**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

Input: The first line consists of a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

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

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

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

using namespace std;

bool isSubsequence(const std::string& a, const std::string& b) {
    int m = a.length();
    int n = b.length();

    int i = 0;
    int j = 0;

    while (i < m && j < n) {
        if (a[i] == b[j])
            i++;
        j++;
    }

    return (i == m);
}

void solve(){
    int n,m;
    cin>>n>>m;
    string a,b;
    cin>>a>>b;
    int k=0;
    string temp="";
    for(int i=0;i<n;i++){
        temp+=a[i];
        if(isSubsequence(temp,b)){
            k++;
        }
        else{
            break;
        }
    }
    cout<<k<<endl;
}

int main() {

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

  return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases where the algorithm might fail. The main issue in the provided code is that it checks if the entire prefix of `a` is a subsequence of `b` in each iteration, which is inefficient and can lead to incorrect results.

Here are some edge cases to consider:
1. **Empty or very short strings**: These can cause the algorithm to fail if it doesn't handle small inputs correctly.
2. **Long strings with many repeated characters**: This can expose issues in the subsequence check.
3. **Strings where the prefix of `a` is a subsequence of `b` but the entire `a` is not**: This can help identify if the algorithm correctly breaks out of the loop at the right point.
4. **Strings where the prefix of `a` is not a subsequence of `b` but parts of it are**: This can help identify if the algorithm correctly identifies the maximum `k`.

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

```plaintext
4
1 1
1
0
5 5
11111
11111
6 5
101010
11111
7 7
1001001
1010101
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`, `m = 1`
   - `a = "1"`, `b = "0"`
   - The prefix of `a` is not a subsequence of `b`, so the expected output is `0`.

2. **Test Case 2**:
   - `n = 5`, `m = 5`
   - `a = "11111"`, `b = "11111"`
   - The entire `a` is a subsequence of `b`, so the expected output is `5`.

3. **Test Case 3**:
   - `n = 6`, `m = 5`
   - `a = "101010"`, `b = "11111"`
   - The prefix of `a` up to the 3rd character ("101") is a subsequence of `b`, but the entire `a` is not. The expected output is `3`.

4. **Test Case 4**:
   - `n = 7`, `m = 7`
   - `a = "1001001"`, `b = "1010101"`
   - The entire `a` is a subsequence of `b`, so the expected output is `7`.

This test input should help expose the bug in the given code by focusing on edge cases where the algorithm might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

Input: The first line consists of a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

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

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int vis[2000],v[2000];
void solve() {
	int n,m,ans=0;
	string a,b;
	cin>>n>>m;
	cin>>a>>b;
	if(b.size()==1){
		if(a[0]!=b[0]) cout<<0<<endl;
		else cout<<1<<endl;
		return;
	}
	for(int i=n; i>0; i--) {
		string c=a.substr(0,i);
		//cout<<c<<endl;
		int l=0,r=0;
		while(l<c.size() && r<b.size()){
			if(c[l]==b[r]) l++,r++;
			if(c[l]!=b[r]) r++;
		}
		//cout<<l<<' '<<r<<endl;
		if(l==a.size()) {
			cout<<l<<endl;
			return;
		}
		if(l==c.size()) ans=max(ans,l);
	}
	cout<<ans<<endl;
}
int main() {
	int t=1;
	cin>>t;
	while(t--) solve();
	return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Let's analyze the code and identify potential issues:

1. **Handling of Substrings**: The code uses a loop to check if a prefix of `a` is a subsequence of `b`. It iterates from the longest possible prefix to the shortest. However, the logic for checking the subsequence might be flawed.

2. **Edge Cases**: The code might fail when:
   - `a` and `b` are very short.
   - `a` and `b` have very different lengths.
   - `a` or `b` contain only one character.
   - `b` contains multiple consecutive characters that are the same.

3. **Specific Issues**:
   - The condition `if(l==a.size())` is incorrect because `l` should be compared to `c.size()`, not `a.size()`.
   - The code might not handle cases where `b` is much longer than `a` efficiently.

Given these points, let's create a test input that can expose these issues:

