PKUWC2018 随机游走
Min-Max容斥（又称最值反演）：
$\max(S)=\sum_{T\subset S}\min(T)(-1)^{|T|+1}$
大概的证明：
假设$S$中元素排序后为$a_1\dots a_n$，则$\max(S)=a_n$，$a_i$对答案的贡献实际上是$[i=n]$，因为每个$i<j\le n$，j在不在T中对答案的贡献会相互抵消。
所以只需要计算x第一次到每个点集$T$的期望步数$f_x$。
这个显然可以高斯消元……
考虑把$f_x$表示成$a_xf_{fa_x}+b_x$的形式，如果以$x$为根dp，则$f_x=b_x$。
这就可以$O(n)$单次了。
至于容斥？
首先取个反，然后 $O(n2^n)$做一次快速莫比乌斯变换就行了。
时间复杂度$O(n2^n)$。
/*
Author: CNYALI_LK
LANG: C++
PROG: 2542.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const signed inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline void read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (char &x) {
x=gc();
}
inline void read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r');
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
}
template<typename A,typename ...B>
inline void read(A &x,B &...y){
read(x);read(y...);
}
// print a signed integer
inline void write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
}
inline void write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
}
inline void write (char x) {
putc(x);
}
inline void write(const char *x){
while(*x){putc(*x);++x;}
}
inline void write(char *x){
while(*x){putc(*x);++x;}
}
template<typename A,typename ...B>
inline void write(A x,B ...y){
write(x);write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int fa[21],fb[21],deg[21],f[1<<18],inv[21];
vector<int> to[21];
int n,q,x,u,v,se;
const int p=998244353;
int sum(int a,int b){return (a+=b)>=p?a-p:a;}
int fpm(int a,int b){int c=1;for(;b;b>>=1,a=(ll)a*a%p)if(b&1)c=(ll)c*a%p;return c;}
void dfs(int x,int f){
// printf("%d!\n",x);
if((1<<(x-1)) & se){fa[x]=fb[x]=0;return;}
fa[x]=0;
fb[x]=0;
int stf=0;
for(auto i:to[x])if(i!=f){
dfs(i,x);
stf=sum(stf,fa[i]);
fb[x]=sum(fb[x],fb[i]);
}
stf=fpm(sum(1,p-(ll)stf*inv[deg[x]]%p),p-2);
fb[x]=((ll)fb[x]*inv[deg[x]]+1)%p*stf%p;
fa[x]=(ll)inv[deg[x]]*stf%p;
}
void calc(){
dfs(x,0);
f[se]=fb[x];
if(!(__builtin_popcount(se)&1))f[se]=(p-f[se])%p;
}
void subset(){
for(int i=1;i<1<<n;i<<=1){
for(int j=0;j<1<<n;j+=i+i){
for(int k=0;k<i;++k)f[j+k+i]=sum(f[j+k+i],f[j+k]);
}
}
}
int main(){
#ifdef cnyali_lk
freopen("2542.in","r",stdin);
freopen("2542.out","w",stdout);
#endif
read(n,q,x);
for(int i=1;i<n;++i){
read(u,v);
++deg[u];++deg[v];
to[u].push_back(v);
to[v].push_back(u);
}
inv[0]=inv[1]=1;
for(int i=2;i<=n;++i)inv[i]=(ll)(p-p/i)*inv[p%i]%p;
for(se=1;se<1<<n;++se){
calc();
}
subset();
for(;q;--q){
read(u);
v=0;
for(;u;--u){read(x);v|=1<<(x-1);}
printf("%d\n",f[v]);
}
return 0;
}