handling of edge cases in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
#define ll long long
#define vi vector<int>
#define fi for(int i=0;i<n;i++)
#include<bits/stdc++.h>
using namespace std;

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int t=0;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        if(n<=3){
            cout<<1<<endl;
            continue;
        }
        ll num=log2(n)/log2(4);
        ll ans=pow(2,num);
        cout<<ans<<endl;

        // ll cnt=0;
        // while(n>3){
        //     n/=4;
        //     cnt+=2;
        // }
        // cout<<cnt<<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 buggy code, we need to focus on edge cases and scenarios where the current logic might fail. The key points to consider are:

1. **Small Values**: The code handles values less than or equal to 3 correctly. We should include a value slightly above 3 to ensure it works for the smallest valid transformation.
2. **Power of 4**: The code uses logarithms to determine the number of transformations. We should include a value that is a power of 4 to check if the logarithm calculation is accurate.
3. **Values Just Below a Power of 4**: These values should be transformed into a specific number of coins, but the logarithm might not handle them correctly.
4. **Large Values**: Large values can also expose issues in the logarithm and power calculations.

Given these considerations, a small-scale test input that includes these edge cases would be:

```plaintext
4
4
5
15
16
```

Explanation:
- **4**: A power of 4. Should transform into 2 coins.
- **5**: Just above 4. Should transform into 2 coins.
- **15**: Just below 16. Should transform into 4 coins.
- **16**: A power of 4. Should transform into 4 coins.

This test input will help verify if the logarithm and power calculations are correct and if the code handles edge cases properly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

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

int main() {
   ios_base::sync_with_stdio(0);
   cin.tie(0);cout.tie(0);
   int t;
   cin>>t;
   while(t--){
       long long int n;
       cin>>n;

       int d = log2(n)/log2(4);
       cout<<(1<<d)<<'\n';
   }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
1
4
5
1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
from math import log2
for _ in range(int(input())):
    n=int(input())
    l=int(log2(n))
    l=l//2
    print(2**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
3
4
8
9
```

### Explanation:
1. **Edge Case: `n = 4`**
   - For `n = 4`, the correct transformation would result in 2 coins of value 1 each. The buggy code calculates `l = int(log2(4)) = 2`, and then `l = l // 2 = 1`, which results in `2**1 = 2`. This is correct, but it's a good test case to ensure the code handles the smallest value where the transformation is possible.

2. **Edge Case: `n = 8`**
   - For `n = 8`, the correct transformation would result in 2 coins of value 2 each. Each of these can be further transformed into 2 coins of value 1, resulting in a total of 4 coins. The buggy code calculates `l = int(log2(8)) = 3`, and then `l = l // 2 = 1`, which results in `2**1 = 2`. This is incorrect, as the correct answer is 4.

3. **Edge Case: `n = 9`**
   - For `n = 9`, the correct transformation would result in 2 coins of value 2 each. Each of these can be further transformed into 2 coins of value 1, resulting in a total of 4 coins. The buggy code calculates `l = int(log2(9)) = 3`, and then `l = l // 2 = 1`, which results in `2**1 = 2`. This is incorrect, as the correct answer is 4.

These test cases will help expose the bug in the code by showing that the transformation logic is not correctly implemented for values where the transformation can be applied multiple times.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
import math
for _ in range(int(input())):
    n = int(input())

    h = int(math.log(n, 4))
    print(2**h)

**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
4
1
5
13
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int l=sc.nextInt();
        for(int t=0;t<l;t++)
        {
            long n=sc.nextLong();
            long c= (long) Math.pow(2,(int)(Math.log(n)/Math.log(4)));
            System.out.println(c);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the transformation logic might fail. The buggy code uses a logarithmic approach to determine the number of coins, which can be incorrect for certain values of `n`.

### Key Points to Consider:
1. **Small Values of `n`**: The code might not handle small values of `n` correctly, especially when `n` is just above 3.
2. **Powers of 4**: The code uses a logarithmic approach to find the number of coins, which might not always yield the correct result for values that are not exact powers of 4.
3. **Large Values of `n`**: The code might fail for very large values of `n` due to precision issues in the logarithmic calculation.

### Chosen Test Input:
Let's choose a test case where `n` is just above 3, a value that is a power of 4, and a value that is not a power of 4 but close to one.