```plaintext
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
```

This test input includes:
1. A case where `a` and `b` are of different lengths and `b` has a repeating pattern.
2. A case where `a` is a subsequence of `b` but not the entire string.
3. A case where `a` is a single character and `b` is a long string.
4. A case where `a` and `b` are of the same length and `a` is a subsequence of `b`.
5. A case where `a` is a subsequence of `b` but `b` is longer.
6. A case where `a` is not a subsequence of `b`.

This input should help identify 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: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

Input: The first line consists of a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

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

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
t = int(input())
for __ in range(t):
    n, m = map(int, input().split())
    a = [int(i) for i in input()]
    b = [int(i) for i in input()]
    x = 0
    for i in range(m):
        try:
            if a[0] == b[0]:
                x+=1
                try:
                    a = a[1:]
                    b = b[1:]
                except IndexError:
                    break
            else:
                b = b[1:]
        except IndexError:
            break
    print(x)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the code might fail to correctly identify the maximum possible number \( k \) such that a prefix of string \( a \) of length \( k \) is a subsequence of string \( b \).

### Analysis of the Buggy Code:
1. **Initial Check**: The code checks if the first character of \( a \) matches the first character of \( b \).
2. **Subsequence Check**: If they match, it increments \( x \) and removes the first character from both \( a \) and \( b \).
3. **Loop Through \( b \)**: It continues to loop through \( b \) until it finds a match or runs out of characters in \( a \) or \( b \).

### Potential Bugs:
1. **Incorrect Subsequence Matching**: The code only checks the first character of \( a \) against the first character of \( b \) and then removes both characters. This can lead to missing valid subsequences.
2. **Premature Termination**: The code might terminate prematurely if it runs out of characters in \( b \) before checking all characters in \( a \).

### Test Input to Expose the Bug:
To expose the bug, we need a test case where the first character of \( a \) matches the first character of \( b \), but the subsequent characters in \( a \) can still form a subsequence in \( b \) even if they are not at the beginning of \( b \).

```plaintext
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
```

### Explanation:
- **Test Case 1**: \( a = "10011" \), \( b = "1110" \)
  - The correct answer is 2, but the code might only find the first '1' and then fail to find the next '0' because it only checks the first character of \( b \).

- **Test Case 2**: \( a = "100" \), \( b = "110" \)
  - The correct answer is 2, but the code might only find the first '1' and then fail to find the next '0'.

- **Test Case 3**: \( a = "1" \), \( b = "111" \)
  - The correct answer is 1, and the code should correctly find the '1'.

- **Test Case 4**: \( a = "1011" \), \( b = "1111" \)
  - The correct answer is 1, but the code might only find the first '1' and then fail to find the next '0'.

- **Test Case 5**: \( a = "100" \), \( b = "11010" \)
  - The correct answer is 3, but the code might only find the first '1' and then fail to find the next '0'.

- **Test Case 6**: \( a = "100" \), \( b = "0" \)
  - The correct answer is 0, and the code should correctly find that there are no matches.

This test input should help expose the bug in the given code by providing cases where the first character of \( a \) matches the first character of \( b \), but the subsequent characters in \( a \) can still form a subsequence in \( b \) even if they are not at the beginning of \( b \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

Input: The first line consists of a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

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

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
t = int(input())
ans = []

def solve(a, b, na, nb):
	k = 0
	pre = ''
	pre_len = 0
	for l_a in a:
		# print(l_a)
		pre += l_a
		pre_len += 1
		# print("Pre:", pre)
		if subseq(b, pre, nb, pre_len):
			k += 1
			continue
		else:
			break
	ans.append(k)

def subseq(a, b, na, nb) -> bool:
	# checks if b is a subsequence of a
	i = 0
	chars_fnd = 0

	for j in range(na):
		if i < nb and b[i] == a[j]:
			chars_fnd += 1
			i += 1
		if chars_fnd == len(b):
			return True
	return False

while t:
	na, nb = map(int, input().split())
	a = input()
	b = input()
	solve(a, b, na, nb)
	t -= 1

for a in ans:
	print(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:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are some considerations:

1. **Empty Strings**: Check if the code handles empty strings correctly.
2. **Single Character Strings**: Check if the code handles single character strings correctly.
3. **Longer Strings**: Check if the code handles longer strings efficiently.
4. **All Zeros or All Ones**: Check if the code handles strings with all zeros or all ones correctly.
5. **Prefixes and Subsequences**: Check if the code correctly identifies the longest prefix that is a subsequence.

Given the constraints, a small-scale test input that can expose potential issues is:

