**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
import java.util.Scanner;

public class VovaXORGame {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int q = sc.nextInt();
        sc.nextLine();

        for (int tc = 0; tc < q; tc++) {
            int n = sc.nextInt();
            sc.nextLine();
            String s = sc.nextLine();
            String t = sc.nextLine();

           int i=0;
           if(s.charAt(0)=='1' ||s.equals(t)){
               System.out.println("YES");
               continue;
           }
           while(i<n){
               if(s.charAt(i)==t.charAt(i))i++;
               else break;
           }
           if(s.charAt(i)=='1'){
               System.out.println("YES");
               continue;
           }
            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:
```plaintext
3
1
1
0
2
10
11
4
1111
0000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

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

int main() {

int t;

cin>>t;

while(t--){
    string ans="yes";
    int l;
    cin>>l;
    int start;
    string s,t;
    cin>>s>>t;
    for(int i=0;i<l;i++){
        if(s[i]=='1'){
            start=i;
            break;
        }

    }
    for(int i=0;i<l;i++){
        if(s[i]=='0'&& t[i]=='1'){
           if(i<=start){
            ans="no";
           }
            break;
           }

        }

        cout<<ans<<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, let's analyze the code's logic and identify potential edge cases or scenarios where it might fail.

### Analysis of the Buggy Code
1. **Initialization of `start`**:
   - The code sets `start` to the index of the first '1' in the string `s`. If `s` does not contain any '1', `start` remains uninitialized, which can lead to undefined behavior.

2. **Checking for '0' and '1'**:
   - The code checks if there is a '0' in `s` and a '1' in `t` at the same index `i` and if `i` is less than or equal to `start`. If such a condition is found, it sets `ans` to "no".

### Edge Cases to Consider
1. **String `s` with no '1'**:
   - If `s` contains no '1', `start` will be uninitialized, which can lead to undefined behavior.

2. **String `s` and `t` of different lengths**:
   - Although the problem guarantees that `s` and `t` are of the same length, it's good to check for this edge case.

3. **String `s` and `t` where `s` can be transformed to `t`**:
   - The code might fail to recognize valid transformations due to the incorrect logic for checking the '0' and '1' conditions.

4. **String `s` and `t` where `s` cannot be transformed to `t`**:
   - The code might incorrectly identify a valid transformation when it should not.

### Challenging Test Input
Given the above analysis, a small-scale test input that can expose the bug is:

```plaintext
3
5
00000
10101
6
010101
010101
7
1001000
1001001
```

### Explanation
1. **Test Case 1**:
   - `s = "00000"` and `t = "10101"`.
   - Since `s` contains no '1', `start` remains uninitialized, and the code might exhibit undefined behavior.
   - Expected output: "No" (since `s` cannot be transformed to `t`).

2. **Test Case 2**:
   - `s = "010101"` and `t = "010101"`.
   - The sequences are already equal, so the game is interesting.
   - Expected output: "Yes".

3. **Test Case 3**:
   - `s = "1001000"` and `t = "1001001"`.
   - The sequences differ by one bit, and it should be possible to transform `s` to `t` using the XOR operation.
   - Expected output: "Yes".

This test input should help 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: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
#pragma GCC optimize("unroll-loops", "unswitch-loops")
#pragma GCC optimize("Ofast", "no-stack-protector")
#pragma GCC optimize("profile-reorder-functions", "tracer")
#pragma GCC optimize("rename-registers")
#pragma GCC optimize("vpt")
#pragma GCC optimize("fast-math")

#include<bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define ld long double

using namespace std;

const ll INF = 1e11;

void solve() {
    ll n;
    string s, t;
    cin >> n >> s >> t;
    if (s[0] != t[0] && s[0] == '0') {
        cout << "NO\n";
        return;
    }
    if (s[0] == '1') {
        cout << "YES\n";
        return;
    }
    ll ss = 0, st = 0;
    for (ll i = 0; i < n; i++) {
        ss += s[i] == '1';
        st += t[i] == '1';
        if (st > ss) {
            cout << "NO\n";
            return;
        }
    }
    cout << "YES\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll q;
    cin >> q;
    while (q--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
1
0
1
2
00
11
3
000
111
4
0000
1111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
def solve():
    n=int(input())
    s=input()
    t=input()

    # find fist 1 idx
    oidx=-1
    for i,c in enumerate(s):
        if c=='1':
            oidx=i
            break
    if oidx==-1:
        possible = True
        for c in enumerate(t):
            if c!='0':
                possible = False
                break
        return possible

    stopping=(2*oidx)-1
    allzeros=True
    for i,c in enumerate(t):
        if i>stopping: break
        if c!='0':
            allzeros=False
            break
    if allzeros: return True

    for i in range(stopping+1):
        if s[i]!=t[i]:
            return False
    return True

for t in range(int(input())):
    ans=solve()
    if ans:
        print("YES")
    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
1
0
1
2
01
10
4
0100
0001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
import sys
input = sys.stdin.readline
import collections
I = lambda: map(int, input().split())

for _ in range(int(input())):
    n,= I()
    a=input().strip()
    b=input().strip()
    c=0
    d=0
    for j in range(n-1):
        if a[j:j+2]=="01":
                c=1
        elif a[j:j+2]=="10":
                d=1
        if c==d==1:
                break
    if c==d==1:
                print("yes")
                continue
    res="yes"
    for j in range(n):
        if a[j]!=b[j] and a[j]=="0":
                res="no"
                break
    print(res)

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

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

Think step by step.

Response:
```plaintext
6
1
0
1
7
0110100
1010100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
import java.util.*;
public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int test = sc.nextInt();
        while (test-->0){
            int n = sc.nextInt();
            String s = sc.next();
            String t = sc.next();
            for (int i = 0; i < n; i++) {
                if (s.charAt(i) == '0' && t.charAt(i) == '1') {
                    System.out.println("NO");
                    break;
                }
                if (s.charAt(i) == '0' && t.charAt(i) == '0')
                {
                    continue;
                }
                else{
                    System.out.println("YES");
                    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
0
1
2
00
11
3
010
101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

