**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upscaling
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. Output a $$$2n \times 2n$$$ checkerboard made of $$$2 \times 2$$$ squares alternating '$$$\texttt{#}$$$' and '$$$\texttt{.}$$$', with the top-left cell being '$$$\texttt{#}$$$'.

The picture above shows the answers for $$$n=1,2,3,4$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 20$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 20$$$) — it means you need to output a checkerboard of side length $$$2n$$$.

Output: For each test case, output $$$2n$$$ lines, each containing $$$2n$$$ characters without spaces — the checkerboard, as described in the statement. Do not output empty lines between test cases.

Examples:
input:
4
1
2
3
4
output:
##
##
##..
##..
..##
..##
##..##
##..##
..##..
..##..
##..##
##..##
##..##..
##..##..
..##..##
..##..##
##..##..
##..##..
..##..##
..##..##

Note:

Buggy Code:
import java.util.*;
public class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int l=s.nextInt();
for(int k=0;k<l;k++){
int t=s.nextInt();
for(int i=0;i<2*t;i++)
{
for(int j=0;j<2*t;j++)
{
    if(i%2==0)
    {
        if(i>2)
        {
        if(j==i||j==i+1||j==i-4||j==i-3)
        {
            System.out.print("#");
        }
        else{
            System.out.print(".");
        }
        }
        else{
            if(j==i||j==i+1||j==i+4||j==i+5){
               System.out.print("#");
            }
             else{
            System.out.print(".");
        }
        }
    }
    else{
        if(i>3)
        {
        if(j==i||j==i-1||j==i-4||j==i-5)
        {
            System.out.print("#");
        }
        else{
            System.out.print(".");
        }
        }
        else{
            if(j==i||j==i-1||j==i+4||j==i+3){
               System.out.print("#");
            }
             else{
            System.out.print(".");
        }
    }
    }

}
 System.out.println();

}
}
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code has issues with how it handles the placement of `#` and `.` in the checkerboard pattern, especially when `i` and `j` are compared in a complex manner.

Let's break down the code and identify potential issues:

1. **Complex Conditionals**: The code uses multiple nested conditionals to determine whether to print `#` or `.`. This can lead to incorrect patterns, especially for larger values of `n`.
2. **Edge Cases**: The code has special handling for `i > 2` and `i > 3`, which might not cover all edge cases correctly.
3. **Checkerboard Pattern**: The correct pattern should alternate `#` and `.` in a consistent manner, but the code's logic is overly complicated and likely incorrect.

Given these points, a small-scale test input that can expose the bug should include:
- A value of `n` that is just above the special handling conditions (e.g., `n = 3`).
- A value of `n` that is exactly at the boundary of the special handling conditions (e.g., `n = 2`).
- A value of `n` that is the minimum (e.g., `n = 1`).

Here is a small-scale test input that includes these edge cases:

```plaintext
3
1
2
3
```

This test input will:
- Test the minimum value of `n` to ensure the basic pattern is correct.
- Test a value just above the minimum to see if the special handling for `i > 2` and `i > 3` works correctly.
- Test a value that is larger but still small enough to be manageable and check if the pattern is consistent.

By running this test input, we can observe if the code fails to produce the correct checkerboard pattern for any of these values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upscaling
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. Output a $$$2n \times 2n$$$ checkerboard made of $$$2 \times 2$$$ squares alternating '$$$\texttt{#}$$$' and '$$$\texttt{.}$$$', with the top-left cell being '$$$\texttt{#}$$$'.

The picture above shows the answers for $$$n=1,2,3,4$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 20$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 20$$$) — it means you need to output a checkerboard of side length $$$2n$$$.

Output: For each test case, output $$$2n$$$ lines, each containing $$$2n$$$ characters without spaces — the checkerboard, as described in the statement. Do not output empty lines between test cases.

Examples:
input:
4
1
2
3
4
output:
##
##
##..
##..
..##
..##
##..##
##..##
..##..
..##..
##..##
##..##
##..##..
##..##..
..##..##
..##..##
##..##..
##..##..
..##..##
..##..##

Note:

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

public class B {
    public static void solve() throws IOException {
        int n = in.nextInt();
        n *= 2;
        boolean start = true;

        for (int i = 0; i < n; i += 2) {
            for (int j = 0; j < 2; j++) {
                if (start) {
                    for (int k = 0; k < n; k++) {
                        out.print('#');
                        out.print('#');
                        k += 2;

                        if (k < n){
                            out.print('.');
                            out.print('.');
                            k += 2;
                        }
                    }
                    if (j + 1 == 2) {
                        start = false;
                    }
                } else {
                    for (int k = 0; k < n; k++) {
                        out.print('.');
                        out.print('.');
                        k += 2;

                        if (k < n){
                            out.print('#');
                            out.print('#');
                            k += 2;
                        }
                    }
                    if (j + 1 == 2) {
                        start = true;
                    }
                }
                out.println();
            }
        }
    }

    public static void main(String[] args) throws IOException {
        int tests = in.nextInt();
        while (tests-- > 0) {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return bf.readLine();
        }

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on the logic of the checkerboard generation. The buggy code has a specific issue where it alternates between printing `#` and `.` in a 2x2 block pattern, but it does so incorrectly, particularly in how it increments the loop variable `k`.

The bug can be exposed by providing a test case where the checkerboard size is large enough to show the pattern issue but small enough to fit within the output limits. Specifically, we need to ensure that the pattern does not align correctly for larger values of `n`.

Let's consider the following test input:

```plaintext
1
5
```

This test case will generate a 10x10 checkerboard. The expected output for `n=5` should be:

```
##..##..##
##..##..##
..##..##..
..##..##..
##..##..##
##..##..##
..##..##..
..##..##..
##..##..##
##..##..##
```

However, the buggy code will likely produce an incorrect pattern due to the incorrect handling of the loop variable `k`.

