**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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 ii pair<ll,ll>
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define lwb lower_bound
#define upb upper_bound
#define ld long double
#define ins insert
#define del erase
#define ull unsigned long long
using namespace std;
const ll big=4e5+9;
const ll sml=1e3+9;
const ll tny=20;
const ll inf=1e15;
const ll hug=1e6+9;
const ll mod=1e9+7;
struct tr{ll fi,se,th;};
struct qu{ll fi,se,th;};
ll mxz(ll &t, ll val){if (t<val){t=val; return 1;} return 0;}
ll mnz(ll &t, ll val){if (t>val){t=val; return 1;} return 0;}
ll qpw(ll n, ll k){ll p=1, t=n%mod; while (k){if (k&1) p=(p*t)%mod; t=(t*t)%mod; k>>=1;} return p;}
ll n,m,p;
ll sz;
ll a[big];
ll vis[big];
vector<ll> con[big];
ll par[big];
ll anc(ll u){return (par[u]==u? u: par[u]=anc(par[u]));}
ll uni(ll u, ll v, ll w){
	if ((u=anc(u))==(v=anc(v))) return 0;
	++sz;
	par[u]=par[v]=par[sz]=sz;
	con[sz].eb(u); con[sz].eb(v);
	a[sz]=w;
	return 1;
}
ll spe[big], cst[big];
ll nxt[big];
ll dfs(ll u, ll p){
//	cout<<u<<" "<<p<<endl;
	cst[u]=inf;
	for (ll v: con[u]){
		dfs(v,u);
		spe[u]+=spe[v];
		if (mnz(cst[u],cst[v])) nxt[u]=v;
	}
	mnz(cst[u],0ll);
	cst[u]+=(a[u]-a[p])*spe[u];
	return 0;
}
ll solve(){
	cin>>n>>m>>p;
	for (ll i=1; i<=p; ++i){
		ll x; cin>>x;
		spe[x]=1;
	}
	sz=n;
	for (ll i=1; i<=n; ++i) par[i]=i;
	vector<tr> edg(m);
	for (ll i=0; i<m; ++i) cin>>edg[i].fi>>edg[i].se>>edg[i].th;
	sort(edg.begin(),edg.end(),[](tr x, tr y){return x.th<y.th;});
	for (ll i=0; i<m; ++i) uni(edg[i].fi, edg[i].se, edg[i].th);
	a[0]=mod;
	dfs(sz,0);
	vector<ii> col;
	for (ll i=1; i<=sz; ++i) col.eb(cst[i],i);
	sort(col.begin(),col.end(),[](ii x, ii y){return x.fi<y.fi;});
//	for (ll i=0; i<sz; ++i) cout<<col[i].se<<": "<<col[i].fi<<"\n";
	ll sol=p*mod;
	for (ll i=0; i<sz; ++i){
		ll u=col[i].se;
		if (vis[u]) continue;
		sol+=cst[u];
		cout<<sol<<" ";
		do {
			vis[u]=1;
			u=nxt[u];
			if (!u) break;
		} while (1);
	}
	for (ll i=1; i<=sz; ++i){
		a[i]=vis[i]=par[i]=spe[i]=cst[i]=nxt[i]=0;
		con[i].clear();
	}
	cout<<"\n";
	return 0;
}
int main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//    freopen("INPUT.TXT","r",stdin);
//    freopen("OUTPUT.TXT","w",stdout);
	ll tst=1;
	cin>>tst;
	for (; tst; --tst) solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that could lead to incorrect behavior. The buggy code involves union-find, graph traversal, and dynamic programming, so we should focus on inputs that can cause issues with these components.

### Step-by-Step Analysis:

1. **Edge Cases:**
   - **Single House:** A graph with only one house.
   - **Complete Graph:** A graph where every house is connected to every other house.
   - **Linear Graph:** A graph where houses are connected in a linear fashion.
   - **Star Graph:** A graph where one central house is connected to all other houses.

2. **Constraints:**
   - Ensure the graph is connected.
   - Ensure the number of houses and cables are within the given constraints.

3. **Specific Scenarios:**
   - **High Latency:** Cables with very high latency values.
   - **Low Latency:** Cables with very low latency values.
   - **Mixed Latency:** A combination of high and low latency cables.
   - **Multiple Servers:** Different configurations of servers and houses that require internet.

### Chosen Test Input:
Let's create a test case that involves a star graph with high and low latency values to see if the code handles the union-find and dynamic programming correctly.

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

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

### Why This Test Case is Challenging:
- **Star Graph:** The central house (House 1) is connected to all other houses.
- **Mixed Latency:** Cables have different latency values, which can test the dynamic programming and union-find logic.
- **Houses Requiring Internet:** Houses 1, 2, and 5 require internet, which means the algorithm needs to handle the central house and a house with low latency correctly.

This test case should help expose any issues in the union-find and dynamic programming parts of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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 <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cassert>
#include <cstdlib>
#include <cctype>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <utility>
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <random>
#include <istream>
#include <numeric>
#include <chrono>
#include <functional>
#include <ranges>

using i64 = long long;
using u64 = unsigned long long;
using u32 = unsigned int;
using i128 = __int128;

std::mt19937_64 rng(std::chrono::system_clock::now().time_since_epoch().count());

struct DSU {
	std::vector<int> f, siz;
	DSU() {}
	DSU(int n_) {
		init(n_);
	}

	void init(int n_) {
		f.resize(n_);
		siz.assign(n_, 1);
		std::iota(f.begin(), f.end(), 0);
	}

	int find(int x) {
		while (x != f[x]) {
			x = f[x] = f[f[x]];
		}
		return f[x];
	}

	bool merge(int x, int y) {
		x = find(x), y = find(y);
		if (x == y) {
			return false;
		}
		siz[x] += siz[y];
		f[y] = x;
		return true;
	}

	int size(int x) {
		return siz[find(x)];
	}
};

void solve() {
	int n, m, p;
	std::cin >> n >> m >> p;
	std::vector<int> c(n);
	DSU dsu(n);
	for (int i = 0; i < p; i++) {
		int s;
		std::cin >> s;
		s--;
		c[s] = 1;
	}
	for (int i = 0; i < n; i++) {
		if (!c[i]) {
			dsu.siz[i] = 0;
		}
	}

	std::vector<std::array<int, 3>> edges(n);
	for (int i = 0; i < m; i++) {
		auto &[w, u, v] = edges[i];
		std::cin >> u >> v >> w;
		u--;
		v--;
	}
	std::sort(edges.begin(), edges.end());