```plaintext
3
4
5
6
```

### Explanation:
1. **n = 4**: This is a power of 4. The correct output should be 2 (since 4 can be transformed into two coins of value 1).
2. **n = 5**: This is just above 3. The correct output should be 2 (since 5 can be transformed into two coins of value 1).
3. **n = 6**: This is not a power of 4 but close to 4. The correct output should be 2 (since 6 can be transformed into two coins of value 1).

This test input will help expose the bug in the logarithmic calculation used in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

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

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;
public class GridDimensionSolver{
        public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t>0) {
            Double n=sc.nextDouble();
            long count=1;
            while(n>3) {
                count=count*2;
                n=Math.floor(((double)n)/4);
            }
            System.out.println(count);
            t--;
        }
    }
}

**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
4
1
5
16
12
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
#include <iostream>
#include <queue>
#include <vector>
#include <stack>
#include <deque>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <cmath>
#define f first
#define s second
#define ll long long
#define ld long double

using namespace std;

int query(int l, int r) {
    cout << "? " << l << ' ' << r << '\n';

    int ans;
    cin >> ans;
    return ans;
}

int query(int len, int n, int idx) {
    int lq = n / 4 * idx + 1;
    int rq = n;

    if (rq - lq + 1 > len) {
        rq = lq + len - 1;
    } else {
        lq = n - len + 1;
    }

    return query(lq, rq);
}

void answer(int p) {
    cout << "! " << p << '\n';
}

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

    int v[4] = {
        query(1, n / 4),
        query(n / 4 + 1, n / 2),
        query(n / 2 + 1, n / 4 * 3),
        query(n / 4 * 3 + 1, n)
    };

    int sum = v[0] + v[1] + v[2] + v[3];
    if (sum == 3) {
        // k < n / 4
        int l = 1, r = n / 4;
        if (v[0]) {
            l = n / 4 + 1;
            r = n / 2;
        }
        int start = l;

        while (l <= r) {
            if (l == r) {
                answer(l - start + 1);
                return;
            }
            int mid = (l + r) / 2;
            int p = query(l, mid);
            if (p) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
    } else {
        // k >= n / 4
        int l = n / 4;
        int r = n;
        int idx = -1;
        for (int i = 0; i < 4; i++) {
            if (v[i]) {
                idx = i;
                break;
            }
        }

        while (l <= r) {
            if (l == r) {
                answer(l);
                return;
            }

            int mid = (l + r) / 2;
            int p = query(mid, n, idx);
            if (p) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
    }
}

int main() {

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

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that maximize the failure probability. The key points to focus on are:

1. **Edge Cases for `k` and `n`**:
   - `k` should be as small as possible (2) or as large as possible (n-1).
   - `n` should be a power of 2, and we should consider small values of `n` to ensure the test input is small-scale.

2. **Position of the 1 in the Array**:
   - The 1 should be placed in positions that are likely to trigger incorrect behavior in the binary search or the query mechanism.

3. **Query Behavior**:
   - The queries should be designed to reveal the incorrect handling of the `k` threshold.

Given these considerations, let's create a test input where `n = 8` and `k = 2`. The 1 in the array will be placed at a position that is not immediately obvious, such as index 3 (0-based index).

### Test Input
```plaintext
2
8

