**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

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

#define assert(expr) {if(!(expr)){while(1);}}

constexpr int A = 33;
constexpr int B = 126;
constexpr int NONE = -123;

struct Node {
    priority_queue<int> id;

    int dp_len;
    int dp_id;

    int parent;
    int children[B - A + 1 + (10)];
};

vector<Node> dictionary;
vector<int> id_where(2'234'567);

void initialize() {
    dictionary.clear();
    Node eps;
    eps.dp_len = NONE;
    eps.parent = NONE;
    for(int i = A; i <= B; ++i)
        eps.children[i - A] = NONE;
    dictionary.push_back(eps);
    fill(id_where.begin(), id_where.end(), NONE);
}

int epsilon() {
    return 0;
}

int go(int u, string word) {
    for(char c : word) {
        assert(A <= c && c <= B);
        assert(0 <= u && u < (int) dictionary.size());
        if(dictionary[u].children[c - A] == NONE) {
            int v = (int) dictionary.size();
            Node sth;
            sth.dp_len = NONE;
            sth.parent = u;
            for(int i = A; i <= B; ++i)
                sth.children[i - A] = NONE;

            dictionary[u].children[c - A] = v;
            dictionary.push_back(sth);
            assert(v < (int) dictionary.size());
            u = v;
        } else u = dictionary[u].children[c - A];
    }
    return u;
}

void repair(int u) {
    while(true) {
        assert(0 <= u && u < (int) dictionary.size());
        dictionary[u].dp_len = NONE;
        dictionary[u].dp_id = NONE;
        while(dictionary[u].id.size() > 0 && NONE == id_where[-dictionary[u].id.top()])
            dictionary[u].id.pop();
        if(dictionary[u].id.size() > 0) {
            dictionary[u].dp_len = 0;
            dictionary[u].dp_id = -dictionary[u].id.top();
        }
        for(int i = A; i <= B; ++i) {
            if(dictionary[u].children[i - A] == NONE) continue;
            if(dictionary[dictionary[u].children[i - A]].dp_len == NONE) continue;
            if(dictionary[u].dp_len == NONE || dictionary[dictionary[u].children[i - A]].dp_len + 1 > dictionary[u].dp_len) {
                dictionary[u].dp_len = dictionary[dictionary[u].children[i - A]].dp_len + 1;
                dictionary[u].dp_id = dictionary[dictionary[u].children[i - A]].dp_id;
            }
        }
        //assert(dictionary[u].dp_len != NONE);
        if(u == epsilon()) break;
        u = dictionary[u].parent;
    }
}

void add(int id, string pattern) {

    int u = epsilon();

    u = go(u, pattern);

    assert(0 <= id && id < (int) id_where.size());
    id_where[id] = u;
    dictionary[u].id.push(-id);

    repair(u);
}

void Delete(int id) {
    assert(0 <= id && id < (int) id_where.size());
    int u = id_where[id];
    assert(0 <= u && u < (int) dictionary.size());
    id_where[id] = NONE;

    repair(u);
}

/*int append(int u, string str) {
    u = go(u, str);

    repair(u);

    return u;
}

int backspace(int u, int cnt) {
    while(u != epsilon() && cnt > 0) {
        u = dictionary[u].parent;
        --cnt;
    }
    return u;
}

int query(int u) {
    if(dictionary[u].dp_len == NONE) return -1;
    else return dictionary[u].dp_id;
}*/

void append_1(int& u, deque<char>& q, string str) {
    for(char c : str)
        q.push_back(c);
}

void backspace_1(int& u, deque<char>& q, int cnt) {
    while(cnt > 0) {
        if(q.size() > 0) {
            q.pop_back();
            --cnt;
        } else if(u != epsilon()) {
            assert(0 <= u && u < (int) dictionary.size());
            u = dictionary[u].parent;
            --cnt;
        } else break;
    }
}

int query_1(int& u, deque<char>& q) {
    while(q.size() > 0) {
        char c = q.front();
        assert(A <= c && c <= B);
        assert(0 <= u && u < (int) dictionary.size());
        if(dictionary[u].children[c - A] == NONE)
            return -1;
        u = dictionary[u].children[c - A];
        q.pop_front();
    }
    assert(0 <= u && u < (int) dictionary.size());
    if(dictionary[u].dp_len == NONE) return -1;
    return dictionary[u].dp_id;
}

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

    int n;
    cin >> n;

    initialize();
    int u = epsilon();
    deque<char> qu;

    while(n--) {
        string what;
        cin >> what;
        int id;
        string pattern;
        string str;
        int cnt;
        switch(what[1]) {
            case 'd': /*add*/       cin >> id >> pattern; add(id, pattern); break;
            case 'e': /*delete*/    cin >> id; Delete(id);                  break;
            case 'p': /*append*/    cin >> str; append_1(u, qu, str);       break;
            case 'a': /*backspace*/ cin >> cnt; backspace_1(u, qu, cnt);    break;
        }

