**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

Examples:
input:
5 1
1 2
1 3
3 4
3 5
1
output:
Ron

Note:

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

//implements Runnable
public class Main {
    static long md=(long)998244353;
    static long Linf=Long.MAX_VALUE/2;
    static int inf=Integer.MAX_VALUE/2;
    static int N=35000;
    static int n=0,m=0;
    static long ans=0;

    static
    class Node{
        int x;
        int y;
        public Node() {

        }
        public Node(int u,int v) {
            x=u;
            y=v;
        }
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Node now = (Node) o;
            return x==now.x&&y==now.y;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }
    static
    class Edge{
        int fr,to,nxt;
        public Edge(int u,int v){
            fr=u;
            to=v;
        }
    }
    static Edge[] e;
    static int[] head;
    static int cnt=0;
    static void addEdge(int fr,int to){
        cnt++;
        e[cnt]=new Edge(fr,to);
        e[cnt].nxt=head[fr];
        head[fr]=cnt;
    }
    static boolean[] Tr;
    static boolean[] Fa;
    static void dfs(int x,int fa){
        if(e[head[x]].nxt==0){
            Tr[x]=true;
            return;
        }

        for(int i=head[x];i>0;i=e[i].nxt){
            int v=e[i].to;
            if(v==fa)continue;
            dfs(v,x);
            if(Tr[v])Fa[x]=true;
            if(Fa[v])Tr[x]=true;
        }

    }

    static void solve() throws Exception{
        AReader input=new AReader();
//        Scanner input=new Scanner(new InputStreamReader(System.in));
//        String fileName="";
//		Scanner input=new Scanner(new FileReader(fileName));
//        BufferedReader input = new BufferedReader(new FileReader(fileName));
//        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        BufferedWriter out=new BufferedWriter(new OutputStreamWriter(System.out));
        String al="abcdefghijklmnopqrstuvwxyz";
        char[] ac=al.toCharArray();
        n=input.nextInt();
        int T=input.nextInt();
        while(T>0){
            T--;
            e=new Edge[2*n+1];
            head=new int[n+1];
            for(int i=1;i<n;++i){
                int u=input.nextInt();
                int v=input.nextInt();
                addEdge(u,v);
                addEdge(v,u);
            }
            int x= input.nextInt();
            Tr=new boolean[n+1];
            Fa=new boolean[n+1];
            dfs(x,0);

            if(Fa[x])out.write("Ron\n");
            else out.write("Hermione\n");
        }

        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        solve();
    }
    //	public static final void main(String[] args) throws Exception {
//		  new Thread(null, new Tx2(), "线程名字", 1 << 27).start();
//	}
//		@Override
//		public void run() {
//			try {
//				//原本main函数的内容
//				solve();
//
//			} catch (Exception e) {
//			}
//		}

    static
    class AReader{
        BufferedReader bf;
        StringTokenizer st;

        public AReader(){
            bf=new BufferedReader(new InputStreamReader(System.in));
            st=new StringTokenizer("");
        }
        public String nextLine() throws IOException{
            return bf.readLine();
        }
        public String next() throws IOException{
            while(!st.hasMoreTokens()){
                st=new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }
        public char nextChar() throws IOException{
            //确定下一个token只有一个字符的时候再用
            return next().charAt(0);
        }
        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException{
            return Long.parseLong(next());
        }
        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }
        public float nextFloat() throws IOException{
            return Float.parseFloat(next());
        }
        public byte nextByte() throws IOException{
            return Byte.parseByte(next());
        }
        public short nextShort() throws IOException{
            return Short.parseShort(next());
        }
        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are some steps to consider:

1. **Tree Structure**: The code uses a depth-first search (DFS) to determine the winning conditions. We need to consider different tree structures, such as:
   - Linear trees (a chain of nodes).
   - Star trees (one central node connected to all other nodes).
   - Balanced trees (a tree with a balanced structure).

2. **Node Initialization**: Ensure that the starting node is one that can lead to a non-trivial game.

3. **DFS Logic**: The DFS logic in the code seems to have some issues, particularly in how it handles leaf nodes and their parents. We need to create a scenario where the DFS logic might misinterpret the tree structure.

4. **Small Scale**: Since the output length is limited, we should focus on a small-scale test input.

Given these considerations, let's create a test input that includes a small, balanced tree and a starting node that could expose the DFS logic flaw.

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

### Explanation
- **Tree Structure**: The tree is balanced with node 1 as the root. Node 1 has two children: node 2 and node 3. Node 2 has two children: node 4 and node 5.
- **Starting Node**: The starting node is node 2, which is a non-leaf node with two children. This should challenge the DFS logic to correctly determine the winning condition.

This test input is small and valid, and it should help expose any issues in the DFS logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main() {
    int n;
    cin >>n;
    for (int i=1;i<=10;i++) {
        int a,b;
        cin >>a >>b;
        if (a>b) cout <<b <<" " <<a <<endl;
        else cout <<a <<" " <<b <<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:
To expose the bug in the provided code, we need to focus on the loop condition. The code uses a fixed loop that runs 10 times, regardless of the value of `t` (the number of test cases). This can cause issues if `t` is less than 10 or more than 10.

### Step-by-Step Reasoning:

1. **Identify the Bug**: The loop condition `for (int i=1;i<=10;i++)` is fixed to run 10 times, but the number of test cases `t` can vary from 1 to 100. If `t` is less than 10, the code will still try to read more inputs than provided, causing an issue. If `t` is more than 10, the code will only process the first 10 test cases and ignore the rest.

2. **Generate Test Input**:
   - **Case 1**: `t` is less than 10. This will cause the code to try to read more inputs than provided.
   - **Case 2**: `t` is more than 10. This will cause the code to ignore the extra test cases.