0

1

0

0

1

0

0
```

### Explanation
- **`n = 8`**: The array length is 8, which is a power of 2.
- **`k = 2`**: The threshold for the device to return the opposite result is 2.
- **Array**: `[0, 0, 1, 0, 0, 0, 0, 0]`
  - The 1 is at index 3 (0-based index).

### Expected Behavior
- The device will return the correct result for ranges of length less than 2.
- The device will return the opposite result for ranges of length 2 or more.

### Why This Input is Challenging
- **Small `k`**: The small value of `k` (2) means that many ranges will trigger the opposite result, making it harder to determine the correct position of the 1.
- **Position of 1**: The 1 is placed at index 3, which is not in the first quarter of the array, making it less likely to be quickly identified by simple binary search.

This test input should help expose the bug in the provided code by forcing it to handle the small `k` value and the position of the 1 in a non-trivial way.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cassert>
#include <memory>
#include <inttypes.h>

int query(int l, int r) {
    std::cout << "? " << l << " " << r << std::endl;
    std::cout.flush();
    int ret;
    std::cin >> ret;
    if (ret == -1) exit(0);
    return ret;
}

void answer(int ans) {
    std::cout << "! " << ans << std::endl;
    std::cout.flush();
    return;
}

void solve() {
    int n;
    std::cin >> n;
    int size = n;

    int q1 = query(1, n / 2);

    auto find_low_k = [&](int start, int end) {
        // find k if k < n / 2.
        // [start, end] is an all 0 interval (size n / 2);
        // 0 0 0 0 0 1 1 1 1 1
        //           ^

        int ans = 0;
        int low = 2;
        int high = n / 2;
        while (low <= high) {
            int mid = (low + high) / 2;
            int q = query(start, start + mid - 1);
            if (q == 0) {
                low = mid + 1;
            } else {
                ans = mid;
                high = mid - 1;
            }
        }
        if (query(start + ans - 1, start + ans - 1) == 0) {
            return ans;
        } else {
            return -1;
        }
    };

    auto find_high_k = [&](int start, int end) {
        // find k if k >= n / 2;
        // [start, end] contains an 1;
        // 1 1 1 1 1 0 0 0 0 0 0 0
        //           ^

        int ans = 0;
        int low = n / 2;
        int high = n;
        while (low <= high) {
            int mid = (low + high) / 2;

            int new_start, new_end;
            if (start == n / 2 + 1) {
                new_start = n - mid + 1;
                new_end = n;
            } else {
                new_start = 1;
                new_end = mid;
            }
            int q = query(new_start, new_end);
            if (q == 0) {
                high = mid - 1;
                ans = mid;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    };

    int start, end;
    if (q1 == 1) {
        start = 1;
        end = n / 2;
    } else {
        start = n / 2 + 1;
        end = n;
    }

    int k_low = find_low_k(start, end);
    int k_high = find_high_k(start, end);

    answer(k_low > 0 ? k_low: k_high);
}

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

#ifdef FILE_INPUT
    std::ifstream file_in("input.txt");
    if (!file_in) {
        std::cerr << "Unable to open the input file" << std::endl;
    }
    std::cin.rdbuf(file_in.rdbuf());
#endif

    int tests;
    std::cin >> tests;

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

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

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

Think step by step.

Response:
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 maximize the failure probability. The code has two main functions: `find_low_k` and `find_high_k`, which handle different ranges of `k`. We need to create a test case that can make these functions fail or behave unexpectedly.

### Step-by-Step Analysis:

1. **Understanding the Code:**
   - The code first queries the first half of the array to determine if the 1 is in the first half or the second half.
   - It then uses binary search to find `k` based on whether `k` is less than or greater than `n/2`.

2. **Edge Cases:**
   - **Small `k` (2 ≤ k < n/2):** The code uses `find_low_k` to find `k` in this range.
   - **Large `k` (n/2 ≤ k < n):** The code uses `find_high_k` to find `k` in this range.
   - **Boundary Conditions:** `k = n/2` and `k = n-1` are critical boundary conditions.

3. **Challenging Scenarios:**
   - **k = n/2:** This is a boundary condition where the code might struggle to correctly identify `k`.
   - **k = n-1:** This is another boundary condition that can be challenging because the query ranges will be very large, and the device will always return the opposite result.

### Test Input:
Let's create a test input for `n = 8` and `k = 4` (a boundary condition where `k = n/2`).

### Test Input:
```plaintext
2
8
0
0
1
0
4
1
0
```

### Explanation:
- **n = 8:** The length of the array.
- **k = 4:** The value of `k` which is a boundary condition (k = n/2).
- **Array `a` = [0, 0, 0, 0, 1, 0, 0, 0]:** The hidden array with a single 1 at index 5.

### Queries and Expected Outputs:
1. **Query 1 4:** The range [1, 4] has length 4, which is equal to `k`. The device should return `1 - s(1, 4) = 1 - 0 = 1`.
2. **Query 5 8:** The range [5, 8] has length 4, which is equal to `k`. The device should return `1 - s(5, 8) = 1 - 1 = 0`.
3. **Query 5 5:** The range [5, 5] has length 1, which is less than `k`. The device should return `s(5, 5) = 1`.
4. **Query 4 4:** The range [4, 4] has length 1, which is less than `k`. The device should return `s(4, 4) = 0`.

This test input should expose the bug in the code by challenging the boundary condition where `k = n/2`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <climits>
#include <cmath>
using namespace std;

#define FAST_IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long
const int MOD = 998244353;
vector<int> seive(int n){
    vector<int> primes(n+1,0);
    for (int i=2;i<=n;i++){
        if (primes[i]==0){
            for (int j=i;j<=n;j+=i){
                primes[j]=i;
            }
        }
    }
    return primes;

}
// Function to compute modular exponentiation
int mod_exp(int base, int exp, int mod) {
    int result = 1;
    while (exp > 0) {
        if (exp % 2 == 1) {
            result = (result * base) % mod;
        }
        base = (base * base) % mod;
        exp /= 2;
    }
    return result;
}
int f1(int n){
    int low = n/2;
    int high = n-1;
    while (high-low>1){
        int mid = (low+high)/2;
        cout<<"? 1 "<<mid<<endl;
        int x;
        cin>>x;
        if (x==0){
            low = mid;
        }
        else{
            high = mid;
        }
    }
    return high;
}
int f2(int n){
    int low = 1;
    int high = n/2;
    while (high-low>1){
        int mid = (low+high)/2;
        cout<<"? "<<n/2+1<<n/2+mid<<endl;
        int x;
        cin>>x;
        if (x==0){
            low = mid;
        }
        else{
            high = mid;
        }
    }
    return high;
}
int f3(int n){
    int low = 1;
    int high = n/2;
    while (high-low>1){
        int mid = (low+high)/2;
        cout<<"? 1 "<<mid<<endl;
        int x;
        cin>>x;
        if (x==0){
            low = mid;
        }
        else{
            high = mid;
        }
    }
    return high;
}
int f4(int n){
    int low = n/2;
    int high = n-1;
    while (high-low>1){
        int mid = (low+high)/2;
        cout<<"? "<<n-mid+1<<' '<<n<<endl;
        int x;
        cin>>x;
        if (x==1){
            low = mid;
        }
        else{
            high = mid;
        }
    }
    return high;
}
void solve() {
    int n;
    cin>>n;
    vector<int> x(2);
    for (int i=0;i<2;i++){
        cout<<'?'<<' '<<1+i*n/4<<' '<<n/4*(i+1)<<endl;
        cin>>x[i];
    }
    bool bo1=0;
    int bo2=0;
    cout<<'?'<<' '<<1<<' '<<n/2<<endl;
    int ans;
    cin>>bo2;
    if (x[0]==x[1]){
        bo1=1;
    }
    if (bo1 && bo2){
        if (n==1073741824){
            cout<<"! 3"<<endl;
            return;
        }
        ans=f3(n);
    }
    else if (bo1 && !bo2){
        if (n==1073741824){
            cout<<"! 4"<<endl;
            return;
        }
       ans= f4(n);
    }
    else if (!bo1 && bo2){
        if (n==1073741824){
            cout<<"! 2"<<endl;
            return;
        }
        ans=f2(n);
    }
    else if (!bo1 && !bo2){
        if (n==1073741824){
            cout<<"! 1"<<endl;
            return;
        }
        ans=f1(n);
    }
    else{
        if (n==1073741824){
            cout<<"! 5"<<endl;
            return;
        }
        cout<<"OIVRE"<<endl;
    }
    cout<<"! "<<ans<<endl;
}

int32_t main() {
    FAST_IO
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
8
0
0
1
0
4
1
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
import sys

def query(l, r):
    print(f"? {l} {r}")
    sys.stdout.flush()
    return int(input())

def solve():
    n = int(input())
    result = query(1, n)
    if result == 1:
        k = n
    else:
        k = 2
        l, r = 2, n
        while l < r:
            mid = (l + r + 1) // 2
            result = query(1, mid)
            if result == 1:
                k = mid
                l = mid + 1
            else:
                r = mid - 1
    print(f"! {k}")
    sys.stdout.flush()

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

# s0eo3a1F7yTmrKMqRdC

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

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

Think step by step.

Response:
```plaintext
1
8
0
0
1
0
4
1
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
import sys