        cout << query_1(u, qu) << "\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
10
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
delete 1
delete 2
delete 3
append pattern
```

10
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
delete 1
delete 2
delete 3
append pattern


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

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

using namespace std;

#define all(x) x.begin(), x.end()
#define ar array
#define pb push_back
#define ln '\n'
#define size(x) (int)(x).size()
//~ #define int long long

using i64 = long long;

template <class F, class _S>
bool chmin(F &u, const _S &v){
    bool flag = false;
    if ( u > v ){
        u = v; flag |= true;
    }
    return flag;
}

template <class F, class _S>
bool chmax(F &u, const _S &v){
    bool flag = false;
    if ( u < v ){
        u = v; flag |= true;
    }
    return flag;
}

using arr = ar <int,94>;

const int inf = 1e9;

const int N = 2e6 + 1007;

int t[N][94], fa[N];

ar <int,2> mx[N];

vector <set<int>> idx(N);

int ct;

void ins(string &s, int &id, int rt, int j, bool del){
	if ( j == size(s) ){
		if ( del ) idx[rt].erase(id);
		else idx[rt].insert(id);

		if ( size(idx[rt]) ){
			mx[rt] = {0, *idx[rt].begin()};
		} else mx[rt] = {-inf, -1};

		return;
	}

	int u = s[j] - 33;

	if ( !t[rt][u] ){
		t[rt][u] = ++ct;
		fa[ct] = rt;
	}

	ins(s, id, t[rt][u], j + 1, del);

	if ( !idx[rt].empty() ){
		mx[rt] = {0, *idx[rt].begin()};
	} else mx[rt] = {-inf, -1};

	for ( int i = 0; i < 94; i++ ){
		if ( t[rt][i] != 0 ){
			int v = t[rt][i];

			if ( mx[v][0] + 1 > mx[rt][0] ){
				mx[rt] = {mx[v][0] + 1, mx[v][1]};
			}
		}
	}
}

void insert(string &s, int &id){ ins(s, id, 0, 0, 0); }

void erase(string &s, int &id){ ins(s, id, 0, 0, 1); }

void move(string &s, int &k){
	for ( auto &u: s ){
		int v = u - 33;

		if ( !t[k][v] ){
			t[k][v] = ++ct;
			fa[ct] = k;
		} k = t[k][v];
	}
}

int qry(int &k){ return mx[k][1]; }

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

    for ( int i = 0; i < N; i++ ) mx[i] = {-inf, -1};

    int n; cin >> n;

    vector <string> p(n + 1);

    int k = 0;

    for ( int i = 0; i < n; i++ ){
		string op; cin >> op;

		if ( op == "add" ){
			int x; string s;
			cin >> x >> s;

			insert(s, x);
			p[x] = s;
		} else if ( op == "delete" ){
			int x; cin >> x;

			erase(p[x], x);
		} else if ( op == "append" ){
			string s; cin >> s;

			move(s, k);
		} else{
			int c; cin >> c;

			while ( c-- ){
				if ( !k ) break;

				k = fa[k];
			}
		}

		cout << qry(k) << ln;
	}

 	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
10
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append a
append b
backspace 1
append c
backspace 2
append d
delete 2
```

10
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append a
append b
backspace 1
append c
backspace 2
append d
delete 2


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
def main():
    import sys
    input = sys.stdin.read
    data = input().splitlines()

    n = int(data[0])
    patterns = {}  # id -> pattern
    current_text = ""
    output = []

    for line in data[1:n+1]:
        parts = line.split()
        action = parts[0]

        if action == 'add':
            i = int(parts[1])
            p_i = parts[2]
            patterns[i] = p_i
        elif action == 'delete':
            i = int(parts[1])
            del patterns[i]
        elif action == 'append':
            s = parts[1]
            current_text = current_text + s
        elif action == 'backspace':
            c = int(parts[1])
            current_text = current_text[:-min(c, len(current_text))]

        candidates = []
        for k, v in patterns.items():
            if v.startswith(current_text):
                candidates.append((v, k))

        if not candidates:
            output.append(-1)
            continue

        candidates.sort(key=lambda x: (-len(x[0]), x[0], x[1]))
        output.append(candidates[0][1])

    for num in output:
        print(num)

if __name__ == "__main__":
    main()

# tGIamVr0UsBqAEbj6uy

**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
10
add 1 a
add 2 aa
add 3 aaa
add 4 aaaa
add 5 aaaaa
append a
append a
append a
append a
backspace 3
```

10
add 1 a
add 2 aa
add 3 aaa
add 4 aaaa
add 5 aaaaa
append a
append a
append a
append a
backspace 3


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:

**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
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
backspace 5
append a
```

### Explanation:
1. **Add 1 pattern1_alice**: The current text is empty, so the suggestion is "pattern1_alice".
2. **Add 2 pattern2_bob**: The current text is still empty, so the suggestion is "pattern1_alice" (it's the longest).
3. **Add 3 pattern3_charlie**: The current text is still empty, so the suggestion is "pattern3_charlie" (it's the longest).
4. **Backspace 5**: The current text is empty, so the suggestion is "pattern3_charlie" (it's the longest).
5. **Append a**: The current text is "a", so the suggestion is "pattern1_alice" (it's the only one that starts with "a").

This test input checks the behavior of the auto-complete feature when the text is cleared and then a new character is appended. It ensures that the auto-complete logic correctly handles the backspace operation and the subsequent append operation.

5
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
backspace 5
append a


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
import java.util.*;

public class Main{
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        scanner.nextLine(); // Consume the newline after the integer

        StringBuilder t = new StringBuilder();
        Map<Integer, String> patterns = new HashMap<>();

        for (int i = 0; i < n; i++) {
            String[] parts = scanner.nextLine().split(" ");
            String operation = parts[0];

            if (operation.equals("add")) {
                int id = Integer.parseInt(parts[1]);
                String pattern = parts[2];
                patterns.put(id, pattern);
            } else if (operation.equals("delete")) {
                int id = Integer.parseInt(parts[1]);
                patterns.remove(id);
            } else if (operation.equals("append")) {
                String s = parts[1];
                t.append(s);
            } else if (operation.equals("backspace")) {
                int c = Integer.parseInt(parts[1]);
                if (c >= t.length()) {
                    t.setLength(0);
                } else {
                    t.setLength(t.length() - c);
                }
            }

            // Find the best matching pattern
            int bestId = -1;
            String bestPattern = "";
            for (Map.Entry<Integer, String> entry : patterns.entrySet()) {
                String p = entry.getValue();
                if (t.length() <= p.length() && p.startsWith(t.toString())) {
                    if (bestId == -1 || p.length() > bestPattern.length() ||
                        (p.length() == bestPattern.length() && p.compareTo(bestPattern) < 0) ||
                        (p.length() == bestPattern.length() && p.compareTo(bestPattern) == 0 && entry.getKey() < bestId)) {
                        bestId = entry.getKey();
                        bestPattern = p;
                    }
                }
            }
            System.out.println(bestId);
        }
        scanner.close();
    }
}

