rt,思路是不难发现黑球落在 2∼n 的概率都一样,然后设 fi 表示考虑操作 i 次后,黑球在 1 的概率
//when you use vector or deque,pay attention to the size of it.
//by OldDirverTree
#include<bits/stdc++.h>
#include<atcoder/all>
#define P pair<int,int>
#define int long long
#define mid (l+r>>1)
using namespace std;
using namespace atcoder;
const int mod=998244353;
using mint=modint998244353;
const mint inv2=mod-mod/2;
const int N=1e5+1;
mint f[N];
int read() {
int x=0; bool f=true; char c=0;
while (!isdigit(c) ) f&=(c!='-'),c=getchar();
while (isdigit(c) ) x=(x<<3)+(x<<1)+(c&15),c=getchar();
return f?x:-x;
}
main()
{
mint n=read(); int m=read();
f[0]=1; mint v=2*(n-1)/n/n;
for (int i=1;i<=m;i++) f[i]=f[i-1]*(1-v)+v*(1-f[i-1]);
mint ans=f[m]+(1-f[m])/(n-1)*(n+2)*inv2;
return !printf("%d",ans.val() );
}