Participated in this event gotta write something .jpg
Consider obtaining just elected a \ (i \) of Article disjoint paths after the first \ (i + 1 \) contribution at coincident with the previous path.
Note \ (F_i \) is selected in the tree \ (I \) number of disjoint paths program bar, then selecting the \ (I \) of the paths \ (i + 1 \) intersecting strips program number is \ (F_i G_i = \ Times \ dbinom {n-2} + {-f_. 1} + {I}. 1 \ Times (I +. 1) \) , \ (G_i \) contribution answer is \ (\ FRAC {G_i \ times I! \ times (I +. 1)} {\ dbinom {n-+. 1} {2} ^ {I +. 1}} \) (because the calculation of \ (G_i \) when we have identified the first \ (i + 1 \) edges, so here we just determined \ (I \) sequence to the edges)
Consider how to calculate \ (F_i \) . Remember \ (f_ {u, i, 0} \) represents the sub-tree \ (U \) have the (I \) \ Number Scheme Intact path, \ (F_ { u, i, 1} \) expressed \ (I \) there are a number of program paths extending upwardly while intact path. In order to facilitate the transfer and then provided assistance state \ (f_ {u, i, 2} \) represents the \ (U \) sub-tree root in the two paths (U \) \ converge into one path at the same time there \ (I \) the number of program paths. Specific transfer may look code, note processing path as well as a start and end point of a path.
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=100000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u];i;i=sq[i].nxt)
#define fir first
#define sec second
#define mp make_pair
#define pb push_back
#define maxd 998244353
#define eps 1e-8
inline int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
namespace My_Math{
#define N 100000
int fac[N+100],invfac[N+100];
int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
int dec(int x,int y) {return x<y?x-y+maxd:x-y;}
int mul(int x,int y) {return 1ll*x*y%maxd;}
ll qpow(ll x,int y)
{
ll ans=1;
while (y)
{
if (y&1) ans=mul(ans,x);
x=mul(x,x);y>>=1;
}
return ans;
}
int inv(int x) {return qpow(x,maxd-2);}
int C(int n,int m)
{
if ((n<m) || (n<0) || (m<0)) return 0;
return mul(mul(fac[n],invfac[m]),invfac[n-m]);
}
int math_init()
{
fac[0]=invfac[0]=1;
rep(i,1,N) fac[i]=mul(fac[i-1],i);
invfac[N]=inv(fac[N]);
per(i,N-1,1) invfac[i]=mul(invfac[i+1],i+1);
}
#undef N
}
using namespace My_Math;
struct node{int to,nxt;}sq[10010];
int all=0,head[5050];
int n,tmp[5050][3],g[5050][3],f[5050][5050][3],siz[5050];
void addedge(int u,int v)
{
all++;sq[all].to=v;sq[all].nxt=head[u];head[u]=all;
}
void dfs(int u,int fu)
{
go(u,i)
{
int v=sq[i].to;
if (v!=fu) dfs(v,u);
}
rep(i,0,n+1) rep(j,0,2) tmp[i][j]=0;tmp[0][0]=1;
go(u,i)
{
int v=sq[i].to;
if (v==fu) continue;
rep(j,0,siz[u]) rep(k,0,siz[v])
{
g[j+k][0]=add(g[j+k][0],mul(tmp[j][0],f[v][k][0]));
g[j+k][1]=add(g[j+k][1],mul(tmp[j][1],f[v][k][0]));
g[j+k][1]=add(g[j+k][1],mul(tmp[j][0],f[v][k][1]));
g[j+k][2]=add(g[j+k][2],mul(tmp[j][2],f[v][k][0]));
g[j+k+1][2]=add(g[j+k+1][2],mul(tmp[j][1],f[v][k][1]));
}
siz[u]+=siz[v];
rep(j,0,siz[u]) rep(k,0,2) {tmp[j][k]=g[j][k];g[j][k]=0;}
}
rep(i,0,siz[u])
{
f[u][i][0]=add(f[u][i][0],add(tmp[i][0],tmp[i][2]));
f[u][i+1][0]=add(f[u][i+1][0],add(tmp[i][0],tmp[i][1]));
f[u][i][1]=add(f[u][i][1],add(tmp[i][0],tmp[i][1]));
}
siz[u]++;
}
int main()
{
read();n=read();
math_init();
rep(i,1,n-1)
{
int u=read(),v=read();
addedge(u,v);addedge(v,u);
}
dfs(1,0);
ll ans=0,cn2=C(n+1,2),inv2=inv(cn2),sum=mul(inv2,inv2);
rep(i,1,n)
{
int tmp=dec(mul(f[1][i][0],cn2),mul(f[1][i+1][0],i+1));
ans=add(ans,mul(mul(tmp,i+1),mul(sum,fac[i])));
sum=mul(sum,inv2);
}
printf("%d",ans);
return 0;
}