// cMqIBbzmgAiy4VuJ0Oa

**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
6
add 1 a
add 2 ab
add 3 abc
append a
append b
append c
```

6
add 1 a
add 2 ab
add 3 abc
append a
append b
append c


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

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

public class Main{
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        int n = Integer.parseInt(br.readLine());
        Map<Integer, String> patterns = new HashMap<>();
        StringBuilder currentText = new StringBuilder();

        for (int i = 0; i < n; i++) {
            String[] input = br.readLine().split(" ");
            if (input[0].equals("add")) {
                int id = Integer.parseInt(input[1]);
                String pattern = input[2];
                patterns.put(id, pattern);
            } else if (input[0].equals("delete")) {
                int id = Integer.parseInt(input[1]);
                patterns.remove(id);
            } else if (input[0].equals("append")) {
                currentText.append(input[1]);
            } else if (input[0].equals("backspace")) {
                int c = Integer.parseInt(input[1]);
                int len = currentText.length();
                if (c >= len) {
                    currentText.setLength(0);
                } else {
                    currentText.setLength(len - c);
                }
            }

            // Suggest the pattern
            String t = currentText.toString();
            String bestPattern = null;
            int bestId = -1;

            for (Map.Entry<Integer, String> entry : patterns.entrySet()) {
                int id = entry.getKey();
                String pattern = entry.getValue();
                if (pattern.startsWith(t) && (bestPattern == null ||
                        pattern.length() > bestPattern.length() ||
                        (pattern.length() == bestPattern.length() &&
                        (pattern.compareTo(bestPattern) < 0 ||
                        (pattern.compareTo(bestPattern) == 0 && id < bestId))))){
                    bestPattern = pattern;
                    bestId = id;
                }
            }

            pw.println(bestId);
        }

        pw.flush();
        pw.close();
        br.close();
    }
}

// C0iOKeLu4RZljQdwmEP

**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 create a challenging test input that exposes bugs in the provided code, we need to consider several edge cases and scenarios that can stress the algorithm:

1. **Add and Delete Operations**: Ensure that adding and deleting patterns works correctly, especially when patterns are added and deleted in a specific order.
2. **Appending and Backspacing**: Test scenarios where the current text is built and then partially or fully removed.
3. **Multiple Matching Patterns**: Test cases where multiple patterns match the current text, and the correct pattern is selected based on the given criteria (longest, lexicographic order, smallest ID).
4. **Empty Patterns and Text**: Test cases where the current text or patterns are empty.

Here's a small-scale test input that covers these scenarios:

```plaintext
10
add 1 aaaa
add 2 bbbb
add 3 cccc
append a
append a
append a
backspace 2
append b
delete 1
append c
```

### Explanation:
1. **Add Operations**:
   - Add patterns "aaaa", "bbbb", and "cccc" with IDs 1, 2, and 3 respectively.

2. **Append Operations**:
   - Append "a" three times, resulting in the current text "aaa".
   - The best match for "aaa" should be "aaaa" (ID 1).

3. **Backspace Operation**:
   - Backspace 2 characters, resulting in the current text "a".
   - The best match for "a" should be "aaaa" (ID 1).

4. **Append Operation**:
   - Append "b", resulting in the current text "ab".
   - The best match for "ab" should be "bbbb" (ID 2).

5. **Delete Operation**:
   - Delete pattern "aaaa" (ID 1).
   - The best match for "ab" should now be "bbbb" (ID 2).

6. **Append Operation**:
   - Append "c", resulting in the current text "abc".
   - The best match for "abc" should be "cccc" (ID 3).

This test input will help expose any issues in the handling of add, delete, append, and backspace operations, as well as the correct selection of the best matching pattern based on the given criteria.

10
add 1 aaaa
add 2 bbbb
add 3 cccc
append a
append a
append a
backspace 2
append b
delete 1
append c


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

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

#define int long long

vector<string> s;
int curr;
int last;

struct compare{
    int depth;
    compare(int d = 0) : depth(d) {}
    bool operator()(int a, int b) const {
        const string& sa = s[a];
        const string& sb = s[b];
        if (sa.length() == sb.length()){
            if (depth==sa.length()) return a < b;
            if (sa[depth]==sb[depth]) return a < b;
            return sa[depth] < sb[depth];
        }
        return sa.length() > sb.length();
    }
};

struct item {
    set<int, compare> ans;
    int prev = 0;
    int depth = 0;
    unordered_map<int, int> next;
}; item trie[2000010];

void add(int i){
    int n = s[i].length();
    int p = 0;
    trie[p].ans.insert(i);
    for(int j=0; j<n; j++){
        int ch = s[i][j]-'!';
        if (trie[p].next.count(ch) == 0){
            trie[p].next[ch] = last;
            trie[last].prev = p;
            trie[last].depth = trie[p].depth + 1;
            last++;
        }
        p = trie[p].next[ch];
        trie[p].ans.insert(i);
    }
}

void update(string& a){
    int n = a.length();
    for(int j=0; j<n; j++){
        int ch = a[j]-'!';
        if (trie[curr].next.count(ch) == 0){
            trie[curr].next[ch] = last;
            trie[last].prev = curr;
            trie[last].depth = trie[curr].depth + 1;
            last++;
        }
        curr = trie[curr].next[ch];
    }
}

void backspace(int n){
    for(int i=0; i<n; i++){
        if(curr==0) break;
        curr = trie[curr].prev;
    }
}

void remove(int i){
    int n = s[i].length();
    int p = 0;
    trie[p].ans.erase(i);
    for(int j=0; j<n; j++){
        int ch = s[i][j]-'!';
        p = trie[p].next[ch];
        trie[p].ans.erase(i);
    }
}

int get(){
    item p = trie[curr];
    if (p.ans.empty()) return -1;
    int idx = *p.ans.begin();
    while (p.depth != s[idx].length()){
        p = trie[p.next[s[idx][p.depth]-'!']];
        idx = *p.ans.begin();
    }
    return idx;
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    string q, query;
    int n, idx;
    cin >> n;
    s.assign(n+1,"");
    curr = 0;
    last = 1;
    while(n--){
        cin >> q;
        if (q=="add"){
            cin >> idx >> query;
            s[idx] = query;
            add(idx);
        }
        else if(q=="append"){
            cin >> query;
            update(query);
        }
        else if(q=="delete"){
            cin >> idx;
            remove(idx);
        }
        else{
            cin >> idx;
            backspace(idx);
        }
        cout << get() << "\n";
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
7
add 1 pattern1
add 2 pattern2
add 3 pattern3
append pattern1
backspace 8
append pattern2
delete 1
```

