**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

Buggy Code:
from collections import deque
import copy

# Dictionary to hold each store's ice cream list as a deque
stores = {1: deque()}

# Function to solve the knapsack problem for query type 4
def max_tastiness_for_budget(ice_creams, max_price):
    # Using dynamic programming to solve the 0/1 knapsack problem
    dp = [0] * (max_price + 1)

    for price, tastiness in ice_creams:
        for j in range(max_price, price - 1, -1):
            dp[j] = max(dp[j], dp[j - price] + tastiness)

    return max(dp)

# Read input
q = int(input().strip())

# Process each query
for _ in range(q):
    query = input().strip().split()
    query_type = int(query[0])

    if query_type == 1:
        # Open a new store, which is a copy of store x
        x = int(query[1])
        new_store_index = len(stores) + 1
        stores[new_store_index] = copy.deepcopy(stores[x])

    elif query_type == 2:
        # Add a new ice cream type to store x
        x = int(query[1])
        price = int(query[2])
        tastiness = int(query[3])
        stores[x].append((price, tastiness))

    elif query_type == 3:
        # Remove the earliest added ice cream from store x
        x = int(query[1])
        stores[x].popleft()

    elif query_type == 4:
        # Find the maximum tastiness for store x with budget p
        x = int(query[1])
        max_price = int(query[2])
        ice_creams = list(stores[x])
        # Use the knapsack algorithm to find the max tastiness
        result = max_tastiness_for_budget(ice_creams, max_price)
        print(result)

**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
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

template<typename T>
struct STACK{
    stack<pair<T, T>> st;
    int get(int p){return st.top().second.arr[p];}
    T get(){return st.top().second;}
    bool empty(){return st.empty();}
    int size(){return st.size();}
    void push(T x){
        T val = x;
        if(!empty()){val = val.compare(get());}
        st.push({x, val});
    }
    void pop(){st.pop();}
    T top(){return st.top().first;}
    void swap(STACK<T> &x){st.swap(x.st);}
};

template<typename T>
struct DEQUE{
    STACK<T> l, r, t;
    void rebalance(){
        bool f = false;
        if(r.empty()){f = true;l.swap(r);}
        int sz = r.size() / 2;
        while(sz--){t.push(r.top());r.pop();}
        while(!r.empty()){l.push(r.top());r.pop();}
        while(!t.empty()){r.push(t.top());t.pop();}
        if(f){l.swap(r);}
    }
    int get(int p){
        if(l.empty()) return r.get(p);
        if(r.empty()) return l.get(p);
        return max(r.get(p), l.get(p));
    }
    T get(){
        if(l.empty()) return r.get();
        if(r.empty()) return l.get();
        return l.get().compare(r.get());
    }
    bool empty(){return l.empty() && r.empty();}
    int size(){return l.size() + r.size();}
    void push_front(T x){l.push(x);}
    void push_back(T x){r.push(x);}
    void pop_front(){if(l.empty()){rebalance();}l.pop();}
    void pop_back(){if(r.empty()){rebalance();}r.pop();}
    T front(){if(l.empty()){rebalance();}return l.top();}
    T back(){if(r.empty()){rebalance();}return r.top();}
    void swap(DEQUE<T> &x){x.l.swap(l);x.r.swap(r);}
};

struct NODE{
    int val;
    int P;
    array<int, 2222> arr = {0};
    NODE(int val, int P): val(val), P(P){
        fill(arr.begin(), arr.end(), 0);
        fill(arr.begin() + P, arr.end(), val);
    }
    NODE(int val, int P, array<int, 2222> arr): val(val), P(P), arr(arr){}
    NODE compare(NODE x){
        NODE c = NODE(0, 0, x.arr);
        for(int i = 2221 - P; i >= 0; i --){
            c.arr[i + P] = max(c.arr[i + P], c.arr[i] + val);
        }
        return c;
    }
};

struct EDGE{
    int to;
    int type;
    int p;
    int val;
};

struct TREE_NODE{
    int id;
    vector<pair<int, int>> query;
    vector<EDGE> next;
};

bool b = false;

void dfs(int head, DEQUE<NODE> &deq, vector<TREE_NODE> &tree, vector<int> &ans){
    if(b){
        cout << head << " " << tree.size() << endl;
    }
    if(!deq.empty()){
        if(!b)
            for(auto [i, j] : tree[head].query){
                ans[i] = deq.get(j);
            }
    }
    for(auto [to, type, p, val] : tree[head].next){
        if(type == 1){
            dfs(to, deq, tree, ans);
        }
        else if(type == 2){
            deq.push_back(NODE(val, p));
            dfs(to, deq, tree, ans);
            deq.pop_back();
        }
        else{
            auto x = deq.front();
            deq.pop_front();
            dfs(to, deq, tree, ans);
            deq.push_front(x);
        }
    }
}

void solve(){
    int Q;
    cin >> Q;
    DEQUE<NODE> deq;
    b = Q == 30000;
    vector<TREE_NODE> tree;
    map<int, int> pos;
    pos[1] = 0;
    tree.push_back({1, {}, {}});
    int cnt = 0;
    int c = 1;
    while(Q --){
        int T;
        cin >> T;
        if(T == 1){
            int x;
            cin >> x;
            c ++;
            tree[pos[x]].next.push_back({(int)tree.size(), 1, 0, 0});
            pos[c] = tree.size();
            tree.push_back({c, {}, {}});
        }
        else if(T == 2){
            int x, p, t;
            cin >> x >> p >> t;
            tree[pos[x]].next.push_back({(int)tree.size(), 2, p, t});
            pos[x] = tree.size();
            tree.push_back({c, {}, {}});
        }
        else if(T == 3){
            int x;
            cin >> x;
            tree[pos[x]].next.push_back({(int)tree.size(), 3, 0, 0});
            pos[x] = tree.size();
            tree.push_back({c, {}, {}});
        }
        else{
            int x, p;
            cin >> x >> p;
            tree[pos[x]].query.push_back({cnt ++, p});
        }
    }
    vector<int> ans(cnt);
    dfs(0, deq, tree, ans);
    for(int i : ans){
        cout << i << endl;
    }
}

signed main(){
    // ios_base::sync_with_stdio(false);
    // cin.tie(0);
    int N = 1;
    while(N--){
        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
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

// 定义最大价格
const int MAX_P = 2000;

// 结构体表示每个店铺
struct Store {
    // 使用 vector 存储冰淇淋，start 指向有效冰淇淋的起始位置
    vector<pair<int, int>> ice_creams;
    int start; // 起始位置
    // dp_cache 存储最大美味度，对于每个价格从 0 到 2000
    // 如果 dp_cache 为空，则表示尚未计算
    vector<int> dp_cache;

    Store() : start(0), dp_cache() {}
};

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

    int q;
    cin >> q;

    // 使用 1-based 索引，stores[1] 是初始店铺
    vector<Store> stores;
    stores.emplace_back(); // stores[0] 未使用
    stores.emplace_back(); // stores[1] 初始店铺，空的

    int current_store_count = 1;

    while(q--){
        int type;
        cin >> type;
        if(type == 1){
            int x;
            cin >> x;
            current_store_count++;
            // 创建新店铺，复制店铺 x 的冰淇淋和 dp_cache
            stores.emplace_back();
            stores[current_store_count].ice_creams = stores[x].ice_creams;
            stores[current_store_count].start = stores[x].start;
            if(!stores[x].dp_cache.empty()){
                stores[current_store_count].dp_cache = stores[x].dp_cache;
            }
        }
        else if(type == 2){
            int x, p, t;
            cin >> x >> p >> t;
            // 向店铺 x 添加新的冰淇淋
            stores[x].ice_creams.emplace_back(p, t);
            // 清除 dp_cache
            stores[x].dp_cache.clear();
        }
        else if(type == 3){
            int x;
            cin >> x;
            // 移除店铺 x 中最早添加的冰淇淋
            if(stores[x].start < (int)stores[x].ice_creams.size()){
                stores[x].start++;
                // 清除 dp_cache
                stores[x].dp_cache.clear();
            }
            // 根据题目约束，不需要处理店铺为空的情况
        }
        else if(type == 4){
            int x, p_limit;
            cin >> x >> p_limit;
            p_limit = min(p_limit, MAX_P); // 限制 p_limit 不超过 2000

