#include <map>
#include <set>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
#define fi first
#define se second
#define u1 (u<<1)
#define u2 (u<<1|1)
#define pb push_back
#define pp pop_back()
#define int long long
#define laile cout<<"laile"<<endl
#define lowbit(x) ((x)&(-x))
#define double long double
#define sf(x) scanf("%lld",&x)
#define sff(x,y) scanf("%lld %lld",&x,&y)
#define sd(x) scanf("%Lf",&x)
#define sdd(x,y) scanf("%Lf %Lf",&x,&y)
#define _for(i,n) for(int i=0;i<(n);++i)
#define _rep(i,a,b) for(int i=(a);i<=(b);++i)
#define _pre(i,a,b) for(int i=(a);i>=(b);--i)
#define all(x) (x).begin(), (x).end()
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
typedef unsigned long long ULL;
typedef pair<int,int>PII;
const int N=1e6+10,INF=4e18;
int n,m;
int fa[N],dep[N],top[N],sz[N],son[N],cnt[N];
vector<int>e[N];
void dfs1(int u, int father) { 
	fa[u] = father, sz[u] = 1, dep[u] = dep[father] + 1;
	for (auto j : e[u]) {
		if (j == father)continue;
		dfs1(j, u);
		sz[u] += sz[j];
		if (sz[son[u]] < sz[j])son[u] = j;
	}
	return;
}
void dfs2(int u, int v) { //找到u结点在重链上深度最小的顶点，所以可以找到每个点的链头O(n)
	top[u] = v; //记录链头
	if (!son[u])return; //没有重儿子,结束
	dfs2(son[u], v); //搜重儿子
	for (auto j : e[u]) {
		if (j == fa[u] || j == son[u])continue;
		dfs2(j, j); //搜轻儿子
	}
	return;
}
int lca(int u, int v) { //log(n)因为每一条链不会被切分成超过logn条链
	while (top[u] != top[v]) {
		if (dep[top[u]] < dep[top[v]])swap(u, v); //保证u的链头深度更深
		u = fa[top[u]];
	}
	return dep[u] < dep[v] ? u : v;
}
void solve()
{
	cin>>n>>m;
	_rep(i,2,n)
	{
		int a,b;
		cin>>a>>b;
		e[a].pb(b);
		e[b].pb(a);
	}
	dfs1(1,0);
	dfs2(1,1);
	while(m--)
	{
		cin>>n;
		bool bl=true;
		int pre=0;
		_rep(i,1,n)
			cin>>son[i];
		_rep(i,1,n)
		{
			cnt[son[i]]++;
			if(pre)
			{
				if(pre==fa[son[i]])
					cnt[pre]++;
				else
				{
					if(!fa[son[i]]||fa[son[i]]!=lca(son[i],pre)||cnt[pre]!=sz[pre])
					{
						bl=false;
						break;
					}
				}
			}
			pre=son[i];
		}
		_rep(i,1,n)cnt[son[i]]=0;
		if(bl)cout<<"Yes\n";
		else cout<<"No\n";
	}
	return ;
}
signed main()
{
	IOS;
	int T=1;
//	cin>>T;
	while(T--)
		solve();
	return 0;
}