**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 java.util.*;

public final class WinningIslands
{
    private static final int MAX_N = 20005;

    public static void helper(Scanner sc)
    {
        int n = sc.nextInt();
        int m = sc.nextInt();

        List<Integer>[] adj = new ArrayList[MAX_N];

        for(int i=0; i<MAX_N; i++)
        adj[i] = new ArrayList<>();

        int[] dp = new int[MAX_N];
        int[] add = new int[MAX_N];

        Arrays.fill(dp, Integer.MAX_VALUE);

        dp[1] = 0;

        for(int i=0; i<m; i++)
        {
            int x = sc.nextInt();
            int y = sc.nextInt();
            adj[x].add(y);
        }

        for(int i=1; i<n; i++)
        {
            for(int x : adj[i])
            dp[x] = Math.min(dp[x], dp[i]+1);

            if(i+1 <= n)
            dp[i+1] = Math.min(dp[i+1], dp[i]+1);
        }

        Arrays.fill(add, 0);

        for(int i=1; i<=n; i++)
        {
            for(int to : adj[i])
            {
                int u = i;
                int v = to;
                if(u+1 <= v-(dp[i]+1)-1)
                {
                    add[u + 1]++;
                    if(v-(dp[i]+1) < MAX_N)
                    add[v-(dp[i]+1)]--;
                }
            }
        }

        for(int i=1; i<=n; i++)
        add[i] += add[i-1];

        StringBuilder result = new StringBuilder();

        for(int i=1; i<n; i++)
        result.append(add[i] > 0 ? '0':'1');

        result.append('\n');

        System.out.print(result.toString());
    }
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0)
        helper(sc);
        sc.close();
    }
}

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

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

Think step by step.

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


**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;

void MonkeyDLuffy() {

  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> remove;
  vector<vector<int>> adj(n + 1);
  vector<int> dis(n + 1, INT_MAX);
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    if (u + 1 != v) {
      adj[u].push_back(v);
      remove.push_back({u, v});
    }
  }

  for (int u = 1; u < n; u++) {
    int v = u + 1;
    adj[u].push_back(v);
  }

  priority_queue <pair<int, int>, vector<pair<int, int>>, greater<>> pq;
  pq.push({1, 0});
  dis[1] = 0;
  while (!pq.empty()) {
    auto[node, dist] = pq.top();
    pq.pop();
    for (auto child : adj[node]) {
      if (dis[child] > dist + 1) {
        dis[child] = dist + 1;
        pq.push({child, dis[child]});
      }
    }
  }

  sort(remove.begin(), remove.end());
  vector<int> intervalRemove(n + 1);
  for (auto [l, r] : remove) {
    if (r - dis[r] > l) {
      intervalRemove[l + 1] += 1;
      intervalRemove[r - dis[r]] -= 1;
    }
  }

  for (int i = 1; i <= n; i++) {
    intervalRemove[i] += intervalRemove[i - 1];
  }

  for (int i = 1; i < n; i++) {
    cout << (intervalRemove[i] ? 0 : 1);
  }

}

int main() {

  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  int tc = 1;
  cin >> tc;
  while (tc--) {
    MonkeyDLuffy();
    if(tc) cout << '\n';
  }

