首先，有
[x^{overline n}=sum_kegin{bmatrix}{n\ k}end{bmatrix}x^{k}\
]
那么我们只需要求出(x^{overline n})即可，考虑倍增
[x^{overline 2n}=x^{overline n}(x+n)^{overline n}
]
假设我们现在已经求出了(x^{overline n})，考虑如何求出((x+n)^{overline n})
开始颓柿子
[egin{aligned}
f(x+n)
&=sum_{i}f_i(x+n)^i\
&=sum_{j}x^jsum_{i}f_i{ichoose j}n^{i-j}\
&=sum_{j}{x^jover j!}sum_{i}f_ii!{n^{i-j}over (i-j)!}\
end{aligned}
]
直接卷就可以了，再把它和原来的多项式卷积来即可
//quming
#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)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
const int P=167772161;
inline void upd(R int &x,R int y){(x+=y)>=P?x-=P:0;}
inline int inc(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))(y&1)?res=mul(res,x):0;
return res;
}
const int N=(1<<19)+5;
int fac[N],ifac[N],lg[N],r[25][N],rt[2][N],inv[25];
int lim,d;
inline void swap(R int &x,R int &y){R int t=x;x=y,y=t;}
inline int C(R int n,R int m){return m>n?0:1ll*fac[n]*ifac[m]%P*ifac[n-m]%P;}
void init(){
fac[0]=ifac[0]=1;fp(i,1,262144)fac[i]=mul(fac[i-1],i);
ifac[262144]=ksm(fac[262144],P-2);fd(i,262143,1)ifac[i]=mul(ifac[i+1],i+1);
fp(d,1,19){
fp(i,1,(1<<d)-1)r[d][i]=(r[d][i>>1]>>1)|((i&1)<<(d-1));
inv[d]=ksm(1<<d,P-2),lg[1<<d]=d;
}
for(R int t=(P-1)>>1,i=1,x,y;i<=262144;i<<=1,t>>=1){
x=ksm(3,t),y=ksm(55924054,t),rt[0][i]=rt[1][i]=1;
fp(k,1,i-1)
rt[1][i+k]=mul(rt[1][i+k-1],x),
rt[0][i+k]=mul(rt[0][i+k-1],y);
}
}
void NTT(int *A,int ty){
fp(i,0,lim-1)if(i<r[d][i])swap(A[i],A[r[d][i]]);
R int t;
for(R int mid=1;mid<lim;mid<<=1)
for(R int j=0;j<lim;j+=(mid<<1))
fp(k,0,mid-1)
A[j+k+mid]=dec(A[j+k],t=mul(rt[ty][mid+k],A[j+k+mid])),
A[j+k]=inc(A[j+k],t);
if(!ty){
t=inv[d];
fp(i,0,lim-1)A[i]=mul(A[i],t);
}
}
int f[N],n;
void solve(int *b,int len){
if(!len)return b[0]=1,void();
solve(b,len>>1);
lim=1,d=0;while(lim<=len)lim<<=1,++d;
int dm=(len>>1);
static int A[N],B[N];
for(R int i=0,c=1;i<=dm;++i,c=mul(c,dm))A[i]=mul(c,ifac[i]);
fp(i,0,dm)B[dm-i]=mul(b[i],fac[i]);
fp(i,dm+1,lim-1)A[i]=B[i]=0;
NTT(A,1),NTT(B,1);
fp(i,0,lim-1)A[i]=mul(A[i],B[i]);
NTT(A,0);
reverse(A,A+dm+1);
fp(i,0,dm)A[i]=mul(A[i],ifac[i]);fp(i,dm+1,lim-1)A[i]=0;
fp(i,0,dm)B[i]=b[i];fp(i,dm+1,lim-1)B[i]=0;
NTT(A,1),NTT(B,1);
fp(i,0,lim-1)A[i]=mul(A[i],B[i]);
NTT(A,0);
fp(i,0,len)b[i]=A[i];
if(len&1){
fd(i,len,1)b[i]=inc(mul(b[i],len-1),b[i-1]);
b[0]=mul(b[0],len-1);
}
}
int main(){
// freopen("testdata.in","r",stdin);
scanf("%d",&n);
init();
solve(f,n);
fp(i,0,n)printf("%d ",f[i]);
return 0;
}