            // 检查是否已经有缓存
            if(stores[x].dp_cache.empty()){
                // 计算 dp_cache
                // 初始化 dp 数组
                vector<int> dp(MAX_P + 1, 0);
                // 遍历所有有效的冰淇淋
                for(int i = stores[x].start; i < (int)stores[x].ice_creams.size(); ++i){
                    int price = stores[x].ice_creams[i].first;
                    int tastiness = stores[x].ice_creams[i].second;
                    if(price > MAX_P) continue;
                    // 0/1 背包，从后向前遍历
                    for(int j = MAX_P; j >= price; --j){
                        if(dp[j - price] + tastiness > dp[j]){
                            dp[j] = dp[j - price] + tastiness;
                        }
                    }
                }
                stores[x].dp_cache = move(dp);
            }

            // 查找最大美味度
            int max_tastiness = 0;
            // 为了加速，使用单循环遍历
            for(int j = 0; j <= p_limit; ++j){
                if(stores[x].dp_cache[j] > max_tastiness){
                    max_tastiness = stores[x].dp_cache[j];
                }
            }
            cout << max_tastiness << "\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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the key points to consider:

1. **Query Type 1 (New Store)**:
   - Ensure that the new store correctly copies the ice creams and the `dp_cache` from the existing store.
   - Test with a store that has a large number of ice creams to see if the copy operation is efficient and correct.

2. **Query Type 2 (Add Ice Cream)**:
   - Test with high prices and tastiness values to ensure the `dp_cache` is cleared and recalculated correctly.
   - Test with multiple ice creams to ensure the `dp` array is updated correctly.

3. **Query Type 3 (Remove Ice Cream)**:
   - Test with a store that has multiple ice creams and ensure the oldest ice cream is removed correctly.
   - Ensure the `dp_cache` is cleared and recalculated correctly after removal.

4. **Query Type 4 (Find Maximum Tastiness)**:
   - Test with a price limit that is very low to ensure the correct handling of empty or near-empty stores.
   - Test with a price limit that is very high to ensure the correct handling of all available ice creams.

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

```plaintext
10
1 1
2 2 5 7
2 2 3 4
1 2
2 3 4 10
3 2
4 2 4
4 3 9
4 1 9
4 3 9
```

Explanation:
- **Query 1 1**: Create a new store that copies store 1 (which is initially empty).
- **Query 2 2 5 7**: Add an ice cream to store 2.
- **Query 2 2 3 4**: Add another ice cream to store 2.
- **Query 1 2**: Create a new store that copies store 2 (which now has two ice creams).
- **Query 2 3 4 10**: Add an ice cream to store 3.
- **Query 3 2**: Remove the oldest ice cream from store 2.
- **Query 4 2 4**: Find the maximum tastiness in store 2 with a price limit of 4.
- **Query 4 3 9**: Find the maximum tastiness in store 3 with a price limit of 9.
- **Query 4 1 9**: Find the maximum tastiness in store 1 with a price limit of 9.
- **Query 4 3 9**: Find the maximum tastiness in store 3 with a price limit of 9 again.

This test input should expose any issues with the copying of ice creams, the addition and removal of ice creams, and the calculation of maximum tastiness.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

public class IceCreamStores {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();
        int q = Integer.parseInt(reader.readLine());

        List<List<int[]>> stores = new ArrayList<>();
        stores.add(new ArrayList<>());

        for (int i = 0; i < q; i++) {
            StringTokenizer st = new StringTokenizer(reader.readLine());
            int type = Integer.parseInt(st.nextToken());

            switch (type) {
                case 1: {
                    int x = Integer.parseInt(st.nextToken()) - 1;
                    // Create a new store with the same ice creams as store x
                    stores.add(new ArrayList<>(stores.get(x)));
                    break;
                }
                case 2: {
                    int x = Integer.parseInt(st.nextToken()) - 1;
                    int p = Integer.parseInt(st.nextToken());
                    int t = Integer.parseInt(st.nextToken());
                    // Add new ice cream type to store x
                    stores.get(x).add(new int[]{p, t});
                    break;
                }
                case 3: {
                    int x = Integer.parseInt(st.nextToken()) - 1;
                    // Remove the oldest ice cream type from store x
                    if (!stores.get(x).isEmpty()) {
                        stores.get(x).remove(0);
                    }
                    break;
                }
                case 4: {
                    int x = Integer.parseInt(st.nextToken()) - 1;
                    int p = Integer.parseInt(st.nextToken());
                    output.append(maxTastiness(stores.get(x), p)).append("\n");
                    break;
                }
            }
        }

        System.out.print(output);
    }

    private static int maxTastiness(List<int[]> iceCreams, int maxPrice) {
        int[] dp = new int[maxPrice + 1];

        for (int[] iceCream : iceCreams) {
            int price = iceCream[0];
            int tastiness = iceCream[1];
            for (int j = maxPrice; j >= price; j--) {
                dp[j] = Math.max(dp[j], dp[j - price] + tastiness);
            }
        }

        return dp[maxPrice];
    }
}

**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
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

public class IceCreamStores {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();
        int q = Integer.parseInt(reader.readLine());

        List<List<int[]>> stores = new ArrayList<>();
        stores.add(new ArrayList<>());

        for (int i = 0; i < q; i++) {
            StringTokenizer st = new StringTokenizer(reader.readLine());
            int type = Integer.parseInt(st.nextToken());

            switch (type) {
                case 1:
                    int x = Integer.parseInt(st.nextToken()) - 1;
                    stores.add(new ArrayList<>(stores.get(x)));
                    break;
                case 2:
                    x = Integer.parseInt(st.nextToken()) - 1;
                    int p = Integer.parseInt(st.nextToken());
                    int t = Integer.parseInt(st.nextToken());
                    stores.get(x).add(new int[]{p, t});
                    break;
                case 3:
                    x = Integer.parseInt(st.nextToken()) - 1;
                    if (!stores.get(x).isEmpty()) {
                        stores.get(x).remove(0);
                    }
                    break;
                case 4:
                    x = Integer.parseInt(st.nextToken()) - 1;
                    p = Integer.parseInt(st.nextToken());
                    output.append(maxTastiness(stores.get(x), p)).append("\n");
                    break;
            }
        }

        System.out.print(output);
    }

    private static int maxTastiness(List<int[]> iceCreams, int maxPrice) {
        int[] dp = new int[maxPrice + 1];

        for (int[] iceCream : iceCreams) {
            int price = iceCream[0];
            int tastiness = iceCream[1];
            for (int j = maxPrice; j >= price; j--) {
                dp[j] = Math.max(dp[j], dp[j - price] + tastiness);
            }
        }

        return dp[maxPrice];
    }
}

**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
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

int main() {
  int q;
  cin >> q;

  struct query {
    int u, v;
    int p;
    int id;
  };
  vector<query> queries;

  vector<vector<int>> g(q);
  struct ice {
    int p, t;
  };
  vector<ice> data(q);
  vector<int> curr_node = {0}, popped = {1};
  int n = 1;

  struct prequery {
    int popped, p;
    int id;
  };
  vector<vector<prequery>> prequeries_above(q);

  for (int i = 0; i < q; ++i) {
    int ty, x;
    cin >> ty >> x;
    x--;

    const int u = curr_node[x];

    if (ty == 1) {
      curr_node.push_back(u);
      popped.push_back(popped[x]);
    } else if (ty == 2) {
      int p, t;
      cin >> p >> t;
      g[u].push_back(n);
      g[n].push_back(u);
      data[n] = {p, t};
      curr_node[x] = n++;
    } else if (ty == 3) {
      popped[x]++;
    } else {
      int p;
      cin >> p;
      prequeries_above[u].push_back({popped[x], p, i});
    }
  }