```plaintext
4
1 1
1
0
3 3
100
101
5 5
10101
01010
2 2
00
11
```

Explanation:
1. **Test Case 1**: `1 1` with `a = "1"` and `b = "0"`. This checks if the code handles a single character string correctly, where the prefix of `a` is not a subsequence of `b`.
2. **Test Case 2**: `3 3` with `a = "100"` and `b = "101"`. This checks if the code correctly identifies that the prefix of length 2 (`"10"`) is a subsequence of `b`, but the prefix of length 3 (`"100"`) is not.
3. **Test Case 3**: `5 5` with `a = "10101"` and `b = "01010"`. This checks if the code correctly identifies that the prefix of length 3 (`"101"`) is a subsequence of `b`, but the prefix of length 4 (`"1010"`) is not.
4. **Test Case 4**: `2 2` with `a = "00"` and `b = "11"`. This checks if the code handles a string with all zeros correctly, where the prefix of `a` is not a subsequence of `b`.

This test input should help in identifying 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: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

Input: The first line consists of a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

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

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
import java.util.*;

public class Main {

    static boolean f(String s1, String s2) {
        int n = s1.length(), m = s2.length();
        int i = 0, j = 0;
        while (i < n && j < m) {
            if (s1.charAt(i) == s2.charAt(j))
                i++;
            j++;
        }
        return i == n;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            scanner.nextLine();

            String a = scanner.nextLine();
            String b = scanner.nextLine();
            String Str="";
            int count=0;
            for(int i=0; i<n; i++) {
                Str = Str + a.charAt(i);
                if(f(Str, b)) {
                    count++;
                }
             else{
break;
}
            }
            System.out.println(count);
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
10101
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

Input: The first line consists of a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

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

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
import java.util.*;

public class Solution {

    static boolean isSubSequence(String str1, String str2,int m, int n){
        if (m == 0)
            return true;
        if (n == 0)
            return false;
        if (str1.charAt(m - 1) == str2.charAt(n - 1))
            return isSubSequence(str1, str2, m - 1, n - 1);
        return isSubSequence(str1, str2, m, n - 1);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int ans = 0;
            String a = sc.next();
            String b = sc.next();
            String prfx = "";
            for(int j = 0;j<n;j++){
                prfx += a.charAt(j);
                if(isSubSequence(prfx, b, j+1, m)){
                    ans = j+1;
                }
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
10101
10100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
int T,n,a[1000009],dp[1000009][3];
int x,y,cnt,fir[1000009],i,vis[1000009],ans;
struct nod{
	int next;
	int u,v;
};
nod edg[4000009];
void add(int u,int v){
	cnt++;
	edg[cnt].next=fir[u];
	edg[cnt].u=u;edg[cnt].v=v;
	fir[u]=cnt;
}
void dfs(int u){
	int i;
	dp[u][0]=0;
	dp[u][1]=a[u];
	for(i=fir[u];i;i=edg[i].next){
		int v=edg[i].v;
		if(vis[v]==0){
			vis[v]=1;
			dfs(v);
			dp[u][0]+=max(dp[v][0],dp[v][1]);
			dp[u][1]+=dp[v][0];
		}
	}
}
signed main(){
	cin>>T;
	for(int o=1;o<=T;o++){
		scanf("%lld",&n);
		cnt=ans=0;
		for(i=0;i<=n;i++){
			a[i]=0;fir[i]=0;vis[i]=0;
			dp[i][0]=dp[i][1]=0;
		}
		for(i=1;i<=n;i++){
			scanf("%lld",&a[i]);
			ans+=a[i];
		}
		for(i=1;i<n;i++){
			scanf("%lld%lld",&x,&y);
			add(x,y);add(y,x);
		}
		vis[1]=1;
		dfs(1);
		ans+=(ans-max(dp[1][0],dp[1][1]));
		printf("%lld\n",ans);
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
1000000000000
5
1 1 1 1 1
1 2
1 3
2 4
2 5
7
1 1 1 1 1 1 1
1 2
1 3
2 4
2 5
3 6
3 7
```

### Explanation:
1. **Test Case 1: Single Monster**
   - This is a simple case with a single monster. The health decrement is equal to the attack points of the monster.
   - Input: `1\n1000000000000`
   - Expected Output: `1000000000000`

2. **Test Case 2: Small Tree with Equal Attack Points**
   - This test case has a small tree with 5 vertices, and each monster has an attack point of 1.
   - The optimal strategy is to kill the monsters in such a way that the total health decrement is minimized.
   - Input: `5\n1 1 1 1 1\n1 2\n1 3\n2 4\n2 5`
   - Expected Output: `5` (since all monsters have equal attack points and the tree structure is simple, the minimum health decrement is the sum of all attack points).

3. **Test Case 3: Larger Tree with Equal Attack Points**
   - This test case has a larger tree with 7 vertices, and each monster has an attack point of 1.
   - The optimal strategy is to kill the monsters in such a way that the total health decrement is minimized.
   - Input: `7\n1 1 1 1 1 1 1\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7`
   - Expected Output: `7` (similar to the previous case, the minimum health decrement is the sum of all attack points).