7
add 1 pattern1
add 2 pattern2
add 3 pattern3
append pattern1
backspace 8
append pattern2
delete 1


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define vi vector<int>
#define pb push_back

using namespace std;

//tree of all prefixes of patterns
//add: add prefixes
//delete: delete prefixes, appearences of matchings
//append: deepen search
//backspace: go back x

const int maxn = 1e6+3;

struct pattern {
    int id;
    string s;
    bool operator <(const pattern& r) const {
        if (this->s.size() < r.s.size()) return true;
        if (this->s.size() > r.s.size()) return false;
        for (int i=0; i<this->s.size(); ++i) {
            if (this->s[i] > r.s[i]) {
                return true;
            }
            if (this->s[i] < r.s[i]) {
                return false;
            }
        }
        return (this->id > r.id) ? true : false;
    }
    pattern(int i, string st) {
        id = i;
        s = st;
    }
};

vector<string> strings;

vector<int> pos;

struct pat {
    int id;
    bool operator <(const pat &r) const {
        return pos[this->id] > pos[r.id];
    }
    pat(int i) {
        id = i;
    }
};

struct node {
    set<pat> match;
    char c;
    int ind;
    node * parent;
    map<char, node*> child;
    node (node * par, int x, char d) {
        parent = par;
        ind = x;
        c = d;
    }
};

void add(node &cur, pat &p, string &s) {
    int idx = cur.ind;
    cur.match.insert(p);
    if (s.size()-1 == idx) {
        return;
    }
    char next = s[idx+1];
    if (cur.child.find(next) == cur.child.end())
        cur.child[next] = new node(&cur, idx+1, next);
    add(*cur.child[next], p, s);
    return;
}

vector<int> nums;
vector<int> type;
vector<string> apps;

int main() {
    int n;
    cin >> n;
    {
        vector<pattern> all;
        for (int i=0; i<n; ++i) {
            pos.pb(0);
            string in;
            cin >> in;
            int num;
            string ad;
            if (in == "add") {
                cin >> num >> ad;
                type.pb(0);
                nums.pb(num);
                all.pb(pattern(num, ad));
            }
            if (in == "delete") {
                cin >> num;
                type.pb(1);
                nums.pb(num);
            }
            if (in == "append") {
                cin >> ad;
                type.pb(2);
                apps.pb(ad);
            }
            if (in == "backspace") {
                cin >> num;
                type.pb(3);
                nums.pb(num);
            }
        }
        sort(all.begin(), all.end());
        for (int i=0; i<all.size(); ++i) {
            pos[all[i].id] = i;
            strings.pb(all[i].s);
        }
        all.clear();
    }
    node start = node(nullptr, -1, 'a');
    int dep = 0, len = 0;
    string text;
    node * cur = &start;
    int numind = 0, appsind = 0;
    for (int i=0; i<n; ++i) {
        int t = type[i];
        if (t == 0) {
            int id = nums[numind];
            ++numind;
            pat newpat = pat(id);
            add(start,newpat,strings[pos[newpat.id]]);
            for (int j=dep; j<len; ++j) {
                if (cur->child.find(text[j]) == cur->child.end()) break;
                cur = cur->child[text[j]];
                ++dep;
            }
        }
        if (t == 1) {
            int id = nums[numind];
            ++numind;
            node * here = &start;
            here->match.erase(id);
            for (char cc : strings[pos[id]]) {
                if (here->child.find(cc) == here->child.end()) break;
                here = here->child[cc];
                here->match.erase(id);
            }
        }
        if (t == 2) {
            string s = apps[appsind];
            ++appsind;
            text.append(s);
            int l = s.size();
            len += l;
            if (len-l == dep) {
                for (int j=0; j<l; ++j) {
                    if (cur->child.find(s[j]) == cur->child.end()) break;
                    cur = cur->child[s[j]];
                    ++dep;
                }
            }
        }
        if (t == 3) {
            int b = nums[numind];
            ++numind;
            for (int j=0; j<b; ++j) {
                if (!text.empty()) text.pop_back();
            }
            for (int j=0; j<b-len+dep; ++j) {
                if (cur->parent != nullptr)
                    cur = cur->parent;
            }
            dep = max(0,min(dep, len-b));
            len = max(0,len-b);
        }
        //now query on current tree
        bool found = false;
        int res = -1;
        while (!found) {
            if (len != dep) break;
            if (!cur->match.empty()) {
                found = true;
                res = cur->match.begin()->id;
            } else if (cur->parent != nullptr) {
                char ch = cur->c;
                cur = cur->parent;
                delete cur->child[ch];
                cur->child.erase(ch);
                --dep;
            } else {
                break;
            }
        }
        cout << res << "\n";
        if (i == 29792) return 0;
    }
    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
add 1 alice
add 2 bob
add 3 charlie
append al
backspace 2
```

5
add 1 alice
add 2 bob
add 3 charlie
append al
backspace 2


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
#define Many_SubTask
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define all(a) begin(a), end(a)
#define pb push_back
#define pii pair<int, int>
#define F first
#define S second
#define mp make_pair
const int mod = 998244353;
const int inf = 1e18;
const int MAXN = 1000005;

struct node {
    int ansID;
    set<int> ids;
    node *nxt[95], *fa;
};
int node_cnt = 0;

node *new_node(node *fa) {
    node_cnt++;
    node *nn = new node;
    nn->ansID = -1;
    for (int i = 0; i < 95; i++) {
        nn->nxt[i] = NULL;
    }
    nn->fa = fa;
    return nn;
}

node *root = new_node(NULL);
node *cur = root;

string pattern[MAXN];

int maxID(int a, int b) {
    if (a == -1) return b;
    if (b == -1) return a;
    if (pattern[a].size() > pattern[b].size()) return a;
    if (pattern[a].size() < pattern[b].size()) return b;
    if (pattern[a] < pattern[b]) return a;
    if (pattern[a] > pattern[b]) return b;
    return min(a, b);
}

void add(int id, string &s) {
    node *p = root;
    for (int i = 0; i < s.size(); i++) {
        if (!p->nxt[s[i]]) p->nxt[s[i]] = new_node(p);
        p = p->nxt[s[i]];
    }
    p->ids.insert(id);
    p->ansID = maxID(p->ansID, id);
    do {
        p = p->fa;
        p->ansID = maxID(p->ansID, id);
    } while (p != root);
}

void updateAnsID(node* p) {
    p->ansID = -1;
    for (auto id_ : p->ids) {
        p->ansID = maxID(p->ansID, id_);
    }
    for (int j = 0; j < 95; j++) {
        if (p->nxt[j]) p->ansID = maxID(p->ansID, p->nxt[j]->ansID);
    }
}

void del(int id, string &s) {
    node *p = root;
    for (int i = 0; i < s.size(); i++) {
        if (!p->nxt[s[i]]) p->nxt[s[i]] = new_node(p);
        p = p->nxt[s[i]];
    }
    p->ids.erase(id);
    if (p->ansID == id) updateAnsID(p);
    do {
        p = p->fa;
        if (p->ansID == id) updateAnsID(p);
    } while (p != root);
}

void append(string &s) {
    int n = s.size();
    for (int i = 0; i < n; i++) {
        if (!cur->nxt[s[i]]) cur->nxt[s[i]] = new_node(cur);

        cur = cur->nxt[s[i]];
    }
}

void backspace(int c) {
    while (cur != root && c--) {
        cur = cur->fa;
    }
}

int N = 1;
void solve() {
    string ope;
    cin >> ope;
    if (ope == "add") {
        int id;
        string s;
        cin >> id >> s;
        for (auto &i : s) i -= 33;
        pattern[id] = s;
        add(id, s);
    } else if (ope == "delete") {
        int id;
        cin >> id;
        del(id, pattern[id]);
    } else if (ope == "append") {
        string s;
        cin >> s;
        for (auto &i : s) i -= 33;
        append(s);
    } else {  // backspace
        int c;
        cin >> c;
        backspace(c);
    }
    if(N != 30000)  cout << cur->ansID << '\n';
}

signed main() {
    cin.sync_with_stdio(0), cin.tie(0);
#ifdef Many_SubTask
    cin >> N;
#endif
    for (int i = 1; i <= N; i++) {
        solve();
        if(N == 30000 && i > 29200) cout << node_cnt << '\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
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
backspace 3
append 2_bobabc
```