Let's verify this:

1. The code initializes `n` to `2 * 5 = 10`.
2. It uses a boolean `start` to alternate between `#` and `.` blocks.
3. For each row, it prints `#` and `.` blocks, but the increment of `k` by 2 inside the loop can cause misalignment.

By using this test input, we can observe if the checkerboard pattern is correct or if there are any misalignments, which would indicate the presence of the bug.

Here is the test input:

```plaintext
1
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
#include <iostream>
#include <string>
using namespace std;

#define ll long long

// Function to print "IMPOSSIBLE" and exit
void imp() {
    cout << "! IMPOSSIBLE" << endl;
    cout.flush();
    exit(0);
}

// Function to send a query and get a response
int qry(int l, int r) {
    cout << "? " << l << " " << r << endl;
    cout.flush();
    int res;
    cin >> res;
    return res;
}

// Function to append a specific character multiple times
void add(string &s, char c, int cnt) {
    while (cnt--) {
        s.push_back(c);
    }
}

// Core function to reconstruct the binary sequence
string recon(int n) {
    string res;
    int p = 0, z = 0;

    for (int i = 1; i < n; i++) {
        int x = qry(1, i + 1);

        if (p == 0) {
            // First significant query
            if (x > 0) {
                if (x > i) imp();
                add(res, '1', i - x);
                add(res, '0', x);
            }
            res.push_back('1');
            z = x;
        } else {
            // Further queries
            if (x > p) {
                if (x - p == z) {
                    res.push_back('1');
                } else {
                    imp();
                }
            } else if (x == p) {
                res.push_back('0');
                z++;
            } else {
                imp();
            }
        }
        p = x;
    }

    // Ensure the result has the correct length
    if (res.size() != n) imp();

    return res;
}

// Main driver function
int main() {
    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        string res = recon(n);
        cout << "! " << res << 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:
```plaintext
2
3
2
0
1
3
3
0
1
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstring>
#include<list>
#include<map>
#include<functional>
#include<set>
#include<numeric>
#include<unordered_map>
#include<unordered_set>
#include<iterator>
#include<deque>
#include<string>
#include<queue>
#include<iomanip>
#include<stack>
#include<fstream>
#include<sstream>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>

using namespace std;

#define ll long long int
#define ld long double

ll count(string s)
{
    // ll n=s.size();
    // ll one[n];
    // // if(s[n-1]=='1') one[n-1]=1;
    // // else one[n-1]=0;
    // one[n-1]=0;
    // for (int i = n-2; i >=0; i--)
    // {
    //     if(s[i+1]=='1')
    //     {
    //         one[i]=one[i+1]+1;
    //     }
    //     else one[i]=one[i+1];
    // }

    // ll cnt=0;
    // ll zero=0;
    // for (int i = 0; i < n; i++)
    // {
    //     if(s[i]=='0') zero++;
    //     cnt+=zero*one[i];
    // }
    // return cnt;
    ll n = s.size();
    ll oneCount = 0;
    ll subseqCount = 0;

    for (ll i = n - 1; i >= 0; --i)
    {
        if (s[i] == '1')
        {
            oneCount++;
        }
        else if (s[i] == '0')
        {
            subseqCount += oneCount;
        }
    }

    return subseqCount;
}

int main(){

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

    // startTime

    /*PROBLEM NO.1*/
    // ll t;
    // cin>>t;

    // while (t--)
    // {
    //     ll n;
    //     cin>>n;

    //     map<ll,ll> mp;
    //     for (int i = 0; i < n; i++)
    //     {
    //         ll x;
    //         cin>>x;

    //         mp[x]++;
    //     }

    //     ll ans=0;
    //     for (auto it:mp)
    //     {
    //         ans+=it.second/2;
    //     }
    //     cout<<ans<<endl;
    // }

    /*PROBLE NO.2*/
    // ll t;
    // cin>>t;

    // while (t--)
    // {
    //     ll n;
    //     cin>>n;

    //     map<ll,ll> mp;
    //     ll a[n];
    //     for (int i = 0; i < n; i++)
    //     {
    //         cin>>a[i];
    //         mp[a[i]]++;
    //     }
    //     ll sz=n-2;
    //     ll x=0,y=0;
    //     for(auto it:mp)
    //     {
    //         auto it2=mp.find(sz/it.first);
    //         if(it2!=mp.end())
    //         {
    //             if((it2->first)*(it.first)==sz)
    //             {
    //                 x=it.first;
    //                 y=it2->first;
    //             }
    //         }
    //     }
    //     cout<<x<<" "<<y<<endl;
    // }

    /*PROBLEM NO.3*/
    // ll t;
    // cin>>t;

    // while (t--)
    // {
    //     ll n;
    //     cin>>n;

    //     if(n<5)
    //     {
    //         cout<<-1<<endl;
    //     }
    //     else{
    //         vector<ll> v;
    //         for (int i = 1; i <=n; i+=2)
    //         {
    //             if(i==1 || i==5) continue;
    //             else v.push_back(i);
    //         }
    //         v.push_back(1);
    //         v.push_back(5);

    //         v.push_back(4);
    //         for (int i = 2; i <= n; i+=2)
    //         {
    //             if(i==4) continue;
    //             v.push_back(i);
    //         }
    //         for(auto it:v) cout<<it<<" ";
    //         cout<<endl;
    //     }
    // }

    /*PROBLEM NO.4*/
    // ll t;
    // cin>>t;

    // while (t--)
    // {
    //     ll n,m,l;
    //     cin>>n>>m>>l;

    //     vector<pair<ll,ll> > v;
    //     for (int i = 0; i < n; i++)
    //     {
    //         ll a,b;
    //         cin>>a>>b;

    //         v.push_back(make_pair(a,b));
    //     }

    //     vector<pair<ll,ll> > v2;
    //     for (int i = 0; i < m; i++)
    //     {
    //         ll a,b;
    //         cin>>a>>b;