  // convert prequeries into queries
  vector<int> st;
  auto dfs = [&](auto &dfs, int u, int p = -1) -> void {
    st.push_back(u);
    for (auto [d, p, id] : prequeries_above[u]) {
      if (d >= st.size()) {
        // "dummy query" which gives right answer for empty path
        queries.push_back({u, u, 0, id});
      } else {
        queries.push_back({u, st[d], p, id});
      }
    }
    for (int v : g[u]) {
      if (v != p) dfs(dfs, v, u);
    }
    st.pop_back();
  };
  dfs(dfs, 0);

  vector<vector<int>> queries_at(n);
  for (int i = 0; i < queries.size(); ++i) {
    queries_at[queries[i].u].push_back(i);
  }

  // centroid decomp
  vector<int> ans(q, -1);
  vector<bool> eaten(n);
  vector<int> sz(n), seen_at(n, -1), side(n);
  vector<int> cc;
  vector<array<int, 2005>> dp(n);
  auto decomp = [&](auto &decomp, int u) -> void {
    if (eaten[u]) return;

    auto dfs_sz = [&](auto &dfs, int u, int p = -1) -> int {
      int sm = 1;
      for (int v : g[u]) {
        if (v == p or eaten[v]) continue;
        sm += dfs(dfs, v, u);
      }
      return sz[u] = sm;
    };

    auto dfs_c = [&](auto &dfs, int u, int p = -1) -> int {
      for (int v : g[u]) {
        if (v == p or eaten[v]) continue;
        if (2 * sz[v] >= sz[u]) return dfs(dfs, v, u);
      }
      return u;
    };

    dfs_sz(dfs_sz, u);
    u = dfs_c(dfs_c, u);

    cc.clear();
    const int time = u;
    auto dfs = [&](auto &dfs, int u, int p = -1, int s = -1) -> void {
      if (p != -1) dp[u] = dp[p];
      else fill(begin(dp[u]), end(dp[u]), 0);

      // exclude root node
      if (p != -1) {
        auto [p, t] = data[u];
        for (int i = size(dp[u]); --i >= p;) {
          dp[u][i] = max(dp[u][i], dp[u][i - p] + t);
        }
      }

      seen_at[u] = time;
      side[u] = s;
      cc.push_back(u);
      for (int v : g[u]) {
        if (v == p or eaten[v]) continue;
        dfs(dfs, v, u, s < 0 ? v : s);
      }
    };
    dfs(dfs, u);

    // solve queries
    auto du = data[u];
    for (int a : cc) {
      for (int i : queries_at[a]) {
        auto [x, y, p, id] = queries[i];
        if (seen_at[x] == u and seen_at[y] == u and
            (x == u or y == u or side[x] != side[y])) {
          int res = 0;
          for (int i = 0; i <= p; ++i) {
            res = max(res, dp[x][i] + dp[y][p - i]);
            if (i >= du.p) {
              res = max(res, dp[x][i - du.p] + dp[y][p - i] + du.t);
            }
          }

          ans[id] = res;
        }
      }
    }

    eaten[u] = true;
    for (int v : g[u]) decomp(decomp, v);
  };
  decomp(decomp, 0);

  for (int x : ans) {
    if (x != -1) cout << x << '\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
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

Buggy Code:
from collections import defaultdict, deque
import sys
input = sys.stdin.read

def process_queries(queries):
    stores = defaultdict(list)
    stores[1] = deque()  # Initialize the first store with an empty deque

    results = []
    for query in queries:
        q_type = query[0]

        if q_type == 1:
            # Open a new store, copy inventory from store x
            x = query[1]
            new_store_id = len(stores) + 1
            stores[new_store_id] = deque(stores[x])

        elif q_type == 2:
            # Add ice cream with price p and tastiness t to store x
            x, p, t = query[1], query[2], query[3]
            stores[x].append((p, t))

        elif q_type == 3:
            # Remove the earliest added ice cream in store x
            x = query[1]
            if stores[x]:
                stores[x].popleft()

        elif q_type == 4:
            # Find maximum tastiness within price limit p in store x
            x, p = query[1], query[2]

            # Knapsack DP for maximum tastiness with price limit p
            dp = [0] + [-float('inf')] * p
            for price, tastiness in stores[x]:
                if price <= p:
                    for j in range(p, price - 1, -1):
                        dp[j] = max(dp[j], dp[j - price] + tastiness)

            # Collect the result for this query
            max_tastiness = max(dp[j] for j in range(p + 1))
            results.append(max(0, max_tastiness))

    return results

# Input processing
queries = input().strip().splitlines()
q = int(queries[0].strip())
queries = [list(map(int, line.split())) for line in queries[1:]]

# Output results
results = process_queries(queries)
print('\n'.join(map(str, results)))

**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
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

Buggy Code:
def solve_knapsack(items, max_price):
    # Решение задачи о рюкзаке
    # items - список пар (цена, вкусность)
    # max_price - максимальная допустимая цена
    dp = [0] * (max_price + 1)

    for price, taste in items:
        # Идем с конца, чтобы не использовать элемент дважды
        for curr_price in range(max_price, price - 1, -1):
            dp[curr_price] = max(dp[curr_price], dp[curr_price - price] + taste)

    return dp[max_price]

def process_queries():
    q = int(input())
    # shops[i] содержит список пар (цена, вкусность) для i-го магазина
    shops = [[]]  # начинаем с пустого магазина с индексом 0
    shops.append([])  # магазин с индексом 1

    for _ in range(q):
        query = list(map(int, input().split()))
        query_type = query[0]

        if query_type == 1:
            x = query[1]
            shops.append(shops[x][:])

        elif query_type == 2:
            x, p, t = query[1:]
            shops[x].append((p, t))

        elif query_type == 3:
            x = query[1]
            if shops[x]:
                shops[x].pop(0)

        else:
            x, p = query[1:]
            result = solve_knapsack(shops[x], p)
            print(result)

# Запускаем обработку запросов
process_queries()

**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
1 1
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

public class IceCreamStores {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();
        int q = Integer.parseInt(reader.readLine());

        Map<Integer, List<int[]>> stores = new HashMap<>();
        stores.put(1, new ArrayList<>());  // Initialize store 1

        int nextStoreId = 2;

        for (int i = 0; i < q; i++) {
            StringTokenizer st = new StringTokenizer(reader.readLine());
            int type = Integer.parseInt(st.nextToken());

            switch (type) {
                case 1: {
                    int x = Integer.parseInt(st.nextToken());
                    // Create a new store with the same ice creams as store x
                    List<int[]> newStoreIceCreams = new ArrayList<>(stores.get(x));
                    stores.put(nextStoreId++, newStoreIceCreams);
                    break;
                }
                case 2: {
                    int x = Integer.parseInt(st.nextToken());
                    int p = Integer.parseInt(st.nextToken());
                    int t = Integer.parseInt(st.nextToken());
                    // Add new ice cream type to store x
                    stores.get(x).add(new int[]{p, t});
                    break;
                }
                case 3: {
                    int x = Integer.parseInt(st.nextToken());
                    // Remove the oldest ice cream type from store x
                    List<int[]> iceCreams = stores.get(x);
                    if (!iceCreams.isEmpty()) {
                        iceCreams.remove(0);
                    }
                    break;
                }
                case 4: {
                    int x = Integer.parseInt(st.nextToken());
                    int p = Integer.parseInt(st.nextToken());
                    output.append(maxTastiness(stores.get(x), p)).append("\n");
                    break;
                }
            }
        }

        System.out.print(output);
    }

    private static int maxTastiness(List<int[]> iceCreams, int maxPrice) {
        int[] dp = new int[maxPrice + 1];

        for (int[] iceCream : iceCreams) {
            int price = iceCream[0];
            int tastiness = iceCream[1];
            for (int j = maxPrice; j >= price; j--) {
                dp[j] = Math.max(dp[j], dp[j - price] + tastiness);
            }
        }

        return dp[maxPrice];
    }
}

**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 lead to incorrect behavior. The main issues in the code could arise from:

1. Incorrect handling of store cloning (type 1 queries).
2. Incorrect handling of adding ice creams to stores (type 2 queries).
3. Incorrect handling of removing the oldest ice cream (type 3 queries).
4. Incorrect calculation of the maximum tastiness (type 4 queries).

### Step-by-Step Analysis

1. **Store Cloning (Type 1 Queries)**:
   - Ensure that the new store correctly clones the ice creams from the specified store.
   - Test with a store that has multiple ice creams.

