**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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.read
data = input().split()

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

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] = w
        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

while t > 0:
    t -= 1
    n = int(data[index])
    m = int(data[index + 1])
    p = int(data[index + 2])
    index += 3
    required_houses = [int(data[index + i]) - 1 for i in range(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)

    min_total_latency = [float('inf')] * (n + 1)
    min_total_latency[0] = 0

    for k in range(1, n + 1):
        for mask in range(1 << n):
            if bin(mask).count('1') == k:
                current_latency = 0
                for house in required_houses:
                    min_dist = float('inf')
                    for server in range(n):
                        if (mask >> server) & 1:
                            min_dist = min(min_dist, dist[house][server])
                    current_latency += min_dist
                min_total_latency[k] = min(min_total_latency[k], current_latency)

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

sys.stdout.write("\n".join(results) + "\n")

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some considerations:

1. **Small-scale Input**: The input should be small enough to be valid but large enough to expose the bug.
2. **Edge Cases**: Consider scenarios where the graph is highly connected, or where the required houses are far apart.
3. **Latency Values**: Use a mix of high and low latency values to ensure the algorithm is tested for both extremes.
4. **Multiple Test Cases**: Include multiple test cases to ensure the code handles various scenarios.

Given these considerations, let's construct a test input:

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

### Explanation of the Test Input:

1. **Test Case 1**:
   - **Graph**: A cycle with 5 houses and 5 edges.
   - **Required Houses**: Houses 1, 3, and 5.
   - **Latency Values**: Mixed, with a high latency of 10 between houses 2 and 3.
   - **Purpose**: To test the algorithm's ability to handle high latency values and ensure it correctly calculates the minimum total latency.

2. **Test Case 2**:
   - **Graph**: A linear graph with 4 houses and 3 edges.
   - **Required Houses**: Houses 1, 2, and 4.
   - **Latency Values**: Low and mixed.
   - **Purpose**: To test the algorithm's ability to handle a linear graph and ensure it correctly calculates the minimum total latency for a simpler graph.

3. **Test Case 3**:
   - **Graph**: A triangle with 3 houses and 3 edges.
   - **Required Houses**: All 3 houses.
   - **Latency Values**: Low and mixed.
   - **Purpose**: To test the algorithm's ability to handle a fully connected graph and ensure it correctly calculates the minimum total latency for a simple, highly connected graph.

This test input should help expose any bugs in the given code, especially in handling high latency values and different graph structures.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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

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()
    k = len(s)
    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][u] = d[v][v] = 0

    # Dijkstra
    for i in range(1, n + 1):
        h = [(0, i)]
        while h:
            dis, node = heapq.heappop(h)
            for to, w in g[node]:
                if d[i][to] > max(dis, w):
                    d[i][to] = max(dis, w)
                    heapq.heappush(h, (d[i][to], to))
    exp, prev, ans = [0] * (n + 1), [], []
    for i in range(1, n + 1):
        if i >= k:
            ans.append(0)
            continue
        cand = [i for i in range(1, n + 1) if not exp[i]]
        best, val = -1, inf
        for v in cand:
            dis = sum([min([d[node][j] for j in prev + [v]]) for node in s])
            if dis <= val:
                best = v
                val = dis
        ans.append(val)
        prev.append(best)
    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 3 2
1 3
1 2 1
2 3 10
3 4 1
5 4 3
1 2 4
1 3 2
3 4 3
4 5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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;
using i64 = long long;
const int N = 1e6 + 10;
const int mod = 998244353;
#define endl '\n'
bool flag = false;
int cnt = 0;
#define all(x) x.begin(), x.end()
// int xx[] = { 1,0,-1,0 };
// int yy[] = { 0,1,0,-1 };
i64 qp(i64 a, i64 b)
{
    i64 res = 1;
    a %= mod;
    while (b)
    {
        if (b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
#define int long long
void solve()
{
    int n, m;
    cin >> n >> m;
    int p;
    cin >> p;
    vector<int> s(p), tag(n);
    for (int i = 0; i < p; i++)
    {
        cin >> s[i];
        s[i]--;
        tag[s[i]] = 1;
    }
    vector<tuple<int, int, int>> edge;
    for (int i = 0; i < m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        u--, v--;
        edge.push_back({w, u, v});
    }
    sort(all(edge));
    vector<int> f(n), siz(n), val(n);
    iota(all(f), 0);
    function<int(int)> find = [&](int x)
    {
        return x == f[x] ? f[x] : f[x] = find(f[x]);
    };
    auto merage = [&](int u, int v)
    {
        u = find(u), v = find(v);
        if (u == v)
            return false;
        f[u] = v;
        siz[v] += siz[u];
        return true;
    };
    vector<int> ans(n + 1);
    for (int i = 0; i < p; i++)
        siz[s[i]]++;
    for (int i = n - 1; i; i--)
    {
        i64 cost = 1e18;
        for (auto [w, u, v] : edge)
        {
            if (find(u) == find(v))
                continue;
            u = find(u), v = find(v);
            i64 res = min(siz[u] * w - val[u], siz[v] * w - val[v]);
            cost = min(cost, res);
        }
        for (auto [w, u, v] : edge)
        {
            if (find(u) == find(v))
                continue;
            u = find(u), v = find(v);
            i64 res = min(siz[u] * w - val[u], siz[v] * w - val[v]);
            if (res == cost)
            {
                merage(u, v);
                val[v] = val[u] + val[v] + res;
                break;
            }
        }
        assert(cost>=0);
        ans[i] = ans[i + 1] + cost;
    }
    for (int i = 1; i <= n; i++)
        cout << ans[i] << " \n"[i == n];
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int T = 1;
    cin >> T;

    while (T--)
        solve();
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that can maximize the failure probability. Here are the key points to consider:

1. **Graph Structure**: The graph should be connected and simple. We need to ensure that the graph is not trivial and has a structure that can challenge the algorithm.
2. **Edge Latency**: The latencies should be chosen to create a scenario where the minimum total latency is not straightforward to compute.
3. **Server Placement**: The number of houses that require internet should be such that the optimal placement of servers is not obvious.

Given these points, let's create a small-scale test input:

- **Number of houses (n)**: 5
- **Number of cables (m)**: 4
- **Number of houses that need internet (p)**: 3
- **Houses that need internet (s)**: 2, 3, 5
- **Cables and latencies (u, v, w)**:
  - (1, 2, 1)
  - (2, 3, 2)
  - (3, 4, 3)
  - (4, 5, 4)

This setup ensures that the graph is connected and the latencies are increasing, which can make the problem more challenging for the algorithm to find the optimal placement of servers.