    //         v2.push_back(make_pair(a,b));
    //     }

    //     vector<ll> len;
    //     for (int i = 0; i < n; i++)
    //     {
    //         len.push_back(v[i].second-v[i].first+1);
    //     }

    //     multiset<ll> ms;
    //     ll j=0;
    //     ll pow=0;
    //     ll mx=0;
    //     ll cnt=0;
    //     bool flag=true;
    //     for (int i = 0; i < n; i++)
    //     {
    //         while (j<m && v2[j].first<v[i].first)
    //         {
    //             ms.insert(v2[j].second);
    //             j++;
    //         }

    //         mx=max(mx,len[i]);
    //         while (pow<mx && ms.size())
    //         {
    //             pow+=*ms.rbegin();
    //             ms.erase(prev(ms.end()));
    //             cnt++;
    //         }
    //         if(pow<mx)
    //         {
    //             flag=false;
    //             break;
    //         }
    //     }
    //     if(flag) cout<<cnt<<endl;
    //     else cout<<-1<<endl;
    // }

    /*PROBLEM NO.5*/
    ll t;
    cin>>t;

    while (t--)
    {
        ll n;
        cin>>n;

        string s="";
        ll prev=0,tot=0;
        bool flag=true;
        for (int i = n; i >=2 && flag; i--)
        {
            cout<<"? "<<1<<" "<<i<<endl;
            cout.flush();
            ll x;
            cin>>x;
            if(i==n)
            {
                prev=x;
                if(x==0) flag=false;
                tot=x;

                continue;
            }
            if(x<prev)
            {
                s+='1';
            }
            else s+='0';
            prev=x;
        }
        if(!flag)
        {
            cout<<"! "<<"IMPOSSIBLE"<<endl;
            continue;
        }
        if(prev==1)
        {
            s+='1';
            s+='0';
            reverse(s.begin(),s.end());
            cout<<"! "<<s<<endl;
            continue;
        }
        string s1=s,s2=s,s3=s,s4=s;
        s1+='0';
        s1+='0';
        s2+='1';
        s2+='1';
        s3+='0';
        s3+='1';
        // s4+='1';
        // s4+='0';
        reverse(s2.begin(),s2.end());
        reverse(s3.begin(),s3.end());
        reverse(s1.begin(),s1.end());
        ll cnt1=count(s1);
        ll cnt2=count(s2);
        ll cnt3=count(s3);
        // cout<<cnt1<<" "<<cnt2<<" "<<cnt3<<endl;
        // cout<<s1<<" "<<s2<<" "<<s3<<endl;
        if(cnt1==tot)
        {
            cout<<"! "<<s1<<endl;
            // cout<<cnt1<<endl;
        }
        else if(cnt2==tot)
        {
            cout<<"! "<<s2<<endl;
            // cout<<cnt2<<endl;
        }
        else if(cnt3==tot){
            cout<<"! "<<s3<<endl;
            // cout<<cnt3<<endl;
        }
        else cout<<"! "<<"IMPOSSIBLE"<<endl;
    }

    // endTime

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios where the algorithm might fail. The code has a few potential issues, particularly in how it handles the determination of the binary string and the conditions under which it declares "IMPOSSIBLE".

### Steps to Generate the Test Input:

1. **Edge Case 1: All Zeros or All Ones**
   - If the string is all zeros or all ones, the code should correctly determine this.
   - However, the code might not handle this correctly if it incorrectly assumes the presence of `01` subsequences.

2. **Edge Case 2: Strings with Few `01` Subsequences**
   - If the string has very few `01` subsequences, the code might incorrectly assume the presence of more `01` subsequences.

3. **Edge Case 3: Strings with Multiple Valid Solutions**
   - If there are multiple valid strings that match the responses to the queries, the code should correctly determine "IMPOSSIBLE".

4. **Edge Case 4: Strings with Specific Patterns**
   - Strings with specific patterns like alternating `01` or `10` can be challenging for the code to handle.

### Chosen Test Input:
Let's choose a test input that covers the edge cases mentioned above. We will use a string that has a few `01` subsequences and another string that has multiple valid solutions.

```plaintext
3
4
2
2
0
5
4
3
1
2
2
0
```

### Explanation:
1. **Test Case 1:**
   - `n = 4`
   - Queries:
     - `f(1, 4) = 2`
     - `f(2, 4) = 2`
     - `f(3, 4) = 0`
   - Possible strings: `0110`, `1110`, etc.
   - The code should correctly determine "IMPOSSIBLE" because there are multiple valid strings.

2. **Test Case 2:**
   - `n = 5`
   - Queries:
     - `f(1, 5) = 4`
     - `f(2, 5) = 3`
     - `f(3, 5) = 1`
     - `f(4, 5) = 2`
   - Possible string: `01010`
   - The code should correctly determine the string.

3. **Test Case 3:**
   - `n = 2`
   - Queries:
     - `f(1, 2) = 0`
   - Possible strings: `00`, `11`
   - The code should correctly determine "IMPOSSIBLE" because there are multiple valid strings.

This test input should help expose the bugs in the provided code by challenging it with different edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
#include <iostream>
#include <array>
#include <cstring>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <utility>
#include <cmath>
#include <string>
#include <ranges>
#include <map>

#define ll long long
#define pb push_back

using namespace std;

ll INF=9223372036854775807;
ll remo=(ll)1e9+7;

// int b_s(int x,vector<int>& v){
//     int n=v.size();
//     int low=0,high=n-1;
//     while(high>low){
//         int mid=low+(high-low)/2;
//         (v[mid]>x?high=mid:low=mid+1);
//     }
//     if(low==n-1)++low;
//     return low;
// }

// int inv(int a) {
//   return a <= 1 ? a : remo - (long long)(remo/a) * inv(remo % a) % remo;
// }

bool isPrime(int n) {
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    }
    return true;
}

void solve(){
    ll n;cin>>n;
    ll mx=n;
    string s;

    auto fnc=[&](int l,int r){
        cout<<"? "<<l<<" "<<r<<endl;
        int temp;cin>>temp;
        return temp;
    };

    int x = fnc(1,n);
    if(x==0) cout<<"! IMPOSSIBLE"<<endl;
    else{
        int ptr=2;
        while(ptr<n){
            int temp=fnc(ptr,n);
            if(temp==0){
                s += string(x, '1');
                s.resize(n, '0');
                break;
            }
            if(temp==x) s+='1';
            else x=temp,s+='0';
            ++ptr;

        }
        if(s.size()!=n) s+="01";
        cout<<"! "<<s<<endl;
    }
}

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

/*   /\_/\
*   (= ._.)
*   / >  \>
*/

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

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

Think step by step.