	std::vector<i64> value(n);
	std::vector<std::multiset<i64>> se(n);
	for (int i = 0; i < n; i++) {
		se[i].emplace(0);
	}
	for (auto [w, u, v] : edges) {
		if (dsu.find(u) != dsu.find(v)) {
			u = dsu.find(u), v = dsu.find(v);
			if (dsu.size(u) < dsu.size(v)) {
				std::swap(u, v);
			}
			for (int t = 0; t < 2; t++, std::swap(u, v)) {
				i64 nv = 1LL * dsu.size(u) * w - value[u];
				i64 delta = *se[u].begin();
				se[u].erase(se[u].begin());
				value[u] += nv;
				se[u].emplace(delta - nv);
			}
			for (auto delta : se[v]) {
				se[u].emplace(delta);
			}
			se[v].clear();
			value[u] += value[v];
			dsu.merge(u, v);
		}
	}

	std::vector<i64> ans;
	int root = dsu.find(0);
	i64 nv = value[root];
	for (auto delta : se[root]) {
		nv += delta;
		ans.emplace_back(nv);
	}
	for (int i = 0; i < n; i++) {
		std::cout << ans[i] << " \n"[i == n - 1];
	}
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int T = 1;
    std::cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Algorithm**:
   - The algorithm uses a Disjoint Set Union (DSU) to manage connected components and their sizes.
   - It sorts the edges by weight and processes them to merge components, updating the latency values.

2. **Identify Potential Bugs**:
   - The code might fail if it does not correctly handle the merging of components and updating of latency values.
   - Edge cases where the number of servers is very small or very large can be challenging.
   - Situations where the graph is highly connected or has many small components can also be problematic.

3. **Construct a Test Case**:
   - Create a graph with a small number of nodes and edges to ensure it fits within the output limits.
   - Ensure the graph is connected and has a mix of small and large components.
   - Include houses that require internet in a way that tests the merging and latency calculation.

Here is a small-scale test input that should challenge the given code:

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

### Explanation:
- **Test Case**:
  - **Number of Test Cases**: 1
  - **Number of Houses (n)**: 5
  - **Number of Cables (m)**: 5
  - **Number of Houses Requiring Internet (p)**: 3
  - **Houses Requiring Internet**: 1, 3, 5
  - **Edges**:
    - Edge 1: (1, 2) with latency 1
    - Edge 2: (2, 3) with latency 2
    - Edge 3: (3, 4) with latency 3
    - Edge 4: (4, 5) with latency 4
    - Edge 5: (1, 5) with latency 5

### Why This Test Case is Challenging:
- **Connected Graph**: The graph is fully connected, ensuring that all houses are reachable.
- **Mixed Latencies**: The latencies are increasing, which can test the algorithm's ability to handle different weights.
- ** houses Requiring Internet**: The houses 1, 3, and 5 are spread out, requiring the algorithm to consider different paths and latencies.
- **Small Number of Houses**: The small number of houses ensures that the output is within the limit, but still complex enough to test the algorithm's logic.

This test case should help expose any issues in the given code, particularly in how it handles the merging of components and the calculation of latency.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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 rep(i, l, r) for(int i = (l); i <= (r); i ++)
#define pre(i, l, r) for(int i = (l); i >= (r); i --)
#define Grp(x, i) for(int i = h[x]; i; i = ne[i])
using namespace std;

typedef long long LL;
const int N = 5009;

inline int read() {
    int x = 0; char fs = 0, c = getchar(); while(! isdigit(c)) fs |= (c == '-'), c = getchar();
    while(isdigit(c)) x = (x << 1) + (x << 3) + (c - '0'), c = getchar();
    return fs ? -x : x;
}

int T;
int n, m, k;
struct edge {
    int x, y, v;
} e[N];
int fa[N], imp[N];
LL f[N][N], tmp[N];

int F(int x) {
    return fa[x] == x ? x : fa[x] = F(fa[x]);
}

int main() {
    T = read();
    while(T --) {
        rep(i, 1, n) memset(f[i], 0, sizeof(LL) * (imp[i] + 1));
        rep(i, 1, n) imp[i] = 0;

        n = read(), m = read(), k = read();
        rep(i, 1, k) imp[read()] = 1;
        rep(i, 1, m) {
            e[i].x = read(), e[i].y = read(), e[i].v = read();
        }
        sort(e + 1, e + m + 1, [](const edge& x, const edge& y) { return x.v < y.v; });
        rep(i, 1, n) fa[i] = i;

        rep(i, 1, m) {
            int x = F(e[i].x), y = F(e[i].y);
            if(x == y) continue;
            if(x > y) swap(x, y);
            memcpy(tmp, f[x], sizeof(LL) * (imp[x] + 1));
            memset(f[x], 0x3f, sizeof(LL) * (imp[x] + imp[y] + 1));
            rep(j, 1, imp[x]) rep(p, 1, imp[y])
                f[x][p + j] = min(f[x][p + j], tmp[j] + f[y][p]);
            rep(j, 1, imp[x]) f[x][j] = min(f[x][j], tmp[j] + 1ll * e[i].v * imp[y]);
            rep(j, 1, imp[y]) f[x][j] = min(f[x][j], f[y][j] + 1ll * e[i].v * imp[x]);
            fa[y] = x;
            imp[x] += imp[y];
        }
        rep(i, 1, n) printf("%lld%c", f[1][i], " \n"[i == n]);
    }
    return 0;
}
*/

#include<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l); i <= (r); i ++)
#define pre(i, l, r) for(int i = (l); i >= (r); i --)
#define Grp(x, i) for(int i = h[x]; i; i = ne[i])
using namespace std;

typedef long long LL;
const int N = 1000009;

inline int read() {
    int x = 0; char fs = 0, c = getchar(); while(! isdigit(c)) fs |= (c == '-'), c = getchar();
    while(isdigit(c)) x = (x << 1) + (x << 3) + (c - '0'), c = getchar();
    return fs ? -x : x;
}

int T;
int n, m, k;
int h[N], ne[N], to[N], tot;
int fa[N];
struct edge {
    int x, y;
    LL v;
} e[N];
int idx;
LL dis[N], ans[N], val[N], imp[N];
int son[N];

inline void add(int x, int y) {
    ne[++ tot] = h[x], h[x] = tot, to[tot] = y;
}

int F(int x) {
    return fa[x] == x ? x : fa[x] = F(fa[x]);
}

void Fyr(int x, int ff) {
    Grp(x, i) {
        int y = to[i];
        Fyr(y, x);
        imp[x] += imp[y];
        if(dis[son[x]] < dis[y]) son[x] = y;
    }
    dis[x] = 1ll * (val[ff] - val[x]) * imp[x] + dis[son[x]];
}

void Zrd(int x, int tp) {
    if(! son[x]) {
        ans[++ idx] = dis[tp];
        return;
    }
    Zrd(son[x], tp);
    Grp(x, i) {
        int y = to[i];
        if(y == son[x]) continue;
        Zrd(y, y);
    }
}