Here is the test input:

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

This input should help expose any issues in the buggy code by requiring careful consideration of the latencies and the optimal placement of servers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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.util.*;
public class Main {

    static List<List<int[]>> adj;
    static int res[],n;
    static List<int[]> newedges;
    static int vis[];
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0) {
             n = sc.nextInt();
            adj = new ArrayList<>();
            for(int i=0;i<=n;i++) adj.add(new ArrayList<>());
            int m = sc.nextInt();
            int p = sc.nextInt();
            res = new int[n+1];
            int a[] = new int[p];
            for(int i=0;i<p;i++) {
                a[i] = sc.nextInt();
                res[a[i]] = 1;
            }
            int edges[][] = new int[m][3];
            for(int i=0;i<m;i++)
            {
                int u = sc.nextInt();
                int v = sc.nextInt();
                int w = sc.nextInt();
                edges[i][0] = u;
                edges[i][1] = v;
                edges[i][2] = w;
            }
            Arrays.sort(edges,new aComparator());
            dsu d = new dsu(n);
            newedges = new ArrayList<>();
            vis = new int[n-1];
            for(int i=0;i<m;i++)
            {
                int u = edges[i][0];
                int v = edges[i][1];
                int w = edges[i][2];
                if(d.union(u,v)) {
                    newedges.add(new int[]{u,v,w});
                    adj.get(u).add(new int[]{v,w,newedges.size()-1});
                    adj.get(v).add(new int[]{u,w,newedges.size()-1});
                }
            }
           long dpt[] = new long[n+1];
            dfs(1,dpt);
            for(int i=1;i<=n;i++)
            {
                System.out.print(dpt[i]+" ");
            }
            System.out.println();
        }
    }
    public static int dfs(int u,long dpt[])
    {
        int index = maxEdge(u,0);
        if(index==-1) {
            dpt[1] = 0;
            if(res[u]==1) return 1;
            return 0;
        }
//        System.out.println(index+" "+newedges.get(index)[0]+" "+newedges.get(index)[1]+" "+newedges.get(index)[2]);
        Arrays.fill(dpt,-1l);
        vis[index] = 1;
        long dpa[] = new long[n+1];long dpb[] = new long[n+1];
        int s1 = dfs(newedges.get(index)[0],dpa);
        int s2 = dfs(newedges.get(index)[1],dpb);
        for(int i=1;i<=n;i++) {

            if (dpa[i] != -1) dpt[i] = dpa[i] + (long) newedges.get(index)[2] * (long) s2;
            if (dpb[i] != -1) dpt[i] = Math.min(dpt[i], dpb[i] + (long) newedges.get(index)[2] * (long) s1);
            for (int j = 1; j < i; j++) {
                if (dpa[j] == -1 || dpb[i - j] == -1) continue;
                dpt[i] = Math.min(dpt[i], dpa[j] + dpb[i - j]);
            }
        }
        return s1+s2;
    }
    public static int maxEdge(int u,int p)
    {
        int ind=  -1;
        for(int v[] : adj.get(u))
        {
            if(vis[v[2]]==1||v[0]==p) continue;
            int i = maxEdge(v[0],u);
            if(i==-1||newedges.get(i)[2]<v[1])
            {
                i = v[2];
            }
            if(ind==-1||newedges.get(ind)[2]<newedges.get(i)[2])
            {
                ind = i;
            }
        }
        return ind;
    }
}
class dsu{
    int par[];
    int size[];
    public dsu(int n)
    {
        par = new int[n+1];
        size = new int[n+1];
        for(int i=1;i<=n;i++)
        {
            par[i] = i;
            size[i] = 1;
        }
    }
    public int find(int u)
    {
        if(u==par[u]) return u;
        return par[u] = find(par[u]);
    }
    public boolean union(int x,int y)
    {
        int px = find(x);
        int py = find(y);
        if(px==py) return false;
        if(size[px]>size[py])
        {
            par[py] = px;
            size[px]+=size[py];
        }
        else {
            par[px] = py;
            size[py]+=size[px];
        }
        return true;
    }
}
class aComparator implements Comparator<int[]> {
    public int compare(int a1[], int a2[])
    {
        return Integer.compare(a1[2],a2[2]);
    }
}

**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 might cause the algorithm to fail. Here are the steps to create such a test input:

1. **Small-scale Input**: Ensure the input is small enough to be manageable and to fit within the constraints.
2. **Connected Graph**: The graph must be connected, and the edges should form a valid simple graph.
3. **Distinct Houses Requiring Internet**: The houses requiring internet should be distinct.
4. **Edge Cases**: Include scenarios where the minimum latency calculation might be tricky, such as:
   - Houses requiring internet are far apart.
   - Houses requiring internet are directly connected by low-latency edges.
   - Houses requiring internet are connected through high-latency edges.

Let's create a test input that includes these considerations:

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

### Explanation:
1. **Test Case 1**:
   - **Number of houses (n)**: 4
   - **Number of cables (m)**: 3
   - **Houses requiring internet (p)**: 3
   - **Houses requiring internet**: 1, 3, 4
   - **Edges**:
     - 1-2 with latency 1
     - 2-3 with latency 10
     - 3-4 with latency 5

2. **Test Case 2**:
   - **Number of houses (n)**: 5
   - **Number of cables (m)**: 4
   - **Houses requiring internet (p)**: 2
   - **Houses requiring internet**: 1, 5
   - **Edges**:
     - 1-2 with latency 1
     - 2-3 with latency 10
     - 3-4 with latency 1
     - 4-5 with latency 1

### Why This Test Input?
- **Test Case 1**: This case includes a mix of low and high-latency edges. The houses requiring internet are not directly connected, which should challenge the algorithm to find the optimal server placement.
- **Test Case 2**: This case includes a simple path where the houses requiring internet are at the endpoints. The algorithm should correctly identify the minimum latency for different values of \( k \).

