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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
for test in range(int(input())):
    L=1; R=999
    cur_l = 333; cur_r = 666

    while L<R:
        print('?',cur_l,cur_r,flush=True)
        a = int(input())

        if a == cur_l*cur_r:
            L = cur_r
        elif a == cur_l*(cur_r+1):
            R = cur_r-1
        elif a == (cur_l+1)*(cur_r+1):
            R = cur_l-1
        gap = (R-L)
        cur_l = L + gap//3; cur_r = R - gap//3

    print('! ' + str(L+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

110889

1
```


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

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

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n;
    int query(int a, int b){
        System.out.println("? "+a+" "+b);
        int ans = i();
        System.out.flush();
        return ans;

    }
    int help(){
        int s = 1, e = 1000;
        while(e-s>=3){
        int m1 = s+ (e-s) / 3;
        int m2 = e - (e-s) / 3;
        int r = query(m1, m2);
        if(r == (m1+1)*(m2+1)) e= m1;
        else if(r == m1*m2) s = m2;
        else{
            s = m1;
            e = m2;
            }
        }
        int r = query(s, s+1);
        if(r == (s)*(s+2)) return s+1;
        return e;
    }

    public void solve() throws IOException {
        t = sc.nextInt();
        while (t-- > 0) {
            System.out.println("! "+ help());
            System.out.flush();
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
2

18

25

9999
```


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
//package com.aakash.Contest.C964;

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

public class G2 {
    static BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));

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

        StringTokenizer s1
                = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(s1.nextToken());
        while (t-- > 0) {
            solve();
        }

    }

    static void solve() throws IOException {
        int s=2,e=999;
        while(s<=e) {
            int m=s+(e-s)/2;
            int s1=s,e1=m,s2=m+1,e2=e;
            int m1=s1+(e1-s1)/2;
            int m2=s2+(e2-s2)/2;
            int a=IR(m1,m2);
            if(a==1) s=m2+1;
            else if(a==2) {
                s=m1+1;
                e=m2-1;
            } else e=m1-1;
        }
        System.out.println("! "+s);

    }
    static int IR(int a,int b) throws IOException {
        System.out.println("? "+a+" "+b);
        System.out.flush();
        StringTokenizer s2
                = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(s2.nextToken());
        int x=a*b;
        int z=(a+1)*(b+1);
        if(n==x) return 1;
        else if(n==z) return 3;
        else return 2;
    }
}

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

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

Think step by step.

Response:
```plaintext
2

25

9999
```

### Explanation:
1. **Test Case 1:**
   - The secret value of \( x \) is 4.
   - The query `? 4 4` will return 25 because the ruler measures 4 as 5, so the area is \( 5 \times 5 = 25 \).
   - The correct answer should be 4.

2. **Test Case 2:**
   - The secret value of \( x \) is 100.
   - The query `? 99 100` will return 9999 because the ruler measures 100 as 101, so the area is \( 99 \times 101 = 9999 \).
   - The correct answer should be 100.

These test cases are designed to challenge the algorithm by providing inputs that are at the boundaries of the possible values for \( x \), thereby maximizing the chance of exposing any bugs in the faulty solution.


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

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

ll ask(ll a,ll b){
    ll res;
    cout<<"? "<<a<<" "<<b<<endl;
    cin>>res;
    return res;
}

void solve(){
    ll l=2,h=999;
    while(l<h){
        ll mid1=l+(h-l+1)/3;
        ll mid2=h-(h-l+1)/3;
        ll res=ask(mid1,mid2);
        if(mid1==mid2){
            cout<<"! "<<mid1<<endl;
            return;
        }
        if(res==mid1*mid2) l=mid2+1;
        else if(res==mid1*(mid2+1)){
            l=mid1+1;
            h=mid2;
        }
        else h=mid1;
    }
    cout<<"! "<<l<<endl;
}

int main(){
    ll t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the algorithm's logic might fail. The key points to consider are:

1. **Edge Cases for \( x \) Values**: Values of \( x \) that are close to the boundaries of the range (2 and 999).
2. **Boundary Conditions for Queries**: Values of \( a \) and \( b \) that are close to \( x \) and can cause the ruler to report incorrect measurements.
3. **Behavior of the Algorithm**: The algorithm uses a ternary search approach, which can be tricky to get right, especially when the values of \( a \) and \( b \) are chosen in a way that can confuse the search.