}

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 3
2 4
5 3
1 3
2 4
3 5
6 4
1 3
2 4
3 5
4 6
```


**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:
// Shivam

#include<bits/stdc++.h>
using namespace std;
constexpr int mod = 1e9 + 7;

#define ff              first
#define ss              second
#define ull 			unsigned long long
#define int             long long
#define pb              push_back
#define mp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define vpii            vector<pair<int,int>>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
mt19937                 rng(chrono::steady_clock::now().time_since_epoch().count());

void c_p_c()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	w(T) {
		int n, m; cin >> n >> m;
		if (m == 0) {
			string s(n - 1, '1');
			cout << s << endl;
			continue;
		}

		// create adj list & reverse adj list
		// also store alternative bridge (extra edges)
		vpii edges;
		vi adj[n + 1], radj[n + 1];
		for (int i = 1; i < n; i++) {
			adj[i].pb(i + 1);
			radj[i + 1].pb(i);
		}
		for (int i = 1; i <= m; i++) {
			int u, v; cin >> u >> v;
			if (u > v) swap(u, v);
			adj[u].pb(v);
			radj[v].pb(u);
			edges.pb({u, v});
		}
		sort(edges.begin(), edges.end());

		vi ans(n, 1);

		// time to reach each node (for elsie)
		vi time(n + 1, inf);
		time[1] = 0;
		for (int i = 2; i <= n; i++) {
			for (auto &it : radj[i]) time[i] = min(time[i], time[it] + 1);
		}

		for (auto &it : edges) {
			int start = it.ff;
			int end = it.ss;

			int i = start + 1;
			while (time[end] < end - i) {
				ans[i] = 0;
				i++;
			}
		}

		for (int i = 1; i < n; i++) cout << ans[i];
		cout << endl;
	}
}

int32_t main()
{
	c_p_c();
	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 scenarios that might cause the code to fail. The primary issues to consider are:

1. **Alternative Bridges**: The code might not correctly handle alternative bridges, especially when they create complex paths that affect the race outcome.
2. **Optimal Strategies**: The code might not correctly simulate the optimal strategies for both cows, especially when Elsie can use alternative bridges to overtake Bessie.
3. **Edge Cases**: The code might fail on edge cases such as when there are no alternative bridges, or when the number of islands is very small or very large.

Given these considerations, let's create a small-scale test input that is complete and valid, and that maximizes the failure probability in the faulty solution.

### Test Input
```plaintext
2
6 1
2 6
10 4
1 3
1 6
2 7
3 8
```

### Explanation
1. **Test Case 1**:
   - **Number of Islands (n)**: 6
   - **Number of Alternative Bridges (m)**: 1
   - **Alternative Bridge**: 2 -> 6
   - **Expected Output**: 11011
   - **Reasoning**: Bessie will lose if she starts on island 3 because Elsie can use the alternative bridge from island 2 to island 6, reaching island 6 first.

2. **Test Case 2**:
   - **Number of Islands (n)**: 10
   - **Number of Alternative Bridges (m)**: 4
   - **Alternative Bridges**: 1 -> 3, 1 -> 6, 2 -> 7, 3 -> 8
   - **Expected Output**: 100001111
   - **Reasoning**: Bessie will lose if she starts on islands 3, 4, 5, and 6 because Elsie can use the alternative bridges to overtake Bessie and reach island 10 first.

This test input is designed to challenge the buggy code by including alternative bridges that create complex paths and affect the race outcome, especially when Bessie starts from certain islands.


**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:
# https://codeforces.com/contest/1998

import sys

input = lambda: sys.stdin.readline().rstrip()  # faster!

def solve_case():
    n, m = map(int, input().split())
    b = []
    for _ in range(m):
        u, v = map(lambda x: int(x) - 1, input().split())
        b += [(u, v)]
    b.sort()

    earliest = list(range(n))
    for u, v in b:
        earliest[v] = min(earliest[v], earliest[u] + 1)

    f = [0] * n
    for u, v in b:
        left = u + 1
        right = v - earliest[u] - 2
        if left <= right:
            f[left] += 1
            f[right + 1] -= 1

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

    ans = ["1"] * (n - 1)
    for i in range(n - 1):
        if f[i] > 0:
            ans[i] = "0"
    return "".join(ans)

ans = []
for _ in range(int(input())):
    ans += [str(solve_case())]
print("\n".join(ans))

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

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

Think step by step.

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


**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 heapq import *

# region fastio
input = lambda: sys.stdin.readline().rstrip()
sint = lambda: int(input())
mint = lambda: map(int, input().split())
ints = lambda: list(map(int, input().split()))
# endregion fastio

# 题做不出来，但过的人很多，请再读一遍题
# 没思路，且没读错题，不妨暴力找找规律

def solve() -> None:
    n, m = mint()
    g = [[] for _ in range(n)]
    for _ in range(m):
        u, v = mint()
        u -= 1
        v -= 1
        g[u].append(v)

    dp = list(range(n))
    for i in range(n):
        for v in g[i]:
            dp[v] = min(dp[v], dp[i] +  1)
    ans = [1] * (n - 1)
    h = []
    for i in range(n - 1):
        while h and h[0] <= i:
            heappop(h)
        if h:
            ans[i] = 0
        for v in g[i]:
            heappush(h, v - dp[v])

    print(*ans, sep='')

for _ in range(sint()):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 3
2 4
5 3
1 3
2 4
3 5
6 2
1 4
2 6
```


**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 static java.lang.Math.*;
import static java.lang.System.*;
import java.util.*;

public class Main {
    public static void main(String... args) {
        Scanner sc = new Scanner(System.in);

        int i, j, t = sc.nextInt();

        test:
        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();

            List<Integer>[] adj = new List[n+1];
            for (i=1; i<n; i++) {
                adj[i] = new ArrayList<>();
                adj[i].add(i+1);
            }
            adj[n] = new ArrayList<>();

            for (i=0; i<m; i++) {
                adj[sc.nextInt()].add(sc.nextInt());
            }

            int[] moves = new int[n+1];
            Queue<Integer> q = new LinkedList<>();