int main() {
    T = read();
    while(T --) {
        memset(imp, 0, sizeof(LL) * (n << 1));
        memset(h, 0, sizeof(int) * (n << 1));
        memset(son, 0, sizeof(int) * (n << 1));
        memset(ans, 0, sizeof(LL) * (idx + 1));
        memset(dis, 0, sizeof(LL) * (n << 1));
        memset(val, 0, sizeof(LL) * (n << 1));
        idx = tot = 0;

        n = read(), m = read(), k = read();
        rep(i, 1, k) imp[read()] ++;
        rep(i, 1, m) {
            e[i].x = read(), e[i].y = read(), e[i].v = read();
        }

        sort(e + 1, e + m + 1, [](const edge& x, const edge& y) { return x.v < y.v; });

        int cnt = n;
        rep(i, 1, n * 2 - 1) fa[i] = i;
        rep(i, 1, m) {
            int x = F(e[i].x), y = F(e[i].y);
            if(x == y) continue;
            fa[x] = fa[y] = ++ cnt;
            val[cnt] = e[i].v;
            add(cnt, x), add(cnt, y);
        }

        memset(fa, 0, sizeof(int) * (cnt + 1));
        Fyr(cnt, cnt);
        Zrd(cnt, cnt);
        sort(ans + 1, ans + idx + 1, [](const int& x, const int& y) { return x > y; });

        LL res = 1ll * imp[cnt] * val[cnt];
        rep(i, 1, n) printf("%lld%c", res -= ans[i], " \n"[i == n]);
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The key aspects to consider are:

1. **Graph Structure**: Ensure the graph is connected but has a structure that can lead to incorrect merging or processing.
2. **Latency Values**: Use latency values that can create confusion in the merging process.
3. **House Requirements**: Select houses that require internet in a way that can cause the algorithm to make suboptimal choices.

Given the constraints, let's create a small-scale test input:

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

### Explanation
1. **Graph Structure**:
   - The graph is a simple linear chain: `1 - 2 - 3 - 4 - 5`.
   - This structure ensures that the graph is connected but has a clear path that can be easily visualized.

2. **Latency Values**:
   - The latencies are in increasing order: `10, 20, 30, 40`.
   - This ensures that the merging process must correctly handle the increasing latency values.

3. **House Requirements**:
   - Houses `1, 3, 5` require internet.
   - This selection ensures that the algorithm must correctly handle the requirement for houses that are not directly connected to each other.

### Expected Output
For this test input, the expected output should be:
```plaintext
100 60 20 0 0
```

### Reasoning
- For `k = 1`: The only server can be placed at house 1, 3, or 5. The minimum total latency is when the server is placed at house 3, and the total latency is `10 + 20 + 30 + 40 = 100`.
- For `k = 2`: Servers can be placed at houses 1 and 5. The minimum total latency is `20 + 40 = 60`.
- For `k = 3`: Servers can be placed at houses 1, 3, and 5. The minimum total latency is `20`.
- For `k = 4` and `k = 5`: All houses have servers, so the total latency is `0`.

This test input should expose any issues in the merging and processing logic of the algorithm, particularly in handling the increasing latency values and the selection of houses requiring internet.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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:
// In the name of Allah

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

typedef		long long int			ll;
typedef		long double				ld;
typedef		pair<int, int>			pii;
typedef		pair<ll, ll>			pll;
typedef		complex<ld>				cld;

#define		all(x)					(x).begin(),(x).end()
#define		len(x)					((ll) (x).size())
#define		F						first
#define		S						second
#define		pb						push_back
#define		sep						' '
#define		endl					'\n'
#define		Mp						make_pair
#define		kill(x)					cout << x << '\n', exit(0)
#define		set_dec(x)				cout << fixed << setprecision(x);
#define		file_io(x,y)			freopen(x, "r", stdin); freopen(y, "w", stdout);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int maxn = 4e5 + 7;

int n, m, q;
vector<pll> adj[maxn];
vector<pair<ll, pii>> E;
int A[maxn], p[maxn], sz;
priority_queue<pll> qu;
ll dp[maxn], ans;

int get(int a) {
	return (p[a] == a) ? a : p[a] = get(p[a]);
}

ll cal(int v) {
	int u1 = adj[v][0].F, u2 = adj[v][1].F;
	return (dp[v] - (dp[u1] + dp[u2]));
}

void addx(int v) {
	while (len(adj[v]) == 1) {
		v = adj[v][0].F;
	}
	ans += dp[v];
	if (len(adj[v]) == 2) qu.push(Mp(cal(v), v));
}

void dfs(int v) {
	for (auto f : adj[v]) {
		int u = f.F; ll w = f.S;
		dfs(u);
	}
	if (len(adj[v]) == 0) dp[v] = 0;
	else if (len(adj[v]) == 1) {
		int u = adj[v][0].F;
		dp[v] = dp[u];
	}
	else {
		ll w = adj[v][0].S;
		int u1 = adj[v][0].F, u2 = adj[v][1].F;
		dp[v] = min(dp[u1] + A[u2] * w, dp[u2] + A[u1] * w);
	}
}

void solve() {
	cin >> n >> m >> q;

	E.clear();
	for (int i = 0; i < 2 * n; i++) {
		adj[i].clear(); A[i] = 0;
	}
	iota(p, p + (2 * n), 0); sz = n;

	for (int i = 0; i < q; i++) {
		int x;
		cin >> x; x--;
		A[x] = 1;
	}

	for (int i = 0; i < m; i++) {
		int u, v; ll w;
		cin >> u >> v >> w; u--; v--;
		E.pb(Mp(w, Mp(u, v)));
	}
	sort(all(E));

	for (int i = 0; i < m; i++) {
		int u = E[i].S.F, v = E[i].S.S; ll w = E[i].F;
		if (get(u) == get(v)) continue;

		int r = sz++; p[r] = r;
		u = get(u); v = get(v);
		p[u] = r; p[v] = r; A[r] = A[u] + A[v];
		if (A[u] != 0) adj[r].pb(Mp(u, w));
		if (A[v] != 0) adj[r].pb(Mp(v, w));
	}

	while (!qu.empty()) qu.pop();
	dfs(sz - 1);

	ans = 0; addx(sz - 1);
	for (int i = 1; i <= n; i++) {
		cout << ans << sep;
		if (!qu.empty()) {
			int v = qu.top().S; qu.pop();
			ans -= dp[v];
			for (auto f : adj[v]) {
				int u = f.F;
				addx(u);
			}
		}
	}
	cout << endl;
}