Let's construct a test input that includes these challenging cases:

1. **Test Case 1**: \( x = 2 \)
2. **Test Case 2**: \( x = 3 \)
3. **Test Case 3**: \( x = 998 \)
4. **Test Case 4**: \( x = 999 \)

For each test case, we will simulate the interaction and provide the correct responses from the jury based on the ruler's behavior.

```plaintext
4

? 1 2
3
! 2

? 2 3
8
! 3

? 997 998
995006
! 998

? 998 999
997007
! 999
```

### Explanation:
1. **Test Case 1: \( x = 2 \)**
   - Query: \( 1 \times 2 \)
   - Response: \( 1 \times 3 = 3 \)
   - Correct output: \( 2 \)

2. **Test Case 2: \( x = 3 \)**
   - Query: \( 2 \times 3 \)
   - Response: \( 2 \times 4 = 8 \)
   - Correct output: \( 3 \)

3. **Test Case 3: \( x = 998 \)**
   - Query: \( 997 \times 998 \)
   - Response: \( 997 \times 999 = 995006 \)
   - Correct output: \( 998 \)

4. **Test Case 4: \( x = 999 \)**
   - Query: \( 998 \times 999 \)
   - Response: \( 998 \times 1000 = 997007 \)
   - Correct output: \( 999 \)

This test input is designed to challenge the ternary search logic and ensure that the algorithm correctly handles edge cases and boundary conditions.


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define ull unsigned long long
#define Mod 998244353
#define PI acos(-1);
#define MAXN 210000
#define debug cout<<"debug:\t"
#define debugn cout<<"debug:\n"
#define enter cout<<"\n"
void fastread() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
};
void solve() {
    ll bel = 1;
    ll ber = 1000;
    ll l = 250;
    ll r = 750;
    ll sum = 0;
    while (1) {
        cout << "? " << l << " " << r << endl;
        cin >> sum;
        if (sum == (l + 1) * (r + 1)) {
            ber = l;
            l = (ber - bel) / 3 + bel;
            r = (ber - bel) / 3 * 2 + bel;
        }
        else if (sum == l * (r + 1)) {
            bel = l;
            ber = r;
            l = (ber - bel) / 3 + bel;
            r = (ber - bel) / 3 * 2 + bel;
        }
        else if (sum == l * r) {
            if (l == r) {
                cout << "! " << l + 1 << endl;
                return;
            }
            bel = r;
            l = (ber - bel) / 3 + bel;
            r = (ber - bel) / 3 * 2 + bel;
        }
    }
}
int main() {
    //fastread();
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ll t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2

18

9999
```


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
from sys import stdout

def solve():
    l = 1
    r = 1000
    while r - l > 2:
        d = (r - l) // 3

        f1 = l + d - 1
        f2 = f1 + d

        print(f'? {f1} {f2}')
        stdout.flush()
        inp = int(input())
        if inp == -1:
            exit()
        if inp == (f1 + 1) * (f2 + 1):
            r = f1
        if inp == f1 * (f2 + 1):
            l = f1
            r = f2
        if inp == f1 * f2:
            l = f2

    while r - l > 1:
        c = (l + r) // 2
        print(f'? {1} {c}')
        stdout.flush()

        inp = int(input())
        if inp == -1:
            exit()
        if inp == c:
            l = c
        else:
            r = c

    print(f'! {r}')
    stdout.flush()

for _ in range(int(input())):
    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

18

24
```

### Explanation:
1. **Test Case 1 (x = 4):**
   - The first query is `? 3 5`, which should return `18` because the ruler measures `3` and `6` (since `x = 4`).
   - The second query is `? 4 4`, which should return `25` because the ruler measures `5` and `5` (since `x = 4`).
   - The solution should correctly determine that `x = 4` and output `! 4`.

2. **Test Case 2 (x = 5):**
   - The first query is `? 3 5`, which should return `18` because the ruler measures `3` and `6` (since `x = 5`).
   - The second query is `? 4 6`, which should return `24` because the ruler measures `4` and `7` (since `x = 5`).
   - The solution should correctly determine that `x = 5` and output `! 5`.

This test input is designed to challenge the buggy code by providing edge cases where the ruler's behavior changes. The first test case is similar to the example provided, while the second test case introduces a slightly different scenario to test the robustness of the solution.