### Chosen Test Input:
- **Test Input 1**: `t = 5` (less than 10)
- **Test Input 2**: `t = 15` (more than 10)

### Test Input:
```plaintext
5
1 9
8 4
1 4
3 4
2 0
```

### Explanation:
- **Test Input 1**:
  - `t = 5` (5 test cases)
  - The code will try to read 10 test cases but only 5 are provided, which will cause an issue.

This test input should expose the bug in the loop condition. The code will either crash or produce incorrect output due to the mismatch between the number of test cases and the fixed loop.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include<iostream>
#include<cstdio>
using namespace std;
int main() {
	int n, a[10][10];
	scanf("%d", &n);

	for (int j = 1; j <= n; j++){
		for (int i = 1; i <=2; i++)
			cin >> a[j][i];
	}
	for (int j = 1; j <= n; j++) {
		if (a[j][2] >= a[j][1])
			cout << a[j][1] << " " << a[j][2];
		else
			cout << a[j][2] << " " << a[j][1];
		printf("\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 provided buggy code, we need to consider the following:

1. **Array Indexing Issue**: The code uses a 2D array `a[10][10]` to store the input values. However, the input values are read into the array starting from index 1, which is unconventional in C++ (arrays typically start at index 0). This could lead to issues if the input values are not correctly indexed.

2. **Input Format**: The input format specifies that each test case contains two integers \(x\) and \(y\) where \(0 \leq x, y \leq 9\).

3. **Edge Cases**: We should consider edge cases where \(x\) and \(y\) are equal, and where they are at the boundary values (0 and 9).

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
3
0 0
9 9
5 5
```

This test input includes:
- Two equal values at the minimum boundary (0).
- Two equal values at the maximum boundary (9).
- Two equal values in the middle (5).

This input will help verify if the code correctly handles equal values and if it correctly indexes the array. If the code is buggy, it might fail to correctly handle these cases, especially if it has issues with array indexing.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

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

int main() {
           int t;
           cin>>t;
           t--;
           while(t--){
            int a,b;
            for(int i=0;i<t;++i){
                cin>>a>>b;
                if(a>b){
                    swap(a,b);
                    cout<<a<<" "<<b<<endl;
                    }else{
                        cout<<a<<" "<<b<<endl;
                        break;
                    }

   }
           }

}

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

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

Think step by step.

Response:
```plaintext
3
1 9
3 3
9 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main(){
    int x[100],y[100],n;
    cin >> n;
    int a,b;
    for (int i = 0;i < n;i++){
        cin >> x[i] >> y[i];
    }
    for (int i = 0; i<=n;i++){
        a=x[i];
        b=y[i];
        if ( a >= 0 and b >= 0 and a <= 9 and b <=9){
            if( a > b){
                cout << b << " " << a << endl;
            }else{
                cout << a <<" " << b << endl;
            }
        }
    }
    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 buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The main bug in the provided code is the loop condition `for (int i = 0; i <= n; i++)`, which should be `for (int i = 0; i < n; i++)` to avoid accessing out-of-bounds array elements.

Here are the steps to generate a test input:

1. **Single Test Case with Equal Values:**
   - This will test the condition where `x` and `y` are the same.
   - Example: `1 1`

2. **Single Test Case with Different Values:**
   - This will test the condition where `x` and `y` are different.
   - Example: `2 3`

3. **Multiple Test Cases with Different Values:**
   - This will test the loop and multiple conditions.
   - Example: `1 2`, `3 4`, `5 6`

4. **Test Case with Maximum and Minimum Values:**
   - This will test the boundary conditions.
   - Example: `0 9`, `9 0`

5. **Test Case with Out-of-Bounds Access:**
   - This will test the loop condition to ensure it doesn't access out-of-bounds array elements.
   - Example: `1 2` (for `n = 1`)

Combining these, we can create a small-scale test input:

```plaintext
3
1 1
2 3
0 9
```

This test input includes:
- A test case with equal values (`1 1`).
- A test case with different values (`2 3`).
- A test case with boundary values (`0 9`).

This should be sufficient to expose the bug in the loop condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int ans=1;
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		ans++;
		int a,b;
		cin>>a>>b;
		if(ans%3!=0)
			cout<<a<<" "<<b<<endl;
		else if(ans%3==0)
			cout<<b<<" "<<a<<endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code first:

### Analysis of the Buggy Code
The code reads the number of test cases, `n`, and then processes each test case. For each test case, it reads two integers, `a` and `b`. The variable `ans` is incremented at the start of each test case. Depending on the value of `ans % 3`, the code outputs the integers in different orders:
- If `ans % 3 != 0`, it outputs `a` and `b` in the original order.
- If `ans % 3 == 0`, it outputs `b` and `a` in reverse order.

### Identifying the Bug
The bug is that the output order is dependent on the value of `ans % 3`, which is not related to the actual values of `a` and `b`. The correct output should always be the minimum of `a` and `b` followed by the maximum of `a` and `b`.

### Challenging Test Input
To expose this bug, we need a test input where the order of `a` and `b` is such that the incorrect output order (when `ans % 3 == 0`) is clearly visible. We can use a small number of test cases to demonstrate this.

### Test Input
Let's use the following test input:
```plaintext
3
1 2
3 4
5 6
```

### Explanation
- For the first test case (`1 2`), `ans` is 2, and `2 % 3 != 0`, so the output will be `1 2` (correct).
- For the second test case (`3 4`), `ans` is 3, and `3 % 3 == 0`, so the output will be `4 3` (incorrect).
- For the third test case (`5 6`), `ans` is 4, and `4 % 3 != 0`, so the output will be `5 6` (correct).

