# Preface

Why can't I do anything? He mang guides the blog to do it.

# Title Solution

We have three sub problems to consider.

## Sub problem 0

Connect the red and blue common edges. Each connected block has the same color, and different connected blocks are independent.

The answer is \ (y ^ {number of connected blocks} \).

## Sub problem 1

For a connection scheme of mangrove tree, suppose that there are some edges on the blue tree, and if \ (i \) edges are connected, then the contribution to the answer is as follows:

$y ^ n / y ^ i$

order

$z = \frac 1 y$

According to binomial theorem

$z ^ a = \sum_{i=0}^a \binom{a}{i} (z-1)^i$

So the contribution is

$\sum_{j=0}^{n-i} \binom{n-i}{j} (z -1) ^ j$

The meaning of combination is to enumerate the subsets of all sides to calculate the answer.

$y ^ n \sum_{i = 0} ^ {n-1} (z-1) ^ j \sum n ^ {n-i-2} \prod_k a_k$

Where \ (a_k \) represents the size of the \ (K \) connected block.

Consider further expanding the meaning of combination:

$$\ prod \\\\\\\\\\\\\\\\. Just transfer it vigorously. Time complexity \ (O(n)$$.

## Sub problem 2

Consider the formula for writing the answer

$ans = y ^ n \sum_{i=1}^ n (z- 1 ) ^ {n-i} \frac{n!}{i!\prod a_j!}\left (\prod a_j^{a_j} \right)(n ^ {i-2}) ^ 2 \\ = y ^ n n ^ {-4} (z-1) ^ n \sum_{i=1}^ n \frac{n!}{i!\prod a_j!}\prod (a_j^{a_j} (z-1)^ {-1}n ^ 2)$

Be aware

$\sum_{i=1}^ n \frac{n!}{i!\prod a_j!}\prod (a_j^{a_j} (z-1)^ {-1}n ^ 2) = [n] exp(\sum_{i\geq 1 } a_j^{a_j} (z-1)^ {-1}n ^ 2\frac{x^i}{i!})$

Therefore, the polynomial exp can be used to solve this problem in the time complexity of \ (O(n\log n) \).