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
def solve():
    l = 0
    r = 1001
    while r - l > 1:
        m1 = (l + r) // 2
        m2 = (l + m1) // 2
        print(f"? {m1} {m2}", flush=True)
        x = int(input())
        if x == m1 * m2:
            l = m1
        if x == (m1+1)*(m2+1):
            r = m2
        if x == m2 * (m1 + 1):
            l = m2
            r = m1
    print(f"! {r}", flush=True)
t = int(input())
for _ in range(t):
    solve()

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

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

Think step by step.

Response:
```plaintext
2

18

25
```


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
import 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 st = 2, end = 999;
   int ans = 0;
   while (st <= end) {
    int mid = (st + end) / 2;
    int x = (mid + st) / 2, y = (end + mid) / 2;
    if (x == y) {
     end = x - 1;
     break;
    }
    System.out.println("? " + x + " " + y);
    int res = sc.nextInt();
    if (res == -1) {
     break;
    }
    if (res == x * y) {
     st = y + 1;
    } else if (res == (x + 1) * (y + 1)) {
     end = x - 1;

    } else if (res == x * (y + 1)) {
     st = x + 1;
     end = y - 1;
    }
   }
   System.out.println("! " + (end + 1));
  }
 }
}

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

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

Think step by step.

Response:
```plaintext
2

18

25
```


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

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

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

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

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

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

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

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

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

Output: None

Examples:
input:
2

18

25

9999
output:
? 3 5

? 4 4

! 4
? 99 100

! 100

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

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

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

Buggy Code:
import java.util.Scanner;

public class problemG2 {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int tests = sc.nextInt();
    while (tests-- > 0) {
      int res = 2;
      int l = 2, r = 999;
      while (r - l > 2) {
        int m1 = (2 * l + r) / 3;
        int m2 = (2 * r + l) / 3;
        System.out.println("? " + m1 + " " + m2);
        int response = sc.nextInt();
        if (response == (m1 + 1) * (m2 + 1)) {
          r = m1;
        } else if (response == (m1 * m2)) {
          l = m2;
        } else {
          l = m1;
          r = m2;
        }
      }
      if (r - l == 2) {
        System.out.println("? 1" + " " + (l + 1));
        int temp = sc.nextInt();
        if (temp == l + 1) {
          l = l + 1;
        } else {
          r = l + 1;
        }
      }
      System.out.println("! " + r);
    }
  }

}

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

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

Think step by step.

Response:
```plaintext
2

18

25
```

This test input is designed to challenge the buggy code by providing specific responses that might cause the algorithm to misinterpret the value of \( x \). Here’s the reasoning:

1. **First Test Case:**
   - The response `18` suggests that the ruler measures a `3 x 5` rectangle as `3 x 6`, indicating that \( x \) could be 4.
   - The response `25` suggests that the ruler measures a `4 x 4` rectangle as `5 x 5`, which is consistent with \( x = 4 \).

2. **Second Test Case:**
   - The response `25` for a `4 x 4` rectangle again suggests \( x = 4 \), but this time we provide a different sequence of queries to ensure the code handles it correctly.

By providing these specific responses, we can test if the code correctly identifies \( x \) and handles edge cases properly. The test input is small and complete, adhering to the constraints and format specified.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

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

typedef long long ll;
#define int ll
typedef pair<int,int> pi;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi,ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0)
#define maxn 200010
#define INF int(1e18)
#define MOD 1000000007
#define DEBUG 0
typedef pair <int, pi> ipi;
typedef pair <pi,pi> pipi;

int N,M,TC;
int d[maxn],ss[maxn];
vi adj[maxn];
int u,v;
void solve(){
	cin >> N >> M;
	vpi edges;
	FOR(i,1,M){
		cin >> u >> v;
		adj[v].pb(u);
		edges.pb(pi(u,v));
	}
	d[1] = 0;
	FOR(i,2,N){
		d[i] = d[i-1] + 1;
		aFOR(j,adj[i]) d[i] = min(d[i], d[j] + 1);
	}
	FOR(i,1,N) ss[i] = 0;
	aFOR(i, edges){
		auto [u,v] = i;
		int l = u + 1;
		int r = v - d[u] - 2;
		if (l <= r){
			ss[l]++; ss[r+1]--;
		}
	}
	FOR(i,2,N) ss[i] += ss[i-1];
	FOR(i,1,N-1) cout << (ss[i] > 0 ? 0 : 1);
	cout << '\n';
}

int32_t main(){
	fast;
	cin >> TC;
	while (TC--) solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The bug in the code seems to be related to how it handles the alternative bridges and the dynamic programming approach used to determine the winning positions.

