**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

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

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef pair<int,int> pii;

const int N = 2e5+10;

int p[N];
int d[N];

void solve()
{
    string a,b;
    cin>>a>>b;
    swap(a,b);
    map<char,int> m;
    for(int i=0;i<b.size();i++)
    {
        int x=10010;
        for(int j=(m.count(b[i]) ? m[b[i]]+1 : 0);j<a.size();j++)
        {
            if(a[j]==b[i])
            {
                x=j;
                break;
            }
        }
        //cout<<x<<"\n";
        p[i]=x;
        m[b[i]]=x;
    }

    /*for(int i=0;i<b.size();i++)
    {
        cout<<p[i]<<" ";
    }
    cout<<"\n";*/

    int ans=0;
    for(int i=0;i<b.size();i++)
    {
        if(p[i]<a.size())
        {
            d[i]=1;
        }
        else
        {
            d[i]=0;
        }
        if(p[i]<a.size())
        {
            for(int j=0;j<i;j++)
            {
                if(p[j]<p[i])
                {
                    d[i]=max(d[i],d[j]+1);
                }
            }
            ans=max(ans,d[i]);
        }
        //ans=max(ans,d[i]);
    }
    //cout<<ans<<"\n";
    cout<<a.size()+b.size()-ans<<"\n";

}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //cout.precision(2);
    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 generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The key aspects to focus on include:

1. **Overlapping Characters**: Ensure that the characters in `a` and `b` overlap in a way that the algorithm might not handle correctly.
2. **Subsequence and Substring Constraints**: Ensure that `b` is a subsequence of a string that contains `a` as a substring.
3. **Edge Cases**: Consider cases where `a` and `b` have the same characters, or where `b` is a substring of `a`.

Given these considerations, let's construct a test input that is small but challenging:

```plaintext
3
abc
cba
aaa
aaa
abc
abc
```

### Explanation:
1. **Test Case 1: `abc` and `cba`**
   - `a` is `abc`.
   - `b` is `cba`.
   - The minimum length string that contains `abc` as a substring and `cba` as a subsequence is `abcba`, which has a length of 5.
   - This test case checks if the algorithm can handle the reverse order of characters in `b`.

2. **Test Case 2: `aaa` and `aaa`**
   - `a` is `aaa`.
   - `b` is `aaa`.
   - The minimum length string that contains `aaa` as a substring and `aaa` as a subsequence is `aaa`, which has a length of 3.
   - This test case checks if the algorithm can handle identical strings.

3. **Test Case 3: `abc` and `abc`**
   - `a` is `abc`.
   - `b` is `abc`.
   - The minimum length string that contains `abc` as a substring and `abc` as a subsequence is `abc`, which has a length of 3.
   - This test case checks if the algorithm can handle the same string for both `a` and `b`.

These test cases are designed to expose potential issues in the buggy code by testing overlapping characters, identical strings, and reverse order of characters.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

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

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
# -*- coding: utf-8 -*-
"""
Created on Thu Aug  8 19:33:37 2024

@author: PRISHIT VORA
"""

def solve(a,b):
    cnt,last = 0,-1
    for i in range(len(b)):
        for j in range(len(a)):
            if b[i] == a[j]:
                if j > last:
                    last = j
                    cnt += 1
                    break

    return len(a)+len(b)-cnt

def main():
    for _ in range(int(input())):
        a = input()
        b = input()
        ans = solve(a,b)
        print(ans)

if __name__ =='__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
3
abc
cba
xyz
zyx
abcde
edcba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

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

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
from bisect import bisect_left

def find_ge(a, key):
    '''Find smallest item greater-than or equal to key.
    Raise ValueError if no such item exists.
    If multiple keys are equal, return the leftmost.

    '''
    i = bisect_left(a, key)
    if i == len(a):
        raise ValueError('No item found with key at or above: %r' % (key,))
    return a[i]

def solve(a, b):
    m = {}
    for i, v in enumerate(a):
        if v not in m:
            m[v] = []
        m[v].append(i)

    i = 0
    j = 0
    ans = 0
    count = 0
    while j < len(b):
        #print(i, j)
        if b[j] in m and m[b[j]][-1] >= i:
            count += 1
            i = find_ge(m[b[j]], i) + 1
        else:
            ans = max(ans, count)
            count = 0
            i = 0
            if count != 0:
                j -= 1

        j += 1
    ans = max(ans, count)
    return ans

t = int(input())
for _ in range(t):
    a = input()
    b = input()
    print(len(a) + len(b) - solve(a, b))

**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
abc
cba
abc
abc
abc
def
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

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

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
import java.util.*;

public class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
	    int t=sc.nextInt();
	    sc.nextLine();
	    while(t-->0){
	        String a=sc.nextLine();
	        String b=sc.nextLine();
	        int count=a.length();
	        HashSet<Character> hs=new HashSet<>();
	        for(int i=0;i<b.length();i++){
	            hs.add(b.charAt(i));
	        }
	        for(Character key:hs){
	            if(a.indexOf(key)==-1)
	            count++;
	        }
	        System.out.println(count);
	    }
	}
}

**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
abc
def
abc
abc
abc
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

