#include #pragma GCC optimize("Ofast") #define _GLIBCXX_DEBUG using namespace std; using std::cout; using std::cin; using std::endl; using ll=long long; using ld=long double; ll ILL=1167167167167167167; const int INF=2100000000; const ll mod=1e9+7; #define rep(i,a) for (ll i=0;i using _pq = priority_queue, greater>; template ll LB(vector &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();} template ll UB(vector &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();} template bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}else return 0;} template bool chmax(T &a,const T &b){if(a void So(vector &v) {sort(v.begin(),v.end());} template void Sore(vector &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});} void yneos(bool a){if(a) cout<<"Yes\n"; else cout<<"No\n";} template void vec_out(vector &p){for(int i=0;i<(int)(p.size());i++){if(i) cout<<" ";cout<> tree_in(int N){ std::vector> G(N); for(int i=0;i>a>>b; a--,b--; G[a].push_back(b); G[b].push_back(a); } return G; } std::tuple,std::vector,std::vector> tree_order_pare_depth(std::vector> &G,int root){ int n=G.size(); std::vector order={root},pare(n,-1),depth(n); pare[root]=-2; for(int i=0;i tree_diameter_path(std::vector> &G){ int n=G.size(); auto r=(std::get<0>(tree_order_pare_depth(G,0))).at(n-1); std::vector order,pare,depth,ans; tie(order,pare,depth)=tree_order_pare_depth(G,r); int ind=order[n-1]; while(ind!=-2){ ans.push_back(ind); ind=pare[ind]; } return ans; } void solve(); // oddloop int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t=1; //cin>>t; rep(i,t) solve(); } void solve(){ int N,M; cin>>N>>M; vector p(M); rep(i,M) cin>>p[i],p[i]--; map,int> m; auto G=tree_in(N); vector order,pare,dp2; tie(order,pare,dp2)=tree_order_pare_depth(G,0); dp2[0]=N; vector table(N+1),dp(N),dp3(N); set s; rep(i,N){ int a=order[N-i-1]; for(auto x:G[a]){ if(pare[x]==a){ table[dp[x]]++; s.insert(dp[x]); } } rep(j,N+1){ if(table[j]==0){ dp[a]=j; break; } } for(auto x:s) table[x]=0; s={}; } rep(i,N){ int a=order[i]; for(auto x:G[a]){ if(pare[x]==a){ table[dp[x]]++; s.insert(dp[x]); } } table[dp2[a]]++; s.insert(dp2[a]); int X=-1; rep(i,N+1){ if(table[i]==0){ X=i; break; } } //cout<<"$"< ban; rep(i,M){ int y=p[i]; if(ban.count(y)) continue; ban.insert(y); if(y!=0&&(ans^dp2[y]^dp3[y])==0){ cout<