6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
backspace 3
append 2_bobabc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
// -*- mode: c++; coding: utf-8-emacs; -*-
#include<bits/stdc++.h>
//#define int long long
using namespace std;
struct NODE{
	int dep;
	int fa;
	int mx;
	int md;
	multiset<int> id;
	map<char,int> c;
	NODE(int d,int f) : dep(d),fa(f),mx(-1),md(1){}
};
struct TREE{
	map<int,string> pattern;
	vector<NODE> a;
	string t;
	int now;
	TREE() : now(0),a{NODE(0,-1)}{};
	void chkmax(int u,string &s){
		while(u!=-1){
			a[u].mx=-1;
			a[u].md=1;
			for(auto [key,v] : a[u].c){
				if(v!=-1 && a[v].mx!=-1){
//					cout<<a[v].md<<'\n';
					if(a[u].mx==-1 || a[v].md+1>a[u].md){
						a[u].md=a[v].md+1;
						a[u].mx=a[v].mx;
					}
				}
			}
			if(a[u].mx==-1 && !a[u].id.empty()){
				a[u].mx=*a[u].id.begin();
			}
//			cout<<' '<<a[u].mx<<endl;
			u=a[u].fa;
		}
	}
	void add(int id,string s){
		pattern[id]=s;
		int u=0;
		for(auto i : s){
			if(a[u].c.find(i)==a[u].c.end()){
				a[u].c[i]=a.size();
				a.emplace_back(a[u].dep+1,u);
			}
			u=a[u].c[i];
		}
		a[u].id.insert(id);
		chkmax(u,s);
	}
	void del(int id){
		string s=pattern[id];
		int u=0;
		for(auto i : s){
			u=a[u].c[i];
		}
		a[u].id.erase(a[u].id.find(id));
		chkmax(u,s);
	}
	void append(string s){
		t+=s;
	}
	void backspace(int n){
		while(!t.empty() && n--){
			t.pop_back();
		}
	}
	int ask(){
//		cout<<"now: "<<now<<' '<<a[now].mx<<'\n';
		while(now>0 && (a[now].mx==-1 || now>(int)t.length())){
			now=a[now].fa;
		}
		while(a[now].dep<(int)t.length()){
			int x=t[a[now].dep];
			if(a[now].c.find(x)==a[now].c.end()){
				return -1;
			}
			int v=a[now].c[x];
			if(a[v].mx==-1){
				return -1;
			}
			now=v;
		}
		return a[now].mx;
	}
};
/*
  6
  add 1 aaa
  add 2 ab
  add 3 accc
  append a
  append babc
  backspace 3
 */
void solve(){
	int qwq;
	cin>>qwq;
	TREE t;
	qwq=min(qwq,29800);
	while(qwq--){
		string o;
		cin>>o;
		if(o[1]=='d'){
			int i;
			string s;
			cin>>i>>s;
			for(auto &c : s){
				c-=33;
			}
			t.add(i,s);
		}else if(o[1]=='p'){
			string s;
			cin>>s;
			for(auto &c : s){
				c-=33;
			}
			t.append(s);
		}else if(o[1]=='a'){
			int i;
			cin>>i;
			t.backspace(i);
		}else{
			int i;
			cin>>i;
			t.del(i);
		}
		cout<<t.ask()<<'\n';
	}
}
signed main(){
	ios::sync_with_stdio(false);
	int qwq=1;
//	cin>>qwq;
	while(qwq--){
		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
6
add 1 aab
add 2 abb
add 3 acc
append a
append b
backspace 1
```

6
add 1 aab
add 2 abb
add 3 acc
append a
append b
backspace 1


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fastio(); ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

char END_CHAR = ' ';

bool compare(pair<int, pair<string, int>>& A, pair<int, pair<string, int>> & B) {
  if (A.first != B.first) {
    return (A.first > B.first);
  }
  if (A.second.first != B.second.first) {
    return (A.second.first < B.second.first);
  }
  return A.second.second < B.second.second;
}

struct Trie {

  unordered_map<char, Trie*> nextPointer;
  set<pair<int, pair<string, int>>> lenStringIDs;
  pair<int, pair<string, int>> maxLenStringID;
  Trie* parent;
  Trie(Trie* p) {
    nextPointer = unordered_map<char, Trie*>();
    maxLenStringID = {0, {"", -1}};
    lenStringIDs = set<pair<int, pair<string, int>>>();
    parent = p;
  }

};

void add(Trie* node, string & s, int index, int ID) {
  if (index == (int) s.size() + 1) {
    // if ((node -> nextPointer).find(END_CHAR) == (node -> nextPointer).end()) {
    //   node->nextPointer[END_CHAR] = new Trie(node);
    // }
    Trie* endNode = node;
    (endNode->lenStringIDs).insert({(int) s.size(), {s, ID}});
    (endNode->maxLenStringID) = *(--((endNode->lenStringIDs).end()));
    return ;
  }

  char ch = (index < ((int)s.size())) ? s[index] : END_CHAR;
  if ((node -> nextPointer).find(ch) == (node -> nextPointer).end()) {
    node->nextPointer[ch] = new Trie(node);
  }

  add(node->nextPointer[ch], s, index + 1, ID);

  if (!compare(node -> maxLenStringID, node -> nextPointer[ch] -> maxLenStringID)) {
    node -> maxLenStringID = node -> nextPointer[ch] -> maxLenStringID;
  }
  // cout << ch << " " << node -> nextPointer[ch] -> maxLenStringID.first << "\t";
}

void remove(Trie* node, string & s, int index, int ID) {
  if (index == (int) s.size() + 1) {
    // if ((node -> nextPointer).find(END_CHAR) == (node -> nextPointer).end()) {
    //   node->nextPointer[END_CHAR] = new Trie(node);
    // }
    Trie* endNode = node;
    auto IT = (endNode->lenStringIDs).find({(int) s.size(), {s, ID}});
    (endNode->lenStringIDs).erase(IT);
    (endNode->maxLenStringID) = {0, {"", -1}};
    if (!(endNode->lenStringIDs).empty()) {
      (endNode->maxLenStringID) = *(--((endNode->lenStringIDs).end()));
    }
    return ;
  }

  char ch = (index < (int) s.size()) ? s[index] : END_CHAR;
  if ((node -> nextPointer).find(ch) == (node -> nextPointer).end()) {
    node->nextPointer[ch] = new Trie(node);
  }

  remove(node->nextPointer[ch], s, index + 1, ID);
  if ((node -> maxLenStringID).second.second == ID) {
    node -> maxLenStringID = {0, {"", -1}};
    for (auto child: node -> nextPointer) {
      if (!compare(node -> maxLenStringID, child.second -> maxLenStringID)) {
        node -> maxLenStringID = child.second -> maxLenStringID;
      }
    }
  }
}

int main() {
  fastio();

  int Q;
  cin >> Q;

  Trie* root = new Trie(NULL);
  Trie* traveller = root;

  map<int, string> idToString;

  while (Q --) {
    string operation;
    cin >> operation;
    if (operation == "add") {
      int ID;
      cin >> ID;
      string s;
      cin >> s;
      idToString[ID] = s;
      add(root, s, 0, ID);

    } else if (operation == "append") {
      string s;
      cin >> s;

      for (int i = 0;i < (int)s.size();i ++) {
        if ((traveller -> nextPointer).find(s[i]) == (traveller -> nextPointer).end()) {
          (traveller -> nextPointer)[s[i]] = new Trie(traveller);
        }

        traveller = (traveller -> nextPointer)[s[i]];
      }

    } else if (operation == "backspace") {
      int cnt = 0;
      cin >> cnt;
      while (traveller -> parent != NULL && cnt > 0) {
        cnt --;
        traveller = traveller -> parent;
      }
    } else if (operation == "delete") {
      int ID;
      cin >> ID;

      string s = idToString[ID];

      remove(root, s, 0, ID);
    }

    cout << (traveller -> maxLenStringID).second.second << "\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
6
add 1 a
add 2 ab
add 3 abc
append a
append b
backspace 2
```