These test cases are designed to expose any issues with the dynamic programming approach in the buggy code, especially in handling small and simple tree structures.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define pii pair<int, int>
#define endl '\n'
#define int ll

using namespace std;

const int MAX_D = 21;

int N;
vector<vector<int>> graph;
vector<vector<pii>> dp;
vector<int> his, val;

void dfs(int v, int p)
{
    for (int u : graph[v])
    {
        if (u == p)
            continue;
        dfs(u, v);
    }

    for (int j = 0; j < MAX_D; j++)
    {
        dp[v][j] = {0, val[v] * (j + 1)};
        int mn_sm1 = 0, mn_sm2 = 0;
        for (int u : graph[v])
        {
            if (u == p)
                continue;
            int cur_mn = dp[u][j].first + val[u] * (j + 2);
            int cur_mn2 = dp[u][j].second;
            for (int k = j + 1; k < MAX_D; k++)
            {
                cur_mn = min(cur_mn, dp[u][k].second);
                cur_mn2 = min(cur_mn2, dp[u][k].second);
            }

            mn_sm1 += cur_mn;
            mn_sm2 += cur_mn2;
        }

        dp[v][j].first += mn_sm2;
        dp[v][j].second += mn_sm1;
    }
}

void solve()
{
    cin >> N;

    graph = vector<vector<int>>(N);
    dp = vector<vector<pii>>(N, vector<pii>(MAX_D));
    his = vector<int>(N, -1);
    val = vector<int>(N);

    for (int i = 0; i < N; i++)
        cin >> val[i];

    for (int i = 0; i < N - 1; i++)
    {
        int a, b;
        cin >> a >> b;
        --a;
        --b;

        graph[a].push_back(b);
        graph[b].push_back(a);
    }
    dfs(0, -1);
    int res = dp[0][0].second;
    for (int i = 1; i < MAX_D; i++)
        res = min(res, dp[0][i].second);

    cout << res << endl;
}

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

    int t;
    cin >> t;

    while (t--)
        solve();
}

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

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

Think step by step.

Response:
```plaintext
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

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

vector<vector<int>> adj;
vector<int> attack;
vector<vector<int>> dp;

void dfs(int node, int parent) {
    dp[node][0] = 0;
    dp[node][1] = attack[node];

    for (int neighbor : adj[node]) {
        if (neighbor == parent) continue;

        dfs(neighbor, node);

        dp[node][0] += max(dp[neighbor][0], dp[neighbor][1]);
        dp[node][1] += dp[neighbor][0];
    }
}

void solve() {
    int n;
    cin >> n;
    adj.assign(n, vector<int>());
    attack.assign(n, 0);
    dp.assign(n, vector<int>(2, 0));

    for(int i = 0; i < n; i++) cin >> attack[i];
    for(int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    dfs(0, -1);

    int total_attack = accumulate(attack.begin(), attack.end(), 0ll);

    int result = max(dp[0][0],dp[0][1]);
    cout << 2*total_attack - result << endl;
}

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on the logic and potential edge cases. The bug in the code lies in the calculation of the minimum health decrement. Specifically, the code incorrectly calculates the total health decrement by subtracting the maximum of `dp[0][0]` and `dp[0][1]` from `2 * total_attack`.

### Analysis of the Bug:
- The `dp[0][0]` value represents the minimum health decrement if the root node is not killed.
- The `dp[0][1]` value represents the minimum health decrement if the root node is killed.
- The code calculates the result as `2 * total_attack - max(dp[0][0], dp[0][1])`.

However, the correct calculation should be the minimum of the two scenarios:
- `total_attack - dp[0][0]` (if the root node is not killed)
- `total_attack - dp[0][1]` (if the root node is killed)

### Challenging Test Input:
We need a test case where the difference between `total_attack - dp[0][0]` and `total_attack - dp[0][1]` is significant, and the incorrect calculation in the code will lead to a wrong result.

### Test Case:
Consider a tree with a root node and two children, where the root node has a high attack value, and one of the children has a very low attack value. This will highlight the bug in the calculation.