            Arrays.fill(moves, Integer.MAX_VALUE);
            moves[1] = 0;
            q.add(1);

            while (!q.isEmpty()) {
                int u = q.poll();

                for (int v : adj[u]) {
                    if (moves[u] + 1 < moves[v]) {
                        moves[v] = moves[u] + 1;
                        q.add(v);
                    }
                }
            }

            List<int[]> zero = new ArrayList<>();

            for (i=1; i<n; i++) {
                for (int v : adj[i]) {
                    if (i+1 < v && i+1 <= v-moves[v]-1)
                        zero.add(new int[]{i+1, v-moves[v]-1});
                }
            }
            zero.add(new int[]{n+2, n+2});

            int z = 0, bgn = 0, end = 0;
            StringBuilder sb = new StringBuilder();

            for (i=1; i<n; i++) {
                if (bgn < i && end < i) {
                    bgn = zero.get(z)[0];
                    end = zero.get(z)[1];
                    while (++z < zero.size() && zero.get(z)[0] <= end) {
                        end = max(end, zero.get(z)[1]);
                    }
                }

                if (i < bgn) {
                    sb.append(1);
                } else if (i <= end) {
                    sb.append(0);
                } else {
                    sb.append(1);
                }
            }

            out.println(sb);
        }

        sc.close();
    }
}

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

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

Think step by step.

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


**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 static java.lang.Math.*;
import static java.lang.System.*;
import java.util.*;

public class Main {
    public static void main(String... args) {
        Scanner sc = new Scanner(System.in);

        int i, j, t = sc.nextInt();

        test:
        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();

            List<Integer>[] adj = new List[n+1];
            for (i=1; i<n; i++) {
                adj[i] = new ArrayList<>();
                adj[i].add(i+1);
            }
            adj[n] = new ArrayList<>();

            for (i=0; i<m; i++) {
                adj[sc.nextInt()].add(sc.nextInt());
            }

            int[] moves = new int[n+1];
            Queue<Integer> q = new LinkedList<>();

            Arrays.fill(moves, Integer.MAX_VALUE);
            moves[1] = 0;
            q.add(1);

            while (!q.isEmpty()) {
                int u = q.poll();

                for (int v : adj[u]) {
                    if (moves[u] + 1 < moves[v]) {
                        moves[v] = moves[u] + 1;
                        q.add(v);
                    }
                }
            }

            int end = 1;
            StringBuilder sb = new StringBuilder();

            for (i=1; i<n; i++) {
                if (end-i > moves[end]) sb.append(0);
                else sb.append(1);

                for (int v : adj[i]) {
                    end = max(end, v);
                }
            }

            out.println(sb);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability in faulty solutions. Let's break down the requirements and think through the steps:

1. **Understand the Bug**: The provided code seems to be using a breadth-first search (BFS) to determine the minimum number of moves Elsie can make to reach each island. However, it does not correctly account for the alternating turns and the collapse of islands and bridges.

2. **Edge Cases**:
   - **Single Alternative Bridge**: This can cause Bessie to lose if Elsie can use it to overtake Bessie.
   - **No Alternative Bridges**: Bessie should always win if there are no alternative bridges.
   - **Multiple Alternative Bridges**: This can create complex scenarios where Elsie can strategically use multiple bridges to overtake Bessie.

3. **Small-Scale Test Input**:
   - We will create a test case with a small number of islands and bridges to ensure it fits within the output limits.
   - We will include a scenario where Elsie can use an alternative bridge to overtake Bessie.

Let's create a test case with 6 islands and 1 alternative bridge:

- **Islands**: 6
- **Alternative Bridge**: From island 2 to island 6

This setup ensures that Bessie will lose if she starts on island 3, as Elsie can use the alternative bridge to reach island 6 before Bessie.

Here is the test input:

```plaintext
1
6 1
2 6
```

This test case will expose the bug in the provided code because:
- Bessie will win if she starts on islands 1, 2, 4, 5.
- Bessie will lose if she starts on island 3.

The expected output for this test case is:

```plaintext
11011
```


**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:
from sys import *
input = stdin.readline
inputlist = lambda: map(int, stdin.readline().split())

# 2 -> 8, 4 -> 9, 8 -> 15
# for (starting points) >= s, min(i - d[i]) >= s  --->  0
# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
# 0 1 2 3 4 5 6 2 3 4  5  6  7  8  3
# 1 1 1 1 1 1 1 6 6 6  6  6  6  6  12
# 1 1 0 0 0 1 1 1 0 0  0  1  1  1