Response:
```plaintext
2
3
2
0
```

**Explanation:**

1. **Test Case 1:**
   - **Input:** `3`
   - **Queries:**
     - `? 1 3` (Expected response: `2`)
     - `? 2 3` (Expected response: `1`)
   - **Output:** `! 101`
   - **Explanation:** The string `101` has exactly 2 subsequences of `01` from index 1 to 3, and 1 subsequence of `01` from index 2 to 3. This test case checks the ability to correctly identify the positions of `0` and `1` in a small string.

2. **Test Case 2:**
   - **Input:** `2`
   - **Queries:**
     - `? 1 2` (Expected response: `0`)
   - **Output:** `! IMPOSSIBLE`
   - **Explanation:** The string can either be `00` or `11`, both of which have 0 subsequences of `01`. This test case checks the ability to correctly identify when it is impossible to determine the string with the given queries.

These test cases are designed to expose the bug in the buggy code by challenging the logic used to determine the binary string and the conditions under which it should report `IMPOSSIBLE`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
from sys import stdout

def query(l: int, r: int) -> int:
    """Make a query for f(l,r) and get response"""
    print(f"? {l} {r}")
    stdout.flush()
    return int(input())

def solve_test_case():
    n = int(input())

    if n == 2:
        count = query(1, 2)
        if count == 1:
            print("! 01")
        else:
            print("! IMPOSSIBLE")
        stdout.flush()
        return

    s = ['?'] * n
    queries_made = 0

    # First get total count
    total = query(1, n)
    queries_made += 1

    if total == 0:
        # Must be all same digit or 1s followed by 0s
        # Try to find first zero by checking sequential pairs
        found_transition = False
        for i in range(1, n):
            if queries_made >= n:
                break
            count = query(i, i+1)  # This ensures l < r
            queries_made += 1
            if count == 0:
                # Could be 00, 11, or 10
                # Check next pair if available
                if i + 2 <= n and queries_made < n:
                    next_count = query(i, i+2)
                    queries_made += 1
                    if next_count == 0:
                        # First i positions must be same digit
                        # Rest must be same digit
                        for j in range(i):
                            s[j] = '1'
                        for j in range(i, n):
                            s[j] = '0'
                        found_transition = True
                        break
        if not found_transition:
            print("! IMPOSSIBLE")
            stdout.flush()
            return
    else:
        # Find a "01" pair
        found_01 = False
        for i in range(1, n):
            if queries_made >= n:
                break
            count = query(i, i+1)  # This ensures l < r
            queries_made += 1
            if count == 1:
                s[i-1] = '0'
                s[i] = '1'
                found_01 = True

                # Fill remaining positions
                # For positions before the 01 pair
                pivot_left = i-1
                for j in range(i-2, -1, -1):
                    if queries_made >= n:
                        break
                    # Must ensure l < r in query
                    count = query(j+1, i+1)
                    queries_made += 1
                    s[j] = '0' if count > 1 else '1'

                # For positions after the 01 pair
                pivot_right = i
                for j in range(i+1, n):
                    if queries_made >= n:
                        break
                    count = query(pivot_right, j+1)  # This ensures l < r
                    queries_made += 1
                    s[j] = '1' if count > 1 else '0'
                break

        if not found_01:
            print("! IMPOSSIBLE")
            stdout.flush()
            return

    # Check if we determined all positions
    if '?' in s:
        print("! IMPOSSIBLE")
    else:
        print(f"! {''.join(s)}")
    stdout.flush()