```plaintext
1
3
1000000000000 1 1
1 2
1 3
```

### Explanation:
- The root node (1) has an attack value of `1000000000000`.
- The two children (2 and 3) have attack values of `1` each.
- The total attack is `1000000000002`.

### Correct Calculation:
- If the root node is not killed: `dp[0][0] = 0 + max(1, 1) = 1`
- If the root node is killed: `dp[0][1] = 1000000000000 + 1 + 1 = 1000000000002`

- The correct result should be:
  - `total_attack - dp[0][0] = 1000000000002 - 1 = 1000000000001` (if the root node is not killed)
  - `total_attack - dp[0][1] = 1000000000002 - 1000000000002 = 0` (if the root node is killed)

The minimum of these two is `0`.

### Incorrect Calculation in the Code:
- The code calculates `2 * total_attack - max(dp[0][0], dp[0][1]) = 2 * 1000000000002 - 1000000000002 = 1000000000002`.

This test case will expose the bug in the code, as the expected output is `0`, but the code will output `1000000000002`.

### Output:
```plaintext
1
3
1000000000000 1 1
1 2
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
import sys; input=sys.stdin.readline; out=[]
from math import inf
for i in range(int(input())):
  n=int(input())
  a=list(map(int,input().split()))
  g=[[] for j in range(n)]
  s1,s2=set(),set()
  for j in range(n-1):
    x,y=map(int,input().split())
    g[x-1].append(y-1)
    g[y-1].append(x-1)
  ans=inf
  for j in range(n):
    v=[0 for j in range(n)]
    s=[j]
    v[j]=0
    s1.add(j)
    while s:
      parent=s.pop()
      u=int(parent in s1)
      for child in g[parent]:
        if not(v[child]):
          v[child]=1
          s.append(child)
          if u:
            s2.add(child)
          else:
            s1.add(child)
    ans=min(sum([a[j] for j in s1]),sum([a[j] for j in s2]),ans)
  out.append(sum(a)+ans)
print('\n'.join(map(str,out)))

**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
4
1 2 3 4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
a=[]
def solve():
    global a
    global e
    global p
    global q
    n=int(input())
    a=list(map(int,input().split()))
    p=[0]
    q=[0]
    e=[[] for _ in range(n)]

    for i in range(n-1):
        x,y=map(int,input().split())
        x-=1
        y-=1
        e[x].append(y)
        e[y].append(x)
    dfs(0,-1,1)
    print(min(p[0],q[0])+p[0]+q[0])
a=[]
e=[]
p=[0]
q=[0]

def dfs(x,par,c):
    global a
    global e
    global p
    global q
    if c==1:

        p[0]+=a[x]
    if c==-1:
        q[0]+=a[x]
    for i in e[x]:
        if par==i:
            continue
        dfs(i,x,c*(-1))

__=int(input())
for ___ in range(__):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
from collections import defaultdict, deque
T = int(input())
for _ in range(T):
    n = int(input())
    health = list(map(int,input().split(" ")))
    neighbours = defaultdict(list)
    for _ in range(n-1):
        a, b = list(map(int,input().split(" ")))
        neighbours[a].append(b)
        neighbours[b].append(a)
    total_healths = sum(health)
    attack_taken = total_healths
    found = True
    while found:
        found = False
        for guy1 in neighbours:
            if health[guy1-1]!=0:
                for guy2 in neighbours[guy1]:
                    if health[guy2-1]!=0:
                        found=True
        def attack_choice(guy):
            queue = deque([(guy,True)])
            visited = set()
            ans = 0
            while queue:
                guy,to_take = queue.popleft()
                if guy in visited:
                    continue
                visited.add(guy)
                if to_take:
                    ans+=health[guy-1]
                for next_guy in neighbours[guy]:
                    queue.append((next_guy,not to_take))
            return ans
        def mark_zero(guy):
            queue = deque([(guy,True)])
            visited = set()
            ans = 0
            while queue:
                guy,to_take = queue.popleft()
                if guy in visited:
                    continue
                visited.add(guy)
                if to_take:
                    health[guy-1]=0
                for next_guy in neighbours[guy]:
                    queue.append((next_guy,not to_take))
            return ans
        if found:
            cost1 = attack_choice(guy1)
            cost2 = attack_choice(guy2)
            if cost1>=cost2:
                mark_zero(guy1)
                attack_taken+=cost2
            else:
                mark_zero(guy2)
                attack_taken+=cost1
    print(attack_taken)

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

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

Think step by step.

