**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Even Positions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp had a regular bracket sequence $s$ of length $n$ ($n$ is even). He even came up with his own way to calculate its cost.

He knows that in a regular bracket sequence (RBS), each opening bracket is paired up with the corresponding closing bracket. So he decided to calculate the cost of RBS as the sum of distances between pairs of corresponding bracket pairs.

For example, let's look at RBS (())(). It has three pairs of brackets:

  * (__)__: the distance between brackets at position $1$ and at $4$ is $4 - 1 = 3$;    * _()___: the distance is $3 - 2 = 1$;    * ____(): the distance is $6 - 5 = 1$.

So the cost of (())() is $3 + 1 + 1 = 5$.

Unfortunately, due to data corruption, Monocarp lost all characters on odd positions $s_1, s_3, \dots, s_{n-1}$. Only characters on even positions ($s_2, s_4, \dots, s_{n}$) remain. For example, (())() turned to _(_)_).

Monocarp wants to restore his RBS by placing brackets on the odd positions. But since the restored RBS may not be unique, he wants to choose one with minimum cost. It's too hard to do for Monocarp alone, so can you help him?

Reminder: A regular bracket sequence is a string consisting of only brackets, such that this sequence, when inserted 1-s and +-s, gives a valid mathematical expression. For example, (), (()) or (()())() are RBS, while ), ()( or ())(() are not.

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Next $t$ cases follow.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$; $n$ is even) — the length of string $s$.

The second line of each test case contains a string $s$ of length $n$, where all characters on the odd positions are '_' and all characters on the even positions are either '(' or ')'.

Additional constraints:

  * $s$ can be restored to at least one regular bracket sequence;    * the total sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print one integer — the minimum cost of the regular bracket sequence that can be obtained from $s$ by replacing '_'-s with brackets.

Examples:
input:
4
6
_(_)_)
2
_)
8
_)_)_)_)
8
_(_)_(_)
output:
5
1
4
8

Note: In the first test case, it's optimal to make $s$ equal to (())(). The cost of $s$ will be equal to $3 + 1 + 1 = 5$.

In the second test case, the only option is to make $s$ equal to () with cost $1$.

In the third test case, the only possible RBS is ()()()() with cost $1 + 1 + 1 + 1 = 4$.

In the fourth test case, it's optimal to make $s$ equal to (())(()) with cost $3 + 1 + 3 + 1 = 8$.

Buggy Code:
import java.util.*;
import java.io.*;
public class EvenPositions {
	public static void main(String [] args) {
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {
			System.err.println("Error");
		}

		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			sc.nextLine();
			String str = sc.nextLine();

			char [] s = str.toCharArray();
			int ob = 0;
			int cb = 0;

			for (char i : s) {
				if (i == '(') {
					ob++;
				} else if (i == ')') {
					cb++;
				}
			}
			String ans = "";

			boolean [] b = new boolean[n];
			// char [] ca = new char[n];
			for (int i = 0 ; i < n ; i++) {
				if (s[i] == '(' || s[i] == ')') {
					b[i] = true;
				}
				// ca[i] = s[i];
			}

			// for (int i = 0 ; i < n ; i++) {
			// 	if (b[i] && s[i] == '(') {
			// 		for (int j = i + 1 ; j < n ; j++) {
			// 			if (s[j] == '_') {
			// 				s[j] = ')';
			// 				break;
			// 			}
			// 		}
			// 	} else if (b[i] && s[i] == ')') {
			// 		for (int j = i - 1 ; j >= 0 ; j--) {
			// 			if (s[j] == '_') {
			// 				s[j] = '(';
			// 				break;
			// 			}
			// 		}
			// 	}
			// }
			s[0] = '(';
			for (int i = n - 2 ; i > 0 ; i -= 2) {
				if (s[i - 1] == '(') {
					s[i] = ')';
				} else {
					s[i] = '(';
				}
			}

			for (int i = 0 ; i < n ; i++) {
				ans += s[i];
			}
			// System.out.println(ans);

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

		sc.close();
	}
	static long point(String str) {
		char [] s = str.toCharArray();
		long ans = 0;

		Stack<Character> st = new Stack<>();
		for (char i : s) {
			if (i == '(') {
				st.push(i);
			} else {
				if (st.size() > 1) {
					ans += st.size() + 1;
				} else {
					ans += st.size();
				}
				if (!st.isEmpty())
					st.pop();
			}
		}
		return ans;
	}

	static int[] arrayGen(int size , int limit) {
		int [] ans = new int[size];
		for (int i = 0 ; i < size ; i++) {
			ans[i] = (int)(Math.random() * limit);
		}
		return ans;
	}

	static List<Integer> seive(int N) {
		boolean[] sieve = new boolean[N];

		for (int i = 0; i < sieve.length; i++) {
			sieve[i] = true;
		}
		for (int i = 2; i < Math.sqrt(N); i++) {
			if (sieve[i] == true) {
				for (int j = (i * i); j < N; j = j + i) {
					sieve[j] = false;
				}
			}
		}
		List<Integer> ls = new ArrayList<>();
		for (int i = 2; i < sieve.length; i++) {
			if (sieve[i] == true) {
				ls.add(i);
			}
		}
		return ls;
	}

	public static void swapInArray(int ind1 , int ind2 , int[] arr) {
		int temp = arr[ind1];
		arr[ind1] = arr[ind2];
		arr[ind2] = temp;
	}

	public static void printArray(int [] a) {
		for (int i : a) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

	public static void printList(List<Integer> ls) {
		for (Integer i : ls) {
			System.out.print(i + " ");
		}
		System.out.println();
	}
	public static int gcd(int a , int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b , a % b);
	}
}

**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 identify potential issues:

### Analysis of the Buggy Code

1. **Input Parsing and Initial Setup:**
   - The code reads the number of test cases and the length of the string.
   - It then reads the string and counts the number of opening and closing brackets.

2. **Restoration of the String:**
   - The code attempts to restore the string by placing brackets on the odd positions.
   - It has a commented-out section that seems to be an attempt to match brackets, but it is not used.
   - The code forces the first character to be an opening bracket and then alternates the brackets on even positions based on the previous character.

3. **Cost Calculation:**
   - The `point` function calculates the cost of the restored string.
   - It uses a stack to track the opening brackets and calculates the cost based on the size of the stack.

### Potential Issues

1. **Incorrect Restoration:**
   - The code forces the first character to be an opening bracket, which might not always be the optimal choice.
   - The alternation of brackets on even positions might not always result in a valid RBS with the minimum cost.

2. **Cost Calculation:**
   - The cost calculation might be incorrect if the stack size is not properly managed.

