简单的题，只要快速幂和乘法逆元。要开O2!!! 代码：

#include<bits/stdc++.h>
using namespace std;
const unsigned long long MOD=998244353;//模数
unsigned long long sumjc=1,sumin=1;//阶乘
unsigned long long f(unsigned long long x,unsigned long long y){ //公约数
    if (y==0){
        return x;
    }
    return f(y,x%y)%MOD;
}
void jc(){//计算阶乘
    sumjc=(sumin*sumjc)%MOD;
    sumin++;
}
unsigned long long Pow(unsigned long long a, unsigned long long b) { //快速幂
  unsigned long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a)%MOD;
    a = (a * a)%MOD;
    b >>= 1;
  }
  return res;
}
unsigned long long inv(unsigned long long a, unsigned long long p) {//乘法逆元
    return Pow(a,p-2,p);
}
 
int main(){
    freopen("math.in","r",stdin);//文件输入
    freopen("math.out","w",stdout);//文件输出
    unsigned long long n,k,abcd;
    cin>>n>>k;
    unsigned long long fs[2]={0,1},fs1[2];
    for (unsigned int i=1;i<=n;i++){
        memset(0,fs1,sizeof fs1);
        fs1[0]=sumjc;
        fs1[1]=Pow(i,k);
        //加法
        fs[0]=(fs[0]*fs1[1]+fs[1]*fs1[0])%MOD;
        fs[1]=(fs[1]*fs1[1])%MOD;
    }
    //约分
    abcd=f(fs[0],fs[1]);
    fs[0]=fs[0]/abcd%MOD;
    fs[1]=fs[1]/abcd%MOD;
    cout<<(fs[0]*inv(fs[1],MOD))%MOD;//计算逆元
    return 0;
}