#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,q;
int dfn[200005];
vector<int>p[200005];
struct node{
	int l,r,sum,lazy;
	void add(int x){
		if(x == 1)sum = r-l,lazy = 1;
		if(x == -1)sum =0 ,lazy =-1;
	}
}tree[1000005];
void push_up(int k){
	tree[k].sum = tree[k<<1].sum+tree[k<<1|1].sum;
}void push_down(int k){
	if(tree[k].lazy!=0){
		tree[k<<1].add(tree[k].lazy),tree[k<<1|1].add(tree[k].lazy);tree[k].lazy = 0;
	}
}void build(int l,int r,int k){
	tree[k].l =l ,tree[k].r = r;
	if(l+1 == r){
		return;
	}build(l,l+r>>1,k<<1),build(l+r>>1,r,k<<1|1);
}void update(int l,int r,int k,int add){
	int ll = tree[k].l,rr = tree[k].r;
	if(l>=rr or ll>=r)return;
	if(l<=ll and rr<=r){
		tree[k].add(add);return;
	}push_down(k);
	update(l,r,k<<1,add),update(l,r,k<<1|1,add);
	push_up(k);
}int query(int l,int r,int k){
	int ll= tree[k].l,rr = tree[k].r;
	if(l>=rr or ll>=r)return 0;
	if(l<=ll and rr<=r)return tree[k].sum;
	push_down(k);
	return query(l,r,k<<1)+query(l,r,k<<1|1);
}
int x[200005],y[200005],fa[200005],dep[200005],sz[200005];int nw=0;
void dfs1(int now,int f){
	fa[now] = f;dep[now] = dep[f]+1;dfn[now] = ++nw;sz[now] = 1;
	for(int i =0;i<p[now].size();i++){
		if(p[now][i] == f)continue;dfs1(p[now][i],now);sz[now]+=sz[p[now][i]];
	}return;
}int k,s;
int a[200005],tt[200005];
bool cmp(int a,int b){
	return dep[a]<dep[b];
}
#define lowbit(x) x&(-x)
int Tree[200005];
void upd(int pos,int add){
	for(int i = pos;i<=n;i+=lowbit(i))Tree[i]+=add;
}int qquery(int pos){
	int res =0;for(int i = pos;i>0;i-=lowbit(i))res+=Tree[i];return res;
}
void solve(){
	cin >> k;
	for(int i = 1;i<=k;i++){
		int aa;cin >> aa;
		if(dep[x[aa]]>dep[y[aa]])a[i] = x[aa];else a[i] = y[aa];
	}sort(a+1,a+1+k,cmp);cin>> s;
	update(1,n+1,1,-1);
	vector<int>pf,ps;
//	for(int i = 1;i<=k;i++){
//		if(query(dfn[a[i]],dfn[a[i]]+sz[a[i]],1) == 0){
//			pf.push_back(a[i]);
//			update(dfn[a[i]],dfn[a[i]]+sz[a[i]],1,1);
//		}
//	}
	update(1,n+1,1,-1);
//	for(int i= 1;i<=n;i++)cout<< query(i,i+1,1) << " ";cout << endl;
	for(int i =k;i>=1;i--){
		if(query(dfn[a[i]],dfn[a[i]]+sz[a[i]],1) == 0){
			ps.push_back(a[i]);
			update(dfn[a[i]],dfn[a[i]]+sz[a[i]],1,1);upd(dfn[a[i]],1);
		}
	}int l = 0,r = s;int mxx =0 ;
	for(int i = 1;i<=k;i++)mxx = max(mxx,qquery(dfn[a[i]]+sz[a[i]]-1)-qquery(dfn[a[i]]-1));
	for(int x:ps)upd(dfn[x],-1);
	//cout << mxx << " " << ps.size() << endl;
	while(l<r){
	//	cout << 111 << endl;
		int mid = l+r>>1;
		/*
			x-(n-x)<=s
			2x-n<=s
			x<=(n+s)/2
			(n-x)-x<=s
			n-2x<=s
			2x>=n-s
			x>=(n-s)/2
		*/
		int mn = (s-mid+1)/2,mx = (s+mid)/2;
	//	cout << " " << s << " " << mid << " " << mn << " " << mx << endl;
		if(mn*mxx>mx or mn*ps.size()>s)l = mid+1;
		else r = mid;
	}cout << r << '\n';
}
signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>> n >> q;
	for(int i = 1;i<n;i++){
		int a,b;cin >> a >> b;
		x[i] = a,y[i] =b;p[a].push_back(b),p[b].push_back(a);
	}dfs1(1,1);build(1,n+1,1);
	while(q--){
		solve();
	} 
	return 0;
}
/*
建虚树然后二分答案。
不想建虚树怎么办？
其实只需要先把极端的点都预处理出来。 
先预处理出叶子节点
我们考虑让子树的size在满足条件的情况下尽可能的小
然后把多余的点全部丢给1
  
*/