int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

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

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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;
const int N = 2e5 + 10;
int T, n, m, p, s[N], fa[N], mnp[N], val[N];
vector<int> grt[N];
int find(int x) {
    if (fa[x] == x) return x;
    return fa[x] = find(fa[x]);
}
int main() {
    ios :: sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    cin >> T;
    while (T--) {
        cin >> n >> m >> p;
        for (int i = 1; i <= n; i++)
            s[i] = 0;
        for (int i = 1, x; i <= p; i++) {
            cin >> x;
            s[x] = 1;
        }
        vector<pair<int, pair<int, int> > > E;
        for (int i = 1; i <= n; i++) {
            fa[i] = mnp[i] = i, val[i] = 0;
            grt[i].clear();
        }
        for (int i = 1; i <= m; i++) {
            int u, v, w;
            cin >> u >> v >> w;
            E.push_back({w, {u, v}});
        }
        sort(E.begin(), E.end());
        for (int i = 0; i < m; i++) {
            int a = E[i].second.first, b = E[i].second.second, w = E[i].first;
            a = find(a), b = find(b);
            if (a == b) continue;
            val[mnp[a]] -= w * s[a], val[mnp[b]] -= w * s[b];
            fa[a] = b;
            if (val[mnp[a]] < val[mnp[b]]) {
                grt[mnp[a]].push_back(mnp[b]);
                mnp[b] = mnp[a];
            } else {
                grt[mnp[b]].push_back(mnp[a]);
            }
            s[b] += s[a];
            val[mnp[b]] += w * s[b];
        }
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>> > Q;
        Q.emplace(val[mnp[find(1)]], mnp[find(1)]);
        int res = 0;
        while (!Q.empty()) {
            auto it = Q.top(); Q.pop();
            res += it.first;
            cout << res << " ";
            for (int v : grt[it.second])
                Q.emplace(val[v], v);
        }
        cout << endl;

    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <array>
#include <algorithm>
#include <random>
#include <chrono>
#include <assert.h>
#include <cstring>
#include <stack>
#include <functional>
#include <bitset>

using namespace std;

// #defines
#define mkuni(v) sort(all(v)); v.erase(unique(all(v)), v.end())
#define all(v) v.begin(), v.end()
#define endl '\n'
#define ll long long
#define vint vector<int>
#define vll vector<ll>
#define vvint vector<vector<int>>
#define ar(x) array<int, x>
#define arl(x) array<ll, x>
#define watch(x) dbg((#x), " : ", (x), '\n')
// #define watch(x) cerr << #x << " : " << (x) << '\n';

#ifndef ONLINE_JUDGE
#define DEBUG true
#endif

// ----------------------------------------------
// print statements
// ----------------------------------------------
template<typename T>
istream& operator>>(istream &in, vector<T> &v) {
    for (auto &x : v) in >> x;
    return in;
}
template<typename T>
ostream& operator<<(ostream &out, vector<T> &v) {
    out << "{";
    for (auto &x : v) out << x << " ";
    out << "}\n";
    return out;
}
template<typename Iterable>
void prnv(const Iterable& iterable, ostream&out = cout) {
    if (iterable.begin() == iterable.end()) {
        out << endl;
        return;
    }
    auto x = iterable.begin();
    out << *x;
    for (++x; x != iterable.end(); ++x) out << ' ' << *x;
    out << endl;
}
void prn(char en = '\n') {
    cout << en;
}
template<typename T, typename... Args>
void prn(T x, Args... args) {
    cout << x << " ";
    prn(args...);
}
void dbg() {
    cerr << endl;
}
#ifdef DEBUG
template<typename T, typename... Args>
void dbg(T x, Args... args) {
    cerr << x << " ";
    dbg(args...);
}
#else
template<typename T, typename... Args>
void dbg(T x, Args... args) {}
#endif

// ----------------------------------------------

/* ----------------------------------------------
 * Juvenile shortcuts
 * ---------------------------------------------- */
template<typename T> void setmin(T &a) {}
template<typename T, typename... Args> void setmin(T &a, T b, Args... args) {a = (a > b ? b : a); setmin(a, args...);}
template<typename T> void setmax(T &a) {}
template<typename T, typename... Args> void setmax(T &a, T b, Args... args) {a = (a < b ? b : a); setmax(a, args...);}
// ----------------------------------------------

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define rand(l, r) uniform_int_distribution<int>(l, r)(rng)

const long long oo = (long long)1e18 + 5;
const int N = (int)2e5 + 5;
const int MOD = 998244353;

int sp[N], par[N], sub[N], vis[N], curW;
ll comp[N], cost;
vector<int> g[N];

int root(int u) {
    while (par[u] != u) {
        par[u] = par[par[u]];
        u = par[u];
    }
    return u;
}

void merge(int uu, int vv) {
    int u = root(uu), v = root(vv);
    if (u == v) return;
    if (sub[u] < sub[v]) swap(u, v);
    sub[u] += sub[v];
    sp[u] += sp[v];
    par[v] = u;
}

vll cost_vector;
void dfs(int u) {
    vis[u] = 1;
    if (sp[u] > 0) cost_vector.push_back(sp[u] * 1LL * curW - comp[u]);
    for (auto v : g[u]) if (!vis[v]) {
        merge(u, v);
        dfs(v);
    }
}

void solve() {
    int n, m, p;
    cin >> n >> m >> p;
    for (int i=1;i<=n;++i) {
        par[i] = i;
        sub[i] = 1;
        sp[i] = 0;
        comp[i] = 0;
        g[i].clear();
        vis[i] = 0;
    }
    for (int i=0;i<p;++i) {
        int x;
        cin >> x;
        sp[x] = 1;
    }
    vector<ar(3)> edges;
    for (int i=0;i<m;++i) {
        int u, v, w;
        cin >> u >> v >> w;
        ar(3) e = ar(3)({w, u, v});
        edges.push_back(e);
    }
    sort(all(edges));
    int curK = p;
    cost = 0;
    vll ans(n + 1, (ll)1e18);
    // prn(edges.back()[0], edges.back()[1], edges.back()[2], " last edge");

    for (int i=p;i<=n;++i) ans[i] = 0;

    auto process_edges = [&](int ell, int arr) {
        // prn("processing edge", edges.back()[2]);
        vint vertices;
        for (int j=ell;j<=arr;++j) {
            int u = root(edges[j][1]), v = root(edges[j][2]);
            // prn(edges[j][1], edges[j][2], u, v, " edge found", j, m, edges.back()[2], " wtf");
            g[u].push_back(v);
            g[v].push_back(u);
            vertices.push_back(u);
            vertices.push_back(v);
        }
        mkuni(vertices);
        vll have;
        for (auto u : vertices) if (!vis[u]) {
            dfs(u);
            auto go = cost_vector;
            sort(all(go));
            // prn(go, " found");
            if (!go.empty()) go.pop_back();
            for (auto x : go) have.push_back(x);
            for (auto x : go) comp[root(u)] += x;
            cost_vector.clear();
        }
        for (auto u : vertices) {
            vis[u] = 0;
            g[u].clear();
        }
        sort(all(have));
        for (auto x : have) {
            curK -= 1;
            cost += x;
            ans[curK] = cost;
        }
    };

    for (int i=0;i<m;++i) {
        int ell = i, arr;
        curW = edges[i][0];
        while (i + 1 < m && edges[i + 1][0] == curW) ++i;
        arr = i;
        process_edges(ell, arr);
    }
    for (int i=1;i<=n;++i) cout << ans[i] << " \n"[i==n];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(15);
    int T = 1;
    cin >> T;
    for (int t=1;t<=T;++t) {
        // cout << "Case #" << t << ": ";
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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
import os
from io import BytesIO, IOBase
from _collections import deque
from heapq import heappop, heappush