### Challenging Test Input

To expose these issues, we need a test input where the optimal placement of brackets on odd positions is not straightforward and where the cost calculation can be easily verified.

### Test Input

Let's consider a test case where the even positions form a pattern that requires careful placement of brackets on odd positions to form a valid RBS with the minimum cost.

```plaintext
1
10
_(_)_(_)_)
```

### Explanation

- **Length of String (n):** 10
- **String (s):** _(_)_(_)_)

### Expected Output

- **Restored String:** (())(())()
- **Cost:** 3 + 1 + 3 + 1 + 1 = 9

### Verification

- **Restored String:** (())(())()
- **Cost Calculation:**
  - (___): 3
  - ___(): 1
  - (___): 3
  - ___(): 1
  - ____(): 1

The total cost is 9.

This test case should challenge the buggy code to correctly place the brackets and calculate the cost accurately.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Even Positions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp had a regular bracket sequence $s$ of length $n$ ($n$ is even). He even came up with his own way to calculate its cost.

He knows that in a regular bracket sequence (RBS), each opening bracket is paired up with the corresponding closing bracket. So he decided to calculate the cost of RBS as the sum of distances between pairs of corresponding bracket pairs.

For example, let's look at RBS (())(). It has three pairs of brackets:

  * (__)__: the distance between brackets at position $1$ and at $4$ is $4 - 1 = 3$;    * _()___: the distance is $3 - 2 = 1$;    * ____(): the distance is $6 - 5 = 1$.

So the cost of (())() is $3 + 1 + 1 = 5$.

Unfortunately, due to data corruption, Monocarp lost all characters on odd positions $s_1, s_3, \dots, s_{n-1}$. Only characters on even positions ($s_2, s_4, \dots, s_{n}$) remain. For example, (())() turned to _(_)_).

Monocarp wants to restore his RBS by placing brackets on the odd positions. But since the restored RBS may not be unique, he wants to choose one with minimum cost. It's too hard to do for Monocarp alone, so can you help him?

Reminder: A regular bracket sequence is a string consisting of only brackets, such that this sequence, when inserted 1-s and +-s, gives a valid mathematical expression. For example, (), (()) or (()())() are RBS, while ), ()( or ())(() are not.

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Next $t$ cases follow.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$; $n$ is even) — the length of string $s$.

The second line of each test case contains a string $s$ of length $n$, where all characters on the odd positions are '_' and all characters on the even positions are either '(' or ')'.

Additional constraints:

  * $s$ can be restored to at least one regular bracket sequence;    * the total sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print one integer — the minimum cost of the regular bracket sequence that can be obtained from $s$ by replacing '_'-s with brackets.

Examples:
input:
4
6
_(_)_)
2
_)
8
_)_)_)_)
8
_(_)_(_)
output:
5
1
4
8

Note: In the first test case, it's optimal to make $s$ equal to (())(). The cost of $s$ will be equal to $3 + 1 + 1 = 5$.

In the second test case, the only option is to make $s$ equal to () with cost $1$.

In the third test case, the only possible RBS is ()()()() with cost $1 + 1 + 1 + 1 = 4$.

In the fourth test case, it's optimal to make $s$ equal to (())(()) with cost $3 + 1 + 3 + 1 = 8$.

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) {
      int n=sc.nextInt();
      String brac=sc.next();
      // int open=0;
      // int close=0;
      // for(int i=0;i<brac.length();i++) {
      //   if(brac.charAt(i)=='(') {
      //     open++;
      //   }else if(brac.charAt(i)==')') {
      //     close++;
      //   }
      // }
      // int temp=close;
      // close=open;
      // open=temp;
      String newly="(";
      for(int i=1;i<brac.length();i++) {
        if(brac.charAt(i)=='_') {
          if(brac.charAt(i-1)=='(') {
            newly+=')';
          }else {
            newly+='(';
          }
        }else {
          newly+=brac.charAt(i);
        }
      }
      ArrayList<Integer> open=new ArrayList<Integer>();
      ArrayList<Integer> close=new ArrayList<Integer>();
      for(int i=0;i<newly.length();i++) {
        if(newly.charAt(i)=='(') {
          open.add(i);
        }else {
          close.add(i);
        }
      }
      int sum=0;
      for(int i=0;i<n/2;i++) {
        sum+=(close.get(i)-open.get(i));
      }
      System.out.println(sum);
    }
  }
}

**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 or scenarios where the code might fail. The buggy code incorrectly handles the placement of brackets, particularly when it tries to balance the brackets by alternating between opening and closing brackets without considering the overall structure of the regular bracket sequence (RBS).

### Steps to Create a Challenging Test Input:
1. **Identify the Bug**: The code incorrectly alternates between opening and closing brackets based on the previous character, which can lead to an invalid RBS.
2. **Create a Test Case**: We need a test case where the buggy code will create an invalid RBS or a suboptimal RBS with a higher cost than the correct solution.

### Challenging Test Input:
Let's create a test case where the input string is such that the buggy code will produce an incorrect RBS.

```plaintext
1
10
_(_)_(_)_(_)
```

### Explanation:
- **Test Case**: The input string is `"_(_)_(_)_(_)"`.
- **Expected Output**: The minimum cost RBS that can be formed is `(()(()(())))`, which has a cost of 9.
- **Buggy Code Output**: The buggy code will likely produce an incorrect RBS by alternating brackets, leading to a higher cost.

### Verification:
- **Correct RBS**: `(()(()(())))`
  - Pairs: (1, 2), (3, 4), (5, 6), (7, 8), (9, 10)
  - Distances: 1, 1, 1, 1, 1
  - Total Cost: 1 + 1 + 1 + 1 + 1 = 5

- **Buggy Code RBS**: `(())(())(())`
  - Pairs: (1, 2), (3, 4), (5, 6), (7, 8), (9, 10)
  - Distances: 1, 1, 1, 1, 1
  - Total Cost: 1 + 1 + 1 + 1 + 1 = 5

However, the buggy code will likely produce a different and incorrect RBS, such as `()()()()()`, which has a higher cost.

```plaintext
1
10
_(_)_(_)_(_)
```

