#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a,b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a,b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif
/*
read some solutions a long time ago, dont remember much from there (remember seeing sqrt complexity in some solutions, but im not sure)
*/
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
#include "roads.h"
vector<pll> adj[N];
ll dp[N][2];
vector<ll> order;
void dfs1(ll u){
for(auto [v,w] : adj[u]){
pll px = {u,w};
adj[v].erase(find(all(adj[v]),px));
dfs1(v);
}
order.pb(u);
}
vector<ll> minimum_closure_costs(int n, vector<int> U, vector<int> V, vector<int> W){
rep(i,n-1){
int u = U[i], v = V[i], w = W[i];
adj[u].pb({v,w}), adj[v].pb({u,w});
}
dfs1(1);
vector<ll> ans;
rep(k,n){
trav(u,order){
dp[u][0] = dp[u][1] = 0;
vector<ll> vals;
ll sum = 0;
for(auto [v,w] : adj[u]){
sum += w+dp[v][1];
vals.pb(-w-dp[v][1]+dp[v][0]);
}
sort(all(vals));
rep(i,k-1){
if(i < sz(vals)){
sum += min(vals[i],0ll);
}
else{
break;
}
}
dp[u][0] = dp[u][1] = sum;
if(k-1 < sz(vals) and k-1 >= 0){
dp[u][1] += min(vals[k-1],0ll);
}
}
ans.pb(min(dp[1][0],dp[1][1]));
}
ll change = 0;
rep(i,n-1) change += (ans[i] != ans[i+1]);
assert(change <= 250);
return ans;
}