# Read number of test cases
t = int(input())
for _ in range(t):
    solve_test_case()

**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
2

3

4
```

### Explanation:
1. **Test Case 1:**
   - **n = 2**
   - This is a straightforward case where the only possible strings are "00", "01", "10", and "11". The only way to uniquely determine the string is if the query `f(1, 2)` returns 1, which means the string is "01". If it returns 0, the string could be "00" or "11", making it impossible to determine.

2. **Test Case 2:**
   - **n = 3**
   - This case is more complex. The possible strings are "000", "001", "010", "011", "100", "101", "110", and "111". The key is to check if the queries can uniquely determine the string. For example, if `f(1, 3)` returns 0, it could be "000" or "111". If it returns 1, it could be "001", "010", "100", or "101". If it returns 2, it could be "011" or "110". This test case checks the ability to handle a small increase in complexity.

3. **Test Case 3:**
   - **n = 4**
   - This case is even more challenging. The possible strings are "0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", and "1111". The key is to check if the queries can uniquely determine the string. For example, if `f(1, 4)` returns 0, it could be "0000" or "1111". If it returns 1, it could be "0001", "0010", "0100", "1000", "1001", "1010", "1100", or "1101". This test case checks the ability to handle a slightly larger input size and the complexity of determining the string.

These test cases are designed to expose potential bugs in the code, especially in handling edge cases and ensuring that the string can be uniquely determined with the given constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
import sys

def reconstruct(s1, f_dict, n):
    c0_prev = (s1 == '0')
    c1_prev = (s1 == '1')
    f_prev =0
    s = [s1]
    for k in range(2, n +1):
        f_k = f_dict.get((1, k), -1)
        if f_k == -1:
            return None
        if f_k - f_prev == c0_prev:
            s.append('1')
            c1_prev +=1
            f_prev += c0_prev
        elif f_k - f_prev ==0:
            s.append('0')
            c0_prev +=1
        else:
            return None
    # Now, compute f_extra
    f_extra = f_dict.get((2, n), -1)
    if f_extra == -1:
        return None
    c0 =0
    f_extra_actual=0
    for m in range(2, n +1):
        if s[m -1] == '1':
            f_extra_actual += c0
        if s[m -1] == '0':
            c0 +=1
    if f_extra_actual == f_extra:
        return ''.join(s)
    else:
        return None

def main():
    import sys

    input = sys.stdin.readline

    t = int(sys.stdin.readline())
    for _ in range(t):
        n = int(sys.stdin.readline())

        if n ==2:
            print("? 1 2")
            sys.stdout.flush()
            f12 = int(sys.stdin.readline())
            if f12 ==1:
                print("! 01")
                sys.stdout.flush()
            else:
                print("! IMPOSSIBLE")
                sys.stdout.flush()
            continue

        f_dict = {}
        for k in range(2, n+1):
            print(f"? 1 {k}")
            sys.stdout.flush()
            f1k = int(sys.stdin.readline())
            f_dict[(1, k)] = f1k
        print(f"? 2 {n}")
        sys.stdout.flush()
        f2n = int(sys.stdin.readline())
        f_dict[(2, n)] = f2n

        possible_s = []
        for s1 in ['0', '1']:
            reconstructed_s = reconstruct(s1, f_dict, n)
            if reconstructed_s is not None:
                possible_s.append(reconstructed_s)
        if len(possible_s)==1:
            print("! " + possible_s[0])
            sys.stdout.flush()
        else:
            print("! IMPOSSIBLE")
            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:
```plaintext
2
5
4
0
1
2
2
0
```

### Explanation:
This test input is designed to expose potential issues in the buggy code by providing a scenario where the algorithm must correctly determine the binary string or conclude that it is impossible to determine.

1. **Test Case 1:**
   - **Input:**
     - `5` (length of the binary string)
     - `4` (f(1, 5))
     - `0` (f(2, 4))
     - `1` (f(4, 5))
     - `2` (f(3, 5))
   - **Expected Output:**
     - `! 01001`
   - **Reasoning:**
     - The first query `? 1 5` returns `4`, indicating there are 4 subsequences of `01` in the entire string.
     - The second query `? 2 4` returns `0`, indicating no `01` subsequences in the substring `s2 s3 s4`.
     - The third query `? 4 5` returns `1`, indicating one `01` subsequence in the substring `s4 s5`.
     - The fourth query `? 3 5` returns `2`, indicating two `01` subsequences in the substring `s3 s4 s5`.
     - Based on these queries, the only possible string is `01001`.

2. **Test Case 2:**
   - **Input:**
     - `4` (length of the binary string)
     - `0` (f(1, 2))
   - **Expected Output:**
     - `! IMPOSSIBLE`
   - **Reasoning:**
     - The first query `? 1 2` returns `0`, indicating no `01` subsequences in the substring `s1 s2`.
     - Since the string length is 4 and only one query is allowed, it is impossible to uniquely determine the string. Both `0000` and `1111` would satisfy the condition of having no `01` subsequences in the first two characters.

This test input is small and valid, ensuring it fits within the constraints and is designed to challenge the buggy code effectively.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
import java.util.*;

public class main {
 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int t = sc.nextInt();
  while (t-- > 0) {
   int n = sc.nextInt();
   // System.out.println("? 1 " + n);
   // long num = sc.nextLong();
   // if (num == 0) {
   // System.out.println("! IMPOSSIBLE");
   // continue;
   // }
   long curr = 0;
   int ch = 0;
   int ans[] = new int[n + 1];
   Arrays.fill(ans, -1);
   for (int i = 2; i <= n; i++) {
    System.out.println("? 1 " + i);
    long num = sc.nextLong();
    long dif = num - curr;
    if (dif < 0) {
     ch = -1;
     break;
    }
    if (dif != 0) {
     ans[i] = 1;
     int in = i - 1;
     while (dif > 0 && ans[in] == -1) {
      ans[in] = 0;
      dif--;
      in--;
     }
    }
    curr = num;
   }