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

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            String s=sc.next();
            String s1=sc.next();
            int c=0;
            int pos=-1;
            int i=0;
            while(i<s1.length()){
                int j=pos+1;
                while(j<s.length()){
                    if(s.charAt(j)==s1.charAt(i)){
                        pos=j;
                        c++;
                        break;
                    }
                    j++;
                }
                i++;
            }
            int temp=s.length()+(s1.length()-c);
            int ans=Math.min(temp,s1.length()+s.length()-c);
            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
5
abc
abc
abc
cba
abc
aaa
abc
xyz
abc
abac
```

5
abc
abc
abc
cba
abc
aaa
abc
xyz
abc
abac


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
#include <bits/stdc++.h>
constexpr int N = 5e3 + 5;

int main() {
    std::cin.tie(nullptr) -> sync_with_stdio(false);
    int T;
    auto solve = []() -> void {
        int n, m, p;
        std::cin >> n >> m >> p;
        std::vector<int> s(p + 1);
        for (int i = 1; i <= p; ++i) std::cin >> s[i];
        class Road {
        public:
            int u, v, w;
        };
        std::vector<Road> r(m + 1);
        for (int i = 1; i <= m; ++i)
            std::cin >> r[i].u >> r[i].v >> r[i].w;
        sort(r.begin() + 1, r.end(), [](Road x,Road y){ return x.w < y.w; });
        std::vector<int> fa(n + 1), sz(n + 1);
        auto find = [&fa](auto &&find, int x) -> int {
            return fa[x] ^ x ? fa[x] = find(find, fa[x]) : x;
        };
        for (int i = 1; i <= n; ++i) fa[i] = i;
        for (int i = 1; i <= p; i++) sz[s[i]] = 1;
        std::vector<std::vector<long long> > f(n + 1, std::vector<long long>(n + 1));
        for (int i = 1; i <= m; ++i) {
            int fx = find(find, r[i].u), fy = find(find, r[i].v);
            if (fx == fy) continue;
            std::vector<long long> g(p + 1, 1e18);
            for (int j = 1; j <= sz[fx]; ++j)
                g[j] = std::min(g[j], f[fx][j] + 1ll * sz[fy] * r[i].w);
            for (int j = 1; j <= sz[fy]; ++j)
                g[j] = std::min(g[j], f[fy][j] + 1ll * sz[fx] * r[i].w);
            for (int j = 1; j <= sz[fx]; ++j)
                for(int k = 1; k <= sz[fy]; ++k)
                    g[j + k] = f[fx][j] + f[fy][k];
            fa[fy] = fx, sz[fx] += sz[fy];
            for (int j = 1; j <= sz[fx]; ++j)
                f[fx][j] = g[j];
        }
        int rt = find(find, 1);
        for (int i = 1; i <= n; ++i)
            std::cout << f[rt][i] << " \n"[i == n];
    };
    for (std::cin >> T; T; --T) solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
// C++ includes used for precompiling -*- C++ -*-

// Copyright (C) 2003-2013 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library.  This library is free
// software; you can redistribute it and/or modify it under the
// terms of the GNU General Public License as published by the
// Free Software Foundation; either version 3, or (at your option)
// any later version.

// This library is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// Under Section 7 of GPL version 3, you are granted additional
// permissions described in the GCC Runtime Library Exception, version
// 3.1, as published by the Free Software Foundation.

// You should have received a copy of the GNU General Public License and
// a copy of the GCC Runtime Library Exception along with this program;
// see the files COPYING3 and COPYING.RUNTIME respectively.  If not, see
// <http://www.gnu.org/licenses/>.

/** @file stdc++.h
 *  This is an implementation file for a precompiled header.
  */

  // 17.4.1.2 Headers

  // C
  #ifndef _GLIBCXX_NO_ASSERT
  #include <cassert>
  #endif
  #include <cctype>
  #include <cerrno>
  #include <cfloat>
  #include <ciso646>
  #include <climits>
  #include <clocale>
  #include <cmath>
  #include <csetjmp>
  #include <csignal>
  #include <cstdarg>
  #include <cstddef>
  #include <cstdio>
  #include <cstdlib>
  #include <cstring>
  #include <ctime>

  #if __cplusplus >= 201103L
  #include <ccomplex>
  #include <cfenv>
  #include <cinttypes>
  #include <cstdbool>
  #include <cstdint>
  #include <ctgmath>
  #include <cwchar>
  #include <cwctype>
  #include <exception>
  #include <stdexcept>
  #endif

  // C++
  #include <algorithm>
  #include <bitset>
  #include <complex>
  #include <deque>
  #include <exception>
  #include <fstream>
  #include <functional>
  #include <iomanip>
  #include <ios>
  #include <iosfwd>
  #include <iostream>
  #include <istream>
  #include <iterator>
  #include <limits>
  #include <list>
  #include <locale>
  #include <map>
  #include <memory>
  #include <new>
  #include <numeric>
  #include <ostream>
  #include <queue>
  #include <set>
  #include <sstream>