This test case should expose the bug in the code by showing that the buggy code produces a suboptimal or incorrect RBS.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
#include <vector>
#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <set>
#include <queue>
#define int long long

using namespace std;

typedef pair<int, int> pr;
typedef pair<pair<int, int>, int> tpl;
int n;
int mod = 1e9 + 7;
vector<int> factorials(n + 1);
int fastpow(int n, int st){
    if(st == 0){
        return 1;
    }
    if(st % 2 == 0){
        int k = fastpow(n, st / 2);
        return (k * k) % mod;
    }
    else{
        return (fastpow(n, st - 1) * n) % mod;
    }
}

int del(int a, int b){
    int k = fastpow(b, mod - 2);
    return (a * k) % mod;
}
int combinations(int n, int k){
    int p = del(factorials[n], ((factorials[k] * factorials[(n - k)]) % mod));
    p %= mod;
    return p;
}

signed main() {
    int tt; cin >> tt;
    while(tt--){
        int l = 2;
        int r = 1000;
        int m1, m2;
        int k;
        int s;
        while(l + 1 != r){
            k = (r - l + 2) / 3;
            m1 = l + k;
            m2 = m1 + k;
            cout << "? " << m1 << ' ' << m2 << '\n';
            cin >> s;
            if(s == (m1 + 1) * (m2 + 1)){
                r = m1;
            }
            else if(s == m1 * m2){
                l = m2;
            }
            else{
                l = m1;
                r = m2;
            }
        }

        cout << "! " << r << '\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
2

25

26
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <string.h>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include<deque>
#include<list>
#include<queue>
#include<set>

using namespace std;
#define LL long long
#define ULL unsigned long long
#define PII pair<int, int>

const int N = 2 * 1e5 + 5;
void init() {

}
void solve() {
	int left = 2, right = 999;
	while (left +1 < right) {
		int m1 = left + (right - left) / 3;
		int m2 = m1 + (right - left) / 3;
		cout << "?  " << m1 << " " << m2 << endl << endl;
		int x;
		cin >> x;
		if (m1 * m2 == x) left = m2 + 1;
		else if ((m2 + 1) * (m1+1) == x) right = m1;
		else left = m1 + 1, right = m2;
	}
	cout << "? " << left << " " << right << endl << endl;
	int x;
	cin >> x;
	if(x==left*left) cout << "! " << right << endl;
	else cout << "! " << left << endl;
}
int main() {
	init();
	//freopen("input.txt", "r", stdin);
	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 the logic of the code and identify scenarios where it might fail. The code uses a ternary search approach to find the missing number \( x \) on the ruler. Let's analyze the code and identify potential edge cases:

1. **Initial Range Selection**: The code initializes the range from 2 to 999.
2. **Ternary Search**: It divides the range into three parts and queries the middle values \( m1 \) and \( m2 \).
3. **Comparison Logic**: It compares the product of the queried values with the expected results to narrow down the range.
4. **Final Query**: It makes a final query to determine the exact value of \( x \).

### Potential Bugs:
1. **Incorrect Range Update**: The code might incorrectly update the range based on the comparison results.
2. **Edge Cases**: The code might fail for values of \( x \) close to the boundaries of the range (2 or 999).
3. **Midpoint Calculation**: The midpoint calculation might not handle cases where the range is not perfectly divisible by 3.

### Challenging Test Input:
To expose these potential bugs, we can use the following test cases:
1. \( x = 2 \) (the lower boundary)
2. \( x = 999 \) (the upper boundary)
3. \( x = 3 \) (a small value)
4. \( x = 998 \) (a value just below the upper boundary)

### Test Input:
```plaintext
4
2
999
3
998
```

### Explanation:
- **Test Case 1 (x = 2)**: This should test the lower boundary. The code should correctly identify that the ruler is missing the number 2.
- **Test Case 2 (x = 999)**: This should test the upper boundary. The code should correctly identify that the ruler is missing the number 999.
- **Test Case 3 (x = 3)**: This should test a small value. The code should correctly identify that the ruler is missing the number 3.
- **Test Case 4 (x = 998)**: This should test a value just below the upper boundary. The code should correctly identify that the ruler is missing the number 998.

This test input will help identify if the code handles edge cases and boundary conditions correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
#include<cstdio>
#include<iostream>
#include<string>
#include<cstdlib>
#include<queue>
#include<stack>
#include<utility>
#include<string>
#include<cstring>
#include<set>
#include<vector>
#include<fstream>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
#define int long long
int mod=1000000007;
int MAX_INT=2147483647;

int32_t main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin>>t;
while(t--){
    int a,b,l=2,r=999;
    int res=-1;
    int prev=0;
    while(l<=r){
        int mid=(l+r)>>1;
        a=l+(r-l)/3;
        b=r-(r-l)/3;
        if(a==prev)
        break;
        prev=a;
        cout<<"? "<<a<<" "<<b<<endl;
        cout.flush();
        int prod;
        cin>>prod;
        if(prod==a*b){
            res=b;
            l=b+1;
        }
        else if(prod==a*(b+1)){
            l=a+1;
            r=b-1;
        }
        else{
            r=a-1;
        }
}
cout<<"! "<<res+1<<endl;
    cout.flush();
}
return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code uses a ternary search approach to find the missing number \( x \). The key points to consider are:

1. **Edge Cases:**
   - The smallest possible value for \( x \) (i.e., \( x = 2 \)).
   - The largest possible value for \( x \) (i.e., \( x = 999 \)).
   - Values of \( x \) that are close to the boundaries of the search range.

2. **Behavior of the Ternary Search:**
   - The code uses the ternary search to narrow down the range of possible values for \( x \). It queries two points \( a \) and \( b \) within the current range and adjusts the search range based on the responses.
   - The code might fail if the queries do not correctly identify the boundaries or if the search range is not properly adjusted.

3. **Challenging Scenarios:**
   - Values of \( x \) that are close to the boundaries of the search range (e.g., \( x = 3 \), \( x = 998 \)).
   - Values of \( x \) that are in the middle of the range (e.g., \( x = 500 \)).
   - Values of \( x \) that cause the search range to be reduced in a way that the code might not handle correctly.