### Steps to Generate the Test Input:
1. **Small-Scale Test Case**: Since the output length is limited, we will use a small number of islands and bridges.
2. **Edge Case**: We need to create a scenario where the alternative bridges can significantly affect the outcome, especially where Elsie can use them to overtake Bessie.
3. **Optimal Strategies**: Ensure that both cows follow optimal strategies to ensure their respective victories.

### Test Case:
- **Number of Islands (n)**: 5
- **Number of Alternative Bridges (m)**: 2
- **Alternative Bridges**:
  - From island 1 to island 3
  - From island 2 to island 4

### Explanation:
- **Island 1 to Island 3**: This bridge allows Elsie to jump directly to island 3, potentially overtaking Bessie if Bessie starts on island 2.
- **Island 2 to Island 4**: This bridge allows Elsie to jump directly to island 4, potentially overtaking Bessie if Bessie starts on island 3.

### Test Input:
```plaintext
1
5 2
1 3
2 4
```

### Expected Output:
```plaintext
1001
```

### Explanation of Expected Output:
- **Bessie starts on island 1**: Bessie can win because she moves first and can reach island 5 before Elsie.
- **Bessie starts on island 2**: Bessie loses because Elsie can use the alternative bridge from island 1 to island 3, then move to island 4 on her next turn, and finally to island 5.
- **Bessie starts on island 3**: Bessie loses because Elsie can use the alternative bridge from island 2 to island 4, then move to island 5.
- **Bessie starts on island 4**: Bessie can win because she is already close to island 5 and can reach it before Elsie.

This test case should expose the bug in the buggy code by challenging the algorithm's ability to handle alternative bridges correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int MAXN=2e5+5;
vector<int> e[MAXN];
int dis[MAXN],vis[MAXN];
int n,m;
void dij()
{
    priority_queue<int,vector<int>,greater<int>> q;
    for(int i=1;i<=n;i++) dis[i]=1e18,vis[i]=0;
    dis[1]=0;
    vis[1]=0;
    q.push(1);
    while(q.size())
    {
        int u=q.top();
        q.pop();
        if(vis[u]) continue;
        vis[u]=1;
        for(auto v:e[u])
        {
            if(dis[v]>dis[u]+1)
            {
                dis[v]=dis[u]+1;
                q.push(v);
            }
        }
    }
}
void solve()
{
    cin>>n>>m;
    for(int i=1;i<n;i++)
    {
        e[i].clear();
        e[i].push_back(i+1);
    }
    vector<pair<int,int>> b;
    for(int i=1;i<=m;i++)
    {
        int u,v;
        cin>>u>>v;
        e[u].push_back(v);
        b.push_back({u,v});
    }
    dij();
    // for(int i=1;i<=n;i++) cout<<dis[i]<<' ';
    // cout<<endl;
    vector<int> ans(n+2,0);
    for(int i=0;i<m;i++)
    {
        int pos=b[i].second-dis[b[i].first]-1;
        // cout<<pos<<endl;
        ans[b[i].first+1]+=1;
        ans[pos]-=1;
    }
    for(int i=1;i<=n;i++) ans[i]=ans[i-1]+ans[i];
    // for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
    // cout<<endl;
    cout<<1;
    for(int i=2;i<n;i++) if(!ans[i]) cout<<1;else cout<<0;
    cout<<endl;
}
signed main()
{
    cin.tie(0)->sync_with_stdio(0);
    int t=1;
    cin>>t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
6 1
2 6
10 2
1 9
2 8
15 3
2 8
4 9
8 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=200010;
int T,n,m,a,b,r,p[N],dis[N],inf=0x3f3f3f3f;
vector<int>fr[N];
signed main(){
	cin>>T;
	while(T--){
		cin>>n>>m;
		r=0;
		for(int i=1;i<=n;i++) {
			p[i]=1;
			dis[i]=i-1;
			fr[i].clear();
		}
		for(int i=1;i<=m;i++){
			cin>>a>>b;
			fr[b].push_back(a);
		}
		for(int i=1;i<=n;i++){
			int a=0,b=i;
			for(int j:fr[i]) {
				dis[i]=min(dis[i],dis[j]+1);
				a=max(a,j);
			}
			b-=dis[i];
			a++,b--;
			a=max(a,r+1);
			if(a<=b){
				for(int j=a;j<=b;j++) p[j]=0;
				r=b;
			}
		}
		for(int i=1;i<n;i++) cout<<p[i];
		puts("");
	}
	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 buggy code, let's consider the following aspects:

1. **Edge Cases**: Test inputs that push the limits of the algorithm, such as the smallest and largest possible values for `n` and `m`.
2. **Critical Scenarios**: Situations where the alternative bridges can significantly affect the outcome, such as when Bessie starts on an island that can be bypassed by Elsie using alternative bridges.
3. **Optimal Strategies**: Ensure that both Bessie and Elsie follow optimal strategies, which means Bessie should win if there are no alternative bridges that Elsie can use to her advantage.

Given these considerations, let's create a test input that includes a scenario where Bessie starts on an island that can be bypassed by Elsie using alternative bridges, but also includes a scenario where Bessie can win.

### Test Input
```plaintext
3
5 0
5 1
1 3
5 2
1 3
3 5
```

### Explanation
1. **Test Case 1:**
   - **n = 5, m = 0**
   - There are no alternative bridges, so Bessie will always win because she always moves first and can reach island 5 without any interference from Elsie.
   - Expected output: `1111`

2. **Test Case 2:**
   - **n = 5, m = 1**
   - There is one alternative bridge from island 1 to island 3.
   - If Bessie starts on island 1, Elsie can use the alternative bridge to bypass Bessie and win.
   - If Bessie starts on island 2 or 3, she can still win because Elsie cannot bypass her with the available alternative bridge.
   - If Bessie starts on island 4, she can win because she is already close to the end.
   - Expected output: `0111`

3. **Test Case 3:**
   - **n = 5, m = 2**
   - There are two alternative bridges: one from island 1 to island 3 and one from island 3 to island 5.
   - If Bessie starts on island 1, Elsie can use the alternative bridge to bypass Bessie and win.
   - If Bessie starts on island 2, Elsie can use the alternative bridge to bypass Bessie and win.
   - If Bessie starts on island 3, Elsie can use the second alternative bridge to bypass Bessie and win.
   - If Bessie starts on island 4, she can win because she is already close to the end.
   - Expected output: `0001`

This test input should expose the bug in the provided buggy code by creating scenarios where the alternative bridges play a critical role in determining the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

Buggy Code:
import sys
from collections import defaultdict

input = sys.stdin.read
data = input().splitlines()

MAXN = 200010

def main():
    testCases = int(data[0])
    idx = 1

    for _ in range(testCases):
        n, m = map(int, data[idx].split())
        idx += 1

        nextIslands = defaultdict(list)
        prevIslands = defaultdict(list)
        startIsland = [0] * (MAXN)
        endIsland = [0] * (MAXN)
        distanceToIsland = [i - 1 for i in range(MAXN)]
        bridgeCount = [0] * (MAXN)

        for i in range(1, m + 1):
            u, v = map(int, data[idx].split())
            idx += 1
            startIsland[i] = u
            endIsland[i] = v
            nextIslands[u].append(v)
            prevIslands[v].append(u)

        for i in range(2, n + 1):
            for prev in prevIslands[i]:
                distanceToIsland[i] = min(distanceToIsland[i], distanceToIsland[prev] + 1)
            distanceToIsland[i] = min(distanceToIsland[i], distanceToIsland[i - 1] + 1)

        for i in range(1, m + 1):
            minDistance = distanceToIsland[startIsland[i]] + 1
            maxValidPosition = endIsland[i] - 1 - minDistance
            if startIsland[i] + 1 <= maxValidPosition:
                bridgeCount[startIsland[i] + 1] += 1
                bridgeCount[maxValidPosition + 1] -= 1

        for i in range(1, n + 1):
            bridgeCount[i] += bridgeCount[i - 1]

        result = []
        for i in range(1, n):
            if bridgeCount[i] > 0:
                result.append('0')
            else:
                result.append('1')

        print(''.join(result))

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
3
6 2
1 3
3 5
10 3
1 4
4 7
7 10
15 4
1 8
2 9
9 12
12 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

Buggy Code:
t = int(input())

for z in range(t):
    n = list(input().split())
    m = int(n[1])
    n = int(n[0])

    most = []
    for i in range(n):
        most.append([])
    for i in range(m):
        j = list(input().split())
        j[0] = int(j[0]) - 1
        j[1] = int(j[1]) - 1
        most[j[1]].append(j[0])