This input should help expose any bugs in the provided buggy code by testing its ability to handle different edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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;
using i64 = long long;
const int N = 1e6 + 10;
const int mod = 998244353;
#define endl '\n'
bool flag = false;
int cnt = 0;
#define all(x) x.begin(), x.end()
// int xx[] = { 1,0,-1,0 };
// int yy[] = { 0,1,0,-1 };
i64 qp(i64 a, i64 b)
{
    i64 res = 1;
    a %= mod;
    while (b)
    {
        if (b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
#define int long long
void solve()
{
    int n, m;
    cin >> n >> m;
    int p;
    cin >> p;
    vector<int> s(p), tag(n);
    for (int i = 0; i < p; i++)
    {
        cin >> s[i];
        s[i]--;
        tag[s[i]] = 1;
    }
    vector<tuple<int, int, int>> edge;
    for (int i = 0; i < m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        u--, v--;
        edge.push_back({w, u, v});
    }
    sort(all(edge));
    vector<int> f(n);
    iota(all(f), 0);
    function<int(int)> find = [&](int x)
    {
        return x == f[x] ? f[x] : f[x] = find(f[x]);
    };
    auto merage = [&](int u, int v)
    {
        u = find(u), v = find(v);
        if (u == v)
            return false;
        f[u] = v;
        return true;
    };
    vector<tuple<int, int, int>> re;
    for (auto [w, u, v] : edge)
    {
        if (merage(u, v))
        {
            re.push_back({w, u, v});
        }
    }
    assert(re.size() == n - 1);
    auto work = [&](auto self, vector<tuple<int, int, int>> &edge, int sz)
    {
        vector<int> dp(sz + 1, 1e18);
        dp[0] = 0;
        if (sz <= 1)
        {
            if (sz == 1)
                dp[1] = 0;
            return dp;
        }
        if (edge.size() == 1)
        {
            auto [w, u, v] = edge[0];
            dp[1] = w;
            if (tag[u] && tag[v])
                dp[1] = w, dp[2] = 0;
            else
                dp[1] = 0;
            return dp;
        }
        vector<vector<pair<int, int>>> e(n);
        int mx = 0;
        int su, sv;
        sort(all(edge));
        for (auto [w, u, v] : edge)
        {
            e[u].push_back({v, w});
            e[v].push_back({u, w});
            su = u, sv = v;
            assert(mx <= w);
            mx = w;
        }
        vector<int> siz(n);
        vector<tuple<int, int, int>> egu, egv;
        function<vector<tuple<int, int, int>>(int x, int fa)> dfs = [&](int x, int fa) -> vector<tuple<int, int, int>>
        {
            vector<tuple<int, int, int>> f;
            for (auto [y, w] : e[x])
            {
                if (y == fa)
                    continue;
                if (!(x == su && y == sv))
                {
                    f.emplace_back(w, x, y);
                }
                auto g = dfs(y, x);
                siz[x] += siz[y];
                if (g.size() > f.size())
                    swap(g, f);
                f.insert(f.end(), all(g));
            }
            if (tag[x])
                siz[x]++;
            if (x == su)
                swap(egu, f);
            if (x == sv)
                swap(egv, f);
            return f;
        };
        dfs(su, -1);
        siz[su] -= siz[sv];
        assert(siz[su] + siz[sv] == sz);
        auto dp1 = self(self, egu, siz[su]);
        auto dp2 = self(self, egv, siz[sv]);
        if (siz[su] == 0)
            return dp2;
        if (siz[sv] == 0)
            return dp1;
        for (int i = 0; i <= siz[su]; i++)
            for (int j = 0; j <= siz[sv]; j++)
            {
                if (i == 0 && j == 0)
                {
                    continue;
                }
                else if (i == 0)
                {
                    dp[i + j] = min(dp[i + j], dp2[j] + mx * siz[su]);
                }
                else if (j == 0)
                {
                    dp[i + j] = min(dp[i + j], dp1[i] + mx * siz[sv]);
                }
                else
                    dp[i + j] = min(dp[i + j], dp1[i] + dp2[j]);
            }

        return dp;
    };
    auto dp = work(work, re, p);
    for (int i = 1; i <= p; i++)
    {
        cout << dp[i] << " ";
    }
    for (int i = p + 1; i <= n; i++)
        cout << 0 << " ";
    cout << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int T = 1;
    cin >> T;

    while (T--)
        solve();
}

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

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

Think step by step.

Response:
```plaintext
1
4 4 3
1 2 3
1 2 1
2 3 2
3 4 3
1 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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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:
//蒟蒻一枚 rp++
//即得易见平凡，仿照上例显然。留作习题答案略，读者自证不难
//反之亦然同理，推论自然成立，略去过程Q.E.D.，由上可知证毕
#include<bits/stdc++.h>
//#pragma GCC optimize("Ofast")
#define re register
#define il inline
#define gc() getchar()
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define repp(i,a,b) for(int i=(a);i<(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define tep(i,x) for(int i=head[x];~i;i=ne[i])
#define ls(x) x<<1
#define rs(x) x<<1|1
#define eps (1e-9)
#define inf 0x3f3f3f3f
#define INF 1e16
#define pii pair<int,int>
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define fi first
#define sc second
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef double db;
namespace IO{
//	#define gc()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<9,stdin),p1==p2)?EOF:*p1++)
//	char buf[1<<9],*p1=buf,*p2=buf;
	template<typename T> inline void read(T &x){
		bool f=1;x=0;char ch=gc();
		while(ch<'0'||ch>'9'){if(ch=='-')f=0;ch=gc();}
		while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+(ch&15),ch=gc();
		x=f?x:-x;
	}
	template<typename T> inline void write(T x){
		if(x<0) putchar('-'),x=-x;
	   	if(x>9) write(x/10);
	   	putchar(char('0'+x%10));
	}
	template <typename T,typename ...Args> inline
	void read(T &x,Args &...args){read(x);read(args...);}
	template<typename T> inline void write(T x,char c){write(x),putchar(c);}
}
using namespace IO;
bool _ST;
const int N=2005,M=2005;
struct EDGE{
	int u,v,w;
	bool operator <(const EDGE&T)const{
		return w<T.w;
	}
}e[N];
int n,m,p,tot,fa[N],cnt[N],siz[N],val[N],ls[N],rs[N];
vector<LL> f[N];
int getf(int x){return fa[x]==x?fa[x]:fa[x]=getf(fa[x]);}
void build(){
	sort(e+1,e+1+m);
	rep(i,1,2*n) fa[i]=i;
	rep(i,1,m){
		int u=getf(e[i].u),v=getf(e[i].v);
		if(u!=v){
			fa[u]=fa[v]=++tot;
			ls[tot]=u,rs[tot]=v;
			val[tot]=e[i].w;
		}
	}
}
void DP(int x){
	if(x<=n) {
		f[x].resize(2,0);
		siz[x]=1;
	}
	else{
		DP(ls[x]),DP(rs[x]);
		siz[x]=siz[ls[x]]+siz[rs[x]],cnt[x]=cnt[ls[x]]+cnt[rs[x]];
		f[x].resize(siz[x]+1,INF);
		f[x][0]=0;
		rep(i,1,siz[ls[x]]) rep(j,1,siz[rs[x]]) f[x][i+j]=min(f[x][i+j],f[ls[x]][i]+f[rs[x]][j]);
		rep(i,1,siz[ls[x]]) f[x][i]=min(f[x][i],f[ls[x]][i]+1ll*cnt[rs[x]]*val[x]);
		rep(i,1,siz[rs[x]]) f[x][i]=min(f[x][i],f[rs[x]][i]+1ll*cnt[ls[x]]*val[x]);
		f[ls[x]].resize(0);f[rs[x]].resize(0);
	}

}
void solve(){
	read(n,m,p);
	rep(i,1,n*2) cnt[i]=0;
	rep(i,1,p){
		int x;read(x);cnt[x]=1;
	}
	rep(i,1,m) read(e[i].u,e[i].v,e[i].w);
	tot=n,build();
	DP(tot);
	rep(i,1,n) write(f[tot][i],' ');puts("");
	f[tot].resize(0);
}
bool _ED;
signed main(){
	fprintf(stderr,"%.20lf MB\n",(&_ST-&_ED)/1048576.0);
	//ios::sync_with_stdio(false);
	//cin.tie(0);cout.tie(0);
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int T;read(T);
	while(T--) solve();
	fprintf(stderr,"%.4lf s\n",1.0*clock()/CLOCKS_PER_SEC);
	return 0;
}
//谨记:
//十年OI一场空，不开longlong见祖宗
//数据千万条，清空第一条。多测不清空，爆零两行泪。清空不规范，TLE总相伴。
//快读要加类型名

**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 for the given algorithm problem, we need to focus on edge cases or scenarios that can maximize the failure probability in faulty solutions. Here are the steps to consider:

1. **Constraints**: The input should be within the given constraints:
   - \(2 \le n \le 5000\)
   - \(n-1 \le m \le 5000\)
   - \(1 \le p \le n\)
   - The sum of \(n\) and the sum of \(m\) over all test cases should not exceed 5000.

2. **Graph Properties**: The graph should be connected and simple. This means there are no self-loops or multiple edges between the same pair of vertices.

3. **Latency Values**: The latency values should be chosen to expose potential bugs in the algorithm, such as handling large latencies or specific edge cases.

4. ** Houses Requiring Internet**: The houses that require internet should be chosen in a way that challenges the algorithm, such as requiring the algorithm to handle a large number of houses needing internet or specific configurations of these houses.

5. **Multiple Test Cases**: The test input should include multiple test cases to ensure the algorithm is robust.

Based on these considerations, here is a small-scale test input that is complete and valid:

```plaintext
3
5 4 3
1 3 5
1 2 10
2 3 1
3 4 2
4 5 5
4 3 2
2 3
1 2 1
2 3 2
3 4 3
7 6 4
1 3 5 7
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1
6 7 1
```

This test input includes three test cases with different configurations to challenge the algorithm:

1. **Test Case 1**:
   - 5 houses, 4 cables, 3 houses requiring internet.
   - The graph is a simple linear chain with varying latencies.
   - The houses requiring internet are at positions 1, 3, and 5.

2. **Test Case 2**:
   - 4 houses, 3 cables, 2 houses requiring internet.
   - The graph forms a simple tree.
   - The houses requiring internet are at positions 2 and 3.

3. **Test Case 3**:
   - 7 houses, 6 cables, 4 houses requiring internet.
   - The graph is a simple linear chain with uniform latencies.
   - The houses requiring internet are at positions 1, 3, 5, and 7.

These test cases are designed to cover different aspects of the problem and should help in exposing potential bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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:
from sys import stdin, stdout

class DisjointSetUnion:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.num_sets = n

    def find(self, a):
        acopy = a
        while a != self.parent[a]:
            a = self.parent[a]
        while acopy != a:
            self.parent[acopy], acopy = a, self.parent[acopy]
        return a

    def union(self, a, b):
        a, b = self.find(a), self.find(b)
        if a != b:
            if self.size[a] < self.size[b]:
                a, b = b, a

            self.num_sets -= 1
            self.parent[b] = a
            self.size[a] += self.size[b]

    def set_size(self, a):
        return self.size[self.find(a)]

    def __len__(self):
        return self.num_sets

t = int(stdin.readline())

for _ in range(t):
    n, m, p = [int(x) for x in stdin.readline().split()]
    s = [int(x)-1 for x in stdin.readline().split()]
    edges = []
    for i in range(m):
        u, v, w = [int(x) for x in stdin.readline().split()]
        u -= 1
        v -= 1
        edges.append((w,u,v))

    edges.sort()

    dsu = DisjointSetUnion(n)
    grid = [[0]*n for i in range(n)]

    for i in range(m):
        w, u, v = edges[i]
        U = dsu.find(u)
        V = dsu.find(v)

        if U != V:
            U_fam = []
            V_fam = []
            for vertex in range(n):
                VERTEX = dsu.find(vertex)
                if VERTEX == U:
                    U_fam.append(vertex)
                elif VERTEX == V:
                    V_fam.append(vertex)

            for i1 in U_fam:
                for i2 in V_fam:
                    grid[i1][i2] = w
                    grid[i2][i1] = w

            dsu.union(u,v)

    best_new = -1
    best_score = 10**12

    for candidate in s:
        score = 0
        for s2 in s:
            score += grid[candidate][s2]

        if score < best_score:
            best_new = candidate
            best_score = score

    answer = [best_score]
    built = set()
    built.add(best_new)
    current_latency = [0]*n
    for s2 in s:
        current_latency[s2] = grid[best_new][s2]

    for bar in range(1,p-1):
        best_savings = -1
        best_new = -1

        for candidate in s:
            if candidate not in built:
                savings = 0
                for s2 in s:
                    if grid[candidate][s2] < current_latency[s2]:
                        savings += current_latency[s2] - grid[candidate][s2]

                if savings > best_savings:
                    best_new = candidate
                    best_savings = savings

        answer.append(answer[-1]-best_savings)
        built.add(best_new)
        for s2 in s:
            current_latency[s2] = min(current_latency[s2], grid[best_new][s2])

    while len(answer) < n:
        answer.append(0)

    stdout.write(' '.join(map(str,answer))+'\n')

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

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

Think step by step.

Response:
```plaintext
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
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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 ll long long
#define nl "\n"
#define all(v) v.begin(),v.end()
#define law_bt_shouf_el_code_dah_ed3ely_atla3_CM_PLZ ios_base::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
using namespace std;
int main() {
    law_bt_shouf_el_code_dah_ed3ely_atla3_CM_PLZ
    ll t;
    cin>>t;
    while(t--){
        ll n,m,k;
        cin>>n>>m>>k;
        vector<ll>ve(n+1,0);
        for(ll i=0;i<k;i++){ll x;cin>>x;ve[x]=1;}
        vector<pair<ll,ll>>adj[n+1];
        while(m--){
            ll u,v,w;
            cin>>u>>v>>w;
            adj[u].push_back({v,w});
            adj[v].push_back({u,w});
        }
        ll best=LLONG_MAX;
        vector<ll>dist(n+1,LLONG_MAX),activated(n+1,0);
        vector<vector<ll>>distB(n+1);
        auto get=[&](ll u){
            ll cost=0;
            vector<ll>d(n+1,LLONG_MAX);
            d[u]=0;
            priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<>>pq;
            pq.push({d[u],u});
            while(!pq.empty()){
                auto [x,node]=pq.top();
                pq.pop();
                if(x>d[node])continue;
                for(auto v:adj[node]){
                    if(max(v.second,x)<d[v.first]){
                        d[v.first]=max(v.second,x);
                        pq.push({d[v.first],v.first});
                    }
                }
            }
            distB[u]=d;
        };
        for(ll i=1;i<=n;i++)get(i);
        for(ll i=1;i<=n;i++){
            for(ll u=1;u<=n;u++){
                if(activated[u])continue;
                ll cost=0;
                for(ll v=1;v<=n;v++)
                    cost+=ve[v]*min(dist[v],distB[u][v]);
                best=min(best,cost);
            }
            for(ll u=1;u<=n;u++){
                if(activated[u])continue;
                ll cost=0;
                for(ll v=1;v<=n;v++)
                    cost+=ve[v]*min(dist[v],distB[u][v]);
                if(best==cost){
                    for(ll v=1;v<=n;v++)
                        dist[v]=min(dist[v],distB[u][v]);
                    activated[u]=1;
                    break;
                }
            }
            cout<<best<<' ';
        }
        cout<<nl;
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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.*;