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

file = sys.stdin
if os.environ.get('USER') == "loic":
    file = open("data.in")

line = lambda: file.readline().split()
ui = lambda: int(line()[0])
mi = lambda: map(int,line())
li = lambda: list(mi())
lsi = lambda: list(map(int,line()[0]))
#######################################################################

class DSU:
    def __init__(self, limit):
        self.par = list(range(limit))
        self.sz = [1] * limit
        self.mx = [0] * limit
        self.s = [0] * limit

    def leader(self,u):
        lead = u
        while lead != self.par[lead]:
            lead = self.par[lead]

        node = u
        while node != lead:
            self.par[node], node = lead, self.par[node]

        return lead

    def union(self, u, v):
        u = self.leader(u)
        v = self.leader(v)
        if u != v:
            if self.sz[u] < self.sz[v]:
                u,v = v,u
            self.par[v] = u
            self.sz[u] += self.sz[v]

    def size(self,a):
        return self.sz[self.leader(a)]

def solve():

    if P == 1:
        return " ".join(str(0) for _ in range(N))

    req = [0] * N
    for val in A:
        req[val-1] = 1

    dsu = DSU(N)

    g = [[] for _ in range(N)]
    E.sort(key = lambda e: e[2])

    for u,v,w in E:
        if dsu.leader(u) != dsu.leader(v):
            dsu.union(u,v)
            g[u].append((v,w))
            g[v].append((u,w))

    start = 0

    seen = [-1] * len(g)
    q = deque([start])
    seen[start] = 0
    par = [None] * N

    h = []
    dp = [[] for _ in range(N)]

    while q:
        n = q[-1]

        chs = g[n]
        nb = seen[n]
        if nb == len(chs):
            q.pop()

            cur = []
            for ch,w in chs:
                if n != start and ch == par[n][0]:
                    continue

                tmp = dp[ch]
                while tmp and w > tmp[0][0]:
                    m,u = heappop(tmp)
                    heappush(tmp, (w,u))

                for m,u in tmp:
                    heappush(dp[n], (m,u))

                if not cur:
                    cur = tmp.copy()
                    continue

                for m,u in cur:
                    for p,v in tmp:
                        heappush(h, (max(m,p), u, v))

                for p,v in tmp:
                    heappush(cur, (p,v))

            if req[n] == 1:
                for w,u in dp[n]:
                    heappush(h, (w, n, u))
                heappush(dp[n], (par[n][1] if par[n] is not None else 0, n))

        else:
            ch,w = chs[nb]
            if seen[ch] == -1:
                seen[ch] = 0
                q.append(ch)
                par[ch] = (n,w)
            seen[n] += 1

    res = [0] * (N + 1)

    dsu = DSU(N)
    j = P - 2
    s = 0

    while h:
        w,u,v = heappop(h)
        lead_u = dsu.leader(u)
        lead_v = dsu.leader(v)
        if lead_u != lead_v:
            sz_u = dsu.size(lead_u)
            sz_v = dsu.size(lead_v)
            s_u = dsu.s[lead_u]
            s_v = dsu.s[lead_v]
            dsu.union(u,v)
            lead = dsu.leader(u)
            dsu.mx[lead] = w
            dsu.s[lead] = min(s_u + w * sz_v, s_v + w * sz_u)
            s += dsu.s[lead] - (s_u + s_v)
            res[j] = s
            j -= 1

    res.pop()

    return " ".join(str(v) for v in res)

for test in range(1,ui()+1):
    N,M,P = mi()
    A = li()
    E = []
    for _ in range(M):
        u,v,w = mi()
        u -= 1
        v -= 1
        E.append((u,v,w))

    print(solve())

file.close()

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

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

Think step by step.

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

1. **Small-scale Test Input**: The input should be small enough to fit within the constraints but large enough to test the algorithm thoroughly.
2. **Edge Cases**:
   - **Single House Requiring Internet**: This can test the base case where only one house needs internet.
   - **Multiple Houses Requiring Internet**: This can test the general case.
   - **High Latency Cables**: This can test how the algorithm handles high latency connections.
   - **Multiple Test Cases**: This can test the algorithm's ability to handle multiple test cases.
3. **Connected Graph**: The graph must be connected, but it can have different structures (e.g., linear, star, tree).

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

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

### Explanation
1. **Test Case 1**:
   - **4 houses, 3 cables, 3 houses requiring internet**.
   - **Houses requiring internet**: 1, 2, 4.
   - **Cables**:
     - House 1 to House 2 with latency 1.
     - House 2 to House 3 with latency 10.
     - House 3 to House 4 with latency 1.

   This test case is designed to test the algorithm's ability to handle a simple linear structure with a high latency cable in the middle.

2. **Test Case 2**:
   - **5 houses, 4 cables, 4 houses requiring internet**.
   - **Houses requiring internet**: 1, 2, 3, 5.
   - **Cables**:
     - House 1 to House 2 with latency 1.
     - House 2 to House 3 with latency 2.
     - House 3 to House 4 with latency 3.
     - House 4 to House 5 with latency 4.

   This test case is designed to test the algorithm's ability to handle a slightly more complex linear structure with increasing latency.

This input should help expose any issues in the algorithm, particularly in how it handles high latency connections and different structures of the graph.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
for __ in range(fg()):
    n, m, p = fgh()
    s = fgh()
    a = [[] for i in range(n)]
    g = []
    for _ in range(m):
        u, v, w = fgh()
        u -= 1
        v -= 1
        g.append((w, u, v))
    g.sort()
    ks = [i for i in range(n)]
    z = [[i] for i in range(n)]
    sz = [1] * n
    ans = [239239239239239] * (n + 1)
    for i in range(p, n + 1):
        ans[i] = 0
    tt = [0] * n
    ans1 = 0
    sz1 = [0] * n
    for i in range(p):
        sz1[s[i] - 1] = 1
    ANS = [0] * n
    last = p
    for _ in range(m):
        u, v = g[_][1], g[_][2]
        if ks[u] != ks[v]:
            if sz[ks[u]] > sz[ks[v]]:
                u, v = v, u
            ttt = ks[u]
            fl = 0
            if sz1[ks[u]] and sz1[ks[v]]:
                fl = 1
                W = g[_][0]
                ans1 += min(ANS[ks[u]] + sz1[ks[v]] * W, ANS[ks[v]] + sz1[ks[u]] * W) - ANS[ks[u]] - ANS[ks[v]]
                ANS[ks[v]] = min(ANS[ks[u]] + sz1[ks[v]] * W, ANS[ks[v]] + sz1[ks[u]] * W)
            if fl == 0 and sz1[ks[u]]:
                ANS[ks[v]] = ANS[ks[u]]
            sz1[ks[v]] += sz1[ks[u]]
            for uu in z[ttt]:
                z[ks[v]].append(uu)
                ks[uu] = ks[v]
                sz[ks[v]] += 1

            if fl:
                ans[last - 1] = ans1
                last -= 1
    print(*ans[1:])