2. **Adding Ice Creams (Type 2 Queries)**:
   - Ensure that ice creams are added to the correct store.
   - Test with multiple ice creams being added to the same store.

3. **Removing the Oldest Ice Cream (Type 3 Queries)**:
   - Ensure that the oldest ice cream is removed from the store.
   - Test with a store that has multiple ice creams and multiple removals.

4. **Calculating Maximum Tastiness (Type 4 Queries)**:
   - Ensure that the maximum tastiness is calculated correctly for different price constraints.
   - Test with a store that has multiple ice creams and different price constraints.

### Test Input

```plaintext
14
2 1 5 7
2 1 3 4
2 1 2 1
1 1
2 2 4 10
2 2 1 2
3 1
3 2
4 1 4
4 1 8
4 1 2
4 2 9
4 2 1
4 2 10
```

### Explanation

1. **Adding Ice Creams to Store 1**:
   - `2 1 5 7`: Add ice cream (5, 7) to store 1.
   - `2 1 3 4`: Add ice cream (3, 4) to store 1.
   - `2 1 2 1`: Add ice cream (2, 1) to store 1.

2. **Cloning Store 1 to Store 2**:
   - `1 1`: Create store 2, which clones the ice creams from store 1.

3. **Adding Ice Creams to Store 2**:
   - `2 2 4 10`: Add ice cream (4, 10) to store 2.
   - `2 2 1 2`: Add ice cream (1, 2) to store 2.

4. **Removing Ice Creams**:
   - `3 1`: Remove the oldest ice cream from store 1.
   - `3 2`: Remove the oldest ice cream from store 2.

5. **Calculating Maximum Tastiness**:
   - `4 1 4`: Calculate maximum tastiness for store 1 with price constraint 4.
   - `4 1 8`: Calculate maximum tastiness for store 1 with price constraint 8.
   - `4 1 2`: Calculate maximum tastiness for store 1 with price constraint 2.
   - `4 2 9`: Calculate maximum tastiness for store 2 with price constraint 9.
   - `4 2 1`: Calculate maximum tastiness for store 2 with price constraint 1.
   - `4 2 10`: Calculate maximum tastiness for store 2 with price constraint 10.

This test input covers various scenarios and should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

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

public class IceCreamStores {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();
        int q = Integer.parseInt(reader.readLine());

        List<List<int[]>> stores = new ArrayList<>();
        stores.add(new ArrayList<>());

        for (int i = 0; i < q; i++) {
            StringTokenizer st = new StringTokenizer(reader.readLine());
            int type = Integer.parseInt(st.nextToken());

            switch (type) {
                case 1:
                    int x = Integer.parseInt(st.nextToken()) - 1;
                    stores.add(new ArrayList<>(stores.get(x)));
                    break;
                case 2:
                    x = Integer.parseInt(st.nextToken()) - 1;
                    int p = Integer.parseInt(st.nextToken());
                    int t = Integer.parseInt(st.nextToken());
                    stores.get(x).add(new int[]{p, t});
                    break;
                case 3:
                    x = Integer.parseInt(st.nextToken()) - 1;
                    if (!stores.get(x).isEmpty()) {
                        stores.get(x).remove(0);
                    }
                    break;
                case 4:
                    x = Integer.parseInt(st.nextToken()) - 1;
                    p = Integer.parseInt(st.nextToken());
                    output.append(maxTastiness(stores.get(x), p)).append("\n");
                    break;
            }
        }

        System.out.print(output);
    }

    private static int maxTastiness(List<int[]> iceCreams, int maxPrice) {
        int[] dp = new int[maxPrice + 1];

        for (int[] iceCream : iceCreams) {
            int price = iceCream[0];
            int tastiness = iceCream[1];
            for (int j = maxPrice; j >= price; j--) {
                dp[j] = Math.max(dp[j], dp[j - price] + tastiness);
            }
        }

        return dp[maxPrice];
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can cause issues. Here are the key points to consider:

1. **Case 1: Opening a store that copies an empty store.**
   - This should not cause any issues, but it's a good sanity check.

2. **Case 2: Adding multiple ice creams to a store and then querying the maximum tastiness.**
   - This should test the dynamic programming part of the code.

3. **Case 3: Removing ice creams from a store and then querying the maximum tastiness.**
   - This should test the removal logic and its impact on the maximum tastiness calculation.

4. **Case 4: Opening a store that copies a store with ice creams, then adding more ice creams, and querying the maximum tastiness.**
   - This should test the copying logic and its interaction with the dynamic programming part.

5. **Case 5: Querying the maximum tastiness for a store with no ice creams.**
   - This should test the edge case where the store has no ice creams available.

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

```plaintext
15
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
3 2
4 2 9
4 1 9
```

This test input includes:
- Adding ice creams to store 1 and querying the maximum tastiness.
- Opening a new store (store 2) that copies store 1.
- Adding more ice creams to store 2 and querying the maximum tastiness.
- Removing ice creams from both stores and querying the maximum tastiness.
- Querying the maximum tastiness for a store with no ice creams after removals.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bermart Ice Cream
time_limit_ms: 2000
memory_limit_mb: 1024
Description: In the Bermart chain of stores, a variety of ice cream is sold. Each type of ice cream has two parameters: price and tastiness.

Initially, there is one store numbered $1$, which sells nothing. You have to process $q$ queries of the following types:

  * $1~x$ — a new store opens, that sells the same types of ice cream as store $x$. It receives the minimum available positive index. The order of the types of ice cream in the new store is the same as in store $x$.    * $2~x~p~t$ — a type of ice cream with price $p$ and tastiness $t$ becomes available in store $x$.    * $3~x$ — a type of ice cream that was available the longest (appeared the earliest) in store $x$ is removed.    * $4~x~p$ — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Input: The first line contains a single integer $q$ ($1 \le q \le 3 \cdot 10^4$) — the number of queries.

Each of the following $q$ lines contains a query in the format described in the statement:

  * $1~x$;    * $2~x~p~t$ ($1 \le p, t \le 2000$);    * $3~x$;    * $4~x~p$ ($1 \le p \le 2000$).

Additional constraints on the input data:

  * $x$ in each query does not exceed the current number of stores (that is, $1$ plus the number of type $1$ queries);    * query type $3$ is not applied to a store that has no types of ice cream;    * there is at least one query of type $4$.

Output: For each query of type $4$, output a single integer — for store $x$, find the maximum total tastiness of a subset of types of ice cream that are sold there, such that the total price does not exceed $p$ (each type can be used in the subset no more than once).

Examples:
input:
12
2 1 5 7
2 1 3 4
4 1 4
4 1 8
4 1 2
1 1
2 2 4 10
4 1 9
4 2 9
3 1
4 1 9
4 2 9
output:
4
11
0
11
17
4
17

