/*input
11 12
0 1 3
0 2 4
2 3 5
3 4 4
4 5 6
0 6 3
6 7 2
6 8 5
8 9 6
8 10 7
2
*/
#include "race.h"
#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;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long
#define ld long double
#define pii pair<int,int>
#define piii pair<int,pii>
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e12;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll mult(ll a,ll b){
return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1){
res=(res*a)%MOD;
}
a=(a*a)%MOD;
b>>=1;
}
return res;
}
const ll maxn=5e5+5;
const ll maxlg=__lg(maxn)+2;
ll pref[maxn];
int n,k;
int ans=INF;
int rt[maxn];
unordered_map<int,int> mp[maxn];
vector<pii> v[maxn];
pii off[maxn];
void dfs(int u,int p){
rt[u]=u;
mp[u][0]=0;
for(auto xx:v[u]){
int vv=xx.f;
if(vv==p) continue;
dfs(vv,u);
int ru=rt[u],rv=rt[vv];
off[rv].f+=xx.s;
off[rv].s++;
if(sz(mp[ru])<sz(mp[rv])) swap(ru,rv);
for(auto x:mp[rv]){
if(mp[ru].count(k-x.f-off[rv].f-off[ru].f)){
MNTO(ans,mp[ru][k-x.f-off[rv].f-off[ru].f]+off[ru].s+off[rv].s+x.s);
}
}
for(auto x:mp[rv]){
if((!mp[ru].count(x.f+off[rv].f-off[ru].f) or mp[ru][x.f+off[rv].f-off[ru].f]>x.s+off[rv].s-off[ru].s)){
mp[ru][x.f+off[rv].f-off[ru].f]=x.s+off[rv].s-off[ru].s;
}
}
rt[u]=ru;
mp[rv].clear();
}
/*cout<<u<<' '<<off[rt[u]].f<<' '<<off[rt[u]].s<<'\n';
for(auto x:mp[rt[u]]) cout<<x.f<<' '<<x.s<<'\n';
cout<<'\n';*/
}
int best_path(int N, int K, int H[][2], int L[]){
n=N;
k=K;
REP(i,n-1){
v[H[i][0]].pb({H[i][1],L[i]});
v[H[i][1]].pb({H[i][0],L[i]});
}
dfs(0,-1);
if(ans>n) return -1;
return ans;
}