This input will clearly show that the second test case is outputted in the wrong order, exposing the bug.

### Output
```plaintext
3
1 2
3 4
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

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

int n,a,b;

int main()
{
    cin>>n;
    for(int i=0;i<10;i++)
    {
        cin>>a>>b;
        cout<<min(a,b)<<" "<<max(a,b)<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
0 0
9 9
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

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

int main() {
    int num;
    int numbers[100] = {0};
    cin >> num;
    if (num <50) {
        for (int i = 0; i < num * 2; i++)
        {
            cin >> numbers[i];
        }
        for (int i = 0; i < num * 2; i += 2)
        {
            if (numbers[i] <= numbers[i + 1]) {
                cout << numbers[i] << " " << numbers[i + 1] << endl;

            }
            else {
                cout << numbers[i + 1] << " " << numbers[i] << endl;
            }
        }

    }
    else {
        cout << "enter 50 only";
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the following points:

1. **Input Validation**: The code has a condition that only allows up to 50 test cases. If the number of test cases is greater than 50, it outputs "enter 50 only" and does not process the test cases. This is a clear bug because the problem description specifies that the number of test cases can be up to 100.

2. **Edge Cases**: We should also consider edge cases where the values of \(x\) and \(y\) are the same, as well as the smallest and largest possible values within the given range (0 and 9).

3. **Input Size**: The test input should be small-scale but complete and valid to ensure it fits within the output length limit.

Given these considerations, a challenging test input that exposes the bug would be:

```plaintext
51
0 9
1 8
2 7
3 6
4 5
5 4
6 3
7 2
8 1
9 0
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
0 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 0
0 2
1 3
2 4
3 5
4 6
5 7
6 8
7 9
8 0
9 1
```

This input includes:
- 51 test cases to trigger the "enter 50 only" message.
- All possible pairs of \(x\) and \(y\) from 0 to 9, including pairs where \(x\) and \(y\) are the same.
- Pairs where \(x\) and \(y\) are in both ascending and descending order.

This input will expose the bug in the code by triggering the incorrect message and ensuring that the code does not process the test cases as expected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

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

int main() {
    int num;
    int numbers[100] = {0};
    cin >> num;
    for (int i = 0; i < num*2; i++)
    {
        cin >> numbers[i];
    }
    for (int i = 0; i < num*2; i+=2)
    {
        if (numbers[i] <= numbers[i + 1]) {
            cout << numbers[i] << " " << numbers[i + 1] << endl;

        }
        else {
            cout<< numbers[i+1] << " " << numbers[i] << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are some key points to consider:

1. **Edge Cases for Input Values**: The input values for \(x\) and \(y\) are constrained to be between 0 and 9. We should test the boundaries of this range.
2. **Duplicate Values**: The code should handle cases where \(x\) and \(y\) are the same.
3. **Order of Input**: The code should correctly handle both \(x < y\) and \(x > y\).

Given these considerations, let's create a small-scale test input that includes these edge cases:

```plaintext
5
0 9
9 0
3 3
5 5
4 4
```

This test input includes:
- The minimum and maximum values (0 and 9).
- Duplicate values (3, 3, 5, 5, 4, 4).
- Both \(x < y\) and \(x > y\) scenarios.

This should help in exposing any potential issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main() {
int n ,n1 ,n2;
int numsmin[10]{};
int numsmax[10]{};
int c=0;
cin >> n;
for(int i=0 ; i<n ; ++i){
    cin >> n1 >> n2;
    if (n1<n2) {
        numsmin[c] = n1;
        numsmax[c] = n2;
    }
    else {
        numsmin[c] = n2;
        numsmax[c] = n1;
    }
    c++;
}
    for (int i=0 ; i<n ; i++){
        cout << numsmin[i] << " " << numsmax[i] << 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:
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 might cause the code to fail. Here are the steps to think through this:

1. **Identify the Bug**: The code seems to correctly handle the basic cases of comparing two integers and outputting the minimum and maximum. However, the bug might be related to how it handles the indexing of the arrays `numsmin` and `numsmax`.

2. **Edge Cases**:
   - **Equal Values**: When `n1` and `n2` are equal, the code should handle it correctly.
   - **Boundary Values**: When `n1` and `n2` are at the boundaries of the allowed range (0 and 9).
   - **Large Number of Test Cases**: Ensure the code handles the maximum number of test cases (100).

3. **Test Input**:
   - We will create a small-scale test input with a few test cases that cover the edge cases and boundary values.

Here is the test input that covers these scenarios:

```plaintext
5
0 0
9 9
1 9
9 1
5 5
```

Explanation:
- The first test case (`0 0`) checks if the code correctly handles equal values at the minimum boundary.
- The second test case (`9 9`) checks if the code correctly handles equal values at the maximum boundary.
- The third and fourth test cases (`1 9` and `9 1`) check if the code correctly identifies the minimum and maximum when the values are at the boundaries and in both possible orders.
- The fifth test case (`5 5`) checks if the code correctly handles equal values in the middle of the range.

This test input is small but comprehensive, covering various edge cases that can help expose any potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main() {
int n ,n1 ,n2;
int numsmin[10]{};
int numsmax[10]{};
int c=0;
cin >> n;
for(int i=0 ; i<n ; ++i){
    cin >> n1 >> n2;
    if (n1<n2) {
        numsmin[c] = n1;
        numsmax[c] = n2;
    }
    else {
        numsmin[c] = n2;
        numsmax[c] = n1;
    }
    c++;
}
    for (int i=0 ; i<n ; i++){
        cout << numsmin[i] << " " << numsmax[i] << 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:
To expose the bug in the given code, we need to focus on edge cases and scenarios where the code might fail. The code provided uses arrays `numsmin` and `numsmax` to store the minimum and maximum values for each test case. The issue with the code is that it does not handle the case where `t` (the number of test cases) is greater than 10, which could lead to out-of-bounds array access.

However, since the problem constraints specify that `1 <= t <= 100`, we should focus on other potential issues. One such issue is the handling of equal values (i.e., when `n1 == n2`), which the code should handle correctly but might still have subtle issues.

Let's consider the following test input to expose the bug:

```plaintext
3
0 0
9 9
1 2
```

This test input includes:
1. Two cases where `n1` and `n2` are equal (0 and 9).
2. One case where `n1` and `n2` are different (1 and 2).

This input will help verify if the code correctly handles equal values and correctly swaps the values when `n1` is greater than `n2`.

Here is the test input:

```plaintext
3
0 0
9 9
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

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

void sort(int loops) {
  for (int i=0; i<loops; i++) {
    int x, y;
    cin >> x >> y;
    if (y < x) {
      cout << y << " " << x << endl;
    } else {
      cout << x << " " << y << endl;
    }
  }
}

int main() {
  int loops;
  cin >> loops;
  sort(10);
}

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

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

Think step by step.

Response:
```plaintext
3
0 9
9 0
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define vl vector<ll>
#define pb push_back
#define endl '\n'
#define pf push_front
#define OO 1e18
#define dql deque<ll>
#define GO  cin.tie(0);cout.tie(0);ios::sync_with_stdio(0)
#define tt ll t;cin>>t
#define nn long long n;cin>>n
#define forn(i,n) for(ll i=0;i<n;i++)
#define all(v)  v.begin(), v.end()
#define allr(v) v.rbegin() , v.rend()
#define ld long double
#define prq priorty_queue
#define R return
#define C continue
#define print(a) for(ll i=0;i<a.size();i++) cout<<a[i]<<" "; cout<<endl;
#define to_int(a) a-48
#define yn(x) if(x) cout<<"YES"; else cout<<"NO"; cout<<endl;
#define len(a,b) sqrtl((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))
#define nod(x) static_cast<ll>(log10(x))+1
#define lg(Fen) (31^__builtin_clz(Fen)
#include <ext/pb_ds/assoc_container.hpp>

int main() {
    GO;
cout<<"1 9\n"
 "4 8\n"
 "1 4\n"
 "3 4\n"
 "0 2\n"
 "2 4\n"
 "6 9\n"
 "3 3\n"
 "0 0\n"
 "9 9";
}

**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 might cause the code to fail. The buggy code in this case is not a functional implementation of the algorithm; it simply prints a hardcoded output. Therefore, any input that does not match the hardcoded output will expose the bug.

Given the constraints:
- \(1 \leq t \leq 100\)
- \(0 \leq x, y \leq 9\)

We can create a small-scale test input that includes:
1. Cases where \(x\) and \(y\) are the same.
2. Cases where \(x\) is less than \(y\).
3. Cases where \(x\) is greater than \(y\).
4. Edge cases where \(x\) or \(y\) is 0 or 9.

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

```plaintext
5
0 0
1 2
2 1
9 9
5 5
```

This input includes:
- Two cases where \(x\) and \(y\) are the same (0 0 and 9 9).
- One case where \(x\) is less than \(y\) (1 2).
- One case where \(x\) is greater than \(y\) (2 1).
- One case with a middle value (5 5).

This test input should expose the bug in the buggy code, as the hardcoded output does not match the expected output for these inputs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
    int t;
    cin>>t;
    int i,j;
    for(int i=0; i<t; i++)
    {
        cin>>i>>j;
        if(i==j)
        {
            cout<<i<<" "<<j<<endl;
        }
        else if(j<i)
        {
            cout<<j<<" "<<i<<endl;
        }
        else
        {
            cout<<i<<" "<<j<<endl;
        }
    }
    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
0 1
1 0
2 2
3 3
4 5
5 4
6 6
7 8
8 7
9 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include<iostream>
using namespace std;
int main(){
    int n;
    cin>>n;
    for(int i=0;i<10;i++){
        int x;
         cin>>x;
         int y;
         cin>>y;
         if(x>y){
            cout<<y<<" "<<x<<endl;
         }
         else {
            cout<<x<<" "<<y<<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
3
0 9
9 0
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int x;
    cin>>x;
    int arr[10][2];
        for(int i=0;i<10;i++)
        {
            for(int j=0;j<2;j++)
                cin>>arr[i][j];
        }
        for(int i=0;i<10;i++)
        {
            cout<<min(arr[i][0],arr[i][1])<<" "<<max(arr[i][0],arr[i][1])<<"\n";
        }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define endl "\n"
#define fx(x) fixed << setprecision(x)
using namespace std;

void speed()
{
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
}

void solve()
{
    int a, b;
    cin >> a >> b;
    int res = a - b;

    if (res == a || res == b)
        cout << b << " " << a << endl;

    else
        cout << a << " " << b << endl;
}

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

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases where the logic in the `solve` function fails. The key issue in the code is the condition `if (res == a || res == b)`, which is incorrect. This condition is always false because `res` is the result of `a - b`, and it will never be equal to `a` or `b` unless `a` and `b` are both zero.

Let's consider the following edge cases:
1. **Equal values (a == b)**: This should output the same value twice.
2. **One value being zero (a == 0 or b == 0)**: This should handle the zero value correctly.
3. **One value being the maximum (a == 9 or b == 9)**: This should handle the maximum value correctly.
4. **One value being the minimum (a == 1 or b == 1)**: This should handle the minimum value correctly.

Given the constraints \(0 \leq x, y \leq 9\), we can create a small test input that includes these edge cases.

Here is the test input:

```plaintext
4
0 0
9 9
1 9
9 1
```

This test input includes:
- Two equal values (0 and 0)
- Two equal values (9 and 9)
- A minimum value and a maximum value (1 and 9)
- A maximum value and a minimum value (9 and 1)

This should effectively expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
t=input()
x = []
y = []
for i in range(10):
    xylist = input().split(" ")
    x.append(int(xylist[0]))
    y.append(int(xylist[1]))

for i in range(10):
    if x[i] < y[i]:
        print(x[i], " ", y[i])
    else:
        print(y[i], " ", x[i])

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on a few key points:

1. **Number of Test Cases**: The code assumes there are always 10 test cases, but the problem statement specifies that the number of test cases can vary from 1 to 100.
2. **Input Handling**: The code reads 10 lines of input regardless of the actual number of test cases, which can lead to incorrect behavior if the number of test cases is less than 10.

Given these points, a small-scale test input that can expose the bug would be one where the number of test cases is less than 10. For example, if we provide only 5 test cases, the code will still try to read 10 lines, leading to an error or incorrect output.

Here is the test input:

```plaintext
5
1 9
8 4
1 4
3 4
2 0
```

This input will cause the code to fail because it will attempt to read 10 lines of input, but only 5 are provided. This will result in an error or incorrect output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
a =input()
for p in range(10):
    q=(input())
    b= int(q[0])
    c =int(q[2])
    if b>c:
        print(f"{c} {b}")
    else:
        print(f"{b} {c}")

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The buggy code has a few issues:

1. It assumes that the input for each test case is always in the format "x y" with a space between the two numbers.
2. It does not handle the case where the input might not have a space or might have extra spaces.
3. It does not handle the case where the input might be a single digit.

Given these issues, we can create a test input that includes:
- Test cases with a single digit.
- Test cases with extra spaces.
- Test cases with no spaces.

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

```plaintext
5
1 9
 8  4
1  4
3 4
9
```

Explanation:
- The first test case "1 9" is a normal input.
- The second test case " 8  4" has extra spaces.
- The third test case "1  4" has an extra space after the first number.
- The fourth test case "3 4" is a normal input.
- The fifth test case "9" is a single digit input, which the code should handle by treating it as both `x` and `y`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
import java.util.Scanner;

public class MyFirstSortingProblem {

    public static void solve() {
        Scanner scanner = new Scanner(System.in);
        if (scanner.hasNextInt()) { // Check if there's next integer
            int x = scanner.nextInt();
            int y = scanner.nextInt();
            //System.out.println(Math.min(x, y) + " " + Math.max(x, y));
        }

    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int testCases = scanner.nextInt();
        for (int i = 1; i <= testCases; i++) {
            solve();
        }
        System.out.println("1 9\n4 8\n1 4\n3 4\n0 2\n2 4\n6 9\n3 3\n0 0\n9 9");
    }
}

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

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

Think step by step.

Response:
```plaintext
3
0 9
9 0
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: My First Sorting Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $x$ and $y$.

Output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases.

The only line of each test case contains two space-separated integers $x$ and $y$ ($0 \leq x, y \leq 9$).

Output: For each test case, output two integers: the minimum of $x$ and $y$, followed by the maximum of $x$ and $y$.

Examples:
input:
10
1 9
8 4
1 4
3 4
2 0
2 4
6 9
3 3
0 0
9 9
output:
1 9
4 8
1 4
3 4
0 2
2 4
6 9
3 3
0 0
9 9

Note:

Buggy Code:
import java.util.Scanner;

public class A {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        A a = new A();
        for (int i = 1; i <= 11; i++) {
            String line = "";
            int num1 = 0;
            int num2 = 0;

            line = scanner.nextLine();
            String[] parts = line.split(" ");

            if (parts.length == 2) {
                num1 = Integer.parseInt(parts[0]);
                num2 = Integer.parseInt(parts[1]);
                a.func(num1, num2);

            }
        }
    }

    void func(int x, int y) {
        if (x < y) {
            System.out.println(x + " " + y);
        } else if (x > y) {
            System.out.println(y + " " + x);
        } else if (x == y) {
            System.out.println(x + " " + y);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
0 0
9 9
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

int a[100];
bool pos[100];

bool check(int a,int b){
	if(b<10){
		while(a>0){
			int temp=a%10;
			if(temp>=b)
				return false;
			a/=10;
		}
		return true;
	}
	else if(b>=10){
		int c=b%10;b/=10;
		while(a>0){
			int temp=a%10;
			if(temp>=b||temp>=c)
				return false;
			a/=10;
		}
		return true;
	}

}
bool check_num(int a){
	int min1=a%10;
	a/=10;
	while(a>0){
		int temp=a%10;
		if(temp>min1)
			return false;
		min1=temp;
		a/=10;
	}
	return true;
}
void solve(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		pos[i]=check_num(a[i]);
	}
	int judge_divide=1;
	for(int i=2;i<=n;i++){
		if(a[i]>a[i-1]){
			if(pos[i-1]&&pos[i]&&check(a[i-1],a[i]))
				continue;
			else {
				judge_divide=0;
				continue;
			}
		}
		if(a[i]<a[i-1]){
			if(judge_divide){
				if(check(a[i-1],a[i])){
					judge_divide=0;
					continue;
				}
			}
			else{
			cout<<"NO"<<endl;
			return;
				}
		}
	}
	cout<<"YES"<<endl;
	return;
}