def find_p(n):
    low = 1
    high = n
    while low < high:
        mid = (low + high) // 2
        print(f'? {mid} {mid}')
        sys.stdout.flush()
        res = int(sys.stdin.readline())
        if res == 1:
            high = mid
        else:
            low = mid + 1
    return low

def find_k(n, p):
    low = 2
    high = n - 1
    while low < high:
        mid = (low + high) // 2
        l = max(1, p - mid + 1)
        r = l + mid - 1
        if r > n:
            r = n
            l = r - mid + 1
        print(f'? {l} {r}')
        sys.stdout.flush()
        res = int(sys.stdin.readline())
        if res == 0:
            high = mid
        else:
            low = mid + 1
    return low

def solve():
    t = int(sys.stdin.readline())
    for _ in range(t):
        n = int(sys.stdin.readline())
        p = find_p(n)
        k = find_k(n, p)
        print(f'! {k}')
        sys.stdout.flush()

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

0

1

0

1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
import sys

def find_p(n):
    left = 1
    right = n
    while left < right:
        mid = (left + right) // 2
        print(f'? {mid} {mid}')
        sys.stdout.flush()
        res = int(sys.stdin.readline())
        if res == 1:
            return mid
        elif res == 0:
            left = mid + 1
        else:
            exit()
    return left

def find_k(n, p):
    left = 2
    right = n - 1
    answer = n  # Initialize with maximum possible k
    while left <= right:
        mid = (left + right) // 2
        l = max(1, p - mid + 1)
        r = l + mid - 1
        if r > n:
            r = n
            l = r - mid + 1
        print(f'? {l} {r}')
        sys.stdout.flush()
        res = int(sys.stdin.readline())
        if res == 0:
            answer = mid
            right = mid - 1
        else:
            left = mid + 1
    return answer

t = int(sys.stdin.readline())
for _ in range(t):
    n = int(sys.stdin.readline())
    p = find_p(n)
    k = find_k(n, p)
    print(f'! {k}')
    sys.stdout.flush()

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

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

Think step by step.

Response:
```plaintext
2
8

0

0

1

0

4

1

0
```

This test input is designed to challenge the buggy code by providing a scenario where the 1 is placed in a position that requires careful handling of the range queries and the inversion logic for ranges of length at least `k`. The input specifies a hidden binary array of length 8 with a hidden `k` value of 6, and the 1 is placed at index 6. This setup ensures that the queries must correctly handle the inversion for ranges of length 6 or more, and the correct identification of `k` will require precise handling of the edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        for (int caseNum = 0; caseNum < t; caseNum++) {
            int n = sc.nextInt();
            int low = 2;
            int high = n - 1;

            while (low < high) {
                int mid = (low + high + 1) / 2;
                System.out.println("? 1 " + mid);
                System.out.flush();
                int sum = sc.nextInt();
                if (sum == 0) {
                    high = mid - 1;
                } else {
                    low = mid;
                }
            }
            int k = low;
            System.out.println("! " + k);
            System.out.flush();
        }

        sc.close();
    }
}

// mIFJsSAV9gi18UhqpMX

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and the problem constraints.

### Problem Analysis
1. **Array and Range Constraints**: The array `a` is a binary array of length `n` (where `n` is a power of 2), and it contains exactly one `1` and all other elements are `0`.
2. **Query Behavior**:
   - For ranges of length less than `k`, the device returns the correct sum.
   - For ranges of length at least `k`, the device returns the opposite sum.
3. **Goal**: Find `k` using at most 33 queries.

### Buggy Code Analysis
The provided code attempts to find `k` by performing a binary search on the range from 2 to `n-1`. It queries the sum from 1 to `mid` and adjusts the search range based on the result.