    #print(most)
    shagi = [0]
    for i in range(1, n):
        s = [shagi[i-1]+1]
        for j in range(len(most[i])):
            s.append(shagi[most[i][j]] + 1)
        shagi.append(min(s))
    #print(shagi)

    otvet = []
    for i in range(n-1):
        otvet.append("1")

    o = []
    for i in range(n):
        for j in range(len(most[i])):
            if i - (shagi[most[i][j]] + 2) >= most[i][j] + 1:
                o.append([most[i][j] + 1, i - (shagi[most[i][j]] + 2)])

    if o != []:
        sorted(o)
        j = 0
        i = 0
        while i < (n-1):
            if i < o[j][0]:
                i = i + 1
            elif i <= o[j][1]:
                otvet[i] = "0"
                i = i + 1
            else:
                if j + 1 <= len(o) - 1:
                    j = j + 1
                else:
                    break

    print(''.join(otvet))

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

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

Think step by step.

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

1. **Alternative Bridges**: The placement of alternative bridges can significantly affect the outcome of the race. We need to place bridges in a way that can confuse the algorithm.
2. **Optimal Strategies**: Both Bessie and Elsie will follow optimal strategies, so the test case should force the algorithm to consider these strategies correctly.
3. **Island Collapse**: The collapse of islands and bridges can change the game dynamics, so we need to ensure that the test case includes scenarios where islands collapse in a critical manner.

Let's create a small-scale test input with 3 islands and 2 alternative bridges to expose the bug:

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

### Explanation:
- **Test Case 1**:
  - **Number of Islands (n)**: 3
  - **Number of Alternative Bridges (m)**: 2
  - **Alternative Bridges**:
    - From island 1 to island 3
    - From island 2 to island 3

### Expected Output:
- **Output**: `10`

### Reasoning:
- **Starting on Island 1**:
  - Bessie starts on island 1.
  - Bessie moves to island 2 (using the main bridge).
  - Elsie starts on island 1.
  - Elsie moves to island 3 (using the alternative bridge).
  - Elsie wins.
- **Starting on Island 2**:
  - Bessie starts on island 2.
  - Bessie moves to island 3 (using the main bridge).
  - Bessie wins.

This test case is designed to expose the bug in the algorithm by creating a scenario where the placement of alternative bridges can significantly affect the outcome, and the algorithm must correctly account for the optimal strategies of both cows.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

Buggy Code:
t = int(input())

for z in range(t):
    n = list(input().split())
    m = int(n[1])
    n = int(n[0])

    most = []
    for i in range(n):
        most.append([])
    for i in range(m):
        j = list(input().split())
        j[0] = int(j[0]) - 1
        j[1] = int(j[1]) - 1
        most[j[1]].append(j[0])

    #print(most)
    shagi = [0]
    for i in range(1, n):
        s = [shagi[i-1]+1]
        for j in range(len(most[i])):
            s.append(shagi[most[i][j]] + 1)
        shagi.append(min(s))
    #print(shagi)

    otvet = []
    for i in range(n-1):
        otvet.append("1")

    o = []
    for i in range(n):
        for j in range(len(most[i])):
            if i - (shagi[most[i][j]] + 2) >= most[i][j] + 1:
                o.append([most[i][j] + 1, i - (shagi[most[i][j]] + 2)])

    if o != []:
        sorted(o)
        j = 0
        i = 0
        while i < (n-1):
            if i < o[j][0]:
                i = i + 1
            elif i <= o[j][1]:
                otvet[i] = "0"
                i = i + 1
            else:
                if j + 1 <= len(o) - 1:
                    j = j + 1
                else:
                    break

    if ''.join(otvet) == "11001":
        st = ""
        for i in range(n):
            if most[i] != []:
                st = st + str(i)
                st = st + str(most[i][0])
        print(s)
    else:
        print(''.join(otvet))

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

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

Think step by step.

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

1. **Edge Cases**:
   - Small values of `n` and `m` to ensure the test case is small-scale.
   - Specific configurations of alternative bridges that can cause the algorithm to fail.

2. **Buggy Code Analysis**:
   - The buggy code seems to have issues with how it handles the alternative bridges and the conditions for determining when Bessie can win.
   - The logic for determining the winning positions might be flawed, especially in cases where Elsie can use alternative bridges to overtake Bessie.

3. **Test Input Design**:
   - We need a configuration where Elsie can use alternative bridges to reach the final island faster than Bessie, but the logic in the buggy code might not correctly handle this.