Given these considerations, let's create a small-scale test input that includes these challenging scenarios:

```plaintext
5

18

25

9999

1

2000
```

### Explanation:
1. **Test Case 1: \( x = 4 \)**
   - The code should correctly identify \( x = 4 \) after a few queries.
   - The expected queries and responses are:
     - Query: `? 3 5` -> Response: `18`
     - Query: `? 4 4` -> Response: `25`
     - Output: `! 4`

2. **Test Case 2: \( x = 100 \)**
   - The code should correctly identify \( x = 100 \).
   - The expected query and response is:
     - Query: `? 99 100` -> Response: `9999`
     - Output: `! 100`

3. **Test Case 3: \( x = 999 \)**
   - The code should correctly identify \( x = 999 \).
   - The expected query and response is:
     - Query: `? 998 999` -> Response: `998000`
     - Output: `! 999`

4. **Test Case 4: \( x = 2 \)**
   - The code should correctly identify \( x = 2 \).
   - The expected query and response is:
     - Query: `? 1 2` -> Response: `2`
     - Output: `! 2`

5. **Test Case 5: \( x = 500 \)**
   - The code should correctly identify \( x = 500 \).
   - The expected queries and responses are:
     - Query: `? 333 666` -> Response: `222222`
     - Query: `? 500 500` -> Response: `250001`
     - Output: `! 500`

This test input includes a variety of challenging scenarios to ensure the code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
import sys

def main():
    t = int(input())

    for _ in range(t):
        query = f"? 99 100"
        print(query)
        sys.stdout.flush()

        response = int(input())
        if response == 9999:
            print("! 100")
        else:
            print("! 4")

        sys.stdout.flush()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's assumptions lead to incorrect outputs. The buggy code always queries the rectangle with dimensions `99 x 100` and makes a decision based on the response. Here are the steps to create a test input:

1. **Identify the Bug**: The buggy code assumes that if the response is `9999`, the missing number \( x \) is `100`, and if the response is anything else, it assumes \( x \) is `4`. This assumption is flawed because there are many other values of \( x \) that can produce a response of `9999` or a different value.

2. **Generate a Test Case**: We need to create a test case where the response to the query `99 x 100` is `9999` but the actual missing number \( x \) is not `100`. For example, if \( x = 50 \), the ruler will measure the rectangle as `99 x 101`, which also results in a response of `9999`.

3. **Test Case Structure**: The test input should include the number of test cases, followed by the responses for each query.

Given these steps, here is a challenging test input:

```plaintext
1
9999
```

This test input will cause the buggy code to incorrectly output `100` when the correct answer should be `50` or any other value that makes the ruler measure `99 x 100` as `9999`.

Here is the complete test input:

```plaintext
1
9999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
import sys

t = int(input())
for _ in range(t):
    print("? 3 5")
    a = int(input())
    print("? 4 4")
    b = int(input())

    if a == 18 and b == 25:
        print("! 4")
    else:
        print("? 99 100")
        input()  # Ignore response
        print("! 100")

**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 create a test case that will cause the code to fail. The buggy code assumes that if the measurements for the queries `? 3 5` and `? 4 4` are 18 and 25 respectively, then the missing number `x` must be 4. However, this assumption is not always correct, and there are other values of `x` that can also result in these measurements.

### Step-by-Step Reasoning:

1. **Identify the Bug:**
   - The code only checks for the specific case where `a == 18` and `b == 25` and concludes that `x` is 4.
   - This assumption fails if there are other values of `x` that can produce the same measurements.

2. **Find an Edge Case:**
   - We need to find a value of `x` other than 4 that can also produce the measurements 18 and 25 for the queries `? 3 5` and `? 4 4`.

3. **Calculate the Measurements:**
   - For `? 3 5`:
     - If `x = 4`, the measurement is `3 * 6 = 18`.
   - For `? 4 4`:
     - If `x = 4`, the measurement is `5 * 5 = 25`.

4. **Check for Other Values:**
   - If `x = 6`, the measurements would be:
     - For `? 3 5`: `3 * 6 = 18` (correct).
     - For `? 4 4`: `4 * 5 = 20` (incorrect, should be 25).

5. **Find a Value that Produces 18 and 25:**
   - If `x = 5`, the measurements would be:
     - For `? 3 5`: `3 * 6 = 18` (correct).
     - For `? 4 4`: `5 * 5 = 25` (correct).

### Test Case:
- The test case should have `x = 5` to expose the bug in the code.

### Output:
```plaintext
1

18

25
```

This test case will cause the buggy code to incorrectly determine that `x` is 4, when in fact `x` is 5.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
import sys

def find_missing_number():
    # Number of test cases
    t = int(input())

    for _ in range(t):
        # Initial query ? 3 5
        print("? 3 5")
        sys.stdout.flush()  # Flush the output to ensure the program continues
        area1 = int(input())  # Input the response from the jury

        # The correct area would have been 3 * 5 = 15, if no side >= x
        if area1 == 18:
            # This means one side is 6, meaning x=4
            print("? 4 4")
            sys.stdout.flush()
            area2 = int(input())
            if area2 == 25:
                print("! 4")
                sys.stdout.flush()
            else:
                print("! 3")
                sys.stdout.flush()
        else:
            # If the area returned was the correct value, continue binary search for a better x
            print("? 99 100")
            sys.stdout.flush()
            area3 = int(input())
            if area3 == 9999:
                print("! 100")
                sys.stdout.flush()

# Call the function
find_missing_number()

**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

18

25

9999
```

### Explanation:
1. **Test Case 1:**
   - The first query is `? 3 5`.
   - The response is `18`, which indicates that one side of the rectangle is measured as 6 instead of 5. This suggests that the missing number \( x \) is 4.
   - The second query is `? 4 4`.
   - The response is `25`, which confirms that both sides are measured as 5 instead of 4. This confirms that \( x \) is 4.
   - The solution correctly outputs `! 4`.

2. **Test Case 2:**
   - The first query is `? 3 5`.
   - The response is `15`, which is the correct area for a `3 x 5` rectangle. This indicates that \( x \) is greater than 5.
   - The second query is `? 99 100`.
   - The response is `9999`, which indicates that one side is measured as 101 instead of 100. This suggests that \( x \) is 100.
   - The solution correctly outputs `! 100`.

### Why This Test Input is Challenging:
- **Edge Cases:**
  - The first test case checks if the solution can correctly identify \( x = 4 \) when the initial query `? 3 5` returns an area of 18.
  - The second test case checks if the solution can correctly identify \( x = 100 \) when the initial query `? 3 5` returns the correct area of 15, and the second query `? 99 100` returns an area of 9999.
- **Faulty Solutions:**
  - If the solution incorrectly handles the initial query and does not correctly identify \( x = 4 \), it will fail the first test case.
  - If the solution does not correctly handle the second query and does not correctly identify \( x = 100 \), it will fail the second test case.

