// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define sz(x) int(x.size())
#define ins insert
using ll = long long;
template<class T> using V = vector<T>;
template<class T, size_t SZ> using AR = array<T, SZ>;
template<class T> using mset = multiset<T>;
using vi = V<int>;
using vl = V<ll>;
const ll MX = 1LL << 47;
unordered_map<ll, ll> AMT, SUM; ll all = 0;
void add(ll p, ll x) {
// cout << "ADD: " << p << " " << x << endl;
ll xi = p * x; all += xi;
for(++p; p <= MX; p += p & -p) { AMT[p - 1] += x; SUM[p - 1] += xi; }
}
ll get(ll k) { // sum of last k
if (k == 0) return 0;
ll p = 0, ans = 0, have = -MX;
for(ll x = MX >> 1; x; x >>= 1) {
const ll &amt = AMT[p + x - 1];
if (amt <= k && p + x <= MX) {
k -= amt;
ans += SUM[p + x - 1];
p += x;
}
if (amt) have = p + x - 1;
}
ans += k * have; // extra room
return ans;
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int N, K; cin >> N >> K;
V<V<AR<int, 2>>> adj(N); for(int i = 0; i < N - 1; i++) {
int u, v, w; cin >> u >> v >> w; --u, --v;
adj[u].pb({v, w});
adj[v].pb({u, w});
}
V<mset<ll>> chd(N); vl todo;
function<ll(int, int)> gen = [&](int u, int p) {
for(auto& [v, w] : adj[u]) if (v != p) {
ll x = gen(v, u) + w;
chd[u].ins(x);
}
int cur = 0;
for(auto& x : chd[u]) {
++cur;
if (cur == sz(chd[u])) break;
todo.pb(x);
}
if (!sz(chd[u])) return 0LL;
return *rbegin(chd[u]);
};
todo.pb(gen(0, 0));
// MX = 1LL << __lg(*max_element(begin(todo), end(todo))); MX *= 2;
for(auto& x : todo) add(x, +1);
int L = 0; for(int i = 0; i < N; i++) if (sz(adj[i]) == 1) {
++L;
chd[i].insert(0);
}
// cout << "LEAF: " << L << endl;
vl ans(N);
function<void(int, int)> answer = [&](int u, int p) {
ans[u] = all - get(max(0, L - K));
// cout << L - K << endl;
// cout << u << " ==> " << ans[u] << endl;
for(auto& [v, w] : adj[u]) if (v != p) {
// cout << sz(chd[v]) << endl;
ll mxv = *rbegin(chd[v]) + w;
chd[u].erase(mxv); add(mxv, -1); // rem v as chd of u
add(*rbegin(chd[v]), +1);
// cout << sz(chd[u]) << endl;
add(*rbegin(chd[u]), -1);
ll mxu = *rbegin(chd[u]) + w;
chd[v].insert(mxu); add(mxu, +1); // add u as chd of v
answer(v, u);
chd[v].erase(mxu); add(mxu, -1); // rem u as chd of v
add(*rbegin(chd[u]), +1);
add(*rbegin(chd[v]), -1);
chd[u].insert(mxv); add(mxv, +1); // add u as chd of v
}
};
answer(0, 0);
for(auto& x : ans) cout << x << nl;
exit(0-0);
}