  #include <stack>
  #include <stdexcept>
  #include <streambuf>
  #include <string>
  #include <typeinfo>
  #include <utility>
  #include <valarray>
  #include <vector>

  #if __cplusplus >= 201103L
  #include <array>
  #include <atomic>
  #include <chrono>
  #include <condition_variable>
  #include <forward_list>
  #include <future>
  #include <initializer_list>
  #include <mutex>
  #include <random>
  #include <ratio>
  #include <regex>
  #include <scoped_allocator>
  #include <system_error>
  #include <thread>
  #include <tuple>
  #include <typeindex>
  #include <type_traits>
  #include <unordered_map>
  #include <unordered_set>
  #endif

using namespace std;

int
gcd (int a, int b)
{
  if (b == 0)
	return a;
  return gcd (b, a % b);
}

/*/---------------------------Defines----------------------/*/
typedef vector < int >vi;
typedef pair < int, int >pii;

#define For(i, n) for (int i = 0; i < (int)n; ++i)
#define Rep(i, n) for (int i = 1; i <= (int)n; ++i)
#define ll long long
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define all(v) (v).begin(), (v).end()
/*/-----------------------Modular Arithmetic---------------/*/

const int mod = 998244353;
#define int long long

void solve(){
    int n,m,p;
    cin>>n>>m>>p;

    vector<tuple<int,int,int>> e;

    vector<int> vp(p);
    for(int &i:vp)cin>>i;

    for(int i=0;i<m;i++){
        int x,y,w;
        cin>>x>>y>>w;
        e.push_back({w,x,y});

    }

    vector<int> parent(n+1);
    for(int i=1;i<=n;i++){
        parent[i]=-i;
    }
    for(int i=0;i<p;i++){
        parent[vp[i]]=vp[i];
    }

    map<int,int> mapp;
    map<int,int> size;
    map<int,int> sizeo;

    for(int i=0;i<p;i++)size[vp[i]]=1;
    for(int i=1;i<=n;i++)sizeo[i]=1;

    sort(e.begin(),e.end());

    priority_queue<int> pq;

    int ans = 0;

    for(auto edge:e){
        int w = get<0>(edge);
        int x = get<1>(edge);
        int y = get<2>(edge);

        while(x!=parent[x] && x!=-parent[x]){x=parent[x];}
        while(y!=parent[y] && y!=-parent[y]){y=parent[y];}

        if(parent[x]>0 && parent[y]>0){
            if(sizeo[x]<sizeo[y])swap(x,y);
            parent[y] = x;
            int val1  = size[x]*w+mapp[y];
            int val2 = size[y]*w+mapp[x];
            if(val1<val2){
                pq.push(size[x]*w-mapp[x]);
            }else{
                pq.push(size[y]*w-mapp[y]);
            }

            mapp[x]= min(val1,val2);
            size[x]+=size[y];
            sizeo[x]+=sizeo[y];

            if(size[x]==p){ans = mapp[x];break;}
        }else if(parent[x]>0 && parent[y]<0){
            if(sizeo[x]<sizeo[y])swap(x,y);
            parent[y] = x;
            sizeo[x]+=sizeo[y];
        }else if(parent[x]<0 && parent[y]>0){
            if(sizeo[x]>sizeo[y])swap(x,y);
            parent[x] = y;

            sizeo[y]+=sizeo[x];
        }else{
            if(sizeo[x]<sizeo[y])swap(x,y);
            parent[y] = x;

            sizeo[x]+=sizeo[y];
        }
    }

    for(int i=0;i<p-1;i++){
        cout<<ans<<" ";
        ans-=pq.top();
        pq.pop();
    }
    for(int i=0;i<n-p+1;i++)cout<<0<<" ";
    cout<<endl;

}

int32_t
main ()
{
    ios_base::sync_with_stdio (false);
  cin.tie (NULL);
  cout.tie (NULL);
  cout << fixed << setprecision (25);
  cerr << fixed << setprecision (10);
  int t;
  cin>>t;
  while(t--)solve();
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes potential bugs in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some key points to consider:

1. **Small Scale**: The test input should be small enough to be manageable but large enough to cover different scenarios.
2. **Connected Graph**: Ensure the graph is connected to avoid trivial disconnections.
3. **Latency Values**: Use a mix of low and high latency values to test different paths.
4. **Server Placement**: Test cases where the optimal server placement is not immediately obvious.
5. **Multiple Houses Requiring Internet**: Ensure there are multiple houses requiring internet to test the algorithm's ability to handle multiple constraints.

Let's construct a small-scale test input based on these considerations:

- **Number of Houses (n)**: 5
- **Number of Cables (m)**: 5
- **Number of Houses Requiring Internet (p)**: 3
- **Houses Requiring Internet**: 1, 3, 5
- **Cables and Latencies**:
  - 1-2: 1
  - 2-3: 2
  - 3-4: 3
  - 4-5: 4
  - 1-5: 10

This setup ensures that the graph is connected and includes a mix of low and high latency values. The optimal placement of servers will not be immediately obvious, making it a good test case for the algorithm.

Here is the test input:

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

This input will test the algorithm's ability to handle a connected graph with varying latencies and multiple houses requiring internet.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+5;
struct node{
    int u,v,w;
}e[N];
bool cmp(node x,node y){