6
add 1 a
add 2 ab
add 3 abc
append a
append b
backspace 2


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
import java.util.*;

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

        int n = scanner.nextInt();
        scanner.nextLine(); // Consume newline

        StringBuilder currentText = new StringBuilder();
        TreeSet<Pattern> patterns = new TreeSet<>(Comparator
            .comparing((Pattern p) -> p.length).reversed()
            .thenComparing(Pattern::toString)
            .thenComparing(p -> p.id));
        Map<Integer, Pattern> patternMap = new HashMap<>();

        for (int i = 0; i < n; i++) {
            String line = scanner.nextLine();
            String[] tokens = line.split(" ", 2); // Split into max 2 parts
            String command = tokens[0];

            if (command.equals("add")) {
                String[] parts = tokens[1].split(" ", 2);
                int id = Integer.parseInt(parts[0]);
                String pattern = parts[1];
                Pattern p = new Pattern(id, pattern);
                patterns.add(p);
                patternMap.put(id, p);
            } else if (command.equals("delete")) {
                int id = Integer.parseInt(tokens[1]);
                Pattern p = patternMap.get(id);
                if (p != null) {
                    patterns.remove(p);
                    patternMap.remove(id);
                }
            } else if (command.equals("append")) {
                currentText.append(tokens[1]);
            } else if (command.equals("backspace")) {
                int c = Integer.parseInt(tokens[1]);
                int len = currentText.length();
                currentText.delete(Math.max(0, len - c), len);
            }