This test input is designed to expose any bugs in the solution by covering both a small and a large value of \( x \) and ensuring that the solution can correctly handle the responses from the jury.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
/*

    ॐ भूर् भुवः स्वः।
    तत् सवितुर्वरेण्यं।
    भर्गो देवस्य धीमहि।
    धियो यो नः प्रचोदयात् ॥

  @Author  : ANKUSH
  @Country : INDIA

*/
import java.io.*;
import java.util.*;

public class Main {// Codechef {
    static class pair {
        int x;
        int y;

        public pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        /*
         * public boolean equals(Object o) {
         * if (this == o)
         * return true;
         * if (o == null || getClass() != o.getClass())
         * return false;
         * pair pair = (pair) o;
         * return val1 == pair.val1 && val2 == pair.val2;
         * }
         *
         * @Override
         * public int hashCode() {
         * return 31 * Long.hashCode(val1) + Long.hashCode(val2);
         * }
         *
         * @Override
         * public String toString() {
         * return "pair{" + "x=" + val1 + ", y=" + val2 + '}';
         * }
         *
         */
    }

    static long MOD = (long) (1e9 + 7);
    static long dp[];
    static long ans;

    public static void main(String[] args) {
        int TTT = 0;
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
            preFactorial();
            int testCases = in.nextInt();
            while (TTT++ != testCases) {
                // write code here
               int low = 2;
               int high = 999;
               while(low+2<high){
                int dis = (high-low+1)/3;
                int m1 = low+dis;
                int m2 = m1+dis;
                System.out.print("? "+m1+" "+m2);
                System.out.print("\n");
                int x = in.nextInt();
                if(x == (m1*m2)){
                    low = m2+1;
                }
                else if(x == ((m1+1)*(m2+1))){
                    high = m1;
                }
                else{
                    low = m1+1;
                    high = m2;
                }
               }
               if(high-low==1){
                    int mid = (low+high)/2;
                    System.out.print("? "+low+" "+mid);
                    System.out.print("\n");
                    int x = in.nextInt();
                    if(x == (low*mid)){
                        low=mid+1;
                    }
                    else{
                        high = mid;
                    }
               }
               else if(high-low==2){
                    int mid = (low+high)/2;
                    System.out.print("? "+low+" "+mid);
                    System.out.print("\n");
                    int x = in.nextInt();
                    if(x == (low*mid)){
                        low = high;
                    }
                    else if(x == ((low*1)+(mid*1))){
                        high = low;
                    }
                    else{
                        high = mid;
                    }
               }
               System.out.print("! "+high);
               System.out.print("\n");
            }
            out.close();
        } catch (Exception e) {
            System.out.println(e);
            return;
        }
    }
    /*-------------------------------------------  MATHS     MATHS     MATHS   ------------------------------------ */

    public static boolean isPrime(long number) {
        if (number <= 1) {
            return false;
        }
        if (number <= 3) {
            return true;
        }
        if (number % 2 == 0 || number % 3 == 0) {
            return false;
        }

        long sqrt = (long) Math.sqrt(number);
        for (long i = 5; i <= sqrt; i += 6) {
            if (number % i == 0 || number % (i + 2) == 0) {
                return false;
            }
        }
        return true;
    }

    public static long power(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1)
                ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    public static long power_mod(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                ans *= a;
                ans %= MOD;
            }
            a *= a;
            a %= MOD;
            b >>= 1;
        }
        return ans;
    }

    static final int SIZE = (int) (1e6);
    private static long[] fact = new long[SIZE];

    public static long add_mod(long a, long b) {
        return (a + b) % MOD;
    }

    public static long sub_mod(long a, long b) {
        return ((a - b) % MOD + MOD) % MOD;
    }

    public static long multiply_mod(long a, long b) {
        return (a * b) % MOD;
    }

    public static long inv(long a) {
        return power_mod(a, MOD - 2);
    }

    public static long divide_mod(long a, long b) {
        return multiply_mod(a, inv(b));
    }

    public static long nCr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], multiply_mod(fact[(int) r], fact[(int) (n - r)]));
    }

    public static long nPr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], fact[(int) (n - r)]);
    }

    public static void preFactorial() {
        fact[0] = 1;
        for (int i = 1; i < SIZE; i++) {
            fact[i] = multiply_mod(i, fact[i - 1]);
        }
    }

    public static void Sieve_Of_Eratosthenes() {
        boolean[] isPrime = new boolean[SIZE + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = false;
        isPrime[1] = false;

        for (int i = 2; i * i <= SIZE; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= SIZE; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    public static long gcd(long x, long y) {
        if (x > y)
            return gcd(y, x);
        if (y == 0) {
            return x;
        }
        if (x == 0) {
            return y;
        }
        return gcd(y % x, x);
    }

    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    /*----------------------------------------  ADVANCED DATA STRUCTURES  ADVANCED DATA STRUCTURES  ------------------------------------ */
    public static class DisjointSet {
        private long[] parent;
        private long[] size;

        public DisjointSet(int n) {
            parent = new long[n];
            size = new long[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        public long findParent(long node) {
            if (node == parent[(int) node]) {
                return node;
            }
            return parent[(int) node] = findParent(parent[(int) node]);
        }

        public void unionSize(long node1, long node2) {
            long p1 = findParent(node1);
            long p2 = findParent(node2);
            if (p1 == p2)
                return;
            if (size[(int) p1] > size[(int) p2]) {
                size[(int) p1] += size[(int) p2];
                parent[(int) p2] = p1;
            } else {
                size[(int) p2] += size[(int) p1];
                parent[(int) p1] = p2;
            }
        }

        public long findSize(long node) {
            long p1 = findParent(node);
            return size[(int) p1];
        }
    }

    static class Segment_Tree {
        long arr[];
        int n;
        long seg[];

        public Segment_Tree(long arr[], int n) {
            this.arr = arr;
            this.n = n;
            this.seg = new long[(4 * n) + 1];
            build(0, 0, n - 1);
        }

        public void build(int ind, int low, int high) {
            if (low == high) {
                seg[ind] = arr[low];
                return;
            }
            int mid = (low + high) / 2;
            build((2 * ind) + 1, low, mid);
            build((2 * ind) + 2, mid + 1, high);
            seg[ind] = (seg[2 * ind + 1] ^ seg[2 * ind + 2]);
        }

        public long query2(int ind, int low, int high, int left, int right) {
            if (low >= left && high <= right) {
                return seg[ind];
            }
            if (low > right || high < left) {
                return 0;
            }
            int mid = (low + high) / 2;
            long l = query2(2 * ind + 1, low, mid, left, right);
            long r = query2(2 * ind + 2, mid + 1, high, left, right);
            return (l ^ r);
        }

        public long query(int l, int r) {
            return query2(0, 0, n - 1, l, r);
        }

    }

    /*-------------------------------------  FAST READER AND WRITER      FAST READER AND WRITER  ------------------------------------ */
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {

        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

}

/*-------------------------------------XXXXXXXX      THE END         THE END        XXXXXXXX------------------------------------ */

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

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

Think step by step.

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

1. **Edge Cases**: Values at the boundaries of the allowed range for `x` (2 to 999).
2. **Critical Points**: Values that are likely to cause the algorithm to make incorrect assumptions, such as values where the ruler's incorrect measurement might be subtle.
3. **Query Distribution**: Values that might cause the algorithm to narrow down the range incorrectly, especially when the algorithm uses a ternary search approach.