Response:
```plaintext
1
3
100 100 100
1 2
2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
import java.util.*;

public class Main {

    Scanner sc = new Scanner(System.in);
	long dp[][];

	void dfs(int u, int p, long[] a, Map<Integer, List<Integer>> tree) {
		// System.out.println(p + "->" + u);
		// initialize dp
		for(int i = 1; i <= 22; ++i) dp[u][i] = a[u] * i;

		// solve for each child
		for(int v: tree.get(u)) {
			if(v != p) {
				dfs(v, u, a, tree);
			}
		}

		// for each round of u
		for(int i = 1; i <= 22; ++i) {
			long sum = 0;
			for(int v: tree.get(u)) {
				long ans = Long.MAX_VALUE;
				if(v != p) {
					// for each round of v
					for(int j = 1; j <= 22; ++j) {
						if(i != j) {
							ans = Math.min(ans, dp[v][j]);
						}
					}
					sum += ans;
				}
			}
			dp[u][i] += sum;
		}
		//for(int i = 1; i <= 5; ++i) System.out.print(dp[u][i] + " ");
		//System.out.println();
	}

    void solve() {
	int n = sc.nextInt();
    	long a[] = new long[n+1];
	for(int i = 1; i <= n; ++i) a[i] = sc.nextLong();
	Map<Integer, List<Integer>> tree = new HashMap<>();
	for(int i = 1; i <= n; ++i) tree.put(i, new ArrayList<>());
	for(int i = 0; i < n-1; ++i) {
		int x = sc.nextInt(), y = sc.nextInt();
		tree.get(x).add(y);
		tree.get(y).add(x);
	}
	dp = new long[n+1][23];
	dfs(1, 0, a, tree);
	//for(int i = 1; i <= n; ++i) for(int j = 0; j < 3; ++j) System.out.print(dp[i][j] +  " ");
	long res = Long.MAX_VALUE;
	for(int i = 1; i <= 22; ++i) res = Math.min(res, dp[1][i]);
	System.out.println(res);
    }

    void run() {
        int t = sc.nextInt();
        while(t-- != 0) {
            solve();
        }
    }

    public static void main(String[] args) {
        new Main().run();
    }
}

**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 test input that exposes the bug in the provided code, let's consider the following:

1. **Understanding the Bug**: The code initializes a 2D array `dp` where `dp[u][i]` is supposed to represent the minimum health decrement if we kill the monster at vertex `u` in the `i`-th round. However, the code limits the number of rounds to 22, which is arbitrary and does not necessarily reflect the optimal solution for all cases.

2. **Edge Cases**:
   - **Single Vertex**: A tree with a single vertex.
   - **Linear Tree**: A tree where each vertex is connected in a linear fashion.
   - **Star Tree**: A tree where one vertex is connected to all other vertices.
   - **Balanced Tree**: A balanced binary tree.
   - **High Attack Points**: Vertices with very high attack points to maximize the health decrement.
   - **Large Number of Vertices**: A tree with the maximum number of vertices allowed.

3. **Test Input**:
   - **Single Vertex**: This should be straightforward and easy to handle.
   - **Linear Tree**: This should test the sequential nature of the tree.
   - **Star Tree**: This should test the central vertex with multiple connections.
   - **Balanced Tree**: This should test the balanced nature of the tree.
   - **High Attack Points**: This should test the handling of large numbers.

Let's create a test input that includes a star tree with high attack points to maximize the health decrement and expose the bug.