# 1 -> 5
# 1 2 3 4 5 6
# 0 1 2 3 2 3
# 1 1 1 1 3 3
# 1 0 0 1 1

def solve():
    n, m = inputlist()
    graph = {}
    for _ in range(m):
        u, v = inputlist()
        if u in graph: graph[u].append(v)
        else: graph[u] = [v]

    shortest = [n+1] * (n+1)
    shortest[1] = 0
    for i in range(1, len(shortest)):
        shortest[i] = min(shortest[i], 1 + shortest[i-1])
        if i in graph:
            for child in graph[i]:
                shortest[child] = min(shortest[child], 1 + shortest[i])
    diff = [i - shortest[i] for i in range(len(shortest))]

    ans  = [1] * (n)
    currentmax = 0
    for i in range(1, n):
        currentmax = max(currentmax, i)
        if diff[currentmax] > i: ans[i] = 0
        if i in graph: currentmax = max(currentmax, max(graph[i]))
    return "".join(str(num) for num in ans[1:])

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

This test input is designed to expose potential issues in the given buggy code by including various edge cases and scenarios that can maximize the failure probability in faulty solutions. Here’s a breakdown of the test cases:

1. **Case 1: No alternative bridges (6 0)**
   - Bessie can always win because Elsie has no alternative bridges to use, and Bessie always moves first.
   - Expected output: `11111`

2. **Case 2: One alternative bridge (6 1)**
   - Bessie can win from most starting points, but she will lose if she starts on island 3 because Elsie can use the alternative bridge to island 6.
   - Expected output: `11011`

3. **Case 3: One alternative bridge (6 1)**
   - Similar to Case 2, but with a different bridge.
   - Expected output: `10011`

4. **Case 4: Multiple alternative bridges (10 4)**
   - Bessie can win from some starting points, but Elsie has more options to use alternative bridges, which can cause Bessie to lose from certain starting points.
   - Expected output: `100001111`

5. **Case 5: Multiple alternative bridges (15 3)**
   - Bessie can win from some starting points, but Elsie has more distant alternative bridges, which can cause Bessie to lose from certain starting points.
   - Expected output: `11000111000111`

These test cases cover a range of scenarios, including no alternative bridges, a single alternative bridge, and multiple alternative bridges, to ensure that the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 heapq import *

# region fastio
input = lambda: sys.stdin.readline().rstrip()
sint = lambda: int(input())
mint = lambda: map(int, input().split())
ints = lambda: list(map(int, input().split()))
# endregion fastio

# 题做不出来，但过的人很多，请再读一遍题
# 没思路，且没读错题，不妨暴力找找规律

def solve() -> None:
    n, m = mint()
    g = [[] for _ in range(n)]
    for _ in range(m):
        u, v = mint()
        u -= 1
        v -= 1
        g[u].append(v)

    dp = list(range(n))
    for i in range(n):
        for v in g[i]:
            dp[v] = min(dp[v], dp[i] +  1)
    ans = [1] * (n - 1)
    h = []
    for i in range(n - 1):
        while h and h[0] <= i:
            heappop(h)
        if h:
            ans[i] = 0
        for v in g[i]:
            heappush(h, v - dp[v])

    print(*ans, sep='')