    return x.w<y.w;
}
int n,m,p;
int s[N];
int fa[N],res;
int cnt[N],num,ans[N];
int find(int x){
    return fa[x]==x?x:fa[x]=find(fa[x]);
}
void unionn(int a,int b,int v){
    int x=find(a),y=find(b);
    if(x==y)return;
    if(cnt[x])num--;
    if(cnt[y])num--;
    if(cnt[x]||cnt[y])num++;
    fa[x]=y;
    res-=ans[x]+ans[y];
    ans[y]=min(ans[y]+cnt[x]*v,ans[x]+cnt[y]*v);
    res+=ans[y];
    cnt[y]+=cnt[x];
}
mt19937 rnd(time(0));
int tmp[N];
void work(){
    cin>>n>>m>>p;
    for(int i=1;i<=n;++i)cnt[i]=ans[i]=0,fa[i]=i;
    res=0;
    for(int i=1;i<=p;++i)cin>>s[i],cnt[s[i]]++;
    for(int i=1;i<=m;++i)cin>>e[i].u>>e[i].v>>e[i].w;
    sort(e+1,e+m+1,cmp);
    int opt=100000;
    for(int i=1;i<p;++i)tmp[i]=1e18;
    for(int i=p;i<=n;++i)tmp[i]=0;
    while(opt--){
        int st=1;
        for(int i=2;i<=m;++i){
            if(e[i].w!=e[i-1].w){
                shuffle(e+st,e+i,rnd);
                st=i;
            }
        }
        shuffle(e+st,e+m+1,rnd);
        for(int i=1;i<=n;++i)cnt[i]=ans[i]=0,fa[i]=i;
        res=0;
        for(int i=1;i<=p;++i)cnt[s[i]]++;
        num=p;
        for(int i=1;i<=m;++i){
            int u=e[i].u,v=e[i].v,w=e[i].w;
            if(find(u)==find(v))continue;
            unionn(u,v,w);
            tmp[num]=min(tmp[num],res);
        }
        for(int i=n-1;i>=1;--i)tmp[i]=max(tmp[i],tmp[i+1]);
    }
    num=p;
    for(int i=1;i<=m;++i){
        int u=e[i].u,v=e[i].v,w=e[i].w;
        if(find(u)==find(v))continue;
        unionn(u,v,w);
        tmp[num]=min(tmp[num],res);
    }
    for(int i=n-1;i>=1;--i)tmp[i]=max(tmp[i],tmp[i+1]);
    for(int i=1;i<=n;++i)cout<<tmp[i]<<' ';
    cout<<'\n';
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)work();
    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
1
5 5 3
1 3 5
1 2 1
2 3 2
3 4 3
4 5 4
5 1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import sys
input = sys.stdin.readline
import heapq
for _ in range(int(input())):
    n,m,p = map(int,input().split())
    G = [[] for _ in range(n)]
    W = set([int(i)-1 for i in input().split()])
    for _ in range(m):
        a,b,c = map(int,input().split())
        a-=1
        b-=1
        G[a].append((b,c))
        G[b].append((a,c))
    SOL = [0] * n
    P = [0] * n
    for i in range(n):
        ma = int(1e16)
        idc = - 1
        for j in range(n):
            if P[j]:continue
            D = [int(1e10)] * n
            tot = 0
            q = []
            for k in range(n):
                if P[k] or k == j:
                    D[k] = 0
                    heapq.heappush(q, (0,k))
            while q:
                cst, idx = heapq.heappop(q)
                if cst > D[idx]:continue
                if idx in W:
                    tot+=cst
                for c,cs in G[idx]:
                    nwcs = max(cs,cst)
                    if nwcs < D[c]:
                        D[c] = nwcs
                        heapq.heappush(q,(nwcs,c))
            if tot < ma:
                ma = tot
                idc = j
        SOL[i] = ma
        P[idc] = 1
    print(*SOL)

**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 5 3
1 3 4
1 2 1
1 3 2
1 4 3
2 3 4
2 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))
def readInt():
    return int(input())
def readInts():
    return map(int, input().split())
def readStr():
    return input().strip()

# BF-OB5, EC, DB, CC, CL
def solve():
    n, m, p = readInts()
    s = readList()
    g = [[] for _ in range(n+1)]
    d = [[inf] * (n + 1) for _ in range(n + 1)]
    for _ in range(m):
        u, v, w = readInts()
        g[u].append([v, w])
        g[v].append([u, w])
        d[u][v] = d[v][u] = w
        d[u][u] = d[v][v] = 0

    # Floyd-Warshall
    for k in range(1, n+1):
        for i in range(1, n+1):
            for j in range(1, n+1):
                d[i][j] = min(d[i][j], max(d[i][k], d[k][j]))

    exp, ans = [0]*(n+1),[0]*n
    v0, v1 = [inf] * (n+1), []
    val = inf
    for i in range(1, p):
        best = -1
        for node in s:
            if exp[node]:
                continue
            dis = sum([min(d[j][node], v0[j]) for j in s])
            if dis < val:
                best = node
                val = dis
        for j in s:
            v0[j] = min(v0[j], d[j][best])
        ans[i-1] = val
        exp[best] = 1
    return " ".join([str(v) for v in ans])

inf = 10**18
for _ in range(int(input())):
    print(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
4 5 3
1 3 4
1 2 1
2 3 2
3 4 3
1 4 5
2 4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))
def readInt():
    return int(input())
def readInts():
    return map(int, input().split())
def readStr():
    return input().strip()

# BF-OB5, EC, DB, CC, CL
def solve():
    n, m, p = readInts()
    s = readList()
    g = [[] for _ in range(n+1)]
    d = [[inf] * (n + 1) for _ in range(n + 1)]
    for _ in range(m):