   if (ch == -1) {
    System.out.println("! IMPOSSIBLE");
    continue;
   }
   StringBuilder sb = new StringBuilder("");
   sb.append("! ");
   for (int i = 1; i <= n; i++) {
    if (ans[i] == -1) {
     ch = -1;
     break;
    }
    sb.append(ans[i]);

   }
   if (ch == 0)
    System.out.println(sb);
   else
    System.out.println("! IMPOSSIBLE");

  }
 }
}

**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
3
0
0
0
4
1
1
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
    {
	static FastScanner fs = null;
	public static void main(String[] args) {
		fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
        while(t-->0){
            int n = fs.nextInt();
            int ans = 0;
            Stack<Pair> stack = new Stack<>();
            for(int i = 2;i<=n;i++){
                System.out.println("? 1 "+i);
                int num = fs.nextInt();
                if(i==2){
                    if(num==1){
                        stack.push(new Pair("01",1));
                    }
                    else{
                        stack.push(new Pair("00",2));
                        stack.push(new Pair("10",1));
                        stack.push(new Pair("11",0));
                    }
                }
                else{
                    int df = num-ans;
                    Stack<Pair> s2 = new Stack<>();
                    while(!stack.isEmpty()){
                        Pair p = stack.pop();
                        String s = p.s;
                        int co = p.co;
                        //System.out.println(s+" -- "+co+" -- "+df+" -- "+co);
                        if((df==0)){
                            if(co==0){
                                s2.push(new Pair((s+'0'),co+1));
                                s2.push(new Pair((s+'1'),co));
                            }
                            else{
                                s2.push(new Pair((s+'0'),co+1));
                            }
                        }
                        else{
                            if(co==df){
                                s2.push(new Pair((s+'1'),co+1));
                            }
                        }
                    }
                    // while(!s2.isEmpty()){
                    //     Pair ss = s2.pop();
                    //     System.out.println(ss.s+" --- "+num+" "+df);
                    //     stack.push(ss);
                    // }
                    stack = s2;
                }
                ans=num;
            }
            //System.out.println(stack.size());
            if(stack.size()==1){
                System.out.println("! "+stack.pop().s);
            }
            else{
                System.out.println("! IMPOSSIBLE");
            }
        }
		out.close();
	}
	static class Pair{
	    String s;
	    int co;
	    Pair(String s,int co){
	        this.s = s;
	        this.co = co;
	    }
	}
	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
2
3
2
0
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

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

// author: Krishna Sharma

int main() {
    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        cout << "? " << n - 1 << " " << n << endl;
        int x; cin >> x;
        if (x == 0 && n == 2) {
            cout << "! IMPOSSIBLE" << endl;
            continue;
        }
        string ans;
        int i = n - 2;
        int one = 0;
        if (x == 0) {
            int hua = 0;
            for (; i >= 1; i--) {
                cout << "? " << i << " " << n << endl;
                int y; cin >> y;
                if (y == 0) continue;
                if (y == (n - i)) {
                    hua = 1;
                    one = (n - i);
                    break;
                }
                else break;
            }
            if (!hua) {
                cout << "! IMPOSSIBLE" << endl;
                continue;

            }
            for (int i = 0; i < one; i++) ans += '1';
            ans += '0';

        }
        else {
            ans += '1'; ans += '0';
            one++;
        }
        int last = 1;
        for (; i >= 1; i--) {
            cout << "? " << i << " " << n << endl;
            int x; cin >> x;
            // agr zero put kiya toh last+=one ho jayega
            if (x == (last + one)) {
                last += one;
                ans += '0';
            }
            else {
                one++;
                ans += '1';
            }
        }
        reverse(ans.begin(), ans.end());
        cout << "! ";
        cout << ans << endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3
0
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

#define pb push_back
#define all(x) (x).begin(), (x).end()
#define debug(x) { cerr << #x << " = " << x << endl; }
#define IO { ios_base::sync_with_stdio(false); cin.tie(0); }
#define endl '\n'
#define int long long
using namespace std;

typedef pair<int, int> ii;
typedef vector<int> vi;

int query(int l, int r) {
    cout << "? " << l << " " << r << endl;
    cout.flush();
    int response;
    cin >> response;
    return response;
}

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

    int prev_rep = query(1, n);
    int last_zero = -1;
    string ans;

    if(prev_rep == 0) {
        cout <<  "! IMPOSSIBLE" << endl;
        return;
    }

    for (int i = 2; i <n && ans.size() < n; i++) {
        int cur_rep = query(i, n);

        if(i==n-1 && cur_rep > 0) ans.pb('0');

        if (prev_rep != cur_rep) {
            ans.pb('0');
            last_zero = i-1;
            prev_rep = cur_rep;
        } else if (cur_rep > 0) {
            ans.pb('1');
            prev_rep = cur_rep;
        } else {
            if (last_zero == -1) {
                cout << "! IMPOSSIBLE" << endl;
                return;
            }

            int cnt = query(last_zero, n);
            ans += string(cnt, '1');
            while (ans.size() < n) ans.pb('0');
        }
    }
    while(ans.size() < n) ans.pb('1');
    cout << "! " << ans << endl;
}

signed main() {
    // IO;
    int tc = 1;
    cin >> tc;
    while (tc--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
5
4
0
1
2
2
0
```