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

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
12 3 10 11
3
9 10 1
2
10 1
```

**Explanation:**

1. **Test Case 1:**
   - Array: `[12, 3, 10, 11]`
   - After splitting the first element, the array becomes `[1, 2, 3, 10, 11]`.
   - The array is sorted, so the output should be `YES`.

2. **Test Case 2:**
   - Array: `[9, 10, 1]`
   - No matter how you split the elements, the array cannot be sorted because `1` cannot be placed before `9` and `10`.
   - The output should be `NO`.

3. **Test Case 3:**
   - Array: `[10, 1]`
   - After splitting the first element, the array becomes `[1, 0, 1]`.
   - The array is sorted, so the output should be `YES`.

These test cases are designed to cover edge cases and scenarios that are likely to expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
//#include <bits/stdc++.h>
#include <iostream>
#include <string>
#include <cmath>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <utility>
#include<cctype>
#include <cstdio>
#include <map>
#include <set>
#include<ctime>
#include <cstdlib>
#define yes cout<<"YES\n"
#define no cout<<"NO\n"
#define ll long long
#define Hanzo ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
using namespace std;
bool isSorted(const std::vector<int>& v) {
    for (size_t i = 1; i < v.size(); ++i) {

        if (v[i] < v[i - 1]) {
            return false;
        }
    }
    return true;
}
int main() {
    Hanzo;
    int t = 0; cin >> t;
    while (t--) {
        ll n = 0; cin >> n;
        vector<int>v1(n);
        vector<int>v2;
        for (int i = 0; i < n; i++)
            cin >> v1[i];
        ll max = 0;
        bool check = true;
        if (!isSorted(v1)) {
            for (int i = 0; i < n; i++) {
                if (v1[i] > 9) {
                    int fd = v1[i] / 10, ld = v1[i] % 10;
                    if (fd >= ld) {
                        v2.push_back(v1[i]);
                        if (v2.back() >= max)  max = v2.back();
                        else {
                            check = false;  break;
                        }
                    }
                    else {
                        if (fd >= max) {
                            v2.push_back(fd);  v2.push_back(ld);
                            if (v2.back() >= max)  max = v2.back();
                            else {
                                check = false;  break;
                            }
                        }
                        else {
                            v2.push_back(v1[i]);
                            if (v2.back() >= max)  max = v2.back();
                            else {
                                check = false;  break;
                            }
                        }
                    }
                }
                else {
                    v2.push_back(v1[i]);
                    if (v2.back() >= max)  max = v2.back();
                    else {
                        check = false;  break;
                    }
                }
            }
            v1.clear(); v2.clear();
            check ? yes : no;
        }
        else
            yes;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
10 10 10 10
3
10 9 8
2
5 55
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

vector<int> getDigits(int n) {
    vector<int> sol;

    while (n > 0) {
        sol.push_back(n%10);
        n = n / 10;
    }

    return sol;
}

string solve(vector<int> arr, int n) {
    int greatest = -1;

    for (int i = 0; i < n-1; i++) {
        if (arr[i] > arr[i+1]) {
            vector<int> digits = getDigits(arr[i]);
            for (int j = 0; j < digits.size(); j++) {
                if (digits[j] > arr[i+1]) return "no";
                if (digits[j] < greatest) return "no";
            }
        }

        greatest = arr[i+1];
    }

    return "yes";
}

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

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

        vector<int> arr (n);
        for (int i = 0; i < n; i++) {
            cin >> arr[i];
        }

        cout << solve(arr, n) << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the array elements are close to the threshold of 10, and the order of elements can be misleading. The code currently checks if the digits of an element are greater than the next element in the array, which might not always be the correct condition to determine if the array can be sorted.

Here's a step-by-step approach to generating a challenging test input:

1. **Identify the Bug**: The bug in the code is that it only checks if the digits of the current element are greater than the next element, but it doesn't consider the possibility of rearranging the digits to form a valid sorted array.

2. **Edge Case**: Create an array where the digits of a number, when split, can form a valid sorted array, but the current logic will incorrectly determine that it cannot be sorted.

3. **Example Input**: Consider an array where an element just above 10 is followed by a single-digit number, and the digits of the element above 10 can be split to form a valid sorted array.

Given these steps, a challenging test input could be:

```plaintext
1
3
12 3 9
```

Explanation:
- The array is `[12, 3, 9]`.
- If we split the element `12`, the array becomes `[1, 2, 3, 9]`.
- The resulting array `[1, 2, 3, 9]` is sorted in non-decreasing order.

The current code will incorrectly return "no" because it checks if the digits of `12` (which are `2` and `1`) are greater than `3`, but the correct answer should be "yes" since the array can be sorted by splitting the element `12`.

This input should effectively expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
a = int(input())

for i in range(a):
    sum = ''
    unof_num = 0
    onesymbol_numbers = 0
    c_unofficial = []
    condition = False
    condition_equals = 0
    b = int(input())
    new_b = b
    c = input()
    c = c.split(" ")

    #Проверка без деления чисел пополам
    for z_without_division in range(b):
        if (z_without_division + 1) != b:
            if not int(c[z_without_division]) < int(c[z_without_division + 1]):
                break
        else:
            condition = True
            break

    for z in range(b):
        if condition:
            print("YES")
            break
        # Проверяем на совпадение числа
        for numbers_equals in range(b):
            if (numbers_equals + 1) < b:
                if int(c[numbers_equals]) == int(c[numbers_equals + 1]):
                    condition_equals += 1
        #Основной блок
        if condition_equals < b:
            if len(str(c[z])) > 1:
                # делим число в массиве
                new_letter = int(c[z]) // 10
                new_letter2 = int(c[z]) % 10
                # пополняем неофициальный массив новыми цифрами
                while unof_num < z:
                    c_unofficial.append(c[int(unof_num)])
                    unof_num += 1
                unof_num += 1
                c_unofficial.append(new_letter)
                c_unofficial.append(new_letter2)
                while unof_num < b:
                    c_unofficial.append(c[int(unof_num)])
                    unof_num += 1
                new_b += 1

            else:
                for numbers in range(b):
                    if len(c[numbers]) == 1:
                        c_unofficial.append(c[numbers])
                        unof_num += 1
                    else:
                        z = numbers
                        # делим число в массиве
                        new_letter = int(c[z]) // 10
                        new_letter2 = int(c[z]) % 10
                        # пополняем неофициальный массив новыми цифрами
                        while unof_num < z:
                            c_unofficial.append(c[int(unof_num)])
                            unof_num += 1
                        unof_num += 1
                        c_unofficial.append(new_letter)
                        c_unofficial.append(new_letter2)
                        while unof_num < b:
                            c_unofficial.append(c[int(unof_num)])
                            unof_num += 1
                        new_b += 1

            for z2 in range(new_b):
                if (z2 + 1) != new_b:
                    if not int(c_unofficial[z2]) < int(c_unofficial[z2 + 1]):
                        unof_num = 0
                        break
                else:
                    condition = True
                    break

        else:
            condition = True #Из-за последней строчки
            print("YES")
            break
    if not condition:
        print("NO")

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

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

Think step by step.

Response:
```plaintext
3
4
10 2 3 4
3
10 9 1
2
9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
import sys
from typing import List

def is_sorted(nums: List[int]):
    for i in range(len(nums) - 1):
        if nums[i] > nums[i + 1]:
            return False
    return True

def get_unsorted_idx(nums: List[int]):
    unsorted_idx = []
    for i in range(len(nums) - 1):
        if nums[i] > nums[i + 1]:
            unsorted_idx.append(i)

    return unsorted_idx

def fix(nums: List[int], idx: int):
    num_to_fix = nums.pop(idx)
    split_nums = [int(x) for x in str(num_to_fix)]
    # [nums.insert(idx, num) for num in split_nums]
    split_nums.reverse()
    for num in split_nums:
        nums.insert(idx, num)

    return len(split_nums) - 1

lines = []
line_idx = 0
for line in sys.stdin:
    if line_idx % 2 == 0:
        lines.append(line)
    line_idx += 1
    if line_idx == (1 + 2 * 7):
        print(line)

lines.pop(0)

for test in lines:
    nums = [int(x) for x in test.strip().split()]
    if is_sorted(nums):
        print("yes")
        continue

    became_sorted = False
    continue_sorting = True
    while continue_sorting:
        unsorted_idx = get_unsorted_idx(nums)
        nums_inserted = 0
        for idx in unsorted_idx:
            if nums[idx] < 10:
                continue_sorting = False  # unsortable
            nums_inserted += fix(nums, idx + nums_inserted)
            if is_sorted(nums):
                became_sorted = True
                continue_sorting = False
                break

    if became_sorted:
        print("yes")
        continue
    print("no")

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

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

Think step by step.

Response:
```plaintext
3
5
10 2 3 4 5
4
9 10 11 12
3
9 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
t = int(input())

def is_sorted(arr):
    for i in range(len(arr) - 1):
        if arr[i] > arr[i + 1]:
            return False
        elif arr[i] == arr[i+1]:
        	return True
    return True

for p in range(t):
    n = int(input())
    arr = list(map(int, input().split()))

    sorted_arr = sorted(arr)
    if sum(arr) == 0:
    	print("YES")
    else:
	    for i in range(n):
	        digits = list(map(int, str(arr[i])))
	        if len(digits) == 1:
	            continue
	        arr.pop(i)
	        arr[i:i] = digits
	        if is_sorted(arr):
	            print("YES")
	            break
	        arr.pop(i)
	        arr.insert(i, arr[i - 1])
	        if is_sorted(arr):
	            print("YES")
	            break

	    else:
	        print("NO")

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

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

Think step by step.

Response:
```plaintext
3
5
10 9 8 7 6
3
99 1 2
4
12 3 45 67
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
// Convert Python code to Java

import java.util.Scanner;
import java.util.ArrayList;

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

        for (int j = 0; j < t; j++) {
            int n = scanner.nextInt();
            ArrayList<Integer> a = new ArrayList<>();
            for (int k = 0; k < n; k++) {
                a.add(scanner.nextInt());
            }

            int i = 0;
            while (i < a.size()) {
                if (a.get(i) >= 10) {
                    char[] digits = String.valueOf(a.get(i)).toCharArray();
                    a.remove(i);
                    for (char digit : digits) {
                        a.add(i, Character.getNumericValue(digit));
                        i++;
                    }
                } else {
                    i++;
                }
            }

            boolean isSorted = true;
            for (int m = 0; m < a.size() - 1; m++) {
                if (a.get(m) > a.get(m + 1)) {
                    isSorted = false;
                    break;
                }
            }

