// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define f first
#define s second
#define mp make_pair
#define pb push_back
#define sz(x) int(x.size())
using ll = long long;
template<class T> using V = vector<T>;
using pl = pair<ll, ll>;
using vi = V<int>;
using vpl = V<pl>;
using vl = V<ll>;
const ll INFL = 1e18;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N, K; cin >> N >> K;
V<vpl> adj(N); vi d(N); for(int i = 0; i < N - 1; i++) {
int u, v, w; cin >> u >> v >> w; --u, --v;
adj[u].pb(mp(v, w));
adj[v].pb(mp(u, w));
}
vi ans; ll k = -1;
function<pl(int, int, ll)> dfs = [&](int u, int p, ll d) {
ll close = -INFL, far = 0;
for(auto& e : adj[u]) {
int v; ll w; tie(v, w) = e; if (v == p) continue;
pl R = dfs(v, u, w);
if (R.f) close = max(close, R.s);
else far = max(far, R.s);
}
pl R;
if (close >= far) R = mp(1, close - d);
else R = mp(0, far + d);
if (R.f && R.s < 0) R = mp(0, 0);
if (!R.f && R.s > k) {
ans.pb(u); R = mp(1, k - d);
}
return R;
};
ll lo = 0, hi = ll(3e15);
while(lo < hi) {
ll mid = (lo + hi) / 2;
ans = {}; k = mid;
dfs(0, -1, INFL);
if (sz(ans) <= K) hi = mid;
else lo = mid + 1;
}
ans = {}; k = lo;
dfs(0, -1, INFL);
set<int> in(begin(ans), end(ans));
int cur = 0;
while(sz(ans) < K) {
if (!in.count(cur)) ans.pb(cur);
cur++;
}
cout << lo << nl;
for(auto& x : ans) cout << x + 1 << " ";
cout << nl;
exit(0-0);
}