Given the constraints and the nature of the problem, a good test input should include values that are likely to cause the algorithm to fail. Here's a small-scale test input that is complete and valid:

```plaintext
3

18

25

9999
```

### Explanation:
1. **First Test Case (x = 4)**:
   - The ruler measures a 3x5 rectangle as 3x6, giving an area of 18.
   - The ruler measures a 4x4 rectangle as 5x5, giving an area of 25.
   - This should correctly identify x = 4.

2. **Second Test Case (x = 100)**:
   - The ruler measures a 99x100 rectangle as 99x101, giving an area of 9999.
   - This should correctly identify x = 100.

3. **Third Test Case (x = 999)**:
   - This is a boundary case. The ruler will measure any rectangle with a side length of 999 or greater incorrectly.
   - This should correctly identify x = 999.

### Why This Input is Challenging:
- **Boundary Values**: The values 4, 100, and 999 are at different points in the range, which can help expose issues with the algorithm's boundary handling.
- **Critical Points**: The values 4 and 100 are critical because they are close to the lower and upper boundaries of the range, respectively.
- **Query Distribution**: The algorithm's ternary search approach can be tested with these values to see if it correctly narrows down the range.

This test input is designed to be small and valid while maximizing the probability of exposing the bug 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: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
/*

    ॐ भूर् भुवः स्वः।
    तत् सवितुर्वरेण्यं।
    भर्गो देवस्य धीमहि।
    धियो यो नः प्रचोदयात् ॥

  @Author  : ANKUSH
  @Country : INDIA

*/
import java.io.*;
import java.util.*;

public class Main {// Codechef {
    static class pair {
        int x;
        int y;

        public pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        /*
         * public boolean equals(Object o) {
         * if (this == o)
         * return true;
         * if (o == null || getClass() != o.getClass())
         * return false;
         * pair pair = (pair) o;
         * return val1 == pair.val1 && val2 == pair.val2;
         * }
         *
         * @Override
         * public int hashCode() {
         * return 31 * Long.hashCode(val1) + Long.hashCode(val2);
         * }
         *
         * @Override
         * public String toString() {
         * return "pair{" + "x=" + val1 + ", y=" + val2 + '}';
         * }
         *
         */
    }

    static long MOD = (long) (1e9 + 7);
    static long dp[];
    static long ans;

    public static void main(String[] args) {
        int TTT = 0;
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
            preFactorial();
            int testCases = in.nextInt();
            while (TTT++ != testCases) {
                // write code here
               int low = 2;
               int high = 999;
               while(low+2<high){
                    int m1 = low + (high-low)/3;
                    int m2 = high - (high-low)/3;
                    System.out.print("? "+m1+" "+m2);
                    System.out.print("\n");
                    int x = in.nextInt();
                    if(x == (m1*m2)){
                        low = m2+1;
                    }
                    else if(x == ((m1+1)*(m2+1))){
                        high = m1-1;
                    }
                    else{
                        low = m1;
                        high = m2;
                    }
               }
               if(high-low==1){
                    int mid = (low+high)/2;
                    System.out.print("? "+low+" "+mid);
                    System.out.print("\n");
                    int x = in.nextInt();
                    if(x == (low*mid)){
                        low=mid+1;
                    }
                    else{
                        high = mid;
                    }
               }
               else if(high-low==2){
                    int mid = (low+high)/2;
                    System.out.print("? "+low+" "+mid);
                    System.out.print("\n");
                    int x = in.nextInt();
                    if(x == (low*mid)){
                        low = high;
                    }
                    else if(x == ((low*1)+(mid*1))){
                        high = low;
                    }
                    else{
                        high = mid;
                    }
               }
               System.out.print("! "+high);
               System.out.print("\n");
            }
            out.close();
        } catch (Exception e) {
            System.out.println(e);
            return;
        }
    }
    /*-------------------------------------------  MATHS     MATHS     MATHS   ------------------------------------ */

    public static boolean isPrime(long number) {
        if (number <= 1) {
            return false;
        }
        if (number <= 3) {
            return true;
        }
        if (number % 2 == 0 || number % 3 == 0) {
            return false;
        }

        long sqrt = (long) Math.sqrt(number);
        for (long i = 5; i <= sqrt; i += 6) {
            if (number % i == 0 || number % (i + 2) == 0) {
                return false;
            }
        }
        return true;
    }

    public static long power(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1)
                ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    public static long power_mod(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                ans *= a;
                ans %= MOD;
            }
            a *= a;
            a %= MOD;
            b >>= 1;
        }
        return ans;
    }

    static final int SIZE = (int) (1e6);
    private static long[] fact = new long[SIZE];

    public static long add_mod(long a, long b) {
        return (a + b) % MOD;
    }

    public static long sub_mod(long a, long b) {
        return ((a - b) % MOD + MOD) % MOD;
    }

    public static long multiply_mod(long a, long b) {
        return (a * b) % MOD;
    }

    public static long inv(long a) {
        return power_mod(a, MOD - 2);
    }

    public static long divide_mod(long a, long b) {
        return multiply_mod(a, inv(b));
    }

    public static long nCr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], multiply_mod(fact[(int) r], fact[(int) (n - r)]));
    }

    public static long nPr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], fact[(int) (n - r)]);
    }

    public static void preFactorial() {
        fact[0] = 1;
        for (int i = 1; i < SIZE; i++) {
            fact[i] = multiply_mod(i, fact[i - 1]);
        }
    }

    public static void Sieve_Of_Eratosthenes() {
        boolean[] isPrime = new boolean[SIZE + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = false;
        isPrime[1] = false;

        for (int i = 2; i * i <= SIZE; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= SIZE; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    public static long gcd(long x, long y) {
        if (x > y)
            return gcd(y, x);
        if (y == 0) {
            return x;
        }
        if (x == 0) {
            return y;
        }
        return gcd(y % x, x);
    }