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int F = 2e3 + 5;
const int Q = 3e4 + 10;
struct icecream{
    int p, t;
};
struct icecreammenu{
    ///maintain dp of item mostly
    ///stack like data structure
    vector <icecream> stk;
    vector <vector<int>> dp;
    int sz;
    int get(int p){
        if(sz == 0)return 0;
        return dp[sz - 1][p];
    }
    void add(icecream x){
        dp.push_back({});
        stk.push_back(x);
        sz++;
        if(sz == 1){
            ///base case
            for(int i = 0;i <= F;i++){
                dp[sz - 1].push_back(0);
            }
            for(int i = x.p;i <= F;i++){
                dp[sz - 1][i] = x.t;
            }
            return;
        }
        dp[sz - 1].resize(F + 1);
        for(int i = F;i >= 0;i--){
            dp[sz - 1][i] = max(dp[sz - 2][i], (i - x.p >= 0?dp[sz - 2][i - x.p] + x.t:0));
        }
    };
    void pop(){
        dp[sz - 1].pop_back();
        stk.pop_back();
        sz--;
    }
    void clr(){
        dp.clear();
        while(!stk.empty())stk.pop_back();
        sz = 0;
    }
    void dbg(){
        for(int i = 0;i < sz;i++){
            for(int j = 0;j <= 10;j++){
                cout<<dp[i][j]<<' ';
            }
            cout<<stk[i].p<<' '<<stk[i].t;
            cout<<'\n';
        }
    }
    icecream top(){
        assert(sz > 0);
        return stk[sz - 1];
    }
};
struct icecreamtruck{
    icecreammenu l,r;
    int sz(){
        return l.sz + r.sz;
    }
    int get(int x){
        int ans = 0;
        for(int i = 0;i <= x;i++){
            ans = max(ans, l.get(i) + r.get(x - i));
        }
        return ans;
    }
    void rebal(int t){
        int sz2 = (sz() + t)/2;
        vector <icecream> v;
        while(l.sz > 0){
            v.push_back(l.top());
            l.pop();
        }
        int sv = (int)v.size();
        while(r.sz > 0){
            v.push_back(r.top());
            r.pop();
        }
        reverse(v.begin() + sv,v.end());
        for(int i = sz2;i < (int)v.size();i++){
            r.add(v[i]);
        }
        for(int i = sz2 - 1;i >= 0;i--){
            l.add(v[i]);
        }
    }
    icecream popfront(){
        assert(sz() > 0);
        if(r.sz == 0)rebal(0);
        icecream y = r.top();
        r.pop();
        return y;
    }
    icecream popback(){
        assert(sz() > 0);
        if(l.sz == 0)rebal(1);
        icecream y = l.top();
        l.pop();
        return y;
    }
    void pushfront(icecream x){
        r.add(x);
    }
    void pushback(icecream x){
        l.add(x);
    }
};
vector <array<int, 4>> e[Q];
vector <array<int, 2>> q2[Q];
int pos[Q];
int ans[Q];
icecreamtruck truck;
int trucks = 1, states = 1, queries = 0;
void dfs(int node){
    for(auto i:q2[node]){
        ans[i[1]] = truck.get(i[0]);
    }
    for(auto i:e[node]){
        if(i[1] == 1){
            ///copy
            dfs(i[0]);
        }else if(i[1] == 2){
            ///add
            truck.pushfront({i[2], i[3]});
            dfs(i[0]);
            truck.popfront();
        }else{
            icecream x = truck.popback();
            dfs(i[0]);
            truck.pushback(x);
        }
    }
}
int main(){
    pos[0] = 0;
    int q;
    cin>>q;
    for(int i = 0;i < q;i++){
        int cer, x, p, t;
        cin>>cer;
        if(cer == 1){
            cin>>x;
            x--;
            pos[trucks] = pos[x];
            trucks++;
            states++;
        }else if(cer == 2){
            cin>>x>>p>>t;
            x--;
            e[pos[x]].push_back({states, 2, p, t});
            pos[x] = states;
            states++;
        }else if(cer == 3){
            cin>>x;
            x--;
            e[pos[x]].push_back({states, 3, 0, 0});
            pos[x] = states;
            states++;
        }else{
            int x, p;
            cin>>x>>p;
            x--;
            q2[pos[x]].push_back({p, queries});
            queries++;
        }
    }
    dfs(0);
    for(int i = 0;i < queries;i++){
        cout<<ans[i]<<'\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
10
2 1 10 10
2 1 5 5
1 1
2 2 5 5
3 1
4 1 15
4 2 15
3 2
4 1 15
4 2 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Stabilization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a matrix of size $n \times m$, where the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The element at the intersection of the $i$-th row and the $j$-th column is denoted by $a_{ij}$.

Consider the algorithm for stabilizing matrix $a$:

  1. Find the cell $(i, j)$ such that its value is strictly greater than the values of all its neighboring cells. If there is no such cell, terminate the algorithm. If there are multiple such cells, choose the cell with the smallest value of $i$, and if there are still multiple cells, choose the one with the smallest value of $j$.   2. Set $a_{ij} = a_{ij} - 1$.    3. Go to step $1$.

In this problem, cells $(a, b)$ and $(c, d)$ are considered neighbors if they share a common side, i.e., $|a - c| + |b - d| = 1$.

Your task is to output the matrix $a$ after the stabilization algorithm has been executed. It can be shown that this algorithm cannot run for an infinite number of iterations.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. This is followed by their description.

The first line of each set of input data contains two integers $n$ and $m$ ($1 \leq n, m \leq 100, n \cdot m > 1$) — the number of rows and columns of matrix $a$.

The next $n$ lines describe the corresponding rows of the matrix. The $i$-th line contains $m$ integers $a_{i1}, a_{i2}, \ldots, a_{im}$ ($1 \leq a_{ij} \leq 10^9$).

It is guaranteed that the sum of $n \cdot m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output $n$ lines with $m$ numbers in each line — the values of the cells of matrix $a$ after the stabilization algorithm.

Examples:
input:
6
1 2
3 1
2 1
1
1
2 2
1 2
3 4
2 3
7 4 5
1 8 10
5 4
92 74 31 74
74 92 17 7
31 17 92 3
74 7 3 92
7 31 1 1
3 3
1000000000 1 1000000000
1 1000000000 1
1000000000 1 1000000000
output:
1 1
1
1
1 2
3 3
4 4 5
1 8 8
74 74 31 31
74 74 17 7
31 17 17 3
31 7 3 3
7 7 1 1
1 1 1
1 1 1
1 1 1

Note: In the first set of input data, the algorithm will select the cell $(1, 1)$ twice in a row and then terminate.

![](CDN_BASE_URL/ce7b57ad25f59927de6d90880ce5a2ba)

In the second set of input data, there is no cell whose value is strictly greater than the values of all neighboring cells.

In the third set of input data, the algorithm will select the cell $(2, 2)$ and then terminate.

![](CDN_BASE_URL/e6f79376ac190defeb9657662c825d80)

In the fourth set of input data, the algorithm will select the cell $(1, 1)$ three times and then the cell $(2, 3)$ twice.

![](CDN_BASE_URL/7651bb6af57f6669236e3dd32a7c1293)

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
	int t;cin>>t;
	while(t--){
	    int n,m;cin>>n>>m;
	    int v[n][m];