This test input is designed to expose the bug in the provided buggy code. Here’s a breakdown of why this input is challenging:

1. **Test Case 1 (n = 5):**
   - The first query asks for `f(1, 5)`, and the response is `4`.
   - The second query asks for `f(2, 4)`, and the response is `0`.
   - The third query asks for `f(4, 5)`, and the response is `1`.
   - The fourth query asks for `f(3, 5)`, and the response is `2`.

   The correct string for this case is `01001`. The buggy code should be able to determine this string correctly if it handles the queries and responses accurately.

2. **Test Case 2 (n = 2):**
   - The first query asks for `f(1, 2)`, and the response is `0`.

   In this case, the string `00` and `11` both have an answer of `0` for `f(1, 2)`, making it impossible to determine the string uniquely. The buggy code should output `IMPOSSIBLE`.

This test input ensures that the buggy code is tested for both a case where the string can be determined and a case where it cannot, thus exposing potential issues in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
/*
「責任を負う人の話は、あなたと話したことがありますよね」
「当時の私はまだわからないけど……今は理解できるようになった」
「大人としての責務。そして、その延長線上にある、あなたの選択」
「それに代表されるアイデアもあります」
「だから、先生」
「信じられる大人のあなたなら、この歪んだ変形の終点とは、違う結果……そこにつながる選択肢……きっとあなたは見つけることができます」
DON'T GET STUCK ON ONE APPROACH!!!
*/
#include <bits/stdc++.h>//喵内～
#define re register//喵内～
#define rep(i,a,b) for (re int i = (a);i <= (b); ++i)
#define debug(x) cout << #x << '=',print(x),putchar(' ')
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
#define pi pair<int,int>
#define mp(a,b) make_pair(a,b)
typedef long long ll;
using namespace std;//喵内～
inline ll read(){
    ll s = 0,f = 1;char c = getchar();
    while (!isdigit(c)){if (c == '-')f = -1;c = getchar();}
    while (isdigit(c)){s = (s<<3) + (s<<1) + (c ^ 48);c = getchar();}
    return s * f;
}//喵内～
void print(__int128 x){if (x < 0) {putchar('-'),print(-x);return ;}if (x >= 10) print(x / 10);putchar(x % 10 + 48);}//喵内～
//const int Mod = 1e9 + 7;//喵内～要填数字哟～
const int Mod = 998244353;//喵内～要填数字哟～
const ll INF = 0x3f3f3f3f;
const int N = 1e4 + 5;//喵内～要填数字哟～
ll qpow(ll x,ll y){
   ll res = 1;
    for (;y;y >>= 1,x = x * x % Mod) if (y & 1) res = res * x % Mod;
    return res;
}
int t,n,a[N],pre,cnt0 = 0;
void solve(){
    n = read(); cnt0 = 0;
    cout << "? 1 " << n << endl;
    pre = read();
    if (pre == 0){cout << "! IMPOSSIBLE" << endl; return ;}
    for (int i = n - 1;i >= 1;--i){
        int val;
        if (i > 1){
            cout << "? 1 " << i << endl;
            val = read();
        } else val = 0;
        if (val == pre){cnt0--; a[i + 1] = 0;}
        else {a[i + 1] = 1;cnt0 = pre - val;}
        pre = val;
    }
    if (cnt0 < 0 || cnt0 > 1){cout << "! IMPOSSIBLE" << endl; return ;}
    if (cnt0 == 0) a[1] = 1;
    else a[1] = 0;
    cout << "! ";
    for (int i = 1;i <= n;++i) cout << a[i];
    cout << endl;
}
signed main(){
    t = read();
    while (t--) solve();
    return 0;
}//喵内～
/*
What's wrong with my code?
1. 小数据？特殊数据？如 n = 1?
2. 最小值，最大值取多少？是否会溢出？
3. 初始值有没有赋值？有没有建树？
4. 数组大小？是否越界？
5. 思考暴力的时候，考虑是否可能是多个连续段？或者是个数不确定无法暴力？
6. 进行详细的分类讨论?
7. 选择的区间是否可以为空？

Trick:
1. 连通 通常带有特殊性质
2. LIS 有另一种求法
3. Brovuka?

About implementation skills:
1. 全局变量多用长变量名，而局部变量，临时变量，和函数传递的参数使用短变量名。
2. 大模拟尽量遵循：怎么方便怎么写。
3. 对于一些数据很小的需要维护的量并且需要大量讨论时，可以考虑把数组拆掉换成变量。
4. 写成多个函数。
*/

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

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

Think step by step.

Response:
```plaintext
2
4
2
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
def f(s):
    n = len(s)
    pre = [0]
    for c in s:
        if c == '0':
            pre.append(pre[-1] + 1)
        else:
            pre.append(pre[-1])
    suf = [0] * (n + 1)
    for i in range(n-1, -1, -1):
        if s[i] == '1':
            suf[i] = suf[i+1] + 1
        else:
            suf[i] = suf[i+1]
    mx = 0
    for i in range(n+1):
        mx = max(mx, pre[i] * suf[i])
    return mx

t = int(input())
for _ in range(t):
    n = int(input())
    a = ['?'] * n
    l, r = 1, n
    print('?', l, r)
    res = int(input())
    if res == 0:
        print('!', 'IMPOSSIBLE')
        continue
    if n == 2 and res == 1:
        print('!', '01')
        continue
    l += 1
    last = res0 = res
    while l < r:
        print('?', l, r)
        res = int(input())
        if res and res < last:
            a[l-2] = '0'
            a[l-1] = "1"
        elif res:
            a[l-2] = "1"
        if l == r-1 and res:
            a[l-1] = '0'
            a[l] = "1"
        elif not res and last > res:
            a[l-2] = '0'
            a[l-1] = '1'
            for i in range(l, r):
                a[i] = "0"
            break
        elif not res:
            a[l-1] = '0'
        last = res
        l += 1
    print('!', ''.join(a))

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

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

Think step by step.

Response:
```plaintext
2
3
2
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

Buggy Code:
for i in range(int(input())):
    n,=map(int,input().strip().split())
    print('?',n-1,n,flush=True)
    m=int(input())
    if m!=1:
        print('!','IMPOSSIBLE',flush=True)
    else:
        i=n-2
        prev=m
        s='10'
        while i!=0:
            print('?',i,n,flush=True)
            m=int(input())
            if m==prev:
                s+='1'
            else:
                s+='0'
                prev=m
            i-=1
        print('!',s[::-1],flush=True)

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

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

Think step by step.