for _ in range(sint()):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
3 1
1 3
5 2
1 5
2 4
7 3
1 7
2 5
3 6
```


**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 java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

public class winningislands {
	static ArrayList<Integer>[] adj;

	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(in.readLine());
		for (int tc = 0; tc < t; tc++) {
			int[] buf = Arrays.stream(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
			int n = buf[0];
			int m = buf[1];
			adj = new ArrayList[n];
			for (int i = 0; i < n; i++)
				adj[i] = new ArrayList<>();
			for (int i = 0; i < m; i++) {
				buf = Arrays.stream(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
				adj[buf[0]-1].add(buf[1]-1);
				adj[buf[1]-1].add(buf[0]-1);
			}

			Queue<int[]> bfs = new LinkedList<>();

			int[] dist = new int[n];
			boolean[] visited = new boolean[n];

			bfs.add(new int[] {0, 0});

			while (!bfs.isEmpty()) {
				int[] a = bfs.poll();
				int node = a[0]; int turns = a[1];

				dist[node] = turns;
				visited[node] = true;

				for (int next: adj[node]) {
					if (!visited[next] && next > node) bfs.add(new int[] {next, turns+1});
				}
				if (node < n-1 && !visited[node+1]) bfs.add(new int[] {node+1, turns+1});
			}

			int[] pfx = new int[n];
			for (int i = 1; i < n; i++) {
				for (int b: adj[i]) {
					if (b >= i) continue;
					int steps = dist[b]+1;

					pfx[b+1] += 1;
					pfx[Math.max(i-steps, b+1)]--;
				}
			}

			int ans = 0;
			StringBuilder res = new StringBuilder();
			res.append("1");
			for (int i = 1; i < n-1; i++) {
				ans += pfx[i];
				if (ans > 0) res.append("0");
				else res.append("1");
			}

			System.out.println(res.toString());
		}
		in.close();
	}
}

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

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

Think step by step.

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


**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;
#define int long long
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pos string::npos
#define endl '\n'
#define inf 0x3f3f3f3f
typedef unsigned long long ull;
const int mod = 1e9+7;
const int N = 5e5+50;
int Inv[N],F[N];//组合数C2
int gcd(int a,int b){ return b == 0 ? a : gcd(b,a%b);}
int lcm(int a,int b){return (a/gcd(a,b)*b);}
int km(int a,int b,int mod)
{int ans=1;while(b){if(b&1) ans=ans*a%mod;b>>=1;a=a*a%mod;}return ans;}
int km1(int a,int b,int mod)//指数取模,2^2^i,单求2^i要取模MOD-1
{int ans=1;mod--;while(b){if(b&1) ans=ans*a%mod;b>>=1;a=a*a%mod;}return ans;}
int C(int a,int b,int p){if(b>a) return 0;int res=1;for(int i=b,j=a;i>=1;i--,j--){res=res*j%p;res=res*km(i,p-2,p)%p;}return res;}
int Lucas(int a,int b,int p){if(a<p&&b<p) return C(a,b,p);return C(a%p,b%p,p)*Lucas(a/p,b/p,p)%p;}
int C2(int a,int b,int p){return F[a]*Inv[a-b]%p*Inv[b]%p;}
void C2init(int n,int p){n-=5;F[0]=F[1]=1;for(int i=2;i<=n;i++)F[i]=F[i-1]*i%p;Inv[n]=km(F[n],p-2,p);for(int i=n-1;i>=0;i--)Inv[i]=Inv[i+1]*(i+1)%p;}
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
typedef pair<int,int> pii;
int n,k,m,d[N],ans[N];
vector<int> h[N];
vector<int> g[N];
vector<int> r[N];
bool vis[N];
// void bfs(int s)
// {
//     queue<int> q;
//     q.push(s);
//     while(q.size())
//     {
//         auto u=q.front();
//         for(auto v:r[u]){

//         }
//     }
// }
void solve()
{
    cin>>n>>m;
    for(int i=0;i<=n+10;i++){
        h[i].clear();g[i].clear();r[i].clear();
        d[i]=i,ans[i]=0;
    }
    for(int i=1;i<n;i++) r[i].push_back(i+1);
    for(int i=1;i<=m;i++){
        int u,v;
        cin>>u>>v;
        h[u].push_back(v);
        g[v].push_back(u);
    }
    for(int u=1;u<=n;u++){
        sort(g[u].begin(),g[u].end());
        for(auto v:g[u]){
            d[u]=min(d[u],d[v]+1);
        }
    }
    for(int i=1;i<n;i++){
        d[i+1]=min(d[i+1],d[i]+1);
    }
    //bfs(1);
    for(int u=1;u<=n;u++){
        for(auto v:h[u]){
            int time=d[u];
            int l=u+1,r=v-time-1;
            if(r<l) continue;
            // if(l>r) swap(l,r);
            // if(l<=0) l=1;
            ans[l]--,ans[r+1]++;
        }
    }
    for(int i=1;i<=n-1;i++){
        ans[i]+=ans[i-1];
        if(i==1||i==n-1){
            cout<<1;
            continue;
        }
        if(ans[i]>=0) cout<<1;
        else cout<<0;
    }
    cout<<endl;
}
signed main()
{
    ios;
    int T=1;
    cin>>T;
    //C2init(N,mod);
    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
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
```


**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;
using ll=long long ;
using pii=pair<int,int>;
const int MAX=200005;

int head[MAX];
int to[MAX<<2];
int net[MAX<<2];
int cnt=1;
int n;
int m;
int vis[MAX];
int nums[MAX];
int S[MAX];

inline void add(int u,int v){
    net[cnt]=head[u];
    to[cnt]=v;
    head[u]=cnt++;
}

void solve(){

    for(int i=1; i<=n; i++) head[i]=vis[i]=S[i]=0;
    cnt=1;
    cin>>n>>m;
    for(int i=1; i<=n-1; i++){
        add(i,i+1);
    }
    for(int i=2 ; i<=n; i++){
        nums[i]=0x3f3f3f3f;
    }
    int u,v;
    for(int i=0; i<m ; i++){
        cin>>u>>v;
        if(u>v) swap(u,v);
        add(u,v);
    }
    nums[1]=1;
    priority_queue<pii> pq;
    pq.push(make_pair(1,1));
    int d;
    while(!pq.empty()){
        tie(d,u)=pq.top();
        pq.pop();
        if(vis[u]) continue;
        vis[u]=true;
        for(int ei=head[u] ;ei!=0; ei=net[ei]){
            int &v=to[ei];
            if(!vis[v]&&nums[v]>d+1){
                pq.emplace(make_pair(d+1,v));
                nums[v]=d+1;
            }
        }
    }
    for(int i=1 ; i<=n ;i++){
        for(int ei=head[i];ei!=0; ei=net[ei]){
            if(i>to[ei]-nums[i]-1) continue;
            S[i+1]++;
            S[to[ei]-nums[i]]--;
        }
    }
    for(int i=1 ;i<n; i++){
        S[i]+=S[i-1];
        cout<<(S[i]==0);
        if(i==n-1) cout<<'\n';
    }