The second line of each test case contains a binary sequence $s$ of length $n$.

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

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

public class Codechef
{
    static long gcd(long a,long b)
    {
        if(a==0)
        return b;
        //at one time a will become 0

        return gcd(b%a,a);
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		FastScanner sc=new FastScanner();
		int t1=sc.nextInt();
		while(t1-->0)
		{
		    int n=sc.nextInt();
		    String s=sc.next();
		    String t=sc.next();
		    StringBuilder sb=new StringBuilder(s);
		    int ind=sb.indexOf("1");
		    boolean ans=true;
		    if(ind==-1)
		    ind=n;
		    int count=0;
		    for(int i=n-1;i>=0;i--)
		    {
		        if(s.charAt(i)!=t.charAt(i))
		        {
		            if(s.charAt(i)=='0')
		            {
		                if(i<ind)
		                {
		                    ans=false;
		                    break;
		                }
		                else
		                sb=sb.replace(i,i+1,"1");
		            }
		            else
		            {
		                sb=sb.replace(i,i+1,"0");
		                ind=sb.indexOf("1");
		            }
		          count++;
		        }

		    }
		    System.out.println(ans?"YES":"NO");

		}

	}
}
class FastScanner {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

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

    String nextLine() throws IOException {
        return br.readLine();
    }

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

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

    public long[] readArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }

    public String[][] read2DArray(int n, int m) {
        String[][] mat = new String[n][];
        for (int i = 0; i < n; i++) {
            mat[i] = new String[m];
            for (int j = 0; j < m; j++) {
                mat[i][j] = next();
            }
        }
        return mat;
    }

    public void printArray(char[][] mat, int n, int m) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                System.out.print(mat[i][j]);
            }
            System.out.println();
        }
    }

    public Long[] readArrayLong(int n) {
        Long[] a = new Long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}

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

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

Think step by step.