Response:
```plaintext
2
3
4
0
1
2
2
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

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

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int test = 0; test < t; ++test) {
            int n = sc.nextInt();
            solve(n);
        }
    }

    private static void solve(int n) {
        int[] f = new int[n + 1];
        for (int i = 2; i <= n; ++i) {
            f[i] = ask(1, i);
        }

        // Assume s1 is '0'
        char[] s = new char[n + 1];
        s[1] = '0';
        int zero_count = 1;
        for (int i = 2; i <= n; ++i) {
            if (f[i] == f[i - 1]) {
                s[i] = '0';
                zero_count++;
            } else {
                s[i] = '1';
                if (zero_count != f[i] - f[i - 1]) {
                    // Inconsistency, but according to problem, jury provides consistent responses
                    // So, this should not happen
                    System.out.println("! IMPOSSIBLE");
                    return;
                }
            }
        }

        // Check if assuming s1 is '1' leads to the same f values
        if (checkImpossible(n, f, s)) {
            System.out.println("! IMPOSSIBLE");
        } else {
            System.out.print("! ");
            for (int i = 1; i <= n; ++i) {
                System.out.print(s[i]);
            }
            System.out.println();
        }
    }

    private static boolean checkImpossible(int n, int[] f, char[] s) {
        int zero_count = 0;
        for (int i = 2; i <= n; ++i) {
            if (s[i] == '0') {
                if (f[i] != f[i - 1]) {
                    return false;
                }
                zero_count++;
            } else {
                if (f[i] != f[i - 1] + zero_count) {
                    return false;
                }
            }
        }
        return true;
    }

    private static int ask(int l, int r) {
        System.out.printf("? %d %d\n", l, r);
        System.out.flush();
        Scanner sc = new Scanner(System.in);
        return sc.nextInt();
    }
}

// wxHRWl8JV9XATC7fNs6

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code lies in its assumption about the consistency of the responses and the way it handles the transition between '0' and '1' in the binary string.

### Step-by-Step Analysis:
1. **Edge Case with All Zeros or All Ones:**
   - If the string is all zeros or all ones, the number of subsequences of "01" will always be zero, which can lead to ambiguity.
   - Example: `0000` or `1111`.

2. **Edge Case with Alternating Zeros and Ones:**
   - If the string alternates between zeros and ones, the number of subsequences of "01" will be consistent but can still lead to ambiguity.
   - Example: `01010`.

3. **Edge Case with a Single '1' or '0':**
   - If the string has only one '1' or '0' in a specific position, the number of subsequences of "01" can be zero for certain ranges, leading to potential issues in the algorithm.
   - Example: `0001000`.

4. **Edge Case with Multiple '01' Subsequences:**
   - If the string has multiple "01" subsequences, the algorithm needs to correctly count and handle these transitions.
   - Example: `010101`.

### Chosen Test Input:
To maximize the failure probability, let's use a string that alternates between zeros and ones and has a length that is small but sufficient to expose the bug. This will ensure that the algorithm has to handle multiple transitions and check its assumptions.

### Test Input:
```plaintext
2
5
4
2
1
2
1
2
0
```

### Explanation:
- **First Test Case:**
  - Length `n = 5`.
  - Queries and responses:
    - `? 1 5` -> `4`
    - `? 2 4` -> `2`
    - `? 3 5` -> `1`
    - `? 4 5` -> `2`
    - `? 1 2` -> `1`
  - The correct string for this input is `01010`.

- **Second Test Case:**
  - Length `n = 2`.
  - Queries and responses:
    - `? 1 2` -> `0`
  - The correct string for this input is `00` or `11` (impossible to determine).

This test input will help expose the bug in the provided code by challenging its assumptions about the consistency of the responses and the handling of transitions between zeros and ones.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kachina's Favorite Binary String
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Kachina challenges you to guess her favorite binary string$$$^{\text{∗}}$$$ $$$s$$$ of length $$$n$$$. She defines $$$f(l, r)$$$ as the number of subsequences$$$^{\text{†}}$$$ of $$$\texttt{01}$$$ in $$$s_l s_{l+1} \ldots s_r$$$. Two subsequences are considered different if they are formed by deleting characters from different positions in the original string, even if the resulting subsequences consist of the same characters.

To determine $$$s$$$, you can ask her some questions. In each question, you can choose two indices $$$l$$$ and $$$r$$$ ($$$1 \leq l < r \leq n$$$) and ask her for the value of $$$f(l, r)$$$.

Determine and output $$$s$$$ after asking Kachina no more than $$$n$$$ questions. However, it may be the case that $$$s$$$ is impossible to be determined. In this case, you would need to report $$$\texttt{IMPOSSIBLE}$$$ instead.

Formally, $$$s$$$ is impossible to be determined if after asking $$$n$$$ questions, there are always multiple possible strings for $$$s$$$, regardless of what questions are asked. Note that if you report $$$\texttt{IMPOSSIBLE}$$$ when there exists a sequence of at most $$$n$$$ queries that will uniquely determine the binary string, you will get the Wrong Answer verdict.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^4$$$) — the length of $$$s$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
2
5

4

0

1

2

2

0
output:
? 1 5

? 2 4

? 4 5

? 3 5

! 01001

? 1 2

! IMPOSSIBLE

Note: In the first test case:

In the first query, you ask Kachina for the value of $$$f(1, 5)$$$, and she responds with $$$4$$$ in the input stream.

In the second query, you ask Kachina for the value of $$$f(2, 4)$$$. Because there are no subsequences of $$$\texttt{01}$$$ in the string $$$\texttt{100}$$$, she responds with $$$0$$$ in the input stream.

After asking $$$4$$$ questions, you report $$$\texttt{01001}$$$ as $$$s$$$, and it is correct.

In the second test case:

In the first query, you ask Kachina for the value of $$$f(1, 2)$$$, and she responds with $$$0$$$ in the input stream. Notice that this is the only distinct question you can ask.

However, notice that the strings $$$00$$$ and $$$11$$$ both have an answer of $$$0$$$, and it is impossible to differentiate between the two. Therefore, we report IMPOSSIBLE.

Please note that this example only serves to demonstrate the interaction format. It is not guaranteed the queries provided are optimal or uniquely determine the answer. However, it can be shown there exists a sequence of at most $$$5$$$ queries that does uniquely determine sample test case $$$1$$$.

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

public class C {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out, true);

        int t = Integer.parseInt(reader.readLine());

        for (int j = 0; j < t; j++) {
            int n = Integer.parseInt(reader.readLine());
            char[] s = new char[n];
            Arrays.fill(s, '?');
            int arr[] = new int[n];
            boolean flag = false;
            for (int i = 0; i < n-1; i++) {
                out.println("? " + (i+1) + " " + n);
                out.flush();

                int temp = Integer.parseInt(reader.readLine());
                arr[i] = temp;
            }
            for (int i = n-1; i >=0; i--) {
                if(arr[i]==0)
                	continue;
                else {
                	flag = true;
                	if(arr[i]!=arr[i+1])
                		s[i]='0';
                	else
                		s[i]='1';
                }
            }
            if(flag) {
            	int index = -1;
            	for (int i = n-1; i >=0; i--) {
                    if(arr[i]!=0) {
                    	if(arr[i]!=n-1-i) {
                    		flag = false;
                    	}
                    	index = i;
                    	break;
                    }
                }
            	for (int i = index+1; i<n; i++) {
                    s[i]='1';
                }
            }
            if (!flag) {
                out.println("! IMPOSSIBLE");
            } else {
                out.println("! " + new String(s));
            }
            out.flush();
        }

        reader.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
2
4
0
3
1
```