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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 speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ll long long
#define YES "Yes"
#define NO "No"
#define endline '\n'

using namespace std;

const ll N = 5e5 + 10, MM = 998244353, MAX = 1e18;
int n,m;
bool house[N];
vector<pair<ll,pair<ll,ll>>>edgs;

int p[N],cnt[N],sum[N];
int get_p(int x){
  if(p[x]==x)return x;
  return p[x]=get_p(p[x]);
}
ll res=0;
ll number;

void join(int x,int y,ll cost){
  x=get_p(x),y=get_p(y);
  if(x==y)return;
  ll mn=cnt[x]*cost-sum[x];
  mn=min(mn,cnt[y]*cost-sum[y]);
  res+=mn;
  if(cnt[x]&&cnt[y])number--;
  cnt[x]+=cnt[y];
  p[y]=x;
  sum[x]=sum[x]+sum[y]+mn;
}

int solve() {
  cin>>n>>m;
  cin>>number;
  for(int i=1;i<=n;i++)house[i]=0;
  for(int i=0;i<number;i++){
    int x;cin>>x;
    house[x]=1;
  }
  res=0;
  edgs.clear();
  for(int i=1;i<=n;i++){
    p[i]=i;
    sum[i]=0;
    cnt[i]=house[i];
  }
  for(int i=0;i<m;i++){
    int a,b,c;cin>>a>>b>>c;
    edgs.push_back({c,{a,b}});
  }
  sort(edgs.begin(),edgs.end());
  vector<ll>ans(n+1,MAX);
  for(int i=number;i<=n;i++)ans[i]=0;
  for(auto [cost,edg]:edgs){
    join(edg.first,edg.second,cost);
    ans[number]=min(ans[number],res);
  }
  for(int i=1;i<=n;i++){
    cout<<ans[i]<<" ";
  }
  cout<<endline;
  return 0;
}

int main() {
	speed;
	int t = 1;
	cin >> t;
	int i = 1;
	while (t--) {
		//cout << "Case #" << i++ << ": ";
		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
1
4 5 3
1 3 4
1 2 1
2 3 2
3 4 3
4 1 4
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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;
int t, n, m, p;
int f[200020];
long long g[200020];
int c[200020];
long long z[200020];
long long zz;
pair<int, pair<int, int> > e[200020];
int F(int x)
{
	return f[x] != x ? f[x] = F(f[x]) : x;
}
int main()
{
	scanf("%d", &t);
	for (int tt = 0; tt < t; tt++)
	{
		scanf("%d%d%d", &n, &m, &p);
		for (int i = 1; i <= n + 5; i++)
		{
			f[i] = i;
			g[i] = 0;
			c[i] = 0;
			z[i] = 0;
		}
		for (int i = 1; i <= p; i++)
		{
			int x;
			scanf("%d", &x);
			c[x] = 1;
		}
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d", &e[i].second.first, &e[i].second.second, &e[i].first);
		}
		sort(e, e + m);
		int zi = n;
		zz = 0;
		for (int i = 0; i < m; i++)
		{
			int x = F(e[i].second.first);
			int y = F(e[i].second.second);
			if (F(x) == F(y))
			{
				continue;
			}
			long long xy = c[x] * e[i].first - g[x];
			long long yx = c[y] * e[i].first - g[y];
			if (xy > yx)
			{
				swap(x, y);
				swap(xy, yx);
			}
			zz += xy;
			g[x] = 0;
			g[y] += c[x] * e[i].first;
			c[y] += c[x];
			f[x] = y;
			z[--zi] = zz;
		}
		for (int i = 1; i <= n; i++)
		{
			printf("%lld ", z[i]);
		}
		puts("");
	}
	return 0;
}

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

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

Think step by step.

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

1. **Identify Potential Bugs**:
   - The code uses a disjoint set (union-find) to manage the connected components.
   - It calculates the total latency for each house and updates the total latency as it merges components.
   - The main issue might be in the way it calculates the total latency and updates the components.

2. **Edge Cases**:
   - **Single Component**: The graph is a single component.
   - **Multiple Components**: The graph has multiple components.
   - **High Latency Edges**: Edges with very high latency values.
   - **Low Latency Edges**: Edges with very low latency values.
   - **Uniform Latency Edges**: All edges have the same latency.
   - **Special Houses**: Houses that need internet are not well connected.

3. **Small-Scale Test Input**:
   - We will create a small graph with a few houses and cables to ensure it is manageable.
   - We will include a scenario where the houses that need internet are not well connected, which could expose the bug in the latency calculation.

Let's create a test input with the following properties:
- 5 houses
- 5 cables
- 3 houses that need internet
- Houses that need internet are connected in a way that requires careful handling of latencies.

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

**Explanation**:
- **Number of Test Cases**: 1
- **Number of Houses (n)**: 5
- **Number of Cables (m)**: 5
- **Houses that Need Internet (p)**: 3 (Houses 1, 3, and 5)
- **Cables**:
  - 1-2 with latency 1
  - 2-3 with latency 10
  - 3-4 with latency 5
  - 4-5 with latency 1
  - 1-4 with latency 2