### Potential Bugs
1. **Binary Search Logic**: The binary search logic might not correctly handle the edge cases where the range length is exactly `k`.
2. **Range Length Calculation**: The code does not explicitly handle the case where the range length is exactly `k`.

### Challenging Test Input
To expose the bug, we need a test input that makes the binary search logic fail. Specifically, we can create a test input where the range length is exactly `k` and the device's behavior changes.

### Test Input
Let's create a test input where `n = 8` and `k = 4`. The 1 in the array is at index 4, so `a = [0, 0, 0, 1, 0, 0, 0, 0]`.

#### Explanation
- **Query 1 4**: Since the range length is 4, the device will return the opposite sum. The sum `s(1, 4)` is 1, so the device returns 0.
- **Query 1 5**: Since the range length is 5, the device will return the opposite sum. The sum `s(1, 5)` is 1, so the device returns 0.
- **Query 1 6**: Since the range length is 6, the device will return the opposite sum. The sum `s(1, 6)` is 1, so the device returns 0.
- **Query 1 7**: Since the range length is 7, the device will return the opposite sum. The sum `s(1, 7)` is 1, so the device returns 0.
- **Query 1 8**: Since the range length is 8, the device will return the opposite sum. The sum `s(1, 8)` is 1, so the device returns 0.

This test input should make the binary search logic in the buggy code fail to correctly identify `k`.

### Test Input
```plaintext
1
8
0
0
0
0
0
1
0
0
```

This test input sets up the array `a` with `n = 8` and `k = 4`, and the 1 is at index 4. The responses to the queries are designed to expose the bug in the binary search logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

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

public class Main{
    static final int MAX_N = 1000000000;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine());
            int pos = findPos(br, n);
            int k = findK(br, n, pos);
            System.out.println("! " + k);
            System.out.flush();
        }
    }

    static int findPos(BufferedReader br, int n) throws IOException {
        int l = 1, r = n;
        while (l < r) {
            int mid = (l + r) / 2;
            System.out.println("? " + l + " " + mid);
            System.out.flush();
            int res = Integer.parseInt(br.readLine());
            if (res == 1) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }

    static int findK(BufferedReader br, int n, int pos) throws IOException {
        int k = 1;
        while (true) {
            int l = Math.max(1, pos - k + 1);
            int r = Math.min(n, pos + k);
            System.out.println("? " + l + " " + r);
            System.out.flush();
            int res = Integer.parseInt(br.readLine());
            if (res == 0) {
                return r - l + 1;
            }
            k++;
        }
    }
}

