首先，因为在(j>i)的时候有(S(i,j)=0)，所以原式可以写成$$Ans=sum_{i=0}^nsum_{j=0}^nS(i,j) imes 2^j imes j!$$
[Ans=sum_{j=0}^n2^j imes j!sum_{i=0}^nS(i,j)
]
根据第二类斯特林数的通项公式代入，有$$Ans=sum_{j=0}^n2^j imes j!sum_{i=0}^nsum_{k=0}^jfrac{(-1)^k}{k!}frac{(j-k)^i}{(j-k)!}$$
[Ans=sum_{j=0}^n2^j imes j!sum_{k=0}^jfrac{(-1)^k}{k!}frac{sum_{i=0}^n(j-k)^i}{(j-k)!}
]
根据等比数列求和公式，知(sum_{i=0}^np^i=frac{p^n-1}{p-1})，于是设(f_i=frac{(-1)^i}{i!},g_i=frac{sum_{k=0}^ni^k}{i!})，则$$Ans=sum_{j=0}^n2^j imes j!(f imes g)(j-k)$$
(NTT)计算即可
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
while(z[++Z]=x%10+48,x/=10);
while(sr[++C]=z[Z],--Z);sr[++C]='
';
}
const int N=5e5+5,P=998244353,Gi=332748118;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
R int res=1;
for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);
return res;
}
int A[N],B[N],O[N],r[N],fac[N],inv[N];
int n,lim,l,res;
void NTT(int *A,int ty){
fp(i,0,lim-1)if(i<r[i])swap(A[i],A[r[i]]);
for(R int mid=1;mid<lim;mid<<=1){
R int I=(mid<<1),Wn=ksm(ty==1?3:Gi,(P-1)/I);O[0]=1;
fp(i,1,mid-1)O[i]=mul(O[i-1],Wn);
for(R int j=0;j<lim;j+=I)for(R int k=0;k<mid;++k){
int x=A[j+k],y=mul(O[k],A[j+k+mid]);
A[j+k]=add(x,y),A[j+k+mid]=dec(x,y);
}
}if(ty==-1)for(R int i=0,inv=ksm(lim,P-2);i<lim;++i)A[i]=mul(A[i],inv);
}
int main(){
// freopen("testdata.in","r",stdin);
scanf("%d",&n);
fac[0]=inv[0]=1;fp(i,1,n)fac[i]=mul(fac[i-1],i);
inv[n]=ksm(fac[n],P-2);fd(i,n-1,1)inv[i]=mul(inv[i+1],i+1);
fp(i,0,n){
A[i]=i&1?P-inv[i]:inv[i];
if(i!=1)B[i]=mul(dec(ksm(i,n+1),1),mul(inv[i],ksm(dec(i,1),P-2)));
else B[i]=n+1;
}lim=1;while(lim<=n+n)lim<<=1,++l;
fp(i,0,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
NTT(A,1),NTT(B,1);
fp(i,0,lim-1)A[i]=mul(A[i],B[i]);
NTT(A,-1);
for(R int i=0,j=1;i<=n;++i,j=add(j,j))res=add(res,mul(j,mul(fac[i],A[i])));
printf("%d
",res);return 0;
}