        u, v, w = readInts()
        g[u].append([v, w])
        g[v].append([u, w])
        d[u][v] = d[v][u] = w
        d[u][u] = d[v][v] = 0

    # Floyd-Warshall
    for i in range(1, n+1):
        for j in range(1, n+1):
            for k in range(1, n+1):
                d[i][k] = min(d[i][k], max(d[i][j], d[j][k]))

    exp, ans = [0]*(n+1),[0]*n
    v0, v1 = [inf] * (n+1), []
    val = inf
    for i in range(1, p):
        best = -1
        for node in s:
            if exp[node]:
                continue
            dis = sum([min(d[j][node], v0[j]) for j in s])
            if dis < val:
                best = node
                val = dis
        for j in s:
            v0[j] = min(v0[j], d[j][best])
        ans[i-1] = val
        exp[best] = 1
    return " ".join([str(v) for v in ans])

inf = 10**18
for _ in range(int(input())):
    print(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
4 4 3
1 2 3
1 2 10
2 3 10
3 4 10
4 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main{
    static long []arr; //存需要联网的房屋号
    static long [][]arr2;//存房屋ij互联的最小延迟
    static long []arr3;//存联网房屋能连上网的最小延迟
    static boolean []flag;
    static int n;
    static int m;
    static int q;
    static Long RES=(long)1e18;
    public static void main(String[] args) throws IOException {
        Scanner sc=new Scanner(System.in);
        int T=sc.nextInt();
        while(T-->0) {
            solver(sc);
        }

    }
    static  void solver(Scanner sc){
            n=sc.nextInt();//n个房屋
            m=sc.nextInt();//m根网线
            q=sc.nextInt();//需要联网的房屋数
            arr=new long[q+1];
            arr2=new long[n+1][n+1];
            arr3=new long[q+1];
            flag=new boolean[n+1];
            for(int i=1;i<=q;i++){
                arr[i]=sc.nextLong();
            }
            for(int i=1;i<=n;i++){
                for(int j=i;j<=n;j++){
                    if(i==j){
                        arr2[i][j]=0;//自通距离为0
                    }else{
                        arr2[i][j]=RES;
                        arr2[j][i]=RES;
                    }
                }
            }
            Arrays.fill(arr3,1, q+1,RES);
            Arrays.fill(flag,1,n+1,false);
            for(int i=1;i<=m;i++){
                int a=sc.nextInt();
                int b=sc.nextInt();
                long c=sc.nextLong();
                arr2[a][b]=Math.min(arr2[a][b],c);
                arr2[b][a]=Math.min(arr2[b][a],c);
            }
            for(int i=1;i<=n;i++){
                for(int k=1;k<=n;k++){
                    for(int j=1;j<=n;j++){
                        arr2[i][j]=Math.min(arr2[i][j],Math.max(arr2[i][k],arr2[k][j]));
                    }
                }
            }
        ArrayList<Long> ans=new ArrayList<>();
            for(int i=1;i<=q;i++){
                int Min=-1;
                long res=RES;
                for(int j=1;j<=n;j++){
                    if(!flag[j]){
                        long tmp=0;
                        for (int k = 1; k <= q; k++) {
                            tmp += Math.min(arr3[k], arr2[j][(int)arr[k]]);
                        }
                        if(tmp<res){
                            res=tmp;
                            Min=j;
                        }
                    }
                }
                flag[Min]=true;
                ans.add(res);
                for(int j=1;j<=q;j++){
                    arr3[j]=Math.min(arr3[j],arr2[Min][(int)arr[j]]);
                }
            }
            while(ans.size()!=n){
                ans.add(0l);
            }
            for (long l: ans) {
                System.out.print(l+" ");
            }
            System.out.println();
        }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - **Small Number of Houses and Cables**: This can help in quickly verifying the correctness of the algorithm.
   - **High Latency**: Cables with very high latency to ensure the algorithm handles large values correctly.
   - **Multiple Houses Requiring Internet**: Ensure that there are multiple houses requiring internet to test the algorithm's ability to handle multiple queries.
   - **Graph Structure**: Ensure the graph is connected but has a structure that can expose bugs in the shortest path calculations.

2. **Constraints**:
   - Ensure the test case adheres to the constraints: $2 \le n \le 400$, $n-1 \le m \le 400$, and $1 \le p \le n$.
   - The sum of $n^3$ and $m^3$ should not exceed $10^8$.

3. **Specific Test Case**:
   - **Number of Test Cases**: Set to 1 for simplicity.
   - **Number of Houses ($n$)**: Set to a small number, e.g., 5.
   - **Number of Cables ($m$)**: Set to $n-1$ to ensure the graph is connected.
   - **Number of Houses Requiring Internet ($p$)**: Set to 3 to ensure multiple queries.
   - **Latencies**: Set to high values to test the algorithm's handling of large numbers.

Here is the test input:

```plaintext
1
5 4 3
1 3 5
1 2 1000000000
2 3 1000000000
3 4 1
4 5 1
```

This test case is designed to:
- Have a small number of houses and cables to keep the input manageable.
- Include high latency values to test the algorithm's ability to handle large numbers.
- Have multiple houses requiring internet to test the algorithm's ability to handle multiple queries.
- Ensure the graph is connected but has a structure that can expose bugs in the shortest path calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

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

public class Main{
    static long []arr; //存需要联网的房屋号
    static long [][]arr2;//存房屋ij互联的最小延迟
    static long []arr3;//存联网房屋能连上网的最小延迟
    static boolean []flag;
    static long n;
    static long m;
    static long q;
    static long RES=Long.MAX_VALUE;
    static ArrayList<Long> ans;
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    static StreamTokenizer st = new StreamTokenizer(br);

    static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    static long nextLong() throws IOException {
        st.nextToken();
        return (long) st.nval;
    }

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

    public static void main(String[] args) throws IOException {
        int T=nextInt();
        while(T-->0){
            n=nextLong();//n个房屋
            m=nextLong();//m根网线
            q=nextLong();//需要联网的房屋数
            arr=new long[(int)(q+1)];
            arr2=new long[(int)(n+1)][(int)(n+1)];
            arr3=new long[(int)(q+1)];
            flag=new boolean[(int)(n+1)];
            ans=new ArrayList<>();
            for(int i=1;i<=q;i++){
                arr[i]=nextLong();
            }
            for(int i=1;i<=n;i++){
               for(int j=i;j<=n;j++){
                   if(i==j){
                       arr2[i][j]=0;//自通距离为0
                   }else{
                       arr2[i][j]=RES;
                       arr2[j][i]=RES;
                   }
               }
            }
            Arrays.fill(arr3,(int)1, (int)(q+1),RES);
//            System.out.println(q);
//            System.out.println(Arrays.toString(arr3));
            Arrays.fill(flag,(int)1,(int)(n+1),false);
            for(int i=1;i<=m;i++){
                int a=nextInt();
                int b=nextInt();
                long c=nextLong();
                arr2[a][b]=Math.min(arr2[a][b],c);
                arr2[b][a]=Math.min(arr2[b][a],c);
            }
            for(int i=1;i<=n;i++){
                for(int k=1;k<=n;k++){
                    for(int j=1;j<=n;j++){
                        arr2[i][j]=Math.min(arr2[i][j],Math.max(arr2[i][k],arr2[k][j]));
                    }
                }
            }
//            System.out.println(Arrays.toString(arr));
//            for(int i=1;i<=n;i++){
//                for(int k=1;k<=n;k++){
//                    System.out.print(arr2[i][k]+" ");
//                }
//                System.out.println();
//            }

            for(int i=1;i<=q;i++){
//                System.out.println("要求服务器个数"+i);
                int Min=-1;
                long res=RES;
                for(int j=1;j<=n;j++){
                    if(!flag[j]){
                        long tmp=0;
                        for (int k = 1; k <= q; k++) {
//                            System.out.print("此时检查"+arr[k]);
                            tmp += Math.min(arr3[k], arr2[j][(int)arr[k]]);
//                            System.out.println("和为:"+tmp+"此时"+arr[k]+"的值为："+arr3[k]+" "+i+"到"+arr[k]+"的距离为："+arr2[i][arr[k]]);
                        }
                        if(tmp<res){
                            res=tmp;
                            Min=j;
                        }
                    }
                }
                flag[Min]=true;
//                System.out.println("此时纳入的节点是"+Min);
                ans.add(res);
                for(int j=1;j<=q;j++){
                    arr3[j]=(int)Math.min(arr3[j],arr2[Min][(int)arr[j]]);
                }
            }
            while(ans.size()!=n){
                ans.add(Long.parseLong("0"));
            }
            for (Long l: ans) {
                pw.print(l+" ");
            }
            pw.flush();
            pw.println();
        }
        pw.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases that might cause the code to fail. The code appears to use a combination of dynamic programming and graph traversal to find the minimum total latency. Some potential issues include:

1. **Cycles in the graph**: The code might not handle cycles correctly.
2. **Multiple paths with different latencies**: The code might not correctly handle multiple paths between the same pair of houses.
3. **Large latencies**: The code might overflow or handle large latencies incorrectly.
4. **Edge cases with small values of \( n \) and \( m \)**: The code might fail for very small graphs.
5. **Special cases with \( k \) values**: The code might not handle the case where \( k \) is very small or very large correctly.

Given these considerations, let's create a small-scale test input that is complete and valid:

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

### Explanation:
1. **Test Case 1**:
   - **4 houses, 4 cables, 3 houses needing internet**: Houses 1, 2, and 4 need internet.
   - **Cables**:
     - 1-2 with latency 1
     - 2-3 with latency 2
     - 3-4 with latency 3
     - 4-1 with latency 4
   - **Expected Output**: The minimum total latency for each \( k \) should be calculated considering the latencies and the paths. The output should be `6 3 0 0`.

2. **Test Case 2**:
   - **3 houses, 3 cables, 3 houses needing internet**: All houses need internet.
   - **Cables**:
     - 1-2 with latency 3
     - 1-3 with latency 2
     - 2-3 with latency 1
   - **Expected Output**: The minimum total latency for each \( k \) should be calculated considering the latencies and the paths. The output should be `3 0 0`.

This test input is designed to challenge the code by including cycles, multiple paths with different latencies, and a small number of houses and cables. It should help expose any issues in the code's handling of these scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
#include <bits/stdc++.h>
#define for0(i,n) for(i=0;i<(n);i++)
#define for1(i,n) for(i=1;i<=(n);i++)
#define forlr(i,l,r) for(i=(l);i<=(r);i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=410;
int T,n,m,p,t[N];
ll a[N][N],ans,ans1,val[N];
bool b[N];
int main(){
    scanf("%d",&T);
    int i,j,k,x,y,z;
    while(T--){
        scanf("%d%d%d",&n,&m,&p);
        for1(i,n) for1(j,n) a[i][j]=1e15;
        for1(i,n) a[i][i]=0;
        for1(i,n) t[i]=0;
        for1(i,p){
            scanf("%d",&x);
            t[x]=1;
        }
        for1(i,m){
            scanf("%d%d%d",&x,&y,&z);