            Pattern bestMatch = findBestMatch(currentText.toString(), patterns);
            if (bestMatch != null) {
                System.out.println(bestMatch.id);
            } else {
                System.out.println(-1);
            }
        }
    }

    private static Pattern findBestMatch(String text, TreeSet<Pattern> patterns) {
        for (Pattern p : patterns) {
            if (p.pattern.startsWith(text)) {
                return p;
            }
        }
        return null;
    }

    static class Pattern {
        int id;
        int length;
        String pattern;

        Pattern(int id, String pattern) {
            this.id = id;
            this.pattern = pattern;
            this.length = pattern.length();
        }

        @Override
        public String toString() {
            return pattern;
        }
    }
}

// 8wrXZand4pEY0I1gWMV

**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
add 1 pattern1_a
add 2 pattern2_ab
add 3 pattern3_abc
append ab
backspace 3
```

5
add 1 pattern1_a
add 2 pattern2_ab
add 3 pattern3_abc
append ab
backspace 3


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const long double PI = acos(-1);
//#define int long long
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll, ll> PLL;
typedef tuple<ll, ll, ll> TLL;
int lowbit(int x) {return x & (- x);};
const int N = 200010, mod = 998244353;
const int Mod = 1e9 + 7, INF = 0x3f3f3f3f;

vector<ll> e[N];

void solve(){
    ll n;
    cin >> n;
    ll l = 1, r = n, L = n + 1, R = 0;
    for(ll i = 1; i <= n; i ++){
        ll x;
        cin >> x;
        l = max(l, i - x + 1);
        r = min(r, i + x - 1);
        e[x].push_back(i);
    }
    for(int i = 1; i <= n; i ++){
        if(e[i].size() == 0) continue;
        L = min(L, e[i][0]), R = max(R, e[i][e[i].size() - 1]);
        e[i].clear();
        if(R - L + 1 > i){
            cout << 0 <<'\n';
            return;
        }
    }
    if(l > r) cout << 0 << '\n';
    else cout << r - l + 1 << '\n';
}

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

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The code seems to have issues with handling the constraints and the logic for determining the valid starting cities. Here are some steps to generate a challenging test input:

1. **Edge Case: All Cities Have the Same Deadline**:
   - If all cities have the same deadline, the code should handle this correctly. However, we can still test it to ensure it works as expected.

2. **Edge Case: Cities with Very Low Deadlines**:
   - If the deadlines are very low, the code should correctly identify that no starting city can meet the deadlines.

3. **Edge Case: Cities with High Deadlines**:
   - If the deadlines are very high, the code should correctly identify that all starting cities can meet the deadlines.

4. **Edge Case: Discontinuous Deadlines**:
   - If the deadlines are not continuous, the code should correctly handle the gaps and determine the valid starting cities.

5. **Edge Case: Single City**:
   - If there is only one city, the code should correctly identify that it can be a starting city.

6. **Edge Case: Multiple Test Cases**:
   - The code should handle multiple test cases correctly.

Based on these considerations, here is a small-scale test input that covers these edge cases:

```plaintext
3
1
1
6
6 3 3 3 5 5
6
5 6 4 1 4 5
```

This test input includes:
1. A single city with a deadline of 1.
2. A case with multiple cities and mixed deadlines (as in the example).
3. A case with a very low deadline that should result in no valid starting cities.

This input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
#include <iostream>
#include <map>
#include <vector>

void solve() {
    int n;
    std::cin >> n;
    std::vector<int> v(n);
    for (int i = 0; i < n; ++i) std::cin >> v[i];

    std::vector<int> first(n + 1);
    int cur = n;
    for (int i = 0; i < n; ++i) {
        while (cur >= v[i]) first[cur] = i, --cur;
    }
    cur = n;
    for (int i = n - 1; i >= 0; --i) {
        while (cur >= v[i]) {
            if (i - first[v[i]] >= v[i]) {
                std::cout << "0\n";
                return;
            }
            --cur;
        }
    }

    std::vector<int> diff(n + 1);
    for (int i = 0; i < n; ++i) {
        int l = std::max(0, i - v[i] + 1);
        int r = std::min(n - 1, i + v[i] - 1);
        ++diff[l];
        --diff[r + 1];
    }
    cur = 0;
    int cnt = 0;
    for (auto el : diff) {
        cur += el;
        cnt += cur == n;
    }
    std::cout << cnt << '\n';
}

int main() {
    int t;
    std::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 that are likely to cause issues. The buggy code has a few potential points of failure:

1. **Incorrect handling of the `first` array**: The `first` array is used to track the earliest position where a certain value can be found. If this array is not updated correctly, it can lead to incorrect results.
2. **Incorrect validation in the second loop**: The second loop checks if the difference between the current position and the earliest position is within the allowed range. If this check is not done correctly, it can lead to incorrect results.
3. **Incorrect calculation of the `diff` array**: The `diff` array is used to calculate the number of valid starting cities. If the boundaries are not correctly calculated, it can lead to incorrect results.

Given these points, we can create a test case that challenges these specific areas. Here is a small-scale test input that should expose the bug:

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

### Explanation:
1. **Test Case 1:**
   - `4 1 2 3 4`
   - This test case checks the basic scenario where the deadlines are in increasing order. The expected output is `1` because only city `1` can be a valid starting city.

2. **Test Case 2:**
   - `4 4 3 2 1`
   - This test case checks the scenario where the deadlines are in decreasing order. The expected output is `1` because only city `4` can be a valid starting city.

3. **Test Case 3:**
   - `5 3 2 1 4 5`
   - This test case checks a mixed scenario where the deadlines are not in a simple increasing or decreasing order. The expected output is `2` because cities `3` and `4` can be valid starting cities.

These test cases should help in exposing any issues in the buggy code by covering different patterns of deadlines and ensuring that the logic for determining valid starting cities is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
/*
    Những mùa thu khác có thể trở về
Nhưng mùa thu của chúng ta đã chết...
*/

// #pragma GCC optimize("Ofast,O3")
// #pragma GCC target("avx,avx2,fma,sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;

#ifdef Debug
#include "./namsee_lib/Debug.h"
#else
#define dbg(...)
#endif

#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)
#define FORD(i,a,b) for(int i=(a),_b=(b); i>=_b; i--)
#define REP(i,a) for(int i=0,_a=(a); i<_a; i++)

#define F first
#define S second
#define MP make_pair
#define PB push_back

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define lb lower_bound
#define ub upper_bound
#define ins insert

#define nl "\n"

#define vi vector<long long>
#define vii vector<pair<long long, long long>>
#define pi pair<ll, ll>

template <class T>
using Tree =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template<class X, class Y> bool ckmin(X& a, const Y& b) {return a > b ? a = b, 1 : 0;}
template<class X, class Y> bool ckmax(X& a, const Y& b) {return a < b ? a = b, 1 : 0;}

// const ll M = 1e9 + 9; // (1ll << 61) - 1
// mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
// const ll B = uniform_int_distribution<ll>(311, M - 1)(rng);

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

    #ifdef Debug
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    #endif

    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        vector<vector<int>> pos(n + 1);
        REP(i, n) {
            int val; cin >> val;
            pos[val].PB(i);
        }
        int ans = 1e9, lf = 1e9, rt = 0;
        FOR(i, 1, n) {
            if (pos[i].empty()) continue;
            int l = pos[i][0], r = pos[i].back();
            if (r - l + 1 > i) {
                ans = 0;
            }
            if (ans == 1e9) {
                ans = r - l + 1;
                rt = r, lf = l;
            }
            else {
                ckmax(r, rt);
                ckmin(l, lf);
                if (r - l + 1 > i) ans = 0;
            }
        }
        cout << ans << nl;
    }
    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 that maximize the failure probability. The code seems to have issues with handling the constraints and the logic for determining the valid starting cities.

Let's consider the following steps:

1. **Edge Case with Single City**: Test with a single city to ensure the code handles the simplest case correctly.
2. **Edge Case with All Cities Having the Same Deadline**: Test with all cities having the same deadline to see if the code can handle uniform deadlines.
3. **Edge Case with Non-Uniform Deadlines**: Test with a sequence of deadlines that are not uniform and have a specific pattern to expose potential logic errors.
4. **Edge Case with a Single Good Starting City**: Test with a sequence of deadlines that have only one valid starting city.
5. **Edge Case with No Good Starting City**: Test with a sequence of deadlines that do not allow any valid starting city.

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

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

Explanation:
1. **Test Case 1**: Single city with a deadline of 1. The only valid starting city is city 1.
2. **Test Case 2**: Three cities with the same deadline of 3. All cities are valid starting cities.
3. **Test Case 3**: Five cities with increasing deadlines. The valid starting city is city 1.
4. **Test Case 4**: Five cities with decreasing deadlines. The valid starting city is city 5.
5. **Test Case 5**: Five cities with a mixed pattern of deadlines. The only valid starting city is city 3.

This test input should help expose any issues in the buggy code by covering a range of edge cases and non-uniform deadline scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

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

public class CodeForcesSolution {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;
    /*static int mod = (int) 1e9+7;*/
    public static void main(String[] args) throws IOException {
        int t = readInt();
        while (t-- > 0) {
            int n = readInt();
            int[] a = new int[n + 1], cnt = new int[n + 1], mx = new int[n + 1], mi = new int[n + 1], df = new int[n + 5];
            for (int i = 1; i <= n; i++) {
                a[i] = readInt();
                if (mi[a[i]] == 0) mi[a[i]] = i;
                mx[a[i]] = i;
                cnt[a[i]]++;
            }
            Arrays.sort(a);
            int ans = 0, l = n, r = 1;
            for (int i = 1; i <= n; i += cnt[a[i]]) {
                l = Math.min(l, mi[a[i]]);
                r = Math.max(r, mx[a[i]]);
                if (a[i] < r - l + 1) {
                    ans = 0; break;
                }
                int nl = Math.max(l, r - a[i] + 1), nr = Math.min(l + a[i] - 1, r);
                df[nl] += cnt[a[i]]; df[nr + 1] -= cnt[a[i]];
            }
            for (int i = 1; i <= n; i++) {
                df[i] += df[i - 1];
                if (df[i] == n) ans++;
            }
            System.out.println(ans);
        }
    }
    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    static long readLong() throws IOException {
        return Long.parseLong(next());
    }
    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter() throws IOException {
        return next().charAt(0);
    }
    static String readLine() throws IOException {
        return br.readLine().trim();
    }
    static int readALotInt() throws IOException{
        int x = 0, c;
        while((c = br.read()) != ' ' && c != '\n')
            x = x * 10 + (c - '0');
        return x;
    }
}

**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 code, we need to consider edge cases and scenarios that can make the algorithm fail. The key points to focus on are:

1. **Deadlines that are very tight**: This can cause the algorithm to incorrectly determine that no starting city is valid.
2. **Multiple cities with the same deadline**: This can cause issues in the `mi` and `mx` arrays.
3. **Cities with deadlines that are very spread out**: This can cause the `df` array to be incorrectly updated.
4. **Single city test cases**: These can help verify the base case handling.
5. **Test cases with all cities having the same deadline**: This can test the algorithm's ability to handle uniform deadlines.

Let's create a test input that combines these scenarios:

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

### Explanation:
1. **Test Case 1: Single City**
   - Input: `1 1`
   - Expected Output: `1`
   - This tests the base case where there is only one city.

2. **Test Case 2: Two Cities with Tight Deadlines**
   - Input: `2 2 1`
   - Expected Output: `1`
   - This tests the scenario where the deadlines are very tight, and only one starting city is valid.

3. **Test Case 3: Three Cities with Decreasing Deadlines**
   - Input: `3 3 2 1`
   - Expected Output: `1`
   - This tests the scenario where the deadlines are decreasing, and only one starting city is valid.

4. **Test Case 4: Four Cities with Decreasing Deadlines**
   - Input: `4 4 3 2 1`
   - Expected Output: `1`
   - This tests a slightly larger scenario with decreasing deadlines, and only one starting city is valid.

This test input is small enough to be easily manageable and should expose potential issues in the given buggy code.