public class DigitalVillageHardVersion {
    // 并查集（Union-Find）结构，带路径压缩和按秩合并
    static class UnionFind {
        int[] parent;
        int[] rank;

        UnionFind(int n){
            parent = new int[n];
            rank = new int[n];
            for(int i=0; i<n; i++) parent[i] = i;
        }

        int findSet(int u){
            while(parent[u] != u){
                parent[u] = parent[parent[u]];
                u = parent[u];
            }
            return u;
        }

        void unionSet(int u, int v){
            int pu = findSet(u);
            int pv = findSet(v);
            if(pu == pv) return;
            if(rank[pu] < rank[pv]){
                parent[pu] = pv;
            }
            else{
                parent[pv] = pu;
                if(rank[pu] == rank[pv]){
                    rank[pu]++;
                }
            }
        }
    }

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st;
        int t = Integer.parseInt(br.readLine());
        while(t-- > 0){
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken()); // 房屋数量
            int m = Integer.parseInt(st.nextToken()); // 电缆数量
            int p = Integer.parseInt(st.nextToken()); // 需要互联网的房屋数量
            int[] required = new int[p];
            st = new StringTokenizer(br.readLine());
            for(int i=0; i<p; i++) {
                required[i] = Integer.parseInt(st.nextToken()) -1; // 0-based
            }
            // 读取所有边
            class Edge implements Comparable<Edge>{
                int w, u, v;
                Edge(int w, int u, int v){
                    this.w = w;
                    this.u = u;
                    this.v = v;
                }
                public int compareTo(Edge other){
                    return Integer.compare(this.w, other.w);
                }
            }
            Edge[] edges = new Edge[m];
            for(int i=0; i<m; i++){
                st = new StringTokenizer(br.readLine());
                int u = Integer.parseInt(st.nextToken()) -1;
                int v = Integer.parseInt(st.nextToken()) -1;
                int w = Integer.parseInt(st.nextToken());
                edges[i] = new Edge(w, u, v);
            }
            // Krusky 算法构建 MST
            Arrays.sort(edges);
            UnionFind uf = new UnionFind(n);
            // 构建 MST 的邻接表
            ArrayList<ArrayList<int[]>> adj = new ArrayList<>();
            for(int i=0; i<n; i++) adj.add(new ArrayList<>());
            for(Edge edge : edges){
                int u = edge.u;
                int v = edge.v;
                int w = edge.w;
                if(uf.findSet(u) != uf.findSet(v)){
                    uf.unionSet(u, v);
                    adj.get(u).add(new int[]{v, w});
                    adj.get(v).add(new int[]{u, w});
                }
            }
            // 计算 c[i][v}，存储为 c[v * p + i}
            // 为了节省内存和提高缓存友好性，使用一维数组
            int INF = 1000000007;
            int[] c = new int[n * p];
            Arrays.fill(c, INF);
            // 高效的 BFS 实现
            for(int i=0; i<p; i++){
                int s = required[i];
                Queue<Integer> q = new ArrayDeque<>();
                q.add(s);
                c[s * p + i] = 0;
                boolean[] visited = new boolean[n];
                visited[s] = true;
                while(!q.isEmpty()){
                    int u = q.poll();
                    int currentLatency = c[u * p + i];
                    for(int[] neighbor : adj.get(u)){
                        int v = neighbor[0];
                        int w = neighbor[1];
                        if(!visited[v]){
                            c[v * p + i] = Math.max(currentLatency, w);
                            visited[v] = true;
                            q.add(v);
                        }
                    }
                }
            }
            // 贪心选择服务器
            // 初始化 current_min 为 INF
            int[] current_min = new int[p];
            Arrays.fill(current_min, INF);
            // 记录选择的服务器节点
            boolean[] selected = new boolean[n];
            // 记录每一步的总延迟
            ArrayList<Long> sumList = new ArrayList<>();
            // 进行最多 p 次选择，因为 k > p 时总延迟为 0
            for(int selection=0; selection < p; selection++){
                int best_v = -1;
                long best_sum = Long.MAX_VALUE;
                for(int v=0; v<n; v++){
                    if(selected[v]) continue;
                    long tmp_sum = 0;
                    int base = v * p;
                    for(int i=0; i<p; i++){
                        tmp_sum += Math.min(current_min[i], c[base + i]);
                    }
                    if(tmp_sum < best_sum){
                        best_sum = tmp_sum;
                        best_v = v;
                    }
                }
                if(best_v == -1){
                    break;
                }
                // 选择 best_v 作为服务器
                selected[best_v] = true;
                // 更新 current_min
                int base = best_v * p;
                for(int i=0; i<p; i++){
                    if(c[base + i] < current_min[i]){
                        current_min[i] = c[base + i];
                    }
                }
                // 计算当前总延迟
                long total_sum = 0;
                for(int i=0; i<p; i++) total_sum += current_min[i];
                sumList.add(total_sum);
            }
            // 构建输出
            StringBuilder sb = new StringBuilder();
            for(int k=1; k<=n; k++){
                if(k <= sumList.size()){
                    sb.append(sumList.get(k-1)).append(' ');
                }
                else{
                    sb.append("0 ");
                }
            }
            // 移除最后一个空格
            if(sb.length() > 0) sb.setLength(sb.length() -1);
            bw.write(sb.toString());
            bw.newLine();
        }
        bw.flush();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes potential bugs in the given code, let's consider a few key points:

1. **Edge Cases**: The code might fail with edge cases such as:
   - A graph with a single house that needs internet.
   - A graph where all houses need internet.
   - A graph with a very specific structure, such as a star graph or a linear chain.

2. **Large Latency Values**: The code might handle large latency values incorrectly.

3. **Complex Graphs**: The code might fail with more complex graphs, such as graphs with multiple connected components (though the problem guarantees a connected graph).

4. **Optimal Server Placement**: The code might fail to find the optimal placement of servers, especially when the graph has a specific structure that makes the greedy approach suboptimal.

Given these considerations, let's create a test case that includes a complex graph with a specific structure and large latency values. Here is a small-scale test input:

```plaintext
2
5 4 3
1 2 3
1 2 100
1 3 100
2 4 1
3 5 1
5 4 4
1 2 3 4
1 2 100
1 3 100
2 4 1
3 5 1
```

### Explanation:
1. **Test Case 1**:
   - **5 houses, 4 cables, 3 houses needing internet (1, 2, 3)**
   - **Cables**:
     - House 1 to House 2 with latency 100
     - House 1 to House 3 with latency 100
     - House 2 to House 4 with latency 1
     - House 3 to House 5 with latency 1

2. **Test Case 2**:
   - **5 houses, 4 cables, 4 houses needing internet (1, 2, 3, 4)**
   - **Cables**:
     - House 1 to House 2 with latency 100
     - House 1 to House 3 with latency 100
     - House 2 to House 4 with latency 1
     - House 3 to House 5 with latency 1

### Expected Output:
```plaintext
101 101 0 0 0
101 101 101 0 0
```

### Reasoning:
- **Test Case 1**:
  - For \( k = 1 \): The best server placement is at house 1, but the total latency is 101 (100 for house 2 and 100 for house 3).
  - For \( k = 2 \): The best server placement is at houses 2 and 3, but the total latency is still 101 (100 for house 1).
  - For \( k \geq 3 \): The total latency is 0 because all required houses can be covered.

- **Test Case 2**:
  - For \( k = 1 \): The best server placement is at house 1, but the total latency is 101 (100 for house 2 and 100 for house 3).
  - For \( k = 2 \): The best server placement is at houses 2 and 3, but the total latency is still 101 (100 for house 1).
  - For \( k = 3 \): The best server placement is at houses 1, 2, and 3, but the total latency is still 101 (100 for house 4).
  - For \( k \geq 4 \): The total latency is 0 because all required houses can be covered.

This test case should help expose any issues in the greedy approach used in the code, especially in handling large latency values and optimal server placement.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $p$ ($2 \le n \le 5000$; $n-1 \le m \le 5000$; $1 \le p \le n$) — 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 $n$ and the sum of $m$ do not exceed $5000$.

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:
from collections import defaultdict
from sys import stdin

class FastIO:
    def __init__(self):
        self.random_seed = 0
        self.flush = False
        self.inf = 1 << 32
        self.dire4 = [(0, -1), (0, 1), (1, 0), (-1, 0)]
        self.dire8 = [(0, -1), (0, 1), (1, 0), (-1, 0)] + [(1, -1), (1, 1), (-1, -1), (-1, 1)]
        return

    @staticmethod
    def read_int():
        return int(stdin.readline().rstrip())

    @staticmethod
    def read_float():
        return float(stdin.readline().rstrip())

    @staticmethod
    def read_list_ints():
        return list(map(int, stdin.readline().rstrip().split()))

    @staticmethod
    def read_list_ints_minus_one():
        return list(map(lambda x: int(x) - 1, stdin.readline().rstrip().split()))

    @staticmethod
    def read_str():
        return stdin.readline().rstrip()

    @staticmethod
    def read_list_strs():
        return stdin.readline().rstrip().split()

    def get_random_seed(self):
        import random
        self.random_seed = random.randint(0, 10 ** 9 + 7)
        return

    def st(self, x):
        return print(x, flush=self.flush)

    def yes(self, s=None):
        self.st("Yes" if not s else s)
        return

    def no(self, s=None):
        self.st("No" if not s else s)
        return

    def lst(self, x):
        return print(*x, flush=self.flush)

    def flatten(self, lst):
        self.st("\n".join(str(x) for x in lst))
        return

    @staticmethod
    def max(a, b):
        return a if a > b else b

    @staticmethod
    def min(a, b):
        return a if a < b else b

    @staticmethod
    def ceil(a, b):
        return a // b + int(a % b != 0)

    @staticmethod
    def accumulate(nums):
        n = len(nums)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] + nums[i]
        return pre

class UnionFindGeneral:
    def __init__(self, n: int) -> None:
        self.root = [i for i in range(n)]
        self.size = [1] * n
        self.part = n
        return

    def find(self, x):
        y = x
        while x != self.root[x]:
            # range_merge_to_disjoint to the direct root node after query
            x = self.root[x]
        while y != x:
            self.root[y], y = x, self.root[y]
        return x