}

signed main(){
    cin.tie(nullptr)->sync_with_stdio(0);
    cout.tie(nullptr)->sync_with_stdio(0);
    int t;
    cin>>t;
    while(t--) solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with correctly handling the dynamics of the race, especially when Elsie can use alternative bridges to overtake Bessie.

### Step-by-Step Analysis:
1. **Initial Setup:**
   - Bessie starts on island `s` and can only use main bridges.
   - Elsie starts on island `1` and can use both main and alternative bridges.
   - The goal is to determine if Bessie can win if she starts on island `s`.

2. **Key Issues in the Buggy Code:**
   - The code might not correctly handle the order of turns and the collapse of islands.
   - The priority queue used for the shortest path might not accurately reflect the optimal strategies of the cows.

3. **Challenging Scenarios:**
   - **Scenario 1:** Elsie can use an alternative bridge to reach the final island before Bessie.
   - **Scenario 2:** Elsie can use an alternative bridge to reach an intermediate island, which collapses before Bessie can use it.
   - **Scenario 3:** Bessie and Elsie can be on the same island, and Elsie can use an alternative bridge to win.

### Test Input Generation:
Let's create a small-scale test case that includes these challenging scenarios.

```plaintext
1
6 2
1 4
3 6
```

### Explanation:
- **Number of Islands (n):** 6
- **Number of Alternative Bridges (m):** 2
- **Alternative Bridges:**
  - From island 1 to island 4
  - From island 3 to island 6

### Expected Output:
- **Bessie's Starting Island (s):** 1 to 5
- **Output:**
  - For `s = 1`: Bessie can win because she starts on island 1 and can use the main bridges to reach island 6.
  - For `s = 2`: Bessie can win because Elsie cannot use the alternative bridge from 1 to 4 to overtake Bessie.
  - For `s = 3`: Bessie can win because Elsie cannot use the alternative bridge from 3 to 6 to overtake Bessie.
  - For `s = 4`: Bessie cannot win because Elsie can use the alternative bridge from 1 to 4 to reach island 4 before Bessie.
  - For `s = 5`: Bessie can win because Elsie cannot use the alternative bridge from 3 to 6 to overtake Bessie.

### Final Test Input:
```plaintext
1
6 2
1 4
3 6
```

This test input should expose the bug in the given code by forcing it to handle the alternative bridges correctly and ensuring that the order of turns and the collapse of islands are accurately simulated.


**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 java.util.*;

public class DeterminingWinningIslandInRace {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        StringBuilder fin_ans = new StringBuilder();
        int t = in.nextInt();
        while (t-- > 0){
            int n = in.nextInt(), m = in.nextInt();
            int[] dist = new int[n+1];
            Arrays.fill(dist,Integer.MAX_VALUE);
            List<List<Integer>> adj = new ArrayList<>();
            for (int i = 0; i < n; i++) adj.add(new ArrayList<>());
            for (int i = 0; i < n-1; i++) adj.get(i).add(i+1);
            for (int i = 0; i < m; i++){
                int u = in.nextInt()-1, v = in.nextInt()-1;
                adj.get(u).add(v);
            }
            BFS(adj,dist);
            int[] arr = new int[n+1];
            for (int i = 0; i < n-1; i++){
                int max = -1;
                for (Integer neigh : adj.get(i)) max = Math.max(max,neigh);
                int first = i+1, last = max - (dist[i]+1) - 1;
                if (last < first) continue;
                arr[first]--;
                arr[last+1]++;
            }
            StringBuilder some = new StringBuilder();
            int sum = 0;
            for (int i = 0; i < n-1; i++){
                sum += arr[i];
                if (sum < 0) some.append(0);
                else some.append(1);
            }
            fin_ans.append(some).append("\n");
        }
        System.out.print(fin_ans);
    }

    private static void BFS(List<List<Integer>> adj, int[] dist){
        Queue<Integer> que = new LinkedList<>();
        que.add(0);
        int curr = 0;
        while (!que.isEmpty()){
            int size =  que.size();
            while (size-- > 0){
                int node = que.poll();
                dist[node] = Math.min(dist[node],curr);
                que.addAll(adj.get(node));
            }
            curr++;
        }
    }
}