            if (isSorted) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
4
12 3 45 67
3
12 28 5
2
0 0
```

This test input is designed to cover a few key scenarios:
1. **Case with multiple operations needed**: The first case `[12, 3, 45, 67]` requires the operation to be applied to the first and third elements to make the array sorted.
2. **Case where the array cannot be sorted**: The second case `[12, 28, 5]` cannot be sorted using the given operation.
3. **Case where the array is already sorted**: The third case `[0, 0]` is already sorted and requires no operations.

These cases should help expose any bugs in the algorithm, particularly issues with the logic for splitting elements and checking the sorted condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
import java.util.*;

public class mat {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long t = sc.nextLong();
        while (t-- > 0) {
            int n = sc.nextInt();
            int arr[] = new int[n];
            ArrayList<Integer> ll = new ArrayList<>();
            boolean sorted = true;
            boolean valid = true;
            boolean ans=true;
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
                if (i > 0 && arr[i] < arr[i - 1]) {
                    sorted = false;
                }

                String s = String.valueOf(arr[i]);
                if(s.length()==1){
                    ll.add(arr[i]);
                }else{
                    if(s.charAt(0)>s.charAt(1)){
                        valid = false;
                    }
                    ll.add(s.charAt(0)-'0');
                    ll.add(s.charAt(1)-'0');
                }

            }

            if (sorted) {
                System.out.println("yes");
            } else {
               for(int i=1;i<ll.size();i++){
                    if(ll.get(i)<ll.get(i-1)){
                        ans = false;

                    }
               }
               if(ans){
                    System.out.println("yes");
               }else{
                    System.out.println("no");
               }
            }
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following points:

1. **Edge Cases:**
   - Arrays where all elements are single digits.
   - Arrays where all elements are two-digit numbers.
   - Arrays where some elements are single digits and some are two-digit numbers.
   - Arrays where the elements are already sorted.
   - Arrays where the elements are not sorted but can be made sorted by the operation.

2. **Buggy Code Analysis:**
   - The code checks if the array is already sorted and prints "yes" if it is.
   - It also checks if the array can be sorted by splitting two-digit numbers and then checks if the resulting list is sorted.
   - The bug might be in the logic that checks the order of digits in two-digit numbers.

3. **Test Input:**
   - A test case where the array is not sorted but can be made sorted by the operation.
   - A test case where the array contains a two-digit number that, when split, results in a sorted array.
   - A test case where the array contains a two-digit number that, when split, results in an unsorted array.

Given these points, a challenging test input could be:

```plaintext
3
4
12 3 45 67
3
12 28 5
2
0 0
```

This input includes:
- A case where the array can be made sorted by splitting the first element (12).
- A case where the array cannot be made sorted by any operation.
- A case where the array is already sorted.

However, to further challenge the code, we can add a more specific test case:

```plaintext
4
4
12 3 45 67
3
12 28 5
2
0 0
4
19 20 21 22
```

This additional test case includes:
- A case where the array contains a two-digit number (19) that, when split, results in an unsorted array (1, 9, 20, 21, 22).

Thus, the complete test input is:

```plaintext
4
4
12 3 45 67
3
12 28 5
2
0 0
4
19 20 21 22
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

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

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

typedef long long int ll;
#define zz 1000000007
#define mz 998244353
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define loop(i, n) for (ll i = 0; i < (n); ++i)
#define pool(i, n) for (ll i = ((n)-1); i >= 0; --i)
#define rep(i, a, b) for (ll i = (a); i <= (b); ++i)
#define in(v) \
    for (auto& i : v) cin >> i
#define out(v)            \
    for (auto& i : v) {   \
        cout << i << " "; \
    }                     \
    cout << "\n"
#define outp(v)                                      \
    for (auto& i : v) {                              \
        cout << "(" << i.fi << "," << i.se << "), "; \
    }                                                \
    cout << "\n"
#define foa(v) for (auto& i : v)
#define vi vector<int>
#define vll vector<ll>
#define vpi vector<pair<int, int> >
#define vpll vector<pair<ll, ll> >
#define LLMAX 9223372036854775807
#define gcd __gcd
#define lcm(a, b) (a * b) / gcd(a, b)

#define ordered_set_int tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

bool comp(pair<ll, ll> p1, pair<ll, ll> p2) {
    if (p1.se != p2.se)
        return (p1.se < p2.se);
    else
        return ((p1.se == p2.se) && (p1.fi < p2.fi));
}

bool compgg(pair<ll, ll> p1, pair<ll, ll> p2) {
    return (p1.fi > p2.fi);
}

struct secintdec {
    bool operator()(const pair<int, int>& a, const pair<int, int>& b) const {
        return a.second > b.second;
    }
};

struct seclldec {
    bool operator()(const pair<ll, ll>& a, const pair<ll, ll>& b) const {
        return a.second > b.second;
    }
};

void myerase_int(ordered_set_int& t, int v) {
    int rank = t.order_of_key(v);                          // Number of elements that are less than v in t
    ordered_set_int::iterator it = t.find_by_order(rank);  // Iterator that points to the (rank+1)th element in t
    t.erase(it);
}

void binsearch(int l, int r, int id) {
    while (l <= r) {
        int m = (l + r) / 2;
        if (1) {
            id = m;
            l = m + 1;
        } else {
            r = m - 1;
        }
    }
}

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

    vll a(n);
    in(a);

    ll cm = a[0] / 10;
    loop(i, n) {
        ll sa = a[i] % 10;
        ll fa = a[i] / 10;

        // cout << "fa: " << fa << " sa: " << sa << " cm: " << cm << "\n";

        if(fa != 0) {
            if (cm <= fa) {
                if (fa <= sa) {
                    cm = sa;
                    continue;
                } else {
                    cout << "NO\n";
                    return;
                }
            } else {
                cout << "NO\n";
                return;
            }
        }
        else
        {
            if (cm <= sa) {
                cm = sa;
                continue;
            } else {
                cout << "NO\n";
                return;
            }
        }
    }

    cout << "YES\n";

    return;
}

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

    int tt;
    cin >> tt;
    while (tt--) solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3
10 10 10
4
1 10 20 30
5
10 20 30 40 50
```