    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x == root_y:
            return False
        if self.size[root_x] >= self.size[root_y]:  # merge to bigger size part
            root_x, root_y = root_y, root_x
        self.root[root_x] = root_y
        self.size[root_y] += self.size[root_x]
        self.size[root_x] = 0
        self.part -= 1
        return True

    def union_right(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x == root_y:
            return False
        if root_x > root_y:  # merge to bigger root number
            root_x, root_y = root_y, root_x
        self.root[root_x] = root_y
        self.size[root_y] += self.size[root_x]
        self.size[root_x] = 0
        self.part -= 1
        return True

    def is_connected(self, x, y):
        return self.find(x) == self.find(y)

    def get_root_part(self):
        part = defaultdict(list)
        n = len(self.root)
        for i in range(n):
            part[self.find(i)].append(i)
        return part

    def get_root_size(self):
        size = defaultdict(int)
        n = len(self.root)
        for i in range(n):
            size[self.find(i)] = self.size[self.find(i)]
        return size

class Solution:
    def __init__(self):
        return

    @staticmethod
    def main(ac=FastIO()):
        """
        url: url of the problem
        tag: algorithm tag
        """
        for _ in range(ac.read_int()):
            n, m, p = ac.read_list_ints()
            s = ac.read_list_ints_minus_one()
            inf = 10 ** 15
            edges = []
            for _ in range(m):
                u, v, w = ac.read_list_ints_minus_one()
                edges.append((w + 1, u, v))
            edges.sort()
            size = [0] * n
            sub = [{0: 0} for _ in range(n)]
            for x in s:
                size[x] = 1
                sub[x][1] = 0
            uf = UnionFindGeneral(n)
            uf.size = size[:]
            for w, i, j in edges:
                i = uf.find(i)
                j = uf.find(j)
                if i != j:
                    dct = dict()
                    for x in sub[i]:
                        for y in sub[j]:
                            if x and y:
                                dct[x + y] = min(dct.get(x + y, inf), sub[i][x] + sub[j][y])
                            elif x:
                                dct[x + y] = min(dct.get(x + y, inf), sub[i][x] + sub[j][y] + w * uf.size[j])
                            elif y:
                                dct[x + y] = min(dct.get(x + y, inf), sub[i][x] + sub[j][y] + w * uf.size[i])
                            else:
                                dct[x + y] = 0
                    sub[i] = None
                    sub[j] = None
                    uf.union(i, j)
                    j = uf.find(j)
                    sub[j] = dct
                    if uf.size[j] == p:
                        ac.lst([dct.get(x, 0) for x in range(1, n + 1)])
                        break
        return

Solution().main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Identify Potential Bugs**:
   - The code uses a union-find data structure to manage the connectivity and merging of houses.
   - The merging process involves updating the `sub` dictionary, which tracks the minimum latency for different sizes of connected components.
   - The code might fail if it incorrectly handles the merging of components, especially when dealing with multiple houses that require internet.

2. **Edge Cases**:
   - **Single House Requires Internet**: If only one house requires internet, the solution should be trivial.
   - **All Houses Require Internet**: If all houses require internet, the code should handle this case efficiently.
   - **Multiple Houses with Different Latencies**: The code should correctly handle cases where different houses have different latencies and need to be connected optimally.
   - **Disconnected Graphs**: The graph is guaranteed to be connected, but we can still test edge cases where the graph is nearly disconnected.

3. **Small-Scale Test Input**:
   - We will create a test case with a small number of houses, cables, and houses that require internet to ensure it is manageable and can be verified easily.

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

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

### Explanation:
1. **Test Case 1**:
   - **5 houses, 4 cables, 3 houses requiring internet (1, 3, 5)**.
   - The cables are:
     - 1-2 with latency 10
     - 2-3 with latency 1
     - 3-4 with latency 10
     - 4-5 with latency 1
   - This setup ensures that the houses requiring internet are spread out, and the latencies are varied, which can help expose issues in the merging process.

2. **Test Case 2**:
   - **3 houses, 3 cables, 2 houses requiring internet (1, 2)**.
   - The cables are:
     - 1-2 with latency 10
     - 2-3 with latency 1
     - 1-3 with latency 10
   - This setup is simpler but still tests the case where multiple houses require internet and the latencies are varied.

This test input should help in identifying any issues in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$q=0$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$q=0$$$) — the number of members, the number of sections and the number of updates.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

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

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 0
1 2 3 4
1 1
3 6 0
1 2 3
1 1 2 3 3 2
4 6 0
3 1 4 2
3 1 1 2 3 4
output:
YA
YA
TIDAK

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line.

For the second test case, the following is a possible way to move members so that the presentation is good:

1. $$$[1,2,3]$$$, do not move member $$$1$$$.
2. $$$[1,2,3]$$$, move member $$$1$$$ after member $$$3$$$.
3. $$$[2,3,1]$$$, move member $$$2$$$ after member $$$3$$$.
4. $$$[3,2,1]$$$, do not move member $$$3$$$.
5. $$$[3,2,1]$$$, move member $$$3$$$ after member $$$1$$$.
6. $$$[2,1,3]$$$, do not move member $$$2$$$.

Buggy Code:
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <bitset>
#include <vector>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define int unsigned long long
#define ll long long
bool FLAG = 0;
#define debug(x) if (FLAG) {  cerr << #x << " = " << x << endl;}
#define debug_vec(x) if (FLAG) {cerr << #x << " = {"; for (auto to x) cout << to << ", "; cout << "}" << endl}
#define F first
#define S second
#define sqr(x) ((x) * (x))
#define make make_pair
#define pb push_back
#define sz(x) (int)(x.size())
#define all(x) x.begin (), x.end ()

using namespace std;

const int inf = (int)1e9;

const int N = (int)2e5 + 10;

int n,m,q,a[N],b[N], used[N];

void solve () {
    cin >> n >> m >> q;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    vector<int> v;
    for(int i = 1; i <= m; i++){
        cin >> b[i];
    }
    used[a[1]] = 1;
    int pos = 2;
    bool ok = 0;
    for (int i = 1; i <= m; ++i) {
   		if (used[b[i]] == 1) {
   			continue;
   	 	}
   	 	if (b[i] == a[pos]) {
   	 		used[a[pos]] = 1;
   	 	    pos ++;
   	 		continue;
   	   	}
   		ok = 1;
	}
	for (int i = 1; i <= n; ++i) {
		used[a[i]] = 0;
		a[i] = 0;
	}
	if (ok == 1) cout << "TIDAK" << endl;
	else cout << "YA" << endl;
}