Response:
```plaintext
5
1
1
0
2
11
00
3
101
010
4
1100
0011
5
10101
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

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

using namespace std;

int getMin(int x, int y) {
	if(x<=y)
		return x;
	return y;
}

int getMax(int x, int y) {
	if(x>=y)
		return x;
	return y;
}

int x1,x2,x3;

void solve() {
	cin>>x1>>x2>>x3;

	int max = getMax(x1,getMax(x2, x3));
    int min = getMin(x1, getMin(x2, x3));
    int ans;
    if(x1!= max && x1!=min) {
    	ans = abs(x1-x2)+abs(x1-x3);
    }
    else if(x2!=max && x2!=min) {
    	ans = abs(x1-x2) + abs(x2-x3);
    }
    else {
    	ans = abs(x1-x3) + abs(x2-x3);
    }

	cout<<ans<<endl;
}

int main() {
	ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
	int tc;
	cin>>tc;
	while(tc) {
		solve();
		tc--;
	}

    //solve(10);

	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 logic in the `solve` function might fail. The current logic in the `solve` function assumes that the median value among the three points is the optimal point `a` to minimize the total distance. However, this assumption is not always correct, especially when the points are not distinct or when the points are very close to each other.

Let's consider the following edge cases:
1. All three points are the same.
2. Two points are the same, and the third point is different.
3. The points are very close to each other but not identical.

### Test Case 1: All three points are the same
This is the simplest edge case. The optimal point `a` should be the same as the points, and the total distance should be 0.

### Test Case 2: Two points are the same, and the third point is different
Here, the optimal point `a` should be the same as the two identical points, and the total distance should be the distance from the third point to the two identical points.

### Test Case 3: Points are very close to each other but not identical
In this case, the optimal point `a` should be the median value, and the total distance should be minimized.

Let's create a small-scale test input that includes these edge cases:

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

### Explanation:
1. **Test Case 1: `1 1 1`**
   - All three points are the same.
   - The optimal point `a` is 1.
   - The total distance is `|1-1| + |1-1| + |1-1| = 0`.

2. **Test Case 2: `2 2 3`**
   - Two points are the same, and the third point is different.
   - The optimal point `a` is 2.
   - The total distance is `|2-2| + |2-2| + |2-3| = 1`.

3. **Test Case 3: `4 5 6`**
   - The points are very close to each other but not identical.
   - The optimal point `a` is 5 (the median).
   - The total distance is `|4-5| + |5-5| + |6-5| = 2`.

This test input should help 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: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
void solve(){
         int x1,x2,x3;
        cin>>x1>>x2>>x3;
        vector<long long>v(10);
        for(int i=0;i<10;i++){
            v[i]=abs(x1-i)+abs(x2-i)+abs(x3-i);
            }
        sort(v.begin(),v.end());
        cout<<v[0]<<endl;
}
int main(){
    int t;
    cin>>t;
    while(t--){
solve();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the code might fail. The main issue with the provided code is that it sorts the entire vector of possible distances and then outputs the smallest value. This approach is inefficient and can be prone to errors, especially when the points are close to the boundaries of the allowed range (1 to 10).

### Step-by-Step Analysis:
1. **Boundary Cases**: Points at the extreme ends of the range (1 and 10) can cause issues if the code does not handle them correctly.
2. **Equal Points**: When all points are the same, the distance should be zero, which is a straightforward case but important to verify.
3. **Symmetric Points**: Points that are symmetric around the middle of the range can also be tricky.
4. **Diverse Points**: Points that are spread out across the range can help identify issues in the distance calculation.

### Test Input:
Let's create a small-scale test input that includes these edge cases:

```plaintext
5
1 1 1
1 10 10
1 5 9
1 6 10
3 7 8
```

### Explanation:
1. **Test Case 1**: `1 1 1` - All points are the same, so the minimum distance is 0.
2. **Test Case 2**: `1 10 10` - Points at the boundaries, the minimum distance is 9.
3. **Test Case 3**: `1 5 9` - Points spread out, the minimum distance is 8.
4. **Test Case 4**: `1 6 10` - Points spread out with a middle point, the minimum distance is 7.
5. **Test Case 5**: `3 7 8` - Points close to the middle of the range, the minimum distance is 4.

This test input should help expose any issues in the code, especially in handling boundary cases and symmetric points.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;
int main() {
int n,result=0;
cin>>n;
while(n--){
    vector <int>v(3);
    cin>>v[0]>>v[1]>>v[2];
    int min=*min_element(v.begin(),v.end());
    int max=*max_element(v.begin(),v.end());
    for(int i=0;i<3;i++){
        if(v[i]!=min||v[i]!=max)
     result= abs(min-v[i])+abs(max-v[i]);
    }
    cout<<result<<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 buggy code, we need to focus on the logic that calculates the minimum total distance. The bug in the code is that it incorrectly calculates the result by only considering the minimum and maximum values, and not the median of the three points. The correct approach is to choose the median of the three points as the value of \(a\), as it minimizes the total distance.

Let's generate a test input that will highlight this issue. We need a test case where the median is different from both the minimum and maximum values, and the distance calculation using the median should be significantly different from the distance calculation using the minimum or maximum values.

Here is a small-scale test input:

```plaintext
3
1 2 10
1 10 2
10 1 2
```

Explanation:
1. For the first test case (1, 2, 10):
   - The median is 2.
   - The correct result is \(|1 - 2| + |2 - 2| + |10 - 2| = 1 + 0 + 8 = 9\).
   - The buggy code will incorrectly calculate it as \(|1 - 1| + |10 - 1| = 0 + 9 = 9\).

2. For the second test case (1, 10, 2):
   - The median is 2.
   - The correct result is \(|1 - 2| + |10 - 2| + |2 - 2| = 1 + 8 + 0 = 9\).
   - The buggy code will incorrectly calculate it as \(|1 - 1| + |10 - 1| = 0 + 9 = 9\).

3. For the third test case (10, 1, 2):
   - The median is 2.
   - The correct result is \(|10 - 2| + |1 - 2| + |2 - 2| = 8 + 1 + 0 = 9\).
   - The buggy code will incorrectly calculate it as \(|10 - 1| + |10 - 10| = 9 + 0 = 9\).

This test input will help highlight the issue in the buggy code where it does not correctly identify the median and use it to calculate the minimum total distance.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
test = int(input())
for test_case in range(test):
    a, b, c = list(map(int, input().split()))
    mini = 179
    for med in range(1, 10):
        if (abs(med - a) + abs(med - b) + abs(med - c) < mini):
            mini = abs(med - a) + abs(med - b) + abs(med - c)
    print(mini)

**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 maximize the failure probability. The buggy code iterates through all possible values of `med` from 1 to 10 and calculates the total distance. However, it does not handle cases where the median value might be the best choice efficiently.

### Step-by-Step Analysis:
1. **Edge Cases:**
   - All points are the same.
   - Points are in a strictly increasing or decreasing order.
   - Points are spread out over the entire range.
   - Points are clustered around a specific value.

2. **Scenarios to Test:**
   - All points are the same (e.g., 1 1 1).
   - Points are in a strictly increasing order (e.g., 1 2 3).
   - Points are in a strictly decreasing order (e.g., 3 2 1).
   - Points are spread out (e.g., 1 5 10).
   - Points are clustered (e.g., 3 3 4).

3. **Challenging Test Cases:**
   - Test case with all points the same.
   - Test case with points in a strictly increasing order.
   - Test case with points in a strictly decreasing order.
   - Test case with points spread out.
   - Test case with points clustered around a specific value.

### Chosen Test Input:
```plaintext
5
1 1 1
1 2 3
3 2 1
1 5 10
3 3 4
```

### Explanation:
- **Test Case 1 (1 1 1):** The smallest value of `f(a)` is 0, achieved when `a = 1`.
- **Test Case 2 (1 2 3):** The smallest value of `f(a)` is 2, achieved when `a = 2`.
- **Test Case 3 (3 2 1):** The smallest value of `f(a)` is 2, achieved when `a = 2`.
- **Test Case 4 (1 5 10):** The smallest value of `f(a)` is 8, achieved when `a = 5`.
- **Test Case 5 (3 3 4):** The smallest value of `f(a)` is 1, achieved when `a = 3`.

This test input covers a variety of scenarios and should help expose any 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: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
testcases = int(input())

def totdist(a, b, c, d):
    tot = abs(d-a) + abs(d-b) + abs(d-c)
    return tot

for _ in range(testcases):
    a, b, c = map(int, input().split())
    print(min([totdist(a,b,c,i) for i in range(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:
To expose the bug in the provided code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The code currently iterates over a fixed range of 10, which might not always find the correct minimum distance. We need to test cases where the points are at the boundaries and where the points are very close to each other.

Here are some test cases to consider:

1. All points are the same.
2. Points are at the extreme ends of the range.
3. Points are very close to each other.
4. Points are evenly distributed.

Let's create a small-scale test input that includes these scenarios:

```plaintext
5
1 1 1
1 2 3
1 10 10
5 5 5
1 1 10
```

Explanation:
1. **All points are the same (1, 1, 1):** The minimum distance should be 0.
2. **Points are very close to each other (1, 2, 3):** The minimum distance should be 2.
3. **Points are at the extreme ends (1, 10, 10):** The minimum distance should be 18.
4. **All points are the same (5, 5, 5):** The minimum distance should be 0.
5. **Points are at the extreme ends with one point in the middle (1, 1, 10):** The minimum distance should be 16.

This test input should help expose any issues in the code that might not handle these edge cases correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
n = int(input())

def func(s):
    return abs(s)

results = []

for _ in range(n):
    x = input().split()
    x1, x2, x3 = map(int, x)

    min_sum = float('inf')
    for j in range(10):
        sum1 = func(x1 - j) + func(x2 - j) + func(x3 - j)
        min_sum = min(min_sum, sum1)

    results.append(min_sum)

for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
3
1 1 10
1 10 10
5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
import java.util.*;
public class A_X_Axis {

    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t>0){
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();

            int x = Math.min(a,b);

            if(Math.min(x,c)==c){
                System.out.println(a+b-2*x+x-c);
            }
            else if(x==a){
                if(Math.min(b,c)==c)
                System.out.println(b-a);
                else
                System.out.println(c-a);
            }
            else if(x==b){
                if(Math.min(a,c)==c)
                System.out.println(a-b);
                else
                System.out.println(a-b);
            }
            t--;
        }

    }

}

**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 where the logic in the code might fail. The code appears to be trying to find the median of the three points, but it does so in a complex and potentially error-prone way. The correct approach to minimize the total distance is to choose the median of the three points.

Let's consider the following scenarios:

1. **All three points are the same**: This is a simple case where the median is obvious.
2. **Two points are the same, and the third is different**: The median will be the point that is repeated.
3. **All three points are different**: The median will be the middle value when the points are sorted.

The buggy code seems to have issues with correctly identifying the median and calculating the distances. Let's create a test case that will likely expose these issues:

### Test Case 1: All three points are the same
- Input: `1 1 1`
- Expected Output: `0`

### Test Case 2: Two points are the same, and the third is different
- Input: `1 1 5`
- Expected Output: `4`

### Test Case 3: All three points are different
- Input: `1 5 9`
- Expected Output: `8`

### Test Case 4: Points in reverse order
- Input: `9 5 1`
- Expected Output: `8`

### Test Case 5: Points in a different order
- Input: `5 1 9`
- Expected Output: `8`

### Test Case 6: Points close to each other
- Input: `1 2 3`
- Expected Output: `2`

### Test Case 7: Points with a large gap
- Input: `1 10 5`
- Expected Output: `9`

### Test Case 8: Points with a medium gap
- Input: `3 7 2`
- Expected Output: `5`

### Test Case 9: Points with a small gap
- Input: `2 3 4`
- Expected Output: `2`

### Test Case 10: Points with a very small gap
- Input: `1 2 2`
- Expected Output: `1`

Now, let's combine these test cases into a single input:

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

This test input should help expose the bugs 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: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class Codeforces {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int[] a = new int[3];
            for(int i=0;i<3;i++){
                a[i]=sc.nextInt();
         }
            Arrays.sort(a);
            int avg = (((a[0]+a[2])/2)+a[1])/2;
            if(Math.abs(avg-a[2])>a[2]-a[1])
                avg = a[1];

            System.out.println(avg-a[0]+Math.abs(a[1]-avg)+a[2]-avg);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