// XfyDYav0zjSL9NmtlRZ

**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
8

0

0

1

0

4

1

0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
#include <bits/stdc++.h>
#define uint unsigned long
#define int long long
#define KAMEHAMEHA freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
#define MADEINHEAVEN ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
#define M_PI 3.14159265358979323846
int bs(int a[], int n, int x){
    int l = 0;
    int r = n-1;
    while(l <= r){
        int mid = (l + r) / 2;
        if(a[mid] == (2*n - x - 1)){
            return mid;
        }
        else if(a[mid] > (2*n - x - 1)){
            r = mid;
        }
        else{
            l = mid + 1;
        }
    }
    return l;
}
bool check(int a, int b, int c){
    int p = 0;
    int q = 0;
    int r = 0;
    if(a=='A'){
        p = 1;
    }
    if(b=='A'){
        q = 1;
    }
    if(c=='A'){
        r = 1;
    }
    if(p+q+r>=2){
        return true;
    }
	else{
        return false;
    }
}
int countABC(const string &s, int i) {
    if (i >= 0 && i + 2 < s.size() && s[i] == 'A' && s[i+1] == 'B' && s[i+2] == 'C') {
        return 1;
    }
    return 0;
}
int ch(int n){
    if(n < 0){
        return INT_MAX;
    }
    int count = 0;
    count += n / 15;
    n %= 15;
    count += n / 6;
    n %= 6;
    count += n / 3;
    n %= 3;
    count += n;
    return count;
}
int distance(int i, int j, int n) {
    return min(abs(i - j), n - abs(i - j));
}
int findPossibleMoves(int mat[][2], int n, int p, int q){
    int X[8] = { 2, 1, -1, -2, -2, -1, 1, 2 };
    int Y[8] = { 1, 2, 2, 1, -1, -2, -2, -1 };
    int count = 0;
    for (int i = 0; i < 8; i++) {
        int x = p + X[i];
        int y = q + Y[i];
        if (x >= 0 && y >= 0 && x < n && y < n
            && mat[x][y] == 0)
            count++;
    }
    return count;
}
void solver(){
    int n, m;
    cin >> n >> m;
    if (n < m){
        m = n;
    }
    if (m == 1) {
        cout << "YES" << endl;
        return;
    }
    if (n % m == 0) {
        cout << "NO" << endl;
        return;
    }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0 && i <= m) {
            cout << "NO" << endl;
            return;
        }
    }
    cout << "YES" << endl;
}
void rec(int i, int j, int k, int H, int W, int K, int &ans, vector<vector<char>> &S, vector<vector<bool>> &visited) {
    if (k == K) {
        ans += 1;
        return;
    }
    visited[i][j] = true;
    int directions[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    for (int d = 0; d < 4; ++d) {
        int ni = i + directions[d][0];
        int nj = j + directions[d][1];
        if (0 <= ni && ni < H && 0 <= nj && nj < W && S[ni][nj] == '.' && !visited[ni][nj]) {
            rec(ni, nj, k + 1, H, W, K, ans, S, visited);
        }
    }
    visited[i][j] = false;
}
bool isP(int x){
    for(int i = 2; i * i <= x; i++){
        if(x % i == 0){
            return false;
        }
    }
    return true;
}
bool subs(const string &s, int i) {
    if (i < 0 || i + 3 >= s.size()) return false;
    return s[i] == '1' && s[i + 1] == '1' && s[i + 2] == '0' && s[i + 3] == '0';
}
int findAns(int vow, int cons, int ng) {
    int ans = 0;
    if (cons >= 2 * vow) {
        ans = 3 * vow;
        ans += min(min(ng, 2 * vow), cons - 2 * vow);
    }
    else {
        ans = (cons / 2) * 3;
        if (cons % 2 == 1) {
            if (ng > 0){
                ans += 1;
            }
        }
    }
    return ans;
}
const int MOD = 1e9;
vector<vector<int>> initMatrix(int rows, int cols) {
    return vector<vector<int>>(rows, vector<int>(cols, 0));
}
vector<vector<int>> multiply(const vector<vector<int>>& A, const vector<vector<int>>& B, int size) {
    vector<vector<int>> result = initMatrix(size, size);
    for (int i = 0; i < size; i++) {
        for (int j = 0; j < size; j++) {
            for (int k = 0; k < size; k++) {
                result[i][j] = (result[i][j] + (A[i][k] * B[k][j]) % MOD) % MOD;
            }
        }
    }
    return result;
}
vector<vector<int>> powerMatrix(vector<vector<int>> mat, int exp, int size) {
    vector<vector<int>> result = initMatrix(size, size);
    for (int i = 0; i < size; i++){
        result[i][i] = 1;
    }
    while (exp > 0) {
        if (exp % 2 == 1) {
            result = multiply(result, mat, size);
        }
        mat = multiply(mat, mat, size);
        exp /= 2;
    }
    return result;
}
const double pi = acos(-1);
bool close(int x1, int y1, int x2, int y2) {
    if(abs(x1 - x2) <= 1 && abs(y1 - y2) <= 1 &&abs(x1 - x2) + abs(y1 - y2) >= 1){
        return 1;
    }
    return 0;
}
bool fix(auto& a, auto b) {
    return a > b ? a = b, 1 : 0;
}
const int MAX_ITER = 1000;
int evaluate_expression(const string& expr) {
    int result = 0;
    int current_number = 0;
    char operation = '+';
    int n = expr.size();
    for(int i = 0; i < n; i++){
        char c = expr[i];
        if(isdigit(c)){
            current_number = current_number * 10 + (c - '0');
        }
        if((!isdigit(c) && c != ' ') || i == n-1){
            if(operation == '+'){
                result += current_number;
            }
            else if(operation == '-'){
                result -= current_number;
            }
            operation = c;
            current_number = 0;
        }
    }
    return result;
}
bool validate(const string& expr) {
    string num;
    for(char c : expr) {
        if(isdigit(c)) {
            num += c;
        }
        else {
            if (!num.empty()) {
                if (num.size() > 1 && num[0] == '0'){
                    return false;
                }
                num.clear();
            }
        }
    }
    if (!num.empty()){
        if (num.size() > 1 && num[0] == '0'){
            return false;
        }
    }
    return true;
}
bool findSubstring(string &txt, string &pat) {
    int n = txt.length();
    int m = pat.length();

    // Iterate through txt
    for (int i = 0; i <= n - m; i++) {

        // Check for substring match
        int j;
        for (j = 0; j < m; j++) {

            // Mismatch found
            if (txt[i + j] != pat[j]) {
                break;
            }
        }

        // If we completed the inner loop, we found a match
        if (j == m) {

            // Return starting index
            return true;
        }
    }

    // No match found
    return false;
}
bool canEscape(long long x) {
    while (x >= 0) {
        if (x == 0) return true;
        if (x % 100 == 33) {
            x /= 100;
        }
        else if (x >= 33) {
            x -= 33;
        }
        else {
            return false;
        }
    }
    return false;
}
int binaryToInt(const string& s) {
    int num = 0;
    for(char c : s) {
        num = (num << 1) | (c - '0');
    }
    return num;
}
long long mod_pow(long long x, long long y, int mod) {
    long long result = 1;
    x = x % mod;
    while (y > 0) {
        if (y % 2 == 1) {
            result = (result * x) % mod;
        }
        y = y / 2;
        x = (x * x) % mod;
    }
    return result;
}
int getMex(vector<int>& arr) {
    set<int> s(arr.begin(), arr.end());
    int mex = 0;
    while (s.count(mex)) mex++;
    return mex;
}

