计数使我快乐（
设 \(T_1,T_2\) 表示两棵树，\(E_1,E_2\) 表示它们的边集，\(k(E)\) 表示 \(E\) 中的边构成的森林的连通块个数。
对于 \({\rm op}=0\)：
这个是 PJ T2 难度吧（
对于 \({\rm op}=1\)：
首先略过 \(y=1\) 的平凡情况。
考虑对 \(E_1 \cap E_2\) 应用子集反演：
\[ \begin{aligned} \sum\limits_{E_2} y^{k(E_1 \cap E_2)} &= \sum\limits_{E_2} y^{n - |E_1 \cap E_2|} \\ &= \sum\limits_{E_2} \sum\limits_{S \subseteq E_1 \cap E_2} \sum\limits_{T \subseteq S} (-1)^{|S| - |T|} y^{n-|T|} \\ &= \sum\limits_{S\subseteq E_1} \left(\sum\limits_{T\subseteq S}(-1)^{|S|-|T|} y^{n-|T|}\right) \sum\limits_{S \subseteq E_2} 1 \\ &= \sum\limits_{S\subseteq E_1} y^{n-|S|} \left(\sum\limits_{T\subseteq S}(-y)^{|S|-|T|}\right) \sum\limits_{S \subseteq E_2} 1 \\ &= \sum\limits_{S\subseteq E_1} y^{n-|S|} (1-y)^{|S|} \sum\limits_{S \subseteq E_2} 1 \\ \end{aligned} \]
设 \(S\) 中的边构成 \(k=n-|S|\) 个连通块，其点数分别为 \(a_1,a_2,\dots,a_k\)。
那么有结论 \[
\sum\limits_{S \subseteq E_2} 1 = n^{k-2} \prod\limits_{i=1}^k a_i
\]
证明可以参见 rqy 的博客或者 shadowice1984 的题解，这里不再赘述了。
（Prufer 序列好！）
于是 \[ \begin{aligned} & \sum\limits_{S\subseteq E_1} y^{n-|S|} (1-y)^{|S|} \sum\limits_{S \subseteq E_2} 1 \\ =& \sum\limits_{S\subseteq E_1} y^{k} (1-y)^{n-k} n^{k-2} \prod\limits_{i=1}^k a_i \\ =& \frac{(1-y)^n}{n^2} \sum\limits_{S\subseteq E_1} \prod\limits_{i=1}^k \frac{ny}{1-y} a_i \end{aligned} \]
设 \(c = \frac{ny}{1-y}\)，考虑后一部分。
相当于在 \(T_1\) 中划分出 \(k\) 个连通块，第 \(i\) 个连通块有 \(ca_i\) 的贡献。
这一问题有显然的 \(O(n^2)\) 的树形 DP 解法。
而可以像 rqy 一样使用生成函数进行优化 DP 做到 \(O(n)\)。
但这里给出一个更为简单的做法。
考虑 \(ca_i\) 的组合意义，相当于在每个连通块中选一个点，每个点有 \(c\) 的贡献。
于是可以简单地设 \(f_{u,0/1}\) 表示 \(u\) 所在的连通块是否选出了这个点的贡献和。
同样是 \(O(n)\) 的。
对于 \({\rm op}=2\)：
类比上一部分，此处略去推导过程，则需要计算 \[
\frac{(1-y)^n}{n^4} \sum\limits_{S} \prod\limits_{i=1}^k \frac{n^2y}{1-y} a_i^2
\]
对于后一部分的和式，考虑枚举所有的 \(k\) 和 \(a_1,a_2,\dots,a_k\)。
\[
\begin{aligned}
& \sum\limits_{k=1}^n \sum\limits_{a_1+a_2+\dots+a_k=n} \frac{n!}{k!\prod_{i=1}^k a_i!} \prod\limits_{i=1}^k \frac{n^2y}{1-y} a_i^2 \cdot a_i^{a_i-2} \\
=& n! \sum\limits_{k=1}^n \frac1{k!} \sum\limits_{a_1+a_2+\dots+a_k=n} \prod\limits_{i=1}^k \frac{n^2y}{(1-y)a_i!} a_i^{a_i}
\end{aligned}
\]
考虑构造生成函数来计算这个东西，设 \[ F(x) = \frac{n^2y}{1-y} \sum\limits_{i=1}^{\infty} \frac{i^i}{i!} x^i \]
则答案等于 \[ \frac{(1-y)^nn!}{n^4} [x^n] \sum\limits_{k=1}^n \frac1{k!} F^k(x) \]
注意到这是一个 exp 的形式，所以答案就是 \[ \frac{(1-y)^nn!}{n^4} [x^n] \exp F(x) \]
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using namespace std;
const int N = 1e5;
const int mod = 998244353;
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
int n,y,op;
namespace Poly
{
const int N = 1 << 18;
const int G = 3;
int lg2[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init()
{
for(register int i = 2;i <= N;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / N);
rt[N >> 1] = 1;
for(register int i = (N >> 1) + 1;i <= N;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (N >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= N;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[N] = fpow(fac[N],mod - 2);
for(register int i = N;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= N;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
struct poly
{
vector<int> a;
inline poly(int x = 0)
{
x && (a.push_back(x),1);
}
inline poly(const vector<int> &o)
{
a = o,shrink();
}
inline void shrink()
{
for(;!a.empty() && !a.back();a.pop_back());
}
inline int size() const
{
return a.size();
}
inline void resize(int x)
{
a.resize(x);
}
inline int operator[](int x) const
{
if(x < 0 || x >= size())
return 0;
return a[x];
}
inline int &operator[](int x)
{
return a[x];
}
inline void clear()
{
vector<int>().swap(a);
}
inline void ntt(int type = 1)
{
int n = size();