Let's create a test case where `n = 5` and `m = 2` with the following alternative bridges:
- An alternative bridge from island 1 to island 3.
- An alternative bridge from island 2 to island 4.

This configuration should make it difficult for Bessie to win if she starts on island 2, as Elsie can use the alternative bridge from island 1 to island 3 and then from island 3 to island 4, reaching the final island before Bessie.

Here is the test input:

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

Explanation:
- `t = 1`: One test case.
- `n = 5`, `m = 2`: 5 islands and 2 alternative bridges.
- Alternative bridges:
  - From island 1 to island 3.
  - From island 2 to island 4.

This test case should expose the bug in the buggy code, as it should correctly determine that Bessie cannot win if she starts on island 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Determine Winning Islands in Race
time_limit_ms: 2000
memory_limit_mb: 256
Description: MOOOOOOOOOOOOOOOOO

— Bessie the Cow, The Art of Racing on Islands

Two of Farmer John's cows, Bessie and Elsie, are planning to race on $n$ islands. There are $n - 1$ main bridges, connecting island $i$ to island $i + 1$ for all $1 \leq i \leq n - 1$. Additionally, there are $m$ alternative bridges. Elsie can use both main and alternative bridges, while Bessie can only use main bridges. All bridges are one way and can only be used to travel from an island with a lower index to an island with a higher index.

Initially, Elsie starts on island $1$, and Bessie starts on island $s$. The cows alternate turns, with Bessie making the first turn. Suppose the cow is on island $i$. During a cow's turn, if there are any bridges connecting island $i$ to island $j$, then the cow can move to island $j$. Then, island $i$ collapses, and all bridges connecting to island $i$ also collapse. Also, note the following:

  * If there are no bridges connecting island $i$ to another island, then island $i$ collapses, and this cow is eliminated from the race.    * If the other cow is also on island $i$, then after this cow moves to another island, the island collapses, and the other cow is eliminated from the race.    * After an island or bridge collapses, no cows may use them.    * If a cow is eliminated, their turn is skipped for the rest of the race.

The race ends once either cow reaches island $n$. It can be shown that regardless of the cows' strategies, at least one cow reaches island $n$. Bessie wins if and only if she reaches island $n$ first.

For each $1 \leq s \leq n - 1$, determine whether Bessie wins if she starts the race on island $s$. Assume both cows follow optimal strategies to ensure their own respective victories.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$) – the number of test cases.

The first line of each test case contains $n$ and $m$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq m \leq 2 \cdot 10^5$) – the number of islands and the number of alternative bridges.

The next $m$ lines of each test case contain $u$ and $v$ ($1 \leq u < v \leq n$) – the islands that the alternative bridge connects. It is guaranteed all alternative bridges are distinct, and they do not coincide with the main bridges.

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^5$.

Output: For each test case, output a binary string of length $n - 1$ on a new line. The $i$'th character is $1$ if it is possible for Bessie to win if she starts on island $i$. Otherwise, it is $0$.

Examples:
input:
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
output:
11111
11011
10011
100001111
11000111000111

Note: In the first test case, there are no alternative bridges for Elsie to overtake Bessie and reach island $n$ first, so Bessie will win on all islands because she always moves first.

In the second case, Bessie will lose if she starts on island $3$ because:

  * Bessie's Turn: Take a main bridge from island $3$ to island $4$.    * Elsie's Turn: Take a main bridge from island $1$ to island $2$.    * Bessie's Turn: Take a main bridge from island $4$ to island $5$.    * Elsie's Turn: Take an alternative bridge from island $2$ to island $6$. Elsie reaches island $n$ first.

Buggy Code:
//package com.aakash.Contest.C965;

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

public class D {
    static BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));

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

        StringTokenizer s1
                = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(s1.nextToken());
        if(t==10000) {
            while (t-- > 0) {

                StringTokenizer s2
                        = new StringTokenizer(br.readLine());
                int n = Integer.parseInt(s2.nextToken());
                int m = Integer.parseInt(s2.nextToken());
                int[][] arr = new int[m][2];

                for (int i = 0; i < m; i++) {
                    StringTokenizer s4
                            = new StringTokenizer(br.readLine());
                    arr[i][0] = Integer.parseInt(s4.nextToken())-1;
                    arr[i][1]=Integer.parseInt(s4.nextToken())-1;
                }
                if(t==10000-26) {
                    System.out.println(n+" "+m);
                    for(int i=0;i<m;i++) {
                        System.out.println(Arrays.toString(arr[i]));
                    }
                }
            }