# Code

#include <bits/stdc++.h>
#define clr(x) memset(x,0,sizeof x)
#define For(i,a,b) for (int i=(a);i<=(b);i++)
#define Fod(i,b,a) for (int i=(b);i>=(a);i--)
#define fi first
#define se second
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define outval(x) cerr<<#x" = "<<x<<endl
#define outtag(x) cerr<<"---------------"#x"---------------"<<endl
#define outarr(a,L,R) cerr<<#a"["<<L<<".."<<R<<"] = ";\
For(_x,L,R)cerr<<a[_x]<<" ";cerr<<endl;
using namespace std;
typedef long long LL;
LL x=0,f=0;
char ch=getchar();
while (!isdigit(ch))
f|=ch=='-',ch=getchar();
while (isdigit(ch))
x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
return f?-x:x;
}
const int mod=998244353;
int Pow(int x,int y){
if (y<0)
x=Pow(x,mod-2),y=-y;
int ans=1;
for (;y;y>>=1,x=(LL)x*x%mod)
if (y&1)
ans=(LL)ans*x%mod;
return ans;
}
if ((x+=y)>=mod)
x-=mod;
}
void Del(int &x,int y){
if ((x-=y)<0)
x+=mod;
}
return x>=mod?x-mod:x;
}
int Del(int x){
return x<0?x+mod:x;
}
const int N=(1<<19)+1;
int Fac[N],Inv[N],Iv[N];
void getFI(){
int n=N-1;
for (int i=Fac[0]=1;i<=n;i++)
Fac[i]=(LL)Fac[i-1]*i%mod;
Inv[n]=Pow(Fac[n],mod-2);
Fod(i,n,1)
Inv[i-1]=(LL)Inv[i]*i%mod;
For(i,1,n)
Iv[i]=(LL)Inv[i]*Fac[i-1]%mod;
}
namespace fft{
int w[N],R[N];
void init(int n){
int d=0;
while ((1<<d)<n)
d++;
For(i,0,n-1)
R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
w[0]=1,w[1]=Pow(3,(mod-1)/n);
For(i,2,n-1)
w[i]=(LL)w[i-1]*w[1]%mod;
}
void FFT(int *a,int n,int flag){
if (flag<0)
reverse(w+1,w+n);
For(i,0,n-1)
if (i<R[i])
swap(a[i],a[R[i]]);
for (int t=n>>1,d=1;d<n;d<<=1,t>>=1)
for (int i=0;i<n;i+=d<<1)
for (int j=0;j<d;j++){
int tmp=(LL)w[t*j]*a[i+j+d]%mod;
a[i+j+d]=Del(a[i+j]-tmp);
}
if (flag<0){
reverse(w+1,w+n);
int inv=Pow(n,mod-2);
For(i,0,n-1)
a[i]=(LL)a[i]*inv%mod;
}
}
}
using fft::FFT;
typedef vector <int> vi;
vi Fix(vi a,int n){
while (a.size()>n)
a.pop_back();
while (a.size()<n)
a.pb(0);
return a;
}
vi operator * (vi a,vi b){
int s=(int)a.size()+b.size()-1,n=1;
while (n<s)
n<<=1;
a=Fix(a,n),b=Fix(b,n);
fft::init(n);
FFT(&a[0],n,1),FFT(&b[0],n,1);
For(i,0,n-1)
a[i]=(LL)a[i]*b[i]%mod;
FFT(&a[0],n,-1);
return Fix(a,s);
}
vi operator + (vi a,vi b){
int s=max(a.size(),b.size());
a=Fix(a,s),b=Fix(b,s);
For(i,0,s-1)
return a;
}
vi operator - (vi a,vi b){
int s=max(a.size(),b.size());
a=Fix(a,s),b=Fix(b,s);
For(i,0,s-1)
Del(a[i],b[i]);
return a;
}
vi pInv(vi a){
if (a.size()==1)
return (vi){Pow(a[0],mod-2)};
int n=a.size();
vi b=pInv(Fix(a,(n+1)>>1));
return Fix(b+b-b*b*a,n);
}
vi Der(vi a){
int n=a.size();
For(i,0,n-2)
a[i]=(LL)a[i+1]*(i+1)%mod;
return Fix(a,n-1);
}
vi Int(vi a){
int n=a.size();
a.pb(0);
Fod(i,n,1)
a[i]=(LL)a[i-1]*Iv[i]%mod;
a[0]=0;
return a;
}
vi Ln(vi a){
return Int(Fix(Der(a)*pInv(a),a.size()-1));
}
vi Exp(vi a){
if (a.size()==1)
return (vi){1};
int n=a.size();
vi b=Fix(Exp(Fix(a,(n+1)>>1)),n);
return Fix(b*((vi){1}-Ln(b)+a),n);
}
int n,z,op;
namespace so0{
map <pair <int,int>,int> Map;
int main(){
Map.clear();
For(i,1,n-1){
if (x>y)
swap(x,y);
Map[mp(x,y)]=1;
}
int c=n;
For(i,1,n-1){
if (x>y)
swap(x,y);
c-=Map[mp(x,y)];
}
cout<<Pow(z,c)<<endl;
return 0;
}
}
namespace so1{
int inv_n,izn;
vector <int> e[N];
int size[N];
int dp[N][2];
void dfs(int x,int pre){
dp[x][0]=dp[x][1]=1;
for (auto y : e[x])
if (y!=pre){
dfs(y,x);
int t0=dp[x][0],t1=dp[x][1];
dp[x][0]=(LL)t0*dp[y][1]%mod;
dp[x][1]=(LL)t1*dp[y][1]%mod;
}
}
int main(){
if (z==1){
cout<<Pow(n,n-2)<<endl;
return 0;
}
inv_n=Pow(n,mod-2);
izn=Del(Pow(z,mod-2)-1);
izn=(LL)izn*inv_n%mod;
For(i,1,n-1){
e[x].pb(y),e[y].pb(x);
}
dfs(1,0);
int ans=(LL)dp[1][1]*Pow(n,n-2)%mod*Pow(z,n)%mod;
cout<<ans<<endl;
return 0;
}
}
namespace so2{
int main(){
if (z==1){
cout<<Pow(n,(n-2)*2)<<endl;
return 0;
}
getFI();
int iz=Del(Pow(z,mod-2)-1),tmp=(LL)Pow(iz,-1)*n%mod*n%mod;
vi a;
a.pb(0);
For(i,1,n)
a.pb((LL)Pow(i,i)*tmp%mod*Inv[i]%mod);
a=Exp(a);
int ans=(LL)a[n]*Fac[n]%mod;
ans=(LL)ans*Pow(z,n)%mod*Pow(iz,n)%mod*Pow(n,-4)%mod;
cout<<ans<<endl;
return 0;
}
}
int main(){
}