	    for(int i=0;i<n;i++){
	        for(int j=0;j<m;j++){
	            cin>>v[i][j];
	        }
	    }
	    //cout<<1;
	    if(n==1){
	        for(int i=0;i<m;i++){
	            if(i==0){
	                if(v[0][i]>v[0][i+1]){
	                    v[0][i] = v[0][i+1];
	                }
	            }
	            else if(i==m-1){
	                if(v[0][i]>v[0][i-1]){
	                    v[0][i] = v[0][i-1];
	                }
	            }
	            else{
	                if(v[0][i]>v[0][i-1]&&v[0][i]>v[0][i+1]){
	                    v[0][i] = max(v[0][i-1],v[0][i+1]);
	                }
	            }
	        }
	        for(int i=0;i<n;i++){
	            for(int j=0;j<m;j++){
	                cout<<v[i][j]<<" ";
	            }
	            cout<<"\n";
	        }
	        continue;
	    }
	    if(m==1){
	        for(int i=0;i<n;i++){
	            if(i==0){
	                if(v[i][0]>v[i+1][0]){
	                    v[i][0] = v[i+1][0];
	                }
	            }
	            else if(i==n-1){
	                if(v[i][0]>v[i-1][0]){
	                    v[i][0] = v[i-1][0];
	                }
	            }
	            else{
	                if(v[i][0]>v[i-1][0]&&v[i][0]>v[i+1][0]){
	                    v[i][0] = max(v[i-1][0],v[i+1][0]);
	                }
	            }
	        }
	        for(int i=0;i<n;i++){
	            for(int j=0;j<m;j++){
	                cout<<v[i][j]<<" ";
	            }
	            cout<<"\n";
	        }
	        continue;
	    }
	    for(int i=0;i<n;i++){
	        for(int j=0;j<m;j++){
	            if(i==0 && j==0){
	                if(v[i][j]>v[i+1][j]&&v[i][j]>v[j+1][i]){
	                    v[i][j] = max(v[i+1][j], v[i][j+1]);
	                }
	            }
	            else if(i==0 && j==m-1){
	                if(v[i][j]>v[i][j-1]&&v[i][j]>v[i+1][j]){
	                    v[i][j] = max(v[i+1][j], v[i][j-1]);
	                }
	            }
	            else if(i==0){
	                if(v[i][j]>v[i][j-1]&&v[i][j]>v[i+1][j]&&v[i][j]>v[i][j+1]){
	                    v[i][j] = max(v[i+1][j], max(v[i][j-1], v[i][j+1]));
	                }
	            }
	            else if(i== n-1 && j==0){
	                if(v[i][j]>v[i-1][j]&&v[i][j]>v[j+1][i]){
	                    v[i][j] = max(v[i-1][j], v[i][j+1]);
	                }
	            }
	            else if(i== n-1 && j==m-1){
	                if(v[i][j]>v[i-1][j]&&v[i][j]>v[j-1][i]){
	                    v[i][j] = max(v[i-1][j], v[i][j-1]);
	                }
	            }
	            else if(i==n-1){
	                if(v[i][j]>v[i][j-1]&&v[i][j]>v[i-1][j]&&v[i][j]>v[i][j+1]){
	                    v[i][j] = max(v[i-1][j], max(v[i][j-1], v[i][j+1]));
	                }
	            }
	            else if(j==0){
	                if(v[i][j]>v[i][j+1]&&v[i][j]>v[i+1][j]&&v[i][j]>v[i-1][j]){
	                    v[i][j] = max(v[i+1][j], max(v[i][j+1], v[i-1][j]));
	                }
	            }
	            else if(j==m-1){
	                if(v[i][j]>v[i][j-1]&&v[i][j]>v[i-1][j]&&v[i][j]>v[i+1][j]){
	                    v[i][j] = max(v[i+1][j], max(v[i][j-1], v[i-1][j]));
	                }
	            }
	            else{
	                if(v[i][j]>v[i][j-1]&&v[i][j]>v[i-1][j]&&v[i][j]>v[i+1][j]&&v[i][j]>v[i][j+1]){
	                    v[i][j] = max(max(v[i+1][j], v[i][j-1]), max(v[i-1][j], v[i][j+1]));
	                }
	            }
	        }
	    }
	    for(int i=0;i<n;i++){
	        for(int j=0;j<m;j++){
	            cout<<v[i][j]<<" ";
	        }
	        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
6
1 2
3 1
2 1
1
1
2 2
1 2
3 4
2 3
7 4 5
1 8 10
5 4
92 74 31 74
74 92 17 7
31 17 92 3
74 7 3 92
7 31 1 1
3 3
1000000000 1 1000000000
1 1000000000 1
1000000000 1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Stabilization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a matrix of size $n \times m$, where the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The element at the intersection of the $i$-th row and the $j$-th column is denoted by $a_{ij}$.

Consider the algorithm for stabilizing matrix $a$:

  1. Find the cell $(i, j)$ such that its value is strictly greater than the values of all its neighboring cells. If there is no such cell, terminate the algorithm. If there are multiple such cells, choose the cell with the smallest value of $i$, and if there are still multiple cells, choose the one with the smallest value of $j$.   2. Set $a_{ij} = a_{ij} - 1$.    3. Go to step $1$.

In this problem, cells $(a, b)$ and $(c, d)$ are considered neighbors if they share a common side, i.e., $|a - c| + |b - d| = 1$.

Your task is to output the matrix $a$ after the stabilization algorithm has been executed. It can be shown that this algorithm cannot run for an infinite number of iterations.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. This is followed by their description.

The first line of each set of input data contains two integers $n$ and $m$ ($1 \leq n, m \leq 100, n \cdot m > 1$) — the number of rows and columns of matrix $a$.

The next $n$ lines describe the corresponding rows of the matrix. The $i$-th line contains $m$ integers $a_{i1}, a_{i2}, \ldots, a_{im}$ ($1 \leq a_{ij} \leq 10^9$).

It is guaranteed that the sum of $n \cdot m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output $n$ lines with $m$ numbers in each line — the values of the cells of matrix $a$ after the stabilization algorithm.

Examples:
input:
6
1 2
3 1
2 1
1
1
2 2
1 2
3 4
2 3
7 4 5
1 8 10
5 4
92 74 31 74
74 92 17 7
31 17 92 3
74 7 3 92
7 31 1 1
3 3
1000000000 1 1000000000
1 1000000000 1
1000000000 1 1000000000
output:
1 1
1
1
1 2
3 3
4 4 5
1 8 8
74 74 31 31
74 74 17 7
31 17 17 3
31 7 3 3
7 7 1 1
1 1 1
1 1 1
1 1 1

Note: In the first set of input data, the algorithm will select the cell $(1, 1)$ twice in a row and then terminate.

![](CDN_BASE_URL/ce7b57ad25f59927de6d90880ce5a2ba)

In the second set of input data, there is no cell whose value is strictly greater than the values of all neighboring cells.

In the third set of input data, the algorithm will select the cell $(2, 2)$ and then terminate.

![](CDN_BASE_URL/e6f79376ac190defeb9657662c825d80)

In the fourth set of input data, the algorithm will select the cell $(1, 1)$ three times and then the cell $(2, 3)$ twice.

![](CDN_BASE_URL/7651bb6af57f6669236e3dd32a7c1293)

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

#define INF (int)1e9
#define PI 3.1415926535897932384626433832795
#define mod 1000000007
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define pb push_back
#define l_b lower_bound
#define u_b upper_bound
#define all(v) v.begin(), v.end()
#define _fastio  ios_base:: sync_with_stdio(false); cin.tie(0); cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef map<int, int> mii;
typedef unordered_map<int, int> umap_ii;
typedef unordered_map<string, int> umap_si;

/**
 * Limits in C++ for reference
 * _____________________________________________________________________________________
 * |Sr| Macro Name | Description                     | Value
 * |No|____________|_________________________________|__________________________________
 * |1.| ULLONG_MAX | Maximum value unsigned long long| 18,446,744,073,709,551,615 (10^20)
 * |2.| LLONG_MAX  | Maximum value long long         | 9,223,372,036,854,775,807 (10^19)
 * |3.| LLONG_MIN  | Minimum value long long         |-9,223,372,036,854,775,808 -1*(10^19)
 * |4.| INT_MAX    | Maximum value int               | 2,147,483,647 (10^10)
 * |5.| INT_MIN    | Minimum value int               |-2,147,483,648 (10^10)
*/

void solve()
{
    // Do your thing here :)
    int n, m;
    cin >> n >> m;
    vector<vector<int>> a(n+1, vector<int> (m+1,0));
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            cin >> a[i][j];
        }
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {

            if(i!=n && j!=m) {

                if(i==1 && j!=1) {
                    if(a[i][j]>a[i+1][j] && a[i][j]>a[i][j+1] && a[i][j-1]) {
                        int tmp = max(a[i+1][j], a[i][j+1]);
                        a[i][j] = max(tmp, a[i][j-1]);
                    }
                }
                else if(j==1 && i!=1) {
                    if(a[i][j]>a[i+1][j] && a[i][j]>a[i][j+1] && a[i][j]>a[i-1][j]) {
                        int tmp = max(a[i+1][j], a[i][j+1]);
                        a[i][j] = max(tmp, a[i-1][j]);
                    }
                }
                else if(i==1 && j==1) {
                    if(a[i][j]>a[i+1][j] && a[i][j]>a[i][j+1]) {
                        a[i][j] = max(a[i+1][j], a[i][j+1]);
                    }
                }
                else if(i!=1 && j!=1) {

                    if(a[i][j]>a[i][j+1] && a[i][j]>a[i][j-1] && a[i][j]>a[i+1][j] && a[i][j]>a[i-1][j]) {
                        int tmp = max(a[i][j+1], a[i][j-1]);
                        int tmp1 = max(a[i-1][j], a[i+1][j]);
                        a[i][j] = max(tmp, tmp1);
                    }
                }
            }
            else if(i==n && j!=m) {

                if(j==1) {
                    if(a[i][j]>a[i][j+1] && a[i][j]>a[i-1][j]) {
                        a[i][j] = max(a[i][j+1], a[i-1][j]);
                    }
                }
                else {
                    if(a[i][j]>a[i][j+1] && a[i][j]>a[i-1][j] && a[i][j]>a[i][j-1]) {
                        int tmp = max(a[i][j+1], a[i-1][j]);
                        a[i][j] = max(tmp, a[i][j-1]);
                    }
                }
            }
            else if(i!=n && j==m) {

                if(i==1) {
                    if(a[i][j]>a[i][j-1] && a[i][j]>a[i+1][j]) {
                        a[i][j] = max(a[i][j-1], a[i+1][j]);
                    }
                }
                else {
                    if(a[i][j]>a[i][j-1] && a[i][j]>a[i+1][j] && a[i][j]>a[i-1][j]) {
                        int tmp = max(a[i][j-1], a[i+1][j]);
                        a[i][j] = max(tmp, a[i-1][j]);
                    }
                }
            }
            else if(i==n && j==m) {
                if(a[i][j]>a[i][j-1] && a[i-1][j]) {
                    a[i][j] = max(a[i][j-1], a[i-1][j]);
                }
            }
        }
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
    //cout << endl;
}

int main()
{

    _fastio;
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

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

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

Think step by step.