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

int mid(int a, int b, int c){
    if ( (a < b && b < c) || (c < b && b < a) )
       return b;
    else if ( (b < a && a < c) || (c < a && a < b) )
       return a;
    else
       return c;
}

int main() {
    int t;
    cin >> t;
    int sum = 0;
    while (t--) {
        int x1,x2,x3;
        cin >> x1 >> x2 >> x3;
        int a = mid(x1, x2, x3);
        sum = sum + abs(x1 - a) + abs(x2 - a) + abs(x3 - a);
        cout << sum << endl;
        sum = 0;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
10
1 1 1
10 10 10
1 2 3
3 2 1
1 10 10
10 1 1
5 5 5
1 5 10
10 5 1
2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

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

int main() {
	int tests; cin >> tests;
	while (tests--) {
		int x, y, z;
		cin >> x >> y >> z;

		int min = INT_MAX;
		for (int i = 1; i < 10; i++) {
			int num;
			num = abs(x - i) + abs(y - i) + abs(z - i);

			if (num < min) min = num;
		}

		cout << min << 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 provided code, we need to focus on edge cases and scenarios that can highlight potential issues. The buggy code has a loop that iterates from 1 to 9 to find the minimum distance, but it misses the point 10. This is a critical oversight because the problem statement specifies that the coordinates can be between 1 and 10, inclusive.

### Challenging Test Input
We will create a test case where the optimal point `a` is 10, and the points `x1`, `x2`, and `x3` are such that the minimum distance is achieved when `a` is 10. This will expose the bug in the code.

### Example Test Case
```plaintext
1
10 10 10
```

### Explanation
- **Test Case 1**:
  - Points: \( x_1 = 10 \), \( x_2 = 10 \), \( x_3 = 10 \)
  - The optimal point `a` is 10, and the total distance \( f(a) \) is:
    \[
    f(10) = |10 - 10| + |10 - 10| + |10 - 10| = 0
    \]
  - The buggy code will not consider the point 10, and it will likely output a higher value, exposing the bug.

This test case should effectively highlight the issue in the buggy code where it fails to consider the point 10.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
test_cases=int(input())
for i in range (test_cases):
    a=input().split()
    for l in range (len(a)):
        a[l]=int(a[l])

    for x in range (len(a)):
        counter=0
        for y in range (len(a)):
            if a[x] == a[y]:
                counter=counter+1
            if counter >= 2:
                big_number=a[x]
    if counter <= 1:
        # Python program to find largest number
        # in a list

        # Removing duplicates from the list
        list2 = list(set(a))

        # Sorting the list
        list2.sort()
        big_number=list2[-2]
    print(abs(a[0] - big_number) + abs(a[1] - big_number) + abs(a[2] - big_number))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
def solve():
    t = int(input())  # Leer el número de casos de prueba

    results = []

    for _ in range(t):
        x4,x2,x3 = map(int, input().split())  # Leer n y k para el caso de prueba actual

        pruebas = []
        for a in range(10):
          n = abs(a-x4) + abs(a-x2) + abs(a-x3)
          pruebas.append(n)

        n = min(pruebas)

        # Almacenar el resultado
        results.append(n)

    # Imprimir todos los resultados, uno por línea
    for result in results:
        print(result)

# Ejecutar la función solve
solve()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the following issues:

1. **Variable Naming Error**: The code incorrectly uses `x4` instead of `x1` when reading the input. This will cause the first coordinate to be read into `x4` instead of `x1`.

2. **Range of `a`**: The code iterates `a` from 0 to 9 (i.e., `range(10)`), but the problem specifies that the coordinates are in the range from 1 to 10. This means the code will not consider `a = 10`.

Given these issues, we can create a test input that will highlight these problems:

1. **Test Case with `x1` Outside the Expected Range**: This will show that the code is not reading `x1` correctly.
2. **Test Case with `a = 10` as the Optimal Point**: This will show that the code is not considering `a = 10`.

Here is a small-scale test input that exposes these issues:

```plaintext
3
10 1 1
1 10 1
1 1 10
```

### Explanation:

1. **Test Case 1: `10 1 1`**
   - The correct optimal point is `a = 10`, but the code will not consider `a = 10` because it only iterates from 0 to 9.
   - The expected output is `18` (since `f(10) = |10 - 10| + |10 - 1| + |10 - 1| = 18`), but the code will output a different value.

2. **Test Case 2: `1 10 1`**
   - The correct optimal point is `a = 10`, but the code will not consider `a = 10`.
   - The expected output is `18` (since `f(10) = |10 - 1| + |10 - 10| + |10 - 1| = 18`), but the code will output a different value.

3. **Test Case 3: `1 1 10`**
   - The correct optimal point is `a = 10`, but the code will not consider `a = 10`.
   - The expected output is `18` (since `f(10) = |10 - 1| + |10 - 1| + |10 - 10| = 18`), but the code will output a different value.

This test input will help expose the 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: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            int x1 = sc.nextInt();
            int x2 = sc.nextInt();
            int x3 = sc.nextInt();
            int diff = 0;
            int min = 99999;
            for (int i = 1; i < 10; i++) {
                diff += Math.abs(x1 - i);
                diff += Math.abs(x2 - i);
                diff += Math.abs(x3 - i);
                min = Math.min(diff, min);
                diff = 0;

            }
            System.out.println(min);
        }
    }
}

**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 2 3
10 10 10
1 10 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
import java.io.BufferedReader;
import java.time.Duration;
import java.time.Instant;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
public class ex{

    // Time complexity = O(sqrt(n))
    public static boolean isPrime(int n) {
        if (n <= 1) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    //GCD OF TWO NUMBERS
    public static long gcd(long a, long b){
        if(a == 0){
            return b;
        }

        return gcd(b % a, a);
    }

    //LCM OF TWO NUMBERS
    public static long LCM(int a, int b){
        return  ((long) a * b) / gcd(a, b);
    }

    //    FOR TAKING FASTER INPUTS
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static long decimalToBinary(long decimal) {
        if (decimal == 0) {
            return 0;
        }
        long binary = 0;
        long place = 1;
        while (decimal > 0) {
            long remainder = decimal % 2;
            binary += remainder * place;
            place *= 10;
            decimal = decimal / 2;
        }
        return binary;
    }

    public static void ha() {
        System.out.println("YES");
    }

    public static void na() {
        System.out.println("NO");
    }

    public static void line() {
        System.out.println();
    }

    public static int mod = 1000000007;

    public static void main(String args[]) {
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();
        while (t-- != 0) {
            // int n = sc.nextInt();
            // int[] a = new int[n];

            // for(int i=0;i<n;i++) {
            //     a[i] = sc.nextInt();
            // }

            int x1 = sc.nextInt();
            int x2 = sc.nextInt();
            int x3 = sc.nextInt();
            int ans = Integer.MAX_VALUE;

            for(int i=0;i<10;i++) {
                ans = Math.min(ans, Math.abs(i-x1)+Math.abs(i-x2)+Math.abs(i-x3));
            }

            System.out.println(ans);

        }
        out.flush();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The main issue in the code is the loop that iterates from 0 to 9, which might not always find the correct minimum distance because the points can be outside this range (1 to 10).

### Step-by-Step Analysis:

1. **Range of Points**: The points \(x_1\), \(x_2\), and \(x_3\) are within the range [1, 10]. The loop in the code iterates from 0 to 9, which means it does not consider the point 10. This can lead to incorrect results if the optimal point \(a\) is 10.

2. **Edge Cases**:
   - All points are the same.
   - Points are spread out.
   - Points are clustered at the boundaries (1 and 10).

3. **Challenging Scenarios**:
   - When the optimal point is 10.
   - When the points are 1, 10, and a middle value.

### Chosen Test Input:

To expose the bug, we will create a test case where the optimal point \(a\) is 10. This will ensure that the loop from 0 to 9 in the code does not consider the correct optimal point.