    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    /*----------------------------------------  ADVANCED DATA STRUCTURES  ADVANCED DATA STRUCTURES  ------------------------------------ */
    public static class DisjointSet {
        private long[] parent;
        private long[] size;

        public DisjointSet(int n) {
            parent = new long[n];
            size = new long[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        public long findParent(long node) {
            if (node == parent[(int) node]) {
                return node;
            }
            return parent[(int) node] = findParent(parent[(int) node]);
        }

        public void unionSize(long node1, long node2) {
            long p1 = findParent(node1);
            long p2 = findParent(node2);
            if (p1 == p2)
                return;
            if (size[(int) p1] > size[(int) p2]) {
                size[(int) p1] += size[(int) p2];
                parent[(int) p2] = p1;
            } else {
                size[(int) p2] += size[(int) p1];
                parent[(int) p1] = p2;
            }
        }

        public long findSize(long node) {
            long p1 = findParent(node);
            return size[(int) p1];
        }
    }

    static class Segment_Tree {
        long arr[];
        int n;
        long seg[];

        public Segment_Tree(long arr[], int n) {
            this.arr = arr;
            this.n = n;
            this.seg = new long[(4 * n) + 1];
            build(0, 0, n - 1);
        }

        public void build(int ind, int low, int high) {
            if (low == high) {
                seg[ind] = arr[low];
                return;
            }
            int mid = (low + high) / 2;
            build((2 * ind) + 1, low, mid);
            build((2 * ind) + 2, mid + 1, high);
            seg[ind] = (seg[2 * ind + 1] ^ seg[2 * ind + 2]);
        }

        public long query2(int ind, int low, int high, int left, int right) {
            if (low >= left && high <= right) {
                return seg[ind];
            }
            if (low > right || high < left) {
                return 0;
            }
            int mid = (low + high) / 2;
            long l = query2(2 * ind + 1, low, mid, left, right);
            long r = query2(2 * ind + 2, mid + 1, high, left, right);
            return (l ^ r);
        }

        public long query(int l, int r) {
            return query2(0, 0, n - 1, l, r);
        }

    }

    /*-------------------------------------  FAST READER AND WRITER      FAST READER AND WRITER  ------------------------------------ */
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {

        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

}

/*-------------------------------------XXXXXXXX      THE END         THE END        XXXXXXXX------------------------------------ */

**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 code uses a ternary search approach to find the missing number \( x \). The key points to consider are:

1. **Edge Cases:**
   - The smallest possible value of \( x \) (i.e., 2).
   - The largest possible value of \( x \) (i.e., 999).
   - Values of \( x \) that are close to the boundaries of the search range (e.g., 3, 998).

2. **Behavior of the Ruler:**
   - The ruler measures lengths correctly for \( y < x \) and incorrectly for \( y \geq x \).

3. **Queries:**
   - The code makes queries of the form \( ? \, a \, b \) and expects the product of the measured lengths.
   - The code uses the results to adjust the search range.

Let's consider the following test cases:

### Test Case 1: \( x = 2 \)
- The smallest possible value of \( x \).
- The ruler will measure any length \( y \geq 2 \) incorrectly as \( y + 1 \).

### Test Case 2: \( x = 999 \)
- The largest possible value of \( x \).
- The ruler will measure any length \( y \geq 999 \) incorrectly as \( y + 1 \).

### Test Case 3: \( x = 3 \)
- A value close to the lower boundary.
- The ruler will measure any length \( y \geq 3 \) incorrectly as \( y + 1 \).

### Test Case 4: \( x = 998 \)
- A value close to the upper boundary.
- The ruler will measure any length \( y \geq 998 \) incorrectly as \( y + 1 \).

### Test Case 5: \( x = 500 \)
- A value in the middle of the range.
- The ruler will measure any length \( y \geq 500 \) incorrectly as \( y + 1 \).

### Test Input:
```plaintext
5

18

9999

15

999900

250000
```

### Explanation:
1. **Test Case 1: \( x = 2 \)**
   - Query: \( ? \, 3 \, 5 \)
   - Expected response: \( 18 \) (since \( 3 \times 6 = 18 \))

2. **Test Case 2: \( x = 999 \)**
   - Query: \( ? \, 99 \, 100 \)
   - Expected response: \( 9999 \) (since \( 99 \times 101 = 9999 \))

3. **Test Case 3: \( x = 3 \)**
   - Query: \( ? \, 2 \, 5 \)
   - Expected response: \( 15 \) (since \( 2 \times 5 = 10 \) and \( 2 \times 6 = 12 \))

4. **Test Case 4: \( x = 998 \)**
   - Query: \( ? \, 997 \, 999 \)
   - Expected response: \( 999900 \) (since \( 997 \times 1000 = 999900 \))

5. **Test Case 5: \( x = 500 \)**
   - Query: \( ? \, 499 \, 500 \)
   - Expected response: \( 250000 \) (since \( 499 \times 501 = 250000 \))

This test input should expose any issues in the ternary search logic and the 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: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

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

using namespace std;
int T,n,s,l,r,lmid,rmid,res;

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

    cin>>T;
    while(T--)
    {
        l=2;r=999;
        while(l<=r)
        {
            lmid=l+(r-l)/3;
            rmid=r-(r-l)/3;
            cout<<"? "<<lmid<<' '<<rmid<< endl;
            cin>>s;
            s-=lmid*rmid;
            if(s==lmid+rmid+1)
            {
                res=lmid;
                r=lmid-1;
            }
            else if(s==lmid)
            {
                res=rmid;
                l=lmid+1;
                r=rmid-1;
            }
            else{
                l=rmid+1;
                r--;
            }
        }
        cout<<"! "<<res<<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:
```plaintext
3

18

25

10000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
// Author: JustKitkat
// Status: WIP

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;

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

#define int long long
#define el "\n"
#define arr array
#define ll long long
#define ld long double
#define ii pair<int, int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vll vector<long long>
#define vii vector<ii>
#define vllll vector<pll>
#define umap unordered_map
#define uset unordered_set
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(a) (a).begin(), (a).end()
#define FOR(a,b) for(auto i=a;i<b;++i)
#define DFOR(a,b) for(auto i=a;i>=b;--i)
#define JFOR(a,b) for(auto j=a;j<b;++j)
#define DJFOR(a,b) for(auto j=a;j>=b;--j)
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define show_vec(a) for(auto &x:a)cerr<<x<<' ';cerr<<endl;
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());

string to_upper(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A'; return a; }
string to_lower(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A'; return a; }
bool prime(ll a) { if (a==1) return 0; for (int i=2;i<=round(sqrt(a));++i) if (a%i==0) return 0; return 1; }
void yes() { cout<<"YES\n"; }
void no() { cout<<"NO\n"; }
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());

#define __time__ { auto duration = chrono::duration<double>( /* Show runtime */ \
chrono::high_resolution_clock::now() - BEG); cout<<"Time: "<<duration.count()<<endl;}
#define __output__ { FILE* file = freopen("../../../Testcases/test.out", "w", stdout); }
#define __input__ { FILE* file = freopen("../../../Testcases/test.in", "r", stdin); }

// PBDS -> initialise using `ordered_set s;` -> change to greater<int> for reversed order.
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>

const int MAX_N = 1e5 + 5;
const ll INF = 1e9;
const double PI = acos(-1);
const auto BEG = std::chrono::high_resolution_clock::now(); //Begining of the program

ll n=0, m=0, k=0, q=0;
void solve(int tc){
    int lo=2,hi=1000,m1,m2;
    while(lo<hi){
        m1=lo+(hi-lo)/3;
        m2=lo+((hi-lo)/3)*2;
        // show2(lo,hi);
        // show2(m1,m2);
        cout<<"? "<<m2<<' '<<m1<<endl;
        int r;
        cin>>r;
        if(r==-1){
            cout<<-1<<endl;
            exit(0);
        }
        if(r==m1*m2){
            lo=m2+1;
        }
        else if(r==m1*(m2-1)){
            hi=m2-1,lo=m1+1;
        }else{
            hi=m1-1;
        }
    }
    cout << "! " << lo<<endl;

}

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

    //__output__ // Redirect output to test.out
    //__input__ // Read test.in for input

    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t << ": ";
        solve(t);
    }

    //__time__ //Runtime
}

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

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

Think step by step.

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

1. **Edge Cases**: The code needs to handle the minimum and maximum values of \( x \) (2 and 999, respectively).
2. **Query Optimization**: The code uses a ternary search to find the value of \( x \). We need to ensure that the queries are not optimized in a way that they might miss certain values of \( x \).
3. **Query Constraints**: The code can make at most 7 queries. We need to ensure that the test input is complex enough to challenge the query logic.

Given these considerations, let's create a test input that includes both edge cases and a value in the middle range to ensure the code handles all scenarios correctly.

### Test Input
```plaintext
3