This test input should challenge the algorithm to correctly handle the high latency between house 2 and house 3, and the low latency between house 4 and house 5, while ensuring the houses that need internet are connected optimally.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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 f(i,x,y) for(int i=x;i<=y;i++)
#define df(i,x,y) for(int i=x;i>=y;i--)
#define int long long
#define pb push_back
using namespace std;
const int N=1.1e4,I=1e9+1;
int n,a[N],o[N],x,y,z,cs,as,m,p,fa[N],tot,si[N];
pair<int,pair<int,int> >w[N];
vector<int>b[N],f[N];
int Gf(int e){return e==fa[e]?e:fa[e]=Gf(fa[e]);}
void D(int e)
{
	si[e]=o[e];
	for(int u:b[e])D(u),si[e]+=si[u];
	f[e].resize(si[e]+2);
	if(e>n)
	{
		x=b[e][0],y=b[e][1];
		f(i,1,si[e])f[e][i]=si[e]*a[e];
		f[x][0]=si[x]*a[e];f[y][0]=si[y]*a[e];
		int sx=0,sy=0;
		f(k,1,si[e])
		{
			if(sx==si[x])++sy;
			else if(sy==si[y])++sx;
			else if(f[x][sx+1]-f[x][sx]<f[y][sy+1]-f[y][sy])++sx;
			else ++sy;
			f[e][k]=f[x][sx]+f[y][sy];
		}
	}
	else if(o[e])f[e][1]=0;
}
signed main()
{
	cin>>cs;
	while(cs--)
	{
		cin>>n>>m>>p;tot=n;
		f(i,1,n*2)
		{
			a[i]=0,fa[i]=i;o[i]=0;
			vector<int>c,g;swap(c,b[i]);swap(g,f[i]);
		}
		f(i,1,p)cin>>x,o[x]=1;
		f(i,1,m)cin>>x>>y>>z,w[i]=make_pair(z,make_pair(x,y));
		sort(w+1,w+m+1);
		f(i,1,m)if((x=Gf(w[i].second.first))!=(y=Gf(w[i].second.second)))
		{
//			if(fa[x]==x)a[x]=w[i].first;if(fa[y]==y)a[y]=w[i].first;
			a[++tot]=w[i].first,fa[fa[x]]=fa[fa[y]]=tot,b[tot].pb(x),b[tot].pb(y);
		}
		D(tot);
		f(i,1,si[tot])cout<<f[tot][i]<<' ';
		f(i,1,n-si[tot])cout<<"0 ";
		cout<<'\n';
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause the code to fail. Here are the steps to think through the process:

1. **Edge Cases**:
   - **Single House**: A graph with only one house.
   - **Linear Graph**: A graph where houses are connected in a linear fashion.
   - **Complete Graph**: A graph where every house is connected to every other house.
   - **Disconnected Components**: A graph with multiple connected components (though this is not applicable here as the graph is guaranteed to be connected).

2. **Constraints**:
   - The graph must be connected.
   - The number of houses and cables should be within the given constraints.
   - The houses that need internet should be distinct and within the valid range.

3. **Specific Scenarios**:
   - **High Latency Edges**: Edges with very high latency to test the algorithm's ability to handle large numbers.
   - **Multiple Houses Requiring Internet**: Multiple houses that need internet to ensure the algorithm can handle multiple queries efficiently.
   - **Optimal Placement of Servers**: Scenarios where the optimal placement of servers is not straightforward, such as when the graph has a complex structure.

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

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

### Explanation:
1. **Test Case 1**:
   - **5 Houses, 4 Cables, 3 Houses Requiring Internet**: This tests a small but non-trivial graph.
   - **Edges with High Latency**: The latencies are 10, 20, 30, and 40, which are high enough to challenge the algorithm.
   - **Houses Requiring Internet**: Houses 1, 3, and 5 require internet. This ensures that the algorithm must consider a non-linear structure and multiple points of interest.

2. **Test Case 2**:
   - **4 Houses, 3 Cables, 2 Houses Requiring Internet**: This tests a smaller graph with fewer houses and cables.
   - **Linear Structure**: The graph is a simple linear structure, which can still be challenging for the algorithm to handle correctly.
   - **Houses Requiring Internet**: Houses 1 and 2 require internet, which tests the algorithm's ability to handle a linear structure with multiple points of interest.

This test input is designed to cover a range of scenarios that can expose potential bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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

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

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

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

So the total latency is $9$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef double db;
#define F(i, a, b) for(int i = a; i <= (b); ++i)
#define F2(i, a, b) for(int i = a; i < (b); ++i)
#define dF(i, a, b) for(int i = a; i >= (b); --i)
template<typename T> void debug(string s, T x) {cerr << "[" << s << "] = [" << x << "]\n";}
template<typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i = 0, b = 0; i < (int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++;else if (s[i] == ')' || s[i] == '}') b--;else if (s[i] == ',' && b == 0) {cerr << "[" << s.substr(0, i) << "] = [" << x << "] | ";debug(s.substr(s.find_first_not_of(' ', i + 1)), args...);break;}}
#ifdef ONLINE_JUDGE
#define Debug(...)
#else
#define Debug(...) debug(#__VA_ARGS__, __VA_ARGS__)
#endif
#define pb push_back
#define fi first
#define se second
#define Mry fprintf(stderr, "%.3lf MB\n", (&Med - &Mbe) / 1048576.0)
#define Try cerr << 1e3 * clock() / CLOCKS_PER_SEC << " ms\n";
typedef long long ll;
// namespace Fread {const int SIZE = 1 << 17; char buf[SIZE], *S, *T; inline char getchar() {if (S == T) {T = (S = buf) + fread(buf, 1, SIZE, stdin); if (S == T) return '\n';} return *S++;}}
// namespace Fwrite {const int SIZE = 1 << 17; char buf[SIZE], *S = buf, *T = buf + SIZE; inline void flush() {fwrite(buf, 1, S - buf, stdout), S = buf;} inline void putchar(char c) {*S++ = c;if (S == T) flush();} struct NTR {~NTR() {flush();}} ztr;}
// #ifdef ONLINE_JUDGE
// #define getchar Fread::getchar
// #define putchar Fwrite::putchar
// #endif
inline int ri() {
	int x = 0;
	bool t = 0;
	char c = getchar();
	while (c < '0' || c > '9') t |= c == '-', c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
	return t ? -x : x;
}
inline void wi(int x) {
	if (x < 0) {
		putchar('-'), x = -x;
	}
	if (x > 9) wi(x / 10);
	putchar(x % 10 + 48);
}
inline void wi(int x, char s) {
	wi(x), putchar(s);
}
bool Mbe;
// mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3f;
const int _ = 2e5 + 5;

int n, m, p;

int g[5005][5005];

int fa[_], sz[_], h[_], sz2[_];

bool s[_];

int find(int x) {
	return x == fa[x] ? x: fa[x] = find(fa[x]);
}

struct edge {
	int u, v, w;
} e[_];

void solve() {
	n = ri(), m = ri(), p = ri();
	if(n <= 5000)
	F(i, 1, n) F(j, 0, n) g[i][j] = infll;
	F(i, 1, n) s[i] = 0;
	F(i, 1, p) s[ri()] = 1;
	F(i, 1, m) {
		e[i].u = ri(), e[i].v = ri(), e[i].w = ri();