```plaintext
1
5
1000000000000 1000000000000 1000000000000 1000000000000 1000000000000
1 2
1 3
1 4
1 5
```

This test input consists of:
- 1 test case.
- 5 vertices, each with an attack point of \(10^{12}\).
- The tree is a star tree with vertex 1 at the center, connected to vertices 2, 3, 4, and 5.

This input should expose the bug in the code by requiring a large number of rounds to minimize the health decrement, which the code may not handle correctly due to the arbitrary limit of 22 rounds.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

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

public class D_The_Omnipotent_Monster_Killer {

    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 void dfs(long[][] mem, int i, int incl, Map<Integer, List<Integer>> adj, int prev, long[] val) {

  if(mem[i][incl] != -1) return;
  long ans = 0;

  if(incl == 1) {
  ans += val[i];
  }
  int nincl = (incl + 1)%2;

  for(int y: adj.getOrDefault(i, new ArrayList<>())) {

    if(y != prev ) {
      if(mem[y][nincl] == -1) {

        dfs(mem, y, nincl, adj, i, val);

      }

        ans += mem[y][nincl];

    }

  }
  mem[i][incl] = ans;

}
    public static void main(String args[]) {

        FastReader s = new FastReader();
        StringBuilder out = new StringBuilder();

        int tc = s.nextInt();
        // int tc = 1;
        while (tc-->0) {
          int n = s.nextInt();
          long[] val = new long[n+1];
          for(int i = 1; i <= n; i++) {
            val[i] = s.nextLong();
          }
          Map<Integer, List<Integer>> adj = new HashMap<>();
          int e = n-1;
          while(e-->0) {
            int x=  s.nextInt(), y = s.nextInt();
            adj.putIfAbsent(x, new ArrayList<>());
            adj.putIfAbsent(y, new ArrayList<>());
            adj.get(x).add(y);
            adj.get(y).add(x);
          }

          boolean odd = true;
          long os = 0, es = 0;

          Queue<Integer> q = new LinkedList<>();
          boolean[] vis = new boolean[n+1];
          long[][] mem = new long[n+1][2];
          long tot = 0;
          for(long v : val) tot += v;
          // System.out.println(os);
          // System.out.println(es);
for(int i = 0; i < mem.length; i++) {
            for(int j = 0; j < mem[i].length; j++) {
              mem[i][j] = -1;
            }
          }
          dfs(mem, 1, 1, adj, -1, val);
          dfs(mem, 1, 0, adj, -1, val);

      //    System.out.println(tot);
     //     System.out.println(mem[1][1]);
      //    System.out.println(mem[1][0]);

          long ans1 = Math.min(tot + mem[1][0], tot + (tot - mem[1][0]));
          long ans2 = Math.min(tot + mem[1][1], tot + (tot - mem[1][1]));

          out.append(Math.min(ans1, ans2)).append('\n');
        }

        System.out.println(out);
    }
}

**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
4
1 2 3 4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int,int>
#define x first
#define y second
#define lowbit(x) x&-x
//#define endl '\n'
const int N=3e5+10;
const int mod=1e9+7;
vector<int> g[N];
int a[N],dep[N];
void dfs(int x,int fa) {
    for(int y:g[x]) {
        if(y==fa) continue;
        dep[y]=dep[x]+1;
        dfs(y,x);
    }
}
void solve() {
    int n;
    cin>>n;
    int ans=0;
    for(int i=1;i<=n;i++) {
        cin>>a[i];
        g[i].clear();
        dep[i]=0;
        ans+=a[i];
    }
    for(int i=1;i<n;i++) {
        int u,v;
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,0);
    int res=0;
    for(int i=1;i<=n;i++) if(dep[i]&1) res+=a[i];
    ans+=min(res,ans-res);
    cout<<ans<<endl;
}
signed main() {
    int T=1;
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    cin>>T;
    while(T--) {
        solve();
    }
    cout<<endl;
}

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

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

Think step by step.

Response:
```plaintext
1
3
100 1 100
1 2
2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

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

using namespace std;

constexpr long long inf = 1e18;

long long get_best(const vector<long long> &v, int i) {
	return min(*min_element(v.begin(), v.begin()+i), *min_element(v.begin()+i+1, v.end()));
}

void dfs(int cur, int prev, vector<vector<long long>> &dp, const vector<vector<int>> &adj, const vector<long long> &a) {
	for (int next: adj[cur]) {
		if (next == prev)
			continue;

		dfs(next, cur, dp, adj, a);
	}

	for (int i = 1; i < 40; i++) {
		long long x {};
		for (int next: adj[cur]) {
			if (next == prev)
				continue;
			x += get_best(dp[next], i);
		}
		dp[cur][i] = min(dp[cur][i], a[cur]*i + x);
	}
}

void test_case() {
	long long n;
	cin >> n;
	vector<long long> a(n);
	for (long long i = 0; i < n; i++)
		cin >> a[i];
	vector<vector<int>> adj(n);
	for (long long i = 0; i < n-1; i++) {
		long long x, y;
		cin >> x >> y;
		x--, y--;
		adj[x].push_back(y);
		adj[y].push_back(x);
	}

	vector<vector<long long>> dp(n, vector<long long>(40, inf));
	dfs(0, -1, dp, adj, a);

	cout << *min_element(dp[0].begin(), dp[0].end()) << '\n';
}

int main() {
	#ifndef ONLINE_JUDGE
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
	#endif

	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed;

	int t {1};
	cin >> t;
	while(t--) test_case();
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases that might cause the algorithm to fail. The key points to consider are:

1. **Tree Structure**: The tree structure can affect the algorithm's performance, especially if there are nodes with a large number of children.
2. **Attack Points**: The attack points of the monsters can be very high, which might cause overflow or incorrect minimum calculations.
3. **Edge Cases**: Small trees with specific structures that might not be handled correctly by the algorithm.

Given the constraints and the nature of the problem, let's create a small-scale test input that might expose the bug:

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

### Explanation
- **Tree Structure**: The tree is a star with node 1 at the center and nodes 2, 3, and 4 as leaves.
- **Attack Points**: The attack points are small and distinct, which should make it easier to verify the correctness of the algorithm.
- **Expected Output**: The optimal strategy is to kill the monsters on nodes 2, 3, and 4 in the first round, and then kill the monster on node 1 in the second round. The total health decrement should be \(1 + 2 + 3 + 4 = 10\).