            a[x][y]=a[y][x]=min(a[x][y],(ll)z);
        }
        for1(k,n) for1(i,n) for1(j,n) a[i][j]=min(a[i][j],max(a[i][k],a[k][j]));
        for1(i,n){
            b[i]=0;
            val[i]=1e15;
        }
        for1(k,n){
            ans=2e18;
            for1(i,n) if(!b[i]){
                ans1=0;
                for1(j,n) if(t[j]) ans1+=min(val[j],a[i][j]);
                if(ans1<ans){
                    x=i;
                    ans=ans1;
                }
            }
            printf("%d ",ans);
            b[x]=1;
            for1(j,n) val[j]=min(val[j],a[x][j]);
        }
        printf("\n");
    }
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to consider scenarios that can cause the algorithm to fail. The code seems to have issues with correctly calculating the minimum total latency, especially when there are multiple possible servers to choose from.

### Key Points to Consider:
1. **Graph Connectivity**: Ensure the graph is connected but has varying latencies to test the algorithm's ability to handle different cable latencies.
2. **Multiple Houses Requiring Internet**: Test with multiple houses requiring internet to ensure the algorithm can handle multiple constraints.
3. **Optimal Server Placement**: Create a scenario where the optimal placement of servers is not straightforward, forcing the algorithm to make suboptimal choices if it has a bug.

### Test Case:
Let's create a test case with 4 houses, 4 cables, and 3 houses requiring internet. We will set up the latencies in a way that the optimal placement of servers is not immediately obvious.

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

### Explanation:
- **Number of Test Cases (t)**: 1
- **Number of Houses (n)**: 4
- **Number of Cables (m)**: 4
- **Number of Houses Requiring Internet (p)**: 3
- **Houses Requiring Internet (s1, s2, s3)**: 1, 3, 4
- **Cables**:
  - House 1 to House 2 with latency 1
  - House 2 to House 3 with latency 10
  - House 3 to House 4 with latency 2
  - House 1 to House 4 with latency 5

### Expected Output:
For each \( k = 1, 2, 3, 4 \), the minimum total latency should be:
- For \( k = 1 \): 12 (placing a server at House 1)
- For \( k = 2 \): 2 (placing servers at House 1 and House 3)
- For \( k = 3 \): 0 (placing servers at House 1, House 3, and House 4)
- For \( k = 4 \): 0 (placing servers at all houses)

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

This test case should expose the bug in the algorithm if it fails to correctly calculate the minimum total latency for the given constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

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

using namespace std;

#define int long long

// int calc_latency(vector<vector<int>> &adj, set<int> &server, set<int> &s, int n){
//     vector<int> dist(n, INT_MAX);
//     vector<bool> visited(n, false);
//     set<pair<int, int>> arr;
//     for(auto p : server){
//         dist[p] = 0;
//         for(int i = 0; i < adj[p].size(); i++){
//             int x = adj[p][i];
//             dist[x] = min(dist[x], w[p][i]);
//         }
//     }
// }
int calc_latency(vector<vector<int>> &adj, vector<vector<int>> &w, set<int> &server, set<int> &arr, int n){
    vector<int> dist(n, INT_MAX);
    set<tuple<int, int, int>> edges;
    vector<bool> visited(n, false);
    for(auto p : server){
        dist[p] = 0;
        visited[p] = true;
        for(int i = 0; i < adj[p].size(); i++){
            edges.insert({w[p][i], p, adj[p][i]});
        }
    }
    while(edges.size() > 0){
        tuple<int, int, int> p = *(edges.begin());
        edges.erase(p);
        int x = get<2>(p);
        dist[x] = min(dist[x], max(get<0>(p), dist[get<1>(p)]));
        if(!visited[x]){
            visited[x] = true;
            for(int i = 0; i < adj[x].size(); i++){
                edges.insert({w[x][i], x, adj[x][i]});
            }
        }
    }
    int ans = 0;
    for(auto p : arr){
        ans += dist[p];
    }
    return ans;

}

int find(vector<int> &parent, int v){
    if(parent[v] == v) return v;
    return parent[v] = find(parent, parent[v]);
}

void graph(vector<vector<int>> &adj, vector<vector<int>> &w, vector<vector<int>> &edges){
    int n = adj.size();
    vector<int> parent(n);
    for(int i = 0; i < n; i++){
        parent[i] = i;
    }
    int a, b, c, x, y;
    for(auto p : edges){
        a = p[0], b = p[1], c = p[2];
        x = find(parent, b);
        y = find(parent, c);
        if(x != y){
            if(x < y) swap(x, y);
            parent[x] = y;
            adj[b].push_back(c);
            adj[c].push_back(b);
            w[b].push_back(a);
            w[c].push_back(a);
        }
    }
}
void work(){
    int n, m, p, x, x2 = 0, k, a, b, c; cin >> n >> m >> x2;
    set<int> arr;
    for(int i = 0; i < x2; i++){
        cin >> x; x--; arr.insert(x);
    }
    vector<vector<int>> adj(n, vector<int> (0));
    vector<vector<int>> w(n, vector<int> (0));
    vector<vector<int>> edges(m, vector<int> (3));
    for(int i = 0; i < m; i++){
        cin >> edges[i][1] >> edges[i][2] >> edges[i][0];
        edges[i][1]--; edges[i][2]--;
    }
    sort(edges.begin(), edges.end());
    graph(adj, w, edges);
    set<int> server;
    for(int i = 1; i < x2; i++){
        int l = n*1000000000, r = -1;
        for(int j = 0; j < n; j++){
            if(server.find(j) == server.end()){
                server.insert(j);
                x = calc_latency(adj, w, server, arr, n);
                if(x <= l){
                    l = x; r = j;
                }
                server.erase(j);
            }
        }
        server.insert(r);
        cout << l << " ";
    }
    for(int i = x2; i <= n; i++){
        cout << "0 ";
    }
    cout << endl;
}

signed main() {
    int t; cin >> t;
    while(t--){
        work();
    }
    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 code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Small Scale but Challenging**:
   - The test input should be small enough to fit within the constraints but large enough to expose potential bugs.
   - Use a small number of test cases, each with a small number of houses and cables.

2. **Edge Cases**:
   - Consider scenarios where the graph is not a simple tree but has multiple paths between nodes.
   - Consider scenarios where the houses requiring internet are not directly connected to each other or to the servers.
   - Consider scenarios where the latencies are high, making it difficult to minimize the total latency.

3. **Specific Test Case**:
   - Let's create a test case with 4 houses, 4 cables, and 2 houses requiring internet.
   - Ensure that the graph is connected but has some complexity in the connections.
   - Set latencies to values that will challenge the algorithm.

Here is a challenging test input:

```plaintext
1
4 4 2
1 3
1 2 10
2 3 5
3 4 1
1 4 15
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Houses**: 4
- **Number of Cables**: 4
- **Houses Requiring Internet**: 1 and 3
- **Cables and Latencies**:
  - House 1 is connected to House 2 with a latency of 10.
  - House 2 is connected to House 3 with a latency of 5.
  - House 3 is connected to House 4 with a latency of 1.
  - House 1 is connected to House 4 with a latency of 15.

### Why This Test Case is Challenging:
- **Multiple Paths**: There are multiple paths between the houses, which can lead to different optimal solutions depending on the algorithm's approach.
- **High Latencies**: The latencies are set to values that can cause the algorithm to make suboptimal choices if it does not handle the maximum latency correctly.
- **Non-Linear Connections**: The houses requiring internet (1 and 3) are not directly connected, which can expose issues in the algorithm's ability to find the optimal servers.

This test case should help in identifying any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import heapq
from collections import defaultdict

def floyd_warshall(n, edges):
    dist = [[float('inf')] * n for _ in range(n)]
    for i in range(n):
        dist[i][i] = 0
    for u, v, w in edges:
        dist[u-1][v-1] = min(dist[u-1][v-1], w)
        dist[v-1][u-1] = min(dist[v-1][u-1], w)
    for k in range(n):
        for i in range(n):
            for j in range(n):
                dist[i][j] = min(dist[i][j], max(dist[i][k], dist[k][j]))
    return dist

def solve():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        m = int(data[index + 1])
        p = int(data[index + 2])
        index += 3

        s = list(map(int, data[index:index + p]))
        index += p

        edges = []
        for _ in range(m):
            u = int(data[index])
            v = int(data[index + 1])
            w = int(data[index + 2])
            edges.append((u, v, w))
            index += 3

        dist = floyd_warshall(n, edges)

        # Calculate the minimum total latency for each k
        min_latency = [float('inf')] * (n + 1)
        min_latency[0] = 0

        for k in range(1, n + 1):
            current_min_latency = float('inf')
            for server_combination in combinations(range(n), k):
                total_latency = 0
                for house in s:
                    house -= 1
                    min_server_latency = float('inf')
                    for server in server_combination:
                        min_server_latency = min(min_server_latency, dist[house][server])
                    total_latency += min_server_latency
                current_min_latency = min(current_min_latency, total_latency)
            min_latency[k] = current_min_latency

        results.append(" ".join(map(str, min_latency[1:])))

    print("\n".join(results))

# Helper function to generate combinations
def combinations(iterable, r):
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = list(range(r))
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)

# Example usage
if __name__ == "__main__":
    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
4 5 3
1 3 4
1 2 1
2 3 2
3 4 3
1 4 4
2 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 400$; $n-1 \le m \le 400$; $1 \le p \le n$) — the number of houses, the number of cables and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n^3$ and the sum of $m^3$ do not exceed $10^8$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))
def readInt():
    return int(input())
def readInts():
    return map(int, input().split())
def readStr():
    return input().strip()

# BF-OB5, EC, DB, CC, CL
def solve():
    n, m, p = readInts()
    s = readList()
    g = [[] for _ in range(n+1)]
    d = [[inf] * (n + 1) for _ in range(n + 1)]
    for _ in range(m):
        u, v, w = readInts()
        g[u].append([v, w])
        g[v].append([u, w])
        d[u][v] = d[v][u] = w
        d[u][u] = d[v][v] = 0

    # Floyd-Warshall
    for i in range(1, n+1):
        for j in range(1, n+1):
            for k in range(1, n+1):
                d[j][k] = min(d[j][k], max(d[j][i], d[i][k]))

    exp, prev, ans = [0]*(n+1), [], [0]*n
    v0, v1 = [], []
    for i in range(1, p+1):
        cand = [j for j in range(1, n+1) if not exp[j]]
        best, val, best_v = -1, inf, [inf] * len(s)
        for v in cand:
            if i == 1:
                v1 = [d[node][v] for node in s]
            else:
                v1 = [min(d[s[j]][v], v0[j]) for j in range(len(s))]
            dis = sum(v1)
            if dis <= val:
                best = v
                val = dis
                best_v = v1
        ans[i-1] = val
        prev.append(best)
        exp[best] = 1
        v0 = best_v
    return " ".join([str(v) for v in ans])

inf = 10**9 + 1
for _ in range(int(input())):
    print(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
2
4 4 2
1 4
1 2 1
2 3 2
3 4 3
1 4 4
5 5 3
1 3 5
1 2 1
2 3 2
3 4 3
4 5 4
```