void solve() {
    int n;
    cin >> n;
    int l = 1, r = n;
    int ans;
    while(l < r) {
        int mid = (l + r) / 2;
        cout << "? " << l << " " << mid << endl;
        cout.flush();
        cin >> ans;
        int len = mid - l + 1;
        int len_indicator = (len >= 2) ? 1 : 0;
        int s = ans ^ len_indicator;
        if(s == 1) {
            r = mid;
        }
        else {
            l = mid + 1;
        }
    }
    int pos = l;
    int low = 1, high = n;
    while(low < high) {
        int mid = (low + high) / 2;
        int ll = max(1LL, pos - mid + 1);
        int rr = min(n, pos + mid - 1);
        int len = rr - ll + 1;
        cout << "? " << ll << " " << rr << endl;
        cout.flush();
        cin >> ans;
        int len_indicator = (len >= mid) ? 1 : 0;
        int s = ans ^ len_indicator;
        if(s == 1) {
            high = mid;
        }
        else {
            low = mid + 1;
        }
    }
    cout << "! " << low << endl;
    cout.flush();
}

int32_t main(){
    MADEINHEAVEN
    #ifndef ONLINE_JUDGE
    KAMEHAMEHA
    #endif
    int t;
    cin >> t;
    while(t--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key aspects to consider are:

1. **Array Length and Power of 2**: The array length \( n \) must be a power of 2.
2. **Hidden Integer \( k \)**: \( k \) should be within the range \( 2 \le k \le n - 1 \).
3. **Position of 1 in the Array**: The position of the single 1 in the array should be chosen to maximize the difficulty in determining \( k \).

Given the constraints and the nature of the problem, a good test case would involve a small array where the position of 1 and the value of \( k \) create a scenario that challenges the binary search logic used in the solution.

### Chosen Test Case
- **Array Length \( n \)**: 8 (a power of 2)
- **Value of \( k \)**: 5 (a value that is not too small and not too large)
- **Position of 1 in the Array**: 4 (to create a challenging range sum scenario)

### Explanation
- **Array**: \( a = [0, 0, 0, 1, 0, 0, 0, 0] \)
- **Queries**:
  - Query 1 4: \( s(1, 4) = 1 \) (length 4 < 5, correct result)
  - Query 5 8: \( s(5, 8) = 0 \) (length 4 < 5, correct result)
  - Query 1 8: \( s(1, 8) = 1 \) (length 8 >= 5, opposite result, returns 0)

This test case is designed to challenge the binary search logic and the handling of the range sum inversion for ranges of length \( k \) or more.

### Output
```plaintext
8
5
4
```

This test input will help expose any issues in the binary search logic and the handling of the range sum inversion in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
// #define endl '\n'
using ll = long long;

int ask(int l, int r) {
	cout << "? " << l << " " << r << endl;
	int ans;
	cin >> ans;
	return ans;
}

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

	vector<int> a(5);
	for (int i = n / 4; i <= n; i += n / 4) {
		a[i / (n / 4)] = ask(i - n / 4 + 1, i);
	}

	int s = a[1] + a[2] + a[3] + a[4];

	int ans = 0;

	if (s == 3) {
		int l, r;
		for (int i = 1; i <= 4; i++) {
			if (a[i] == 1) {
				l = (i - 1) * (n / 4) + 1;
				r = i * (n / 4);
			}
		}

		int lo = 1, hi = n / 4;
		while (lo <= hi) {
			int mid = lo + hi >> 1;
			if (ask(l + mid - 1, r) == 1) {
				ans = mid;
				hi = mid - 1;
			}
			else {
				lo = mid + 1;
			}
		}
	}
	else {
		int x = 0;
		for (int i = 1; i < 4; i++) {
			if (a[i] == 0 && a[i + 1] == 0) {
				x = ask((i - 1) * (n / 4) + 1, (i + 1) * (n / 4));
				break;
			}
		}

		if (x == 1) {
			int l, r;
			for (int i = 1; i < 4; i++) {
				if (a[i] == 0 && a[i + 1] == 0) {
					l = (i - 1) * (n / 4) + 1;
					r = (i + 1) * (n / 4);
				}
			}

			int lo = n / 4 + 1, hi = n / 2;
			while (lo <= hi) {
				int mid = lo + hi >> 1;
				if (ask(l + mid - 1, r) == 0) {
					ans = mid;
					hi = mid - 1;
				}
				else {
					lo = mid + 1;
				}
			}
		}
		else {
			int lo = n / 2 + 1, hi = n;
			if (a[1] + a[2] == 1) {
				while (lo <= hi) {
					int mid = lo + hi >> 1;
					if (ask(1, mid) == 0) {
						ans = mid;
						hi = mid - 1;
					}
					else {
						lo = mid + 1;
					}
				}
			}
			else {
				while (lo <= hi) {
					int mid = lo + hi >> 1;
					if (ask(n - mid + 1, n) == 0) {
						ans = mid;
						hi = mid - 1;
					}
					else {
						lo = mid + 1;
					}
				}
			}
		}
	}

	cout << "! " << ans << endl;
}

int main() {
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
    int tt = 1;
    cin >> tt;
    while (tt--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
8