```plaintext
1
1 10 10
```

### Explanation:
- **Test Case**: The points are \(1, 10, 10\).
- **Optimal Point**: The optimal point \(a\) is 10, because the total distance \(f(10)\) is:
  - \(|1 - 10| + |10 - 10| + |10 - 10| = 9 + 0 + 0 = 9\).
- **Expected Output**: The smallest value of \(f(a)\) is 9.

This test case will expose the bug in the code because the loop from 0 to 9 does not consider the point 10, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define fastio()                  \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);
#define nl << "\n";
#define rep(i, n) for (ll i = 0; i < n; i++)

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    fastio();

    int test;
    cin >> test;
    while (test--)
    {
        ll n1,n2,n3;
        cin >> n1>>n2>>n3;
        if(n1==n2==n3){
            cout<<"0" nl;
        }else if((n1==n2) || (n2==n3)||(n1==n3)){
            if(n1==n2) {cout<<abs(n3-n1) nl;
            }else if(n2==n3){
                cout<<abs(n1-n2) nl;
            }else{
                cout<<abs(n1-n2) nl;
            }
        }else{
           vector<int>v;
           v.push_back(n1);
           v.push_back(n2);v.push_back(n3);
         sort(v.begin(),v.end());
         cout<<(abs(v[1]-v[0])+abs(v[1]-v[2])) nl;
        }

    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. The code has several logical conditions, and we need to test each of these conditions carefully.

### Step-by-Step Analysis:
1. **All Points Are the Same:**
   - The code correctly handles the case where all three points are the same by outputting `0`.

2. **Two Points Are the Same:**
   - The code has a condition to handle cases where two points are the same. However, it might fail if the third point is significantly different from the other two.

3. **All Points Are Different:**
   - The code sorts the points and calculates the total distance from the median point. This should generally work, but we need to ensure that the sorting and distance calculation are correct.

### Challenging Test Cases:
1. **Two Points Are the Same, Third Point is Far:**
   - Example: `1 1 10`
   - Expected Output: `9` (When `a = 1`, the total distance is `|1-1| + |1-1| + |1-10| = 9`).

2. **All Points Are Different, but Median is Not the Middle Value:**
   - Example: `1 10 5`
   - Expected Output: `9` (When `a = 5`, the total distance is `|5-1| + |5-10| + |5-5| = 9`).

3. **Points Are in Descending Order:**
   - Example: `10 9 8`
   - Expected Output: `3` (When `a = 9`, the total distance is `|9-10| + |9-9| + |9-8| = 3`).

4. **Points Are in Ascending Order:**
   - Example: `1 2 3`
   - Expected Output: `2` (When `a = 2`, the total distance is `|2-1| + |2-2| + |2-3| = 2`).

### Test Input:
```plaintext
4
1 1 10
1 10 5
10 9 8
1 2 3
```

This test input covers a variety of edge cases and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

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

int main() {
	int t;
	cin >>t;
	while(t--) {
		int x1,x2,x3;
		cin >> x1 >> x2 >> x3;
		int f[10];
		for(int i = 1 ; i <= 10 ; ++i) {
			f[i] = abs(i - x1) + abs(i - x2) + abs(i - x3);
		}
		cout << *min_element(f,f + 10) << 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
5
1 1 1
10 10 10
1 10 10
5 5 5
1 2 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
for _ in range(int(input())):
    x1, x2, x3 = map(int, input().split())
    best = x1 + x2 + x3
    for i in range(10):
        best = min(best, abs(x1 - i) + abs(x2 - i) + abs(x3 - i))
    print(best)

**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 1 1
10 10 10
1 10 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Axis
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given three points with integer coordinates $x_1$, $x_2$, and $x_3$ on the $X$ axis ($1 \leq x_i \leq 10$). You can choose any point with an integer coordinate $a$ on the $X$ axis. Note that the point $a$ may coincide with $x_1$, $x_2$, or $x_3$. Let $f(a)$ be the total distance from the given points to the point $a$. Find the smallest value of $f(a)$.

The distance between points $a$ and $b$ is equal to $|a - b|$. For example, the distance between points $a = 5$ and $b = 2$ is $3$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases. Then follows their descriptions.

The single line of each test case contains three integers $x_1$, $x_2$, and $x_3$ ($1 \leq x_i \leq 10$) — the coordinates of the points.

Output: For each test case, output the smallest value of $f(a)$.

Examples:
input:
8
1 1 1
1 5 9
8 2 8
10 9 3
2 1 1
2 4 1
7 3 5
1 9 4
output:
0
8
6
7
1
3
4
8

Note: In the first test case, the smallest value of $f(a)$ is achieved when $a = 1$: $f(1) = |1 - 1| + |1 - 1| + |1 - 1| = 0$.

In the second test case, the smallest value of $f(a)$ is achieved when $a = 5$: $f(5) = |1 - 5| + |5 - 5| + |9 - 5| = 8$.

In the third test case, the smallest value of $f(a)$ is achieved when $a = 8$: $f(8) = |8 - 8| + |2 - 8| + |8 - 8| = 6$.

In the fourth test case, the smallest value of $f(a)$ is achieved when $a = 9$: $f(10) = |10 - 9| + |9 - 9| + |3 - 9| = 7$.

Buggy Code:
# -*- coding:utf-8 -*-
def x_axis(x1:int, x2:int, x3:int)->int:
    # write code here
    dis_sum = 0
    dis_min = 30
    # a = 0
    for i in range(1,10):
        dis_sum = abs(x1-i) + abs(x2-i) +abs(x3-i)
        if dis_sum < dis_min:
            dis_min = dis_sum
            # a = i
    return dis_min

def main():
    t = int(input())
    for i in range(t):
        x1, x2, x3 = map(int, input().split())
        out = x_axis(x1, x2, x3)
        print(out)

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios where the code might fail. The code iterates over all possible values of `a` from 1 to 10 and calculates the total distance for each value. It then finds the minimum distance. However, the code initializes `dis_min` to 30, which is a reasonable value but might not cover all edge cases.

Let's consider the following scenarios:
1. All points are the same.
2. Points are very close to each other.
3. Points are spread out.

### Step-by-Step Analysis

1. **All points are the same:**
   - If `x1 = x2 = x3`, the minimum distance should be 0.
   - Example: `1 1 1`

2. **Points are very close to each other:**
   - If `x1`, `x2`, and `x3` are close, the minimum distance should be small.
   - Example: `2 3 4`

3. **Points are spread out:**
   - If `x1`, `x2`, and `x3` are spread out, the minimum distance should be larger.
   - Example: `1 5 10`

4. **Edge case with initialization:**
   - If the points are such that the minimum distance is exactly 30, the code might not update `dis_min` correctly.
   - Example: `1 10 10` (This is an unlikely scenario, but we can test it to ensure the code handles large distances correctly.)

### Test Input
Let's create a test input that covers these scenarios:

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

### Explanation
1. **Test Case 1: `1 1 1`**
   - All points are the same, so the minimum distance is 0.
2. **Test Case 2: `2 3 4`**
   - Points are close, so the minimum distance is small.
3. **Test Case 3: `1 5 10`**
   - Points are spread out, so the minimum distance is larger.
4. **Test Case 4: `1 10 10`**
   - Points are spread out with one point far from the others, testing the initialization of `dis_min`.

This test input should help expose any issues in the code, particularly with the initialization of `dis_min` and the handling of different point distributions.