**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
6 2
1 4
4 5
10 3
1 3
3 5
5 7
```

### Explanation:
1. **Test Case 1:**
   - **Islands (n):** 6
   - **Alternative Bridges (m):** 2
   - **Bridges:** (1, 3) and (3, 5)
   - **Expected Output:** "11011"
   - **Reason:** Bessie can win from islands 1, 2, 4, and 5, but not from island 3. Elsie can use the alternative bridge from 1 to 3 and then from 3 to 5, outpacing Bessie.

2. **Test Case 2:**
   - **Islands (n):** 6
   - **Alternative Bridges (m):** 2
   - **Bridges:** (1, 4) and (4, 5)
   - **Expected Output:** "11011"
   - **Reason:** Similar to the first test case, but with a different set of bridges. Bessie can win from islands 1, 2, 4, and 5, but not from island 3. Elsie can use the alternative bridge from 1 to 4 and then from 4 to 5, outpacing Bessie.

3. **Test Case 3:**
   - **Islands (n):** 10
   - **Alternative Bridges (m):** 3
   - **Bridges:** (1, 3), (3, 5), and (5, 7)
   - **Expected Output:** "110001111"
   - **Reason:** Bessie can win from islands 1, 2, 6, 7, 8, and 9, but not from islands 3, 4, and 5. Elsie can use the alternative bridges to outpace Bessie by moving from 1 to 3, then from 3 to 5, and finally from 5 to 7.

These test cases are designed to expose the bug in the BFS function and the logic for determining the winning islands. The alternative bridges are placed to create scenarios where Elsie can outpace Bessie, leading to a different outcome based on the starting island.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

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

class Node{
  public:
    int data;
    Node* left;
    Node* right;

    Node(int d){
        data = d;
        left = NULL;
        right = NULL;
    }
};

int getHeight(Node* root){
    if(root == NULL){
        return 0;
    }

    int left = getHeight(root->left);
    int right = getHeight(root->right);

    return (1 + max(left, right));
}

bool buildFromLevelOrder(Node* &root, int a, int b, int c){
    queue<Node*> q ;
    root = NULL ;
    if(a){
        root = new Node(2) ;
        a--;
    }
    else if(b){
        root = new Node(1) ;
        b--;
    }
    else{
        root = new Node(0) ;
        c--;
    }
    q.push(root) ;

    while(!q.empty()){
        Node* temp = q.front() ;
        q.pop() ;

        if(temp->data == 2){
            if(a){
                temp->left = new Node(2) ;
                a--;
                q.push(temp->left) ;
            }
            else if(b){
                temp->left = new Node(1) ;
                b--;
                q.push(temp->left) ;
            }
            else if(c){
                temp->left = new Node(0) ;
                c--;
            }
            else return false;

            if(a){
                temp->right = new Node(2) ;
                a--;
                q.push(temp->right) ;
            }
            else if(b){
                temp->right = new Node(1) ;
                b--;
                q.push(temp->right) ;
            }
            else if(c){
                temp->right = new Node(0) ;
                c--;
            }
            else return false;
        }
        else if(temp->data == 1){
            if(b){
                temp->left = new Node(1) ;
                b--;
                q.push(temp->left) ;
            }
            else if(c){
                temp->left = new Node(0) ;
                c--;
                if(q.empty() && c) return false;
            }
            else return false;
        }
    }

    return true;
}

int main(){
    int a,b,c;
    int t;
    cin>>t;
    for(int i=0 ; i<t ; i++){
        cin >> a >> b >> c;
        int minHeight;
        Node* root = NULL;
        if(buildFromLevelOrder(root, a, b, c)){
            minHeight = getHeight(root);
        }
        else{
            minHeight = 0;
        }
        cout << minHeight-1 << endl;
    }
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose a potential bug in the code where the tree has only one vertex with 0 children. The expected output is `0` because a tree with a single vertex and no children has a height of 0.

### Why This Test Case is Challenging:
- **Edge Case:** This is an edge case where the tree has the minimum number of vertices (1) and no children. The code needs to handle this case correctly.
- **Initialization:** The `buildFromLevelOrder` function initializes the root node and then immediately returns `true` because there are no more nodes to add. The height of the tree should be 0, but the code incorrectly subtracts 1 from the height, leading to an incorrect output of `-1`.

This test case will help identify the issue with the height calculation and the logic in the `buildFromLevelOrder` function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
#include <algorithm>
#include <iostream>
#include <string>
#include <utility>
#include <vector>
#include <cmath>
using namespace std;

class Solver {
public:
	void solve() {
		int a, b, c;
		cin >> a >> b >> c;

		if (a != c - 1) {
			cout << -1 << endl;
			return;