This test input is small and simple, but it can help identify if the algorithm correctly handles the star tree structure and the optimal killing strategy. If the algorithm fails to produce the correct output, it indicates a potential bug in the implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
from collections import deque

def main():
    n = int(input())
    arr = list(map(int,input().split()))

    neigh = [[] for _ in range(n)]
    for _ in range(n-1):
        u,v = map(int,input().split())
        neigh[u-1].append(v-1)
        neigh[v-1].append(u-1)

    children = [[] for _ in range(n)]

    queue = deque()

    queue.append(0)

    parent = [-2]*n

    parent[0] = -1

    seq = []

    while queue:
        index = queue.popleft()
        seq.append(index)
        for nextindex in neigh[index]:
            if parent[nextindex] > -2:  continue
            parent[nextindex] = index
            children[index].append(nextindex)
            queue.append(nextindex)

    k = 10
    inf = 10**20

    dp = [[inf]*(k+1) for _ in range(n)]
    dpl = [[inf]*(k+1) for _ in range(n)]
    dpr = [[inf]*(k+1) for _ in range(n)]

    for index in seq[::-1]:

        for d in range(k):
            dp[index][d] = d * arr[index]

        for d in range(k):
            for nextindex in children[index]:
                currmin = min(dpl[nextindex][d-1], dpr[nextindex][d+1])
                dp[index][d] += currmin
        for d in range(k):
            dpl[index][d] = min(dpl[index][d-1], dp[index][d])
        for d in range(k-1,-1,-1):
            dpr[index][d] = min(dpr[index][d+1], dp[index][d])

    ans = min(dp[0]) + sum(arr)

    print(ans)

T = int(input())
t = 1
while t <= T:
    main()
    t += 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
import sys

# sys.setrecursionlimit(300005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = -1-(-1 << 62)
# inf = -1-(-1 << 31)
md = 10**9+7
# md = 998244353

def solve():
    def dfs(root=0):
        uu, stack = [], [root]
        while stack:
            u = stack.pop()
            uu.append(u)
            for v in to[u]:
                if v == parent[u]: continue
                parent[v] = u
                depth[v] = depth[u]+1
                stack.append(v)
        return uu

    n = II()
    aa = LI()
    to = [[] for _ in range(n)]
    for _ in range(n-1):
        u, v = LI1()
        to[u].append(v)
        to[v].append(u)

    parent, depth = [-1]*n, [0]*n
    dfs()
    ans = sum(aa)
    s = sum(aa[u]*(depth[u] & 1) for u in range(n))
    ans += min(s, ans-s)
    print(ans)

# ***************************************

testnumber = II()
# testnumber = 1
for testcase in range(testnumber):
    solve()
    # print("YES" if solve() else "NO")
    # print("Yes" if solve() else "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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic fails. The primary issue in the buggy code is the way it handles the tree structure and the calculation of the minimum health decrement. Specifically, the code does not correctly account for the restriction that you cannot kill two directly connected monsters in the same round.

### Steps to Generate the Test Input:
1. **Tree Structure**: Create a tree where the nodes have a specific arrangement that can expose the bug. For example, a tree with a long path or a tree with many branches.
2. **Attack Points**: Assign attack points to the nodes in such a way that the optimal solution is not immediately obvious.
3. **Edge Cases**: Consider edge cases like a tree with a single node, a tree with a linear structure, and a tree with multiple branches.

### Example Test Input:
Let's create a test input with a tree that has a linear structure, which is a common edge case. This will help us see if the code correctly handles the restriction of not killing two directly connected monsters in the same round.

```plaintext
1
5
10 20 30 40 50
0 1
1 2
2 3
3 4
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Nodes**: 5
- **Attack Points**: [10, 20, 30, 40, 50]
- **Edges**: (0, 1), (1, 2), (2, 3), (3, 4)

### Expected Output:
The optimal sequence of operations for this test case would be:
1. **First Round**: Receive attacks from all monsters (10 + 20 + 30 + 40 + 50 = 150). Kill monsters on nodes 0, 2, and 4.
2. **Second Round**: Receive attacks from monsters on nodes 1 and 3 (20 + 40 = 60). Kill monsters on nodes 1 and 3.
3. **Subsequent Rounds**: All monsters are killed, so no further attacks.

The total health decrement is 150 + 60 = 210.

### Test Input:
```plaintext
1
5
10 20 30 40 50
0 1
1 2
2 3
3 4
```

This test input should expose the bug in the provided code by ensuring that the restriction of not killing directly connected monsters in the same round is correctly handled.