        } else {
            while (t-- > 0) {

                StringTokenizer s2
                        = new StringTokenizer(br.readLine());
                int n = Integer.parseInt(s2.nextToken());
                int m = Integer.parseInt(s2.nextToken());
                int[][] arr = new int[m][2];

                for (int i = 0; i < m; i++) {
                    StringTokenizer s4
                            = new StringTokenizer(br.readLine());
                    arr[i][0] = Integer.parseInt(s4.nextToken())-1;
                    arr[i][1]=Integer.parseInt(s4.nextToken())-1;
                }

                solve(arr, n,m);
            }

        }

    }

    static void solve(int[][] arr, int n,int m) {
        //now we do some pre processing
        int[][] dp=new int[n][2];

        preprocess(dp,arr,n,m);
        //now our preprocess has been done now we will try to find res;
//        for(int i=0;i<n;i++) {
//            System.out.println(Arrays.toString(dp[i]));
//        }
        StringBuilder sb=new StringBuilder();
        sb.append("0".repeat(n-1));
        Stack<Integer> s=new Stack<>();
        Stack<Integer> temp=new Stack<>();
        if(dp[n-1][1]!=n-1) s.push(n-1);
        for(int i=n-2;i>=0;i--) {
            while(!s.isEmpty()) {
                int ind=s.peek();
                int minpar=dp[ind][0];
                int step=dp[ind][1];
                if(minpar>=i ||ind-i<=step) {
                    if(minpar<i)  temp.push(s.pop());
                    else s.pop();
                }
                else break;
            }
            if(s.isEmpty()) sb.setCharAt(i,'1');
            while(!temp.isEmpty()) s.push(temp.pop());
            if(dp[i][1]!=i) s.push(i);
        }

        System.out.println(sb);
    }
    static void preprocess(int[][] dp,int[][]arr,int n,int m) {
        //finding the min index from which it can be reachable
        //dp[0]-> storing the min index from which it is reachable
        for(int i=0;i<n;i++) dp[i][0]=i;
        for(int i=0;i<m;i++) {
            int s=arr[i][0],e=arr[i][1];
            dp[e][0]=Math.min(s,dp[e][0]);
        }
        //now need dijkstra algorithm to find the shortest path between source and dest with fixed source
        //dp[1]-> storing the minimum time at which it can be reachable
        ArrayList<Integer>[] list=new ArrayList[n];
        for(int i=0;i<n;i++) list[i]=new ArrayList<>();
        for(int i=0;i<n-1;i++) list[i].add(i+1);
        for(int i=0;i<m;i++) list[arr[i][0]].add(arr[i][1]);
        Dijkstra dj=new Dijkstra(list,n);
        int[] dist=dj.dist;
        for(int i=0;i<n;i++) dp[i][1]=dist[i];
    }
}
class Dijkstra {
    ArrayList<Integer>[] list;
    int n;
    int[] dist;
    Dijkstra(ArrayList<Integer>[] list,int n) {
        this.list=list;
        this.n=n;
        this.dist=new int[n];
        Arrays.fill(dist,Integer.MAX_VALUE);
        dist[0]=0;
        find();
    }
    void find() {
        PriorityQueue<Pair> pq=new PriorityQueue<>(new PC());
        pq.add(new Pair(0,0));
        while(!pq.isEmpty()) {
            Pair p=pq.remove();
            int curr=p.v;
            int w=p.w;
            if(w>dist[curr]) continue;
            ArrayList<Integer> internal=list[curr];
            for(int i=0;i<internal.size();i++) {
                int next=internal.get(i);
                int nw=w+1;
                if(dist[next]>nw) {
                    dist[next]=nw;
                    pq.add(new Pair(next,nw));
                }
            }

        }
    }
}
class PC implements Comparator<Pair>{
    public int compare(Pair a,Pair b) {
        return Integer.compare(a.w, b.w);
    }
}
class Pair{
    int v;
    int w;
    Pair(int v,int w) {
        this.v=v;
        this.w=w;
    }
}

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

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

Think step by step.

Response:
```plaintext
5
6 0
6 1
2 6
6 1
1 5
10 4
1 3
1 6
2 7
3 8
15 3
2 8
4 9
8 15
```