signed main () {
    #ifdef Maga
         freopen("input.txt", "r", stdin);
    #endif
    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:
```plaintext
3
4 2 0
1 2 3 4
4 4
3 6 0
1 2 3
3 2 1 2 3 1
4 6 0
3 1 4 2
4 3 2 1 4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$q=0$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$q=0$$$) — the number of members, the number of sections and the number of updates.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

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

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 0
1 2 3 4
1 1
3 6 0
1 2 3
1 1 2 3 3 2
4 6 0
3 1 4 2
3 1 1 2 3 4
output:
YA
YA
TIDAK

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line.

For the second test case, the following is a possible way to move members so that the presentation is good:

1. $$$[1,2,3]$$$, do not move member $$$1$$$.
2. $$$[1,2,3]$$$, move member $$$1$$$ after member $$$3$$$.
3. $$$[2,3,1]$$$, move member $$$2$$$ after member $$$3$$$.
4. $$$[3,2,1]$$$, do not move member $$$3$$$.
5. $$$[3,2,1]$$$, move member $$$3$$$ after member $$$1$$$.
6. $$$[2,1,3]$$$, do not move member $$$2$$$.

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include<numeric>
#include <cstring>
#include<utility>
#include<set>
#include<array>
#include<bitset>
using namespace std;
const int nMax = 2e5 + 7;
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	long long tst = 1;
	cin >> tst;
	while(tst--){
		set <long long> mp[nMax];
		long long an[nMax];
		long long b[nMax];
		long long n , m , x , q , l , r;
		cin >> n >> m >> q;
		vector <long long> arr;
		for(long long i = 0; n > i; i++){
			cin >> x;
			an[x] = i;
			arr.push_back(x);
		}
		long long piv = 0;
		long long flag = 0;
		for(long long i = 0; m > i; i++){
			cin >> x;
			b[i] = x;
			mp[x].insert(i);
		}
		for(long long i = 1; n >= i; i++){
			mp[i].insert(1e13);
		}
		for(long long i = 0; (n - 1) > i; i++){
			if(*mp[arr[i]].begin() > *mp[arr[i + 1]].begin()){
				flag++;
			}
		}

		if(flag){
			cout << "TIDAK\n";
		}
		else{
			cout << "YA\n";
		}
		for(long long i = 0; q > i; i++){
			cin >> l >> r;
			l--;

			long long k = an[r];
			if((n - 1) > k && *mp[arr[k]].begin() > *mp[arr[k + 1]].begin()){
				flag--;
			}
			if(k > 0 && *mp[arr[k]].begin() < *mp[arr[k - 1]].begin()){
				flag--;
			}
			k = an[b[l]];
			if((n - 1) > k && *mp[arr[k]].begin() > *mp[arr[k + 1]].begin()){
				flag--;
			}
			if(k > 0 && *mp[arr[k]].begin() < *mp[arr[k - 1]].begin()){
				flag--;
			}
			if(an[b[l]] - an[r] == 1){
				k = an[b[l]];
				if(k > 0 && *mp[arr[k]].begin() < *mp[arr[k - 1]].begin()){
					flag++;
				}
			}
			if(an[b[l]] - an[r] == -1){
				k = an[r];
				if(k > 0 && *mp[arr[k]].begin() < *mp[arr[k - 1]].begin()){
					flag++;
				}
			}
			mp[b[l]].erase(l);
			mp[r].insert(l);
			k = an[b[l]];
			//cout << an[b[l]] << " " << an[r] << " " << flag << ".\n";
			if((n - 1) > k && *mp[arr[k]].cbegin() > *mp[arr[k + 1]].cbegin()){
				flag++;
			}
			if(k > 0 && *mp[arr[k]].cbegin() < *mp[arr[k - 1]].cbegin()){
				flag++;
			}
			k = an[r];
			if((n - 1) > k && *mp[arr[k]].cbegin() > *mp[arr[k + 1]].cbegin()){
				flag++;
			}
			if(k > 0 && *mp[arr[k]].cbegin() < *mp[arr[k - 1]].cbegin()){
				flag++;
			}
			if(an[b[l]] - an[r] == 1){
				k = an[b[l]];
				if(k > 0 && *mp[arr[k]].begin() < *mp[arr[k - 1]].begin()){
					flag--;
				}
			}
			if(an[b[l]] - an[r] == -1){
				k = an[r];
				if(k > 0 && *mp[arr[k]].begin() < *mp[arr[k - 1]].begin()){
					flag--;
				}
			}
			//cout << an[b[l]] << " " << an[r] << " " << flag << ".\n";
			if(flag){
				cout << "TIDAK\n";
			}
			else{
				cout << "YA\n";
			}
			b[l] = r;

		}

	}

}

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

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

Think step by step.

Response:
```plaintext
3
4 2 0
1 2 3 4
2 1
3 6 0
1 2 3
3 2 1 2 3 1
4 6 0
3 1 4 2
4 3 2 1 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$q=0$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$q=0$$$) — the number of members, the number of sections and the number of updates.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

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

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 0
1 2 3 4
1 1
3 6 0
1 2 3
1 1 2 3 3 2
4 6 0
3 1 4 2
3 1 1 2 3 4
output:
YA
YA
TIDAK

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line.

For the second test case, the following is a possible way to move members so that the presentation is good:

1. $$$[1,2,3]$$$, do not move member $$$1$$$.
2. $$$[1,2,3]$$$, move member $$$1$$$ after member $$$3$$$.
3. $$$[2,3,1]$$$, move member $$$2$$$ after member $$$3$$$.
4. $$$[3,2,1]$$$, do not move member $$$3$$$.
5. $$$[3,2,1]$$$, move member $$$3$$$ after member $$$1$$$.
6. $$$[2,1,3]$$$, do not move member $$$2$$$.

Buggy Code:
// 1300
// https://codeforces.com/problemset/problem/2021/C1
#include <iostream>
#include <vector>
using namespace std;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  int cnt = 1;
  while (t--) {
    int n, m, q;
    cin >> n >> m >> q;
    int big = max(n, m);
    vector<int> a(big, 0), b(big, 0);
    vector<bool> visited(big + 1, false);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (cnt == 8238) {
        cout << a[i] << ".";
      }
    }
    if (cnt == 8238) {
      cout << "/";
    }
    int increment = 0;
    for (int i = 0; i < m; i++) {
      int member;
      cin >> member;
      if (cnt == 8238) {
        cout << member << ".";
      } else {
        if ((increment == 0) ||
            (increment != 0 && member != b[increment - 1])) {
          b[increment] = member;
          increment++;
        }
      }
    }
    if (cnt != 8238) {
      bool frontMatch = true;
      for (int i = 0; i < big; i++) {
        if (b[i] == 0) {
          break;
        }
        if (a[i] != 0) {
          visited[a[i]] = true;
        }
        if (!visited[b[i]]) {
          frontMatch = false;
          break;
        }
      }
      string res = (frontMatch) ? "ya" : "tidak";
      cout << res << endl;
    }
    cnt++;
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 3 0
1 2 3 4
1 2 3
3 3 0
1 2 3
3 2 1
5 5 0
1 2 3 4 5
5 4 3 2 1
```