0

0

1

0

4

1

0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
step = [1]
for i in range(30):
    step.append(step[-1] * 2)
for __ in range(fg()):
    n = fg()
    m = step.index(n)
    print('?', 1, step[m - 1])
    tt()
    a = fg()
    b = 1 - a
    print('?', 1, step[m - 2])
    tt()
    x1 = fg()
    print('?', step[m - 2] + 1, step[m - 1])
    tt()
    x2 = fg()

    print('?', step[m - 1] + 1, 3 * step[m - 2])
    tt()
    x3 = fg()
    print('?', 3 * step[m - 2] + 1, step[m])
    tt()
    x4 = fg()
    if a != x1 + x2:
        l = 1
        r = step[m - 1] + 1
        fl = 1
    else:
        l = step[m - 1]
        r = step[m]
        fl = 0
    while r - l > 1:
        mi = (l + r) // 2
        if fl:
            if a == 1:
                print('?', 1, mi)
                tt()
                if fg():
                    r = mi
                else:
                    l = mi
            else:
                print('?', step[m], step[m] + mi)
                tt()
                if fg():
                    r = mi
                else:
                    l = mi
        else:
            if a == 1:
                print('?', 1, mi)
                tt()
                if fg() == 0:
                    r = mi
                else:
                    l = mi
            else:
                print('?', step[m] - mi + 1, step[m])
                tt()
                if fg() == 0:
                    r = mi
                else:
                    l = mi
    print('!', r)
    tt()

**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
8

0

0

1

0

4

1

0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Broken Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You, a wizard whose creation was destroyed by a dragon, are determined to hunt it down with a magical AOE tracker. But it seems to be toyed with...

This is an interactive problem.

There is a hidden binary array $$$a$$$ of length $$$n$$$ ($$$\mathbf{n}$$$ is a power of 2) and a hidden integer $$$k\ (2 \le k \le n - 1)$$$. The array $$$a$$$ contains exactly one 1 (and all other elements are 0). For two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), define the range sum $$$s(l, r) = a_l + a_{l+1} + \cdots + a_r$$$.

You have a magical device that takes ranges and returns range sums, but it returns the opposite result when the range has length at least $$$k$$$. Formally, in one query, you can give it a pair of integers $$$[l, r]$$$ where $$$1 \le l \le r \le n$$$, and it will return either $$$0$$$ or $$$1$$$ according to the following rules:

- If $$$r - l + 1 < k$$$, it will return $$$s(l, r)$$$.
- If $$$r - l + 1 \ge k$$$, it will return $$$1 - s(l, r)$$$.

Find $$$k$$$ using at most $$$33$$$ queries.

The device is not adaptive. It means that the hidden $$$a$$$ and $$$k$$$ are fixed before the interaction and will not change during the interaction.

Input: None

Output: None

Examples:
input:
2
8

0

0

1

0

4

1

0
output:
? 3 5

? 1 8

? 4 8

? 3 8

! 6

? 3 3

? 3 4

! 2

Note: In the first test case, $$$k = 6$$$ and the 1 in the hidden array is at index 6, so $$$a = [0, 0, 0, 0, 0, 1, 0, 0]$$$.

- For the query 3 5, since $$$5-3+1 = 3 < k$$$, the device answers correctly. Since 6 is not contained in the range $$$[3, 5]$$$, the device answers $$$0$$$.
- For the query 1 8, since $$$8 - 1 + 1 = 8 \ge k$$$, the device answers $$$0$$$ incorrectly.
- For the query 4 8, since $$$8 - 4 + 1 = 5 < k$$$, the device answers $$$1$$$ correctly.
- For the query 3 8, since $$$8 - 3 + 1 = 6 \ge k$$$, the device answers $$$0$$$ incorrectly.

In the second test case, $$$k = 2$$$ and the 1 in the hidden array is at index 3, so $$$a = [0, 0, 1, 0]$$$.

Note that the example solution may not have enough information to determine $$$k$$$ above; this is only an example.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
step = [1]
for i in range(30):
    step.append(step[-1] * 2)
for __ in range(fg()):
    n = fg()
    m = step.index(n)
    print('?', 1, step[m - 1])
    tt()
    a = fg()
    b = 1 - a
    print('?', 1, step[m - 2])
    tt()
    x1 = fg()
    print('?', step[m - 2] + 1, step[m - 1])
    tt()
    x2 = fg()

    print('?', step[m - 1] + 1, 3 * step[m - 2])
    tt()
    x3 = fg()
    print('?', 3 * step[m - 2] + 1, step[m])
    tt()
    x4 = fg()
    if a != x1 + x2 and b != x3 + x4:
        l = 1
        r = step[m - 1]
        fl = 1
    else:
        l = step[m - 1]
        r = step[m] - 1
        fl = 0
    while r - l > 1:
        mi = (l + r) // 2
        if fl:
            if a == 1:
                print('?', 1, mi)
                tt()
                if fg():
                    r = mi
                else:
                    l = mi
            else:
                print('?', step[m - 1], step[m - 1] + mi - 1)
                tt()
                if fg():
                    r = mi
                else:
                    l = mi
        else:
            if a == 1:
                print('?', 1, mi)
                tt()
                if fg() == 0:
                    r = mi
                else:
                    l = mi
            else:
                print('?', step[m] - mi + 1, step[m])
                tt()
                if fg() == 0:
                    r = mi
                else:
                    l = mi
    print('!', r)
    tt()

**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
8

0

0

1

0

4

1

0
```