type == -1 && (reverse(a.begin() + 1,a.end()),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
friend inline poly operator+(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = add(a[i],b[i]);
return poly(ret);
}
friend inline poly operator-(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = dec(a[i],b[i]);
return poly(ret);
}
friend inline poly operator*(poly a,poly b)
{
if(a.a.empty() || b.a.empty())
return poly();
int lim = 1,tot = a.size() + b.size() - 1;
for(;lim < tot;lim <<= 1);
a.resize(lim),b.resize(lim);
a.ntt(),b.ntt();
for(register int i = 0;i < lim;++i)
a[i] = (long long)a[i] * b[i] % mod;
a.ntt(-1),a.shrink();
return a;
}
poly &operator+=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = add(a[i],o[i]);
return *this;
}
poly &operator-=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = dec(a[i],o[i]);
return *this;
}
poly &operator*=(poly o)
{
return (*this) = (*this) * o;
}
poly deriv() const
{
if(a.empty())
return poly();
vector<int> ret(size() - 1);
for(register int i = 0;i < size() - 1;++i)
ret[i] = (long long)(i + 1) * a[i + 1] % mod;
return poly(ret);
}
poly integ() const
{
if(a.empty())
return poly();
vector<int> ret(size() + 1);
for(register int i = 0;i < size();++i)
ret[i + 1] = (long long)a[i] * inv[i + 1] % mod;
return poly(ret);
}
inline poly modxn(int n) const
{
n = min(n,size());
return poly(vector<int>(a.begin(),a.begin() + n));
}
inline poly inver(int m) const
{
poly ret(fpow(a[0],mod - 2));
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (2 - modxn(k) * ret)).modxn(k);
return ret.modxn(m);
}
inline poly log(int m) const
{
return (deriv() * inver(m)).integ().modxn(m);
}
inline poly exp(int m) const
{
poly ret(1);
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (1 - ret.log(k) + modxn(k))).modxn(k);
return ret.modxn(m);
}
};
}
using Poly::init;
using Poly::poly;
namespace Task0
{
unordered_map<int,unordered_set<int>> vis;
int ans,k;
void main()
{
k = n;
int u,v;
for(register int i = 2;i <= n;++i)
scanf("%d%d",&u,&v),u > v && (swap(u,v),1),vis[u].insert(v);
for(register int i = 2;i <= n;++i)
scanf("%d%d",&u,&v),u > v && (swap(u,v),1),k -= vis.count(u) && vis[u].count(v);
ans = fpow(y,k);
printf("%d\n",ans);
}
}
namespace Task1
{
int to[(N << 1) + 5],pre[(N << 1) + 5],first[N + 5];
inline void add_edge(int u,int v)
{
static int tot = 0;
to[++tot] = v,pre[tot] = first[u],first[u] = tot;
}
int fa[N + 5];
int c,ans,f[N + 5][2];
void dfs(int p)
{
f[p][0] = 1,f[p][1] = c;
for(register int i = first[p];i;i = pre[i])
if(to[i] ^ fa[p])
fa[to[i]] = p,
dfs(to[i]),
f[p][1] = ((long long)f[p][0] * f[to[i]][1] + (long long)f[p][1] * f[to[i]][0] + (long long)f[p][1] * f[to[i]][1]) % mod,
f[p][0] = ((long long)f[p][0] * f[to[i]][0] + (long long)f[p][0] * f[to[i]][1]) % mod;
}
void main()
{
if(y == 1)
{
printf("%d\n",fpow(n,n - 2));
return ;
}
int u,v;
for(register int i = 2;i <= n;++i)
scanf("%d%d",&u,&v),add_edge(u,v),add_edge(v,u);
c = (long long)n * y % mod * fpow(dec(1,y),mod - 2) % mod;
dfs(1);
ans = (long long)fpow(dec(1,y),n) * fpow(n,mod - 3) % mod * f[1][1] % mod;
printf("%d\n",ans);
}
}
namespace Task2
{
poly f;
int c,ans;
void main()
{
if(y == 1)
{
printf("%d\n",fpow(n,2 * n - 4));
return ;
}
init();
c = (long long)n * n % mod * y % mod * fpow(dec(1,y),mod - 2) % mod;
f.resize(n + 1);
for(register int i = 1;i <= n;++i)
f[i] = (long long)c * fpow(i,i) % mod * Poly::ifac[i] % mod;
f = f.exp(n + 1),ans = (long long)fpow(dec(1,y),n) * Poly::fac[n] % mod * fpow(n,mod - 5) % mod * f[n] % mod;
printf("%d\n",ans);
}
}
int main()
{
scanf("%d%d%d",&n,&y,&op);
if(!op)
Task0::main();
else if(op == 1)
Task1::main();
else
Task2::main();
}