	}
	sort(e + 1, e + m + 1, [&](edge a, edge b) { return a.w <b.w; });
	F(i, 1, n) fa[i] = i, sz[i] = s[i], sz2[i] = 1;
	F(i, 1, n) if(n <= 5000) g[i][0] = 0, g[i][1] = 0;
	F(i, 1, m) {
		int u = find(e[i].u), v = find(e[i].v);
		if(u == v) continue;
		Debug(i, u, v);
		F(j, 0, sz2[u] + sz2[v]) h[j] = infll;
		F(a, 0, sz2[u]) F(b, 0, sz2[v]) {
			if(!a && b && sz[u]) h[a + b] = min(h[a + b], g[u][a] + g[v][b] + sz[u] * e[i].w);
			else if(a && !b && sz[v]) h[a + b] = min(h[a + b], g[u][a] + g[v][b] + sz[v] * e[i].w);
			else {
				h[a + b] = min(h[a + b], g[u][a] + g[v][b]);
			}
		}
		fa[u] = v;
		sz2[v] += sz2[u], sz2[u] = 0;
		sz[v] += sz[u], sz[u] = 0;
		F(j, 1, sz2[v] - 2) {
			Debug(h[j], h[j + 1], h[j + 2]);
			assert(h[j] - h[j + 1] >= h[j + 1] - h[j + 2]);
		}
		if(n <= 5000)
		F(j, 0, sz2[v]) {
			g[v][j] = h[j];
		}
	}
	int rt = find(1);
	if(n <= 5000)
	F(i, 1, n) cout << g[rt][i] << ' ';
	cout << '\n';
}
bool Med;
signed main() {
	// Mry;
	int T = ri();
	while(T--) {
		solve();
	}
	// Try;
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are the steps to create such a test input:

1. **Small Scale**: Since the output length is limited, we need a small-scale test input. This means using smaller values for \( n \), \( m \), and \( p \).

2. **Edge Cases**:
   - **Single House Requiring Internet**: This can test if the code handles the simplest case correctly.
   - **Multiple Houses Requiring Internet**: Test if the code can handle multiple houses requiring internet.
   - **High Latency Edges**: Test if the code can handle high latency values correctly.
   - **Connected Components**: Ensure the graph is connected but has a structure that can expose bugs.

3. **Constraints**:
   - \( 2 \le n \le 2 \cdot 10^5 \)
   - \( n-1 \le m \le 2 \cdot 10^5 \)
   - \( 1 \le p \le n \)

4. **Example Test Input**:
   - **Test Case 1**: Single house requiring internet.
   - **Test Case 2**: Multiple houses requiring internet with high latency edges.
   - **Test Case 3**: A connected graph with a structure that can expose bugs.

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

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

### Explanation of the Test Input

1. **Test Case 1**:
   - **n = 3**, **m = 2**, **p = 1**
   - **Houses requiring internet**: 1
   - **Edges**:
     - 1-2 with latency 1
     - 2-3 with latency 1
   - **Expected Output**: 0 0 0
   - **Reason**: Only one house requires internet, and it can be a server itself.

2. **Test Case 2**:
   - **n = 4**, **m = 3**, **p = 2**
   - **Houses requiring internet**: 1, 4
   - **Edges**:
     - 1-2 with latency 10
     - 2-3 with latency 1
     - 3-4 with latency 10
   - **Expected Output**: 20 10 0 0
   - **Reason**: The minimum total latency is achieved by placing servers at houses 1 and 4, which results in a total latency of 0. Placing a server at house 2 or 3 with high latency edges can expose bugs.

3. **Test Case 3**:
   - **n = 5**, **m = 4**, **p = 3**
   - **Houses requiring internet**: 1, 2, 3
   - **Edges**:
     - 1-2 with latency 1
     - 2-3 with latency 1
     - 3-4 with latency 1
     - 4-5 with latency 1
   - **Expected Output**: 3 1 0 0 0
   - **Reason**: The minimum total latency is achieved by placing servers at houses 1, 2, and 3, which results in a total latency of 0. Placing servers at other houses can expose bugs in handling multiple houses requiring internet.

This test input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme 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 $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — 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 $2 \cdot 10^5$.

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;
typedef long long LL;
const int N=5e3+3;
LL INF=1e15;
int n,m,p,x,fa[N],sz[N];
LL sum[N];
multiset<LL,greater<LL>>s[N];//从大到小排序的可重set，符合数组顺序情况
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
struct node{
	int x,y,z;
}b[N];
bool cmp(node X,node Y){
	return X.z<Y.z;
}
void merge(int x,int y,int z){
	for(auto it=s[y].begin();it!=s[y].end();it++)
		s[x].insert(*it);
	s[y].clear();
	LL fx=1ll*z*sz[x]-sum[x],fy=1ll*z*sz[y]-sum[y];
	if(fx<=fy) s[x].insert(fx),sum[x]+=fx;
	else s[x].insert(fy),sum[x]+=fy;
	sz[x]+=sz[y],sum[x]+=sum[y],fa[y]=x;
}
int Main()
{
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++) fa[i]=i,sz[i]=sum[i]=0;
    for(int i=1;i<=p;i++) scanf("%d",&x),sz[x]=1;
    for(int i=1;i<=m;i++)
    	scanf("%d%d%d",&b[i].x,&b[i].y,&b[i].z);
    sort(b+1,b+m+1,cmp);
    for(int i=1;i<=m;i++){
    	int x=find(b[i].x),y=find(b[i].y),z=b[i].z;
    	//cout<<x<<' '<<y<<' '<<z<<endl;
		if(x==y) continue;
		if(sz[x]>sz[y]) merge(x,y,z);
		else merge(y,x,z);
	}
	int rt=find(1),now=0;
	for(auto it=s[rt].begin();it!=s[rt].end();it++) printf("%lld ",sum[rt]),sum[rt]-=*it;puts("0");
	s[rt].clear();
	return 0;
}
int main()
{
    int Case=1;
    scanf("%d",&Case);
    while(Case--) Main();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the probability of failure. Here are the steps to create such a test input:

1. **Small Scale**: Since the output length is limited, we need to keep the test input small but effective.
2. **Edge Cases**: Consider scenarios where the graph is highly connected, or where the number of houses requiring internet is small or large.
3. **Latency Values**: Use latency values that can cause the algorithm to make incorrect decisions, such as large differences in latencies or latencies that are very close to each other.
4. **Multiple Test Cases**: Include multiple test cases to cover different scenarios.

Let's create a test input that includes a connected graph with a small number of houses, a few houses requiring internet, and a mix of latencies that can expose the bug in the merging logic.

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

### Explanation:
1. **Test Case 1**:
   - **Graph**: A small cycle graph with 5 houses and 5 cables.
   - **Houses Requiring Internet**: Houses 1, 3, and 5.
   - **Latencies**: Cables with latencies 1, 2, 3, 4, and 5.
   - **Expected Output**: This test case will check how the algorithm handles a cycle and ensures that the minimum total latency is calculated correctly for each value of \( k \).

2. **Test Case 2**:
   - **Graph**: A simple line graph with 3 houses and 2 cables.
   - **Houses Requiring Internet**: Houses 1 and 3.
   - **Latencies**: Cables with latencies 1 and 2.
   - **Expected Output**: This test case will check how the algorithm handles a simple connected graph and ensures that the minimum total latency is calculated correctly for each value of \( k \).

This test input is designed to be small and complete, while still challenging the buggy code to handle different scenarios and edge cases.