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

### Explanation:
This test input is designed to challenge the buggy code in the following ways:

1. **Symmetry and Equal Values**: The matrix is symmetric around the center, which means that the center cell (2, 2) is the only cell that is strictly greater than all its neighbors. This will test the code's ability to correctly identify and process the center cell.

2. **Multiple Iterations**: The center cell (2, 2) has a value of 3, which is greater than all its neighbors (2). The algorithm should repeatedly select this cell and decrement its value until it is no longer greater than its neighbors. This will test the code's ability to handle multiple iterations.

3. **Edge Cases**: The matrix is small (3x3), which makes it easier to verify the output manually. The values are also small and manageable, ensuring that the test input is simple and clear.

The correct output for this test input should be:
```
1 2 1
2 2 2
1 2 1
```

This output reflects the stabilization process where the center cell (2, 2) is decremented from 3 to 2, and the algorithm terminates since no cell is strictly greater than all its neighbors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Stabilization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a matrix of size $n \times m$, where the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The element at the intersection of the $i$-th row and the $j$-th column is denoted by $a_{ij}$.

Consider the algorithm for stabilizing matrix $a$:

  1. Find the cell $(i, j)$ such that its value is strictly greater than the values of all its neighboring cells. If there is no such cell, terminate the algorithm. If there are multiple such cells, choose the cell with the smallest value of $i$, and if there are still multiple cells, choose the one with the smallest value of $j$.   2. Set $a_{ij} = a_{ij} - 1$.    3. Go to step $1$.

In this problem, cells $(a, b)$ and $(c, d)$ are considered neighbors if they share a common side, i.e., $|a - c| + |b - d| = 1$.

Your task is to output the matrix $a$ after the stabilization algorithm has been executed. It can be shown that this algorithm cannot run for an infinite number of iterations.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. This is followed by their description.

The first line of each set of input data contains two integers $n$ and $m$ ($1 \leq n, m \leq 100, n \cdot m > 1$) — the number of rows and columns of matrix $a$.

The next $n$ lines describe the corresponding rows of the matrix. The $i$-th line contains $m$ integers $a_{i1}, a_{i2}, \ldots, a_{im}$ ($1 \leq a_{ij} \leq 10^9$).

It is guaranteed that the sum of $n \cdot m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output $n$ lines with $m$ numbers in each line — the values of the cells of matrix $a$ after the stabilization algorithm.

Examples:
input:
6
1 2
3 1
2 1
1
1
2 2
1 2
3 4
2 3
7 4 5
1 8 10
5 4
92 74 31 74
74 92 17 7
31 17 92 3
74 7 3 92
7 31 1 1
3 3
1000000000 1 1000000000
1 1000000000 1
1000000000 1 1000000000
output:
1 1
1
1
1 2
3 3
4 4 5
1 8 8
74 74 31 31
74 74 17 7
31 17 17 3
31 7 3 3
7 7 1 1
1 1 1
1 1 1
1 1 1

Note: In the first set of input data, the algorithm will select the cell $(1, 1)$ twice in a row and then terminate.

![](CDN_BASE_URL/ce7b57ad25f59927de6d90880ce5a2ba)

In the second set of input data, there is no cell whose value is strictly greater than the values of all neighboring cells.

In the third set of input data, the algorithm will select the cell $(2, 2)$ and then terminate.

![](CDN_BASE_URL/e6f79376ac190defeb9657662c825d80)

In the fourth set of input data, the algorithm will select the cell $(1, 1)$ three times and then the cell $(2, 3)$ twice.

![](CDN_BASE_URL/7651bb6af57f6669236e3dd32a7c1293)

Buggy Code:
/******************************************************************************

                              Online C++ Compiler.
               Code, Compile, Run and Debug C++ program online.
Write your code in this editor and press "Run" button to compile and execute it.

*******************************************************************************/

#include <iostream>
#include <algorithm>
#include <climits>
#include <cstring>
#include <cmath>
#define lit long long
int binarySearch(lit nums[][0], lit left, lit right, lit target) {

    return -1;
}
using namespace std;
int main()
{
lit i;
cin>>i;
    while(i--)
    {
        lit n,k;
        cin>>n>>k;

        lit a[n][k];

        for(lit i=0;i<n;i++)
        {
            for(lit j=0;j<k;j++)
            {
                cin>>a[i][j];
            }
        }
        if(n*k==2)
        {
            if(n==2)
            {
                lit d=min(a[0][0],a[1][0]);
                a[0][0]=d;
                a[1][0]=d;
            }
            else
            {
                lit d=min(a[0][0],a[0][1]);
                a[0][0]=d;
                a[0][1]=d;
            }

        }
        else
        {
        for(lit i=0;i<n;i++)
        {
            for(lit j=0;j<k;j++)
            {
                if(i>0 && i<n-1 && j>0 && j<k-1)
                {
                    if(a[i+1][j]<a[i][j] && a[i][j+1]<a[i][j] && a[i-1][j]<a[i][j] && a[i][j-1]<a[i][j])
                    {
                        a[i][j]=max(max(a[i+1][j],a[i][j+1]),max(a[i-1][j],a[i][j-1]));
                    }
                }
                else if(i>0 && i<n-1 && j==0)
                {
                    if(a[i+1][j]<a[i][j] && a[i][j+1]<a[i][j] && a[i-1][j]<a[i][j])
                    {
                        a[i][j]=max(max(a[i+1][j],a[i][j+1]),a[i-1][j]);
                    }

                }
                else if(i>0 && i<n-1 && j==k-1)
                {
                    if(a[i+1][j]<a[i][j] && a[i][j-1]<a[i][j] && a[i-1][j]<a[i][j])
                    {
                        a[i][j]=max(max(a[i+1][j],a[i][j-1]),a[i-1][j]);
                    }
                }
                else if(i==0 && j>0 && j<k-1)
                {
                    if(a[i+1][j]<a[i][j] && a[i][j-1]<a[i][j] && a[i][j+1]<a[i][j])
                    {
                        a[i][j]=max(max(a[i+1][j],a[i][j-1]),a[i][j+1]);
                    }
                }
                else if(i==n-1 && j>0 && j<k-1)
                {
                    if(a[i-1][j]<a[i][j] && a[i][j-1]<a[i][j] && a[i][j+1]<a[i][j])
                    {
                        a[i][j]=max(max(a[i-1][j],a[i][j-1]),a[i][j+1]);
                    }
                }
                else if(i==0 && j==0)
                {
                    if(a[i+1][j]<a[i][j] && a[i][j+1]<a[i][j])
                    {
                        a[i][j]=max(a[i+1][j],a[i][j+1]);
                    }
                }

                else if(i==n-1 && j==0)
                {
                    if(a[i-1][j]<a[i][j] && a[i][j+1]<a[i][j])
                    {
                        a[i][j]=max(a[i-1][j],a[i][j+1]);
                    }
                }
                else if(i==0 && j==k-1)
                {
                    if(a[i][j-1]<a[i][j] && a[i+1][j]<a[i][j])
                    {
                        a[i][j]=max(a[i][j-1],a[i+1][j]);
                    }
                }
                else if(i==n-1 && j==k-1)
                {
                    if(a[i-1][j]<a[i][j] && a[i][j-1]<a[i][j])
                    {
                        a[i][j]=max(a[i-1][j],a[i][j-1]);
                    }
                }

            }
        }
        }
        for(lit i=0;i<n;i++)
        {
            for(lit j=0;j<k;j++)
            {
                cout<<a[i][j]<<" ";
            }
            cout<<endl;
        }

    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Stabilization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a matrix of size $n \times m$, where the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The element at the intersection of the $i$-th row and the $j$-th column is denoted by $a_{ij}$.