		}

		int res = ceil(log2(c));
		// cout << res << endl;

		int d = 0;
		for (int i = 1; i < c; i *= 2) {
			d += ((c - 1) & i) != 0;
		}

		b -= (res - d);
		while (b > 0) {
			b -= c;
			res++;
		}

		cout << res << endl;
	}

public:
};

int main() {
	ios::sync_with_stdio(false);
  	cin.tie(0);

	int t;
	cin >> t;

	while (t--) {
		Solver().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 given buggy code, let's analyze the conditions under which the code might fail:

1. **Incorrect Condition for Valid Tree:**
   - The code checks if `a != c - 1` and outputs `-1` if this condition is not met. However, this condition is incorrect. A valid tree can exist even if `a != c - 1`.

2. **Height Calculation:**
   - The code uses a logarithmic function to calculate the height, but it does not correctly account for the number of vertices with 1 child (`b`).

3. **Adjustment for `b` Vertices:**
   - The code attempts to adjust the height based on the number of `b` vertices, but it does so in a way that might not always be correct.

Given these points, we can construct a test case that satisfies the conditions but fails the buggy code's checks.

### Test Case:
- **Input:**
  - `a = 2`
  - `b = 2`
  - `c = 3`

### Explanation:
- **Vertices:**
  - 2 vertices with 2 children.
  - 2 vertices with 1 child.
  - 3 vertices with 0 children.

- **Tree Structure:**
  - A possible structure for this tree is:
    - Root (2 children)
      - Child 1 (2 children)
        - Child 1.1 (0 children)
        - Child 1.2 (0 children)
      - Child 2 (1 child)
        - Child 2.1 (0 children)
    - An additional vertex with 1 child:
      - Child (0 children)

- **Height:**
  - The height of this tree is 2.

### Expected Output:
- The correct output should be `2`.

### Buggy Code Output:
- The buggy code will incorrectly output `-1` because it checks `a != c - 1` and fails this condition.

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

This test input should expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#define endl '\n'
using namespace std;
#define ll unsigned long long
#define mod 1000000007

void solve() {
   int a,b,c;
        cin >> a >> b >> c;
        vector<int>vis(4*1e5);
        string s;
        int ndx = 1;
        int size = a + b;
        int ans = 0;
        vis[0] = 1;
        if (c != a + 1) {
            cout << -1 << '\n';
            return;
        }
        for(int i = 0; i < size;i++)
        {
            if(a)
            {
                if(size - ndx + c < 2)
                {
                    ndx = 0;
                    break;
                }
                else
                {
                    vis[ndx] = vis[i] + 1;
                    vis[ndx+1] = vis[i] + 1;
                    ndx+=2;
                }
                a--;
            }
            else if(b)
            {
                if(size - ndx + c < 1)
                {
                    cout << -1 << endl;
                }
                else
                {
                    vis[ndx] = vis[i] + 1;
                    ndx++;

                }
                b--;
            }
        }
        cout << vis[size + c - 1] - 1 << endl;

}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The primary issues in the code include:

1. Incorrect handling of the condition where the number of leaves (vertices with 0 children) does not equal the number of vertices with 2 children plus one.
2. Incorrect calculation of the height of the tree.

Let's construct a test input that will expose these issues:

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

### Explanation
1. **Test Case 1: 2 1 3**
   - This is a valid case where the tree can be formed. The expected output is `2`.
   - The code should correctly handle this case, but we will check if it does.

2. **Test Case 2: 1 0 1**
   - This is a valid case where the tree can be formed. The expected output is `1`.
   - The code should correctly handle this case, but we will check if it does.

3. **Test Case 3: 0 2 2**
   - This is an invalid case because the number of leaves (2) does not equal the number of vertices with 2 children plus one (0 + 1 = 1).
   - The code should output `-1`, but we will check if it does.

### Expected Output
```plaintext
2
1
-1
```

### Why This Test Input is Challenging
1. **Test Case 1: 2 1 3**
   - This case is straightforward and should be handled correctly by the code. However, it serves as a baseline to ensure the code is working for a simple valid case.

2. **Test Case 2: 1 0 1**
   - This case is also straightforward and should be handled correctly by the code. It tests the scenario where there is only one vertex with 2 children and one leaf.

3. **Test Case 3: 0 2 2**
   - This case is invalid and should be caught by the code. It tests the scenario where the number of leaves does not match the expected number (0 + 1 = 1).

By using these test cases, we can ensure that the code correctly handles both valid and invalid scenarios, and we can identify any issues in the logic or calculations.