18

25

9999
```

### Explanation
1. **First Test Case (x = 4)**:
   - The first query is for a \( 5 \times 3 \) rectangle, which should return 18 (since \( 5 \times 4 \)).
   - The second query is for a \( 4 \times 4 \) rectangle, which should return 25 (since \( 5 \times 5 \)).
   - The code should correctly determine that \( x = 4 \).

2. **Second Test Case (x = 500)**:
   - The first query is for a \( 500 \times 500 \) rectangle, which should return a value that is not immediately obvious, forcing the code to make more queries.
   - The code should correctly determine that \( x = 500 \) after a few more queries.

3. **Third Test Case (x = 999)**:
   - The first query is for a \( 999 \times 999 \) rectangle, which should return a very large value, indicating that \( x \) is close to 1000.
   - The code should correctly determine that \( x = 999 \).

This test input is small and complete, covering edge cases and a middle value to ensure the code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ruler (hard version)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is that in this version, you can make at most $$$\mathbf{7}$$$ queries.

This is an interactive problem. If you are unsure how interactive problems work, then it is recommended to read the guide for participants.

We have a secret ruler that is missing one number $$$x$$$ ($$$2 \leq x \leq 999$$$). When you measure an object of length $$$y$$$, the ruler reports the following values:

- If $$$y < x$$$, the ruler (correctly) measures the object as having length $$$y$$$.
- If $$$y \geq x$$$, the ruler incorrectly measures the object as having length $$$y+1$$$.

The ruler above is missing the number $$$4$$$, so it correctly measures the first segment as length $$$3$$$ but incorrectly measures the second segment as length $$$6$$$ even though it is actually $$$5$$$.

You need to find the value of $$$x$$$. To do that, you can make queries of the following form:

- $$$\texttt{?}~a~b$$$ — in response, we will measure the side lengths of an $$$a \times b$$$ rectangle with our ruler and multiply the results, reporting the measured area of the rectangle back to you. For example, if $$$x=4$$$ and you query a $$$3 \times 5$$$ rectangle, we will measure its side lengths as $$$3 \times 6$$$ and report $$$18$$$ back to you.

Find the value of $$$x$$$. You can ask at most $$$\mathbf{7}$$$ queries.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

Note: In the first test, the interaction proceeds as follows.

SolutionJuryExplanation$$$\texttt{2}$$$There are 2 test cases.$$$\texttt{? 3 5}$$$$$$\texttt{18}$$$Secretly, the jury picked $$$x=4$$$. The solution requests the $$$3 \times 5$$$ rectangle, and the jury responds with $$$3 \times 6 = 18$$$, as described in the statement.$$$\texttt{? 4 4}$$$$$$\texttt{25}$$$The solution requests the $$$4 \times 4$$$ rectangle, which the jury measures as $$$5 \times 5$$$ and responds with $$$25$$$.$$$\texttt{! 4}$$$The solution has somehow determined that $$$x=4$$$, and outputs it. Since the output is correct, the jury continues to the next test case.$$$\texttt{? 99 100}$$$$$$\texttt{1}$$$Secretly, the jury picked $$$x=100$$$. The solution requests the $$$99 \times 100$$$ rectangle, which the jury measures as $$$99 \times 101$$$ and responds with $$$9999$$$.$$$\texttt{! 100}$$$The solution has somehow determined that $$$x=100$$$, and outputs it. Since the output is correct and there are no more test cases, the jury and the solution exit.

Note that the line breaks in the example input and output are for the sake of clarity, and do not occur in the real interaction.

Buggy Code:
X_MAX = 999
X_MIN = 2
X_REAL = 0
MAX_TRIES = 7
SIMU = False
LOG = False

def get_input():
    return input()

def send_output(msg:str):
    print(msg)
    #sys.stdout.flush()

def query(a,b):
    global SIMU, X_REAL
    send_output("? "+str(a)+" "+str(b))
    if SIMU:
        a = a if a<X_REAL else a+1
        b = b if b<X_REAL else b+1
        res = a*b
        print(res)
    else:
        res = int(get_input())
    return res

cases = int(get_input())

for c in range(cases):
    # Initialize x
    x0 = X_MIN
    x1 = X_MAX
    if SIMU:
        # x_real
        X_REAL = random.randint(X_MIN, X_MAX)
        print("#", X_REAL)
    cont = 0
    while cont <= MAX_TRIES:
        cont += 1
        if cont>MAX_TRIES:
            print("ISSUE!!!")
            sys.exit(None)
        x = int((x1-x0)/2+x0)
        if LOG: print(f"cont {cont}, x={x}, x0={x0}, x1={x1}")
        a = int(-(x-x0)/3+x)
        b = int((x1-x)/3+x)
        res = query(a, b)
        if res == -1:
            sys.exit(None)
        if res == a*b:
            x0 = b
        elif res == (a+1)*(b+1):
            x1 = a
        else:
            x0 = a
            x1 = b
        if (x1-x0)<=1:
            x = x1
            if LOG: print(f"x={x} (cont={cont})")
            print("!", x)
            break

**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

18

25

9999
```