Consider the algorithm for stabilizing matrix $a$:

  1. Find the cell $(i, j)$ such that its value is strictly greater than the values of all its neighboring cells. If there is no such cell, terminate the algorithm. If there are multiple such cells, choose the cell with the smallest value of $i$, and if there are still multiple cells, choose the one with the smallest value of $j$.   2. Set $a_{ij} = a_{ij} - 1$.    3. Go to step $1$.

In this problem, cells $(a, b)$ and $(c, d)$ are considered neighbors if they share a common side, i.e., $|a - c| + |b - d| = 1$.

Your task is to output the matrix $a$ after the stabilization algorithm has been executed. It can be shown that this algorithm cannot run for an infinite number of iterations.

Input: Each test consists of multiple sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. This is followed by their description.

The first line of each set of input data contains two integers $n$ and $m$ ($1 \leq n, m \leq 100, n \cdot m > 1$) — the number of rows and columns of matrix $a$.

The next $n$ lines describe the corresponding rows of the matrix. The $i$-th line contains $m$ integers $a_{i1}, a_{i2}, \ldots, a_{im}$ ($1 \leq a_{ij} \leq 10^9$).

It is guaranteed that the sum of $n \cdot m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output $n$ lines with $m$ numbers in each line — the values of the cells of matrix $a$ after the stabilization algorithm.

Examples:
input:
6
1 2
3 1
2 1
1
1
2 2
1 2
3 4
2 3
7 4 5
1 8 10
5 4
92 74 31 74
74 92 17 7
31 17 92 3
74 7 3 92
7 31 1 1
3 3
1000000000 1 1000000000
1 1000000000 1
1000000000 1 1000000000
output:
1 1
1
1
1 2
3 3
4 4 5
1 8 8
74 74 31 31
74 74 17 7
31 17 17 3
31 7 3 3
7 7 1 1
1 1 1
1 1 1
1 1 1

Note: In the first set of input data, the algorithm will select the cell $(1, 1)$ twice in a row and then terminate.

![](CDN_BASE_URL/ce7b57ad25f59927de6d90880ce5a2ba)

In the second set of input data, there is no cell whose value is strictly greater than the values of all neighboring cells.

In the third set of input data, the algorithm will select the cell $(2, 2)$ and then terminate.

![](CDN_BASE_URL/e6f79376ac190defeb9657662c825d80)

In the fourth set of input data, the algorithm will select the cell $(1, 1)$ three times and then the cell $(2, 3)$ twice.

![](CDN_BASE_URL/7651bb6af57f6669236e3dd32a7c1293)

Buggy Code:
from sys import stdin
def input(): return stdin.readline()

def solve():
    x = list(map(int, input().split()))
    a=[0]*x[0]
    i=0
    while(i<x[0]):
        a[i]=list(map(int, input().split()))
        i=i+1
    j=0
    i=0
    while(i<x[0]):
        while(j<x[1]):
            if(i==0):
                if(j==0):
                    if(x[1]>1 and x[0]>1):
                        if(a[i][j+1]<a[i][j] and (a[i+1][j]<a[i][j])):
                            a[i][j]=max([a[i][j+1],a[i+1][j]])
                            i=0
                            j=0
                            continue
                    elif(x[1]>1):
                        if(a[i][j+1]<a[i][j] ):
                            a[i][j]=a[i][j+1]
                            i=0
                            j=0
                            continue
                    elif(x[0]>1):
                        if(a[i+1][j]<a[i][j] ):
                            a[i][j]=a[i+1][j]
                            i=0
                            j=0
                            continue
                elif(j==x[1]-1):
                    if(x[0]>1):
                        if(a[i][j-1]<a[i][j] and (a[i+1][j]<a[i][j])):
                            a[i][j]=max([a[i][j-1],a[i+1][j]])
                            i=0
                            j=0
                            continue
                    elif(x[0]==1):
                        if(a[i][j-1]<a[i][j] ):
                            a[i][j]=a[i][j-1]
                            i=0
                            j=0
                            continue
                else:
                    if(x[0]>1):
                        if(a[i][j+1]<a[i][j] and (a[i+1][j]<a[i][j]) and a[i][j-1]<a[i][j]):
                            a[i][j]=max([a[i][j+1],a[i][j-1],a[i+1][j]])
                            i=0
                            j=0
                            continue
                    else:
                        if(a[i][j+1]<a[i][j] and a[i][j-1]<a[i][j]):
                            a[i][j]=max([a[i][j+1],a[i][j-1]])
                            i=0
                            j=0
                            continue
            elif(i==x[0]-1):
                if(j==0):
                    if(x[1]>1):
                        if(a[i][j+1]<a[i][j] and a[i-1][j]<a[i][j]):
                            a[i][j]=max([a[i-1][j],a[i][j+1]])
                            i=0
                            j=0
                            continue
                    else:
                        if(a[i-1][j]<a[i][j]):
                            a[i][j]=a[i-1][j]
                            i=0
                            j=0
                            continue
                elif(j==x[1]-1):
                    if(a[i][j-1]<a[i][j] and (a[i-1][j]<a[i][j])):
                        a[i][j]=max([a[i-1][j],a[i][j-1]])
                        i=0
                        j=0
                        continue
                else:
                    if(a[i][j+1]<a[i][j] and (a[i-1][j]<a[i][j]) and a[i][j-1]<a[i][j]):
                        a[i][j]=max([a[i-1][j],a[i][j-1],a[i][j+1]])
                        i=0
                        j=0
                        continue
            else:
                if(j!=x[1]-1):
                    if(a[i][j+1]<a[i][j] and (a[i-1][j]<a[i][j]) and a[i][j-1]<a[i][j] and (a[i+1][j]<a[i][j])):
                            a[i][j]=max([a[i-1][j],a[i][j-1],a[i+1][j],a[i][j+1]])
                            i=0
                            j=0
                            continue
                else:
                    if((a[i-1][j]<a[i][j]) and a[i][j-1]<a[i][j] and (a[i+1][j]<a[i][j])):
                            a[i][j]=max([a[i-1][j],a[i][j-1],a[i+1][j]])
                            i=0
                            j=0
                            continue
            if(i==x[0]-1 and j==x[1]-1):
                i=1000
                j=1000
            j=j+1
        i=i+1
        j=0
    i=0
    j=0
    while(i<x[0]):
        while(j<x[1]):
            print(a[i][j],end=' ')
            j=j+1
        print()
        i=i+1
        j=0

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

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

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

Think step by step.

Response:
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 probability of the algorithm failing. Here are the key considerations:

1. **Boundary Conditions**: Test the behavior at the edges of the matrix (top, bottom, left, right).
2. **Uniform Values**: Test with uniform values to see how the algorithm handles identical cells.
3. **Local Maximums**: Test with local maximums that are strictly greater than their neighbors.
4. **Mixed Values**: Test with a mix of values to see how the algorithm handles different configurations.

Given the constraints and the need for a small-scale test input, let's construct a test case that includes a local maximum and some boundary conditions.

### Test Input
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```

### Explanation
- **Matrix Size**: 3x3 to keep it small and manageable.
- **Values**: The values are arranged such that the center cell (5) is a local maximum but not the global maximum. The cell (9) is the global maximum and is at the bottom-right corner.
- **Expected Behavior**: The algorithm should stabilize the matrix by reducing the value of the local maximum (5) until it is no longer greater than its neighbors. The global maximum (9) should also be reduced as it is greater than its neighbors.

This test case will help expose issues in the algorithm's logic for handling local and global maximums, as well as boundary conditions.