思路：

RT

懶得寫了

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=(1<<18)+5,mod=1004535809;
int tmp[N],R[N],fac[N],A[N],B[N],C[N],niB[N];
int pow(ll x,ll y){
    ll res=1;
    while(y){
        
 
if(y&1)res=res*x%mod;
        x=x*x%mod,y>>=1;
    }return (int)res;
}
void NTT(int *a,int n,int f){
    int m=1,L=0;
    for(;m<n;m<<=1)L++;
    for(int i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    for(int i=0;i<n;i++)if(i<R[i])swap(a[i],a[R[i]]);
    
 
for(int l=1;l<n;l<<=1){
        int wn=pow(3,((mod-1)/(l<<1)*f+mod-1)%(mod-1));
        for(int j=0;j<n;j+=(l<<1)){
            int w=1;
            for(int k=0;k<l;k++,w=1ll*w*wn%mod){
                int x=a[j+k],y=1ll*w*a[j+k+l]%mod;
                a[j+k]=(x+y)%mod,a[j+k+l]=(x-y+mod)%mod;
            }
        }
    }
    
 
if(f==-1){
        int ni=pow(n,mod-2);
        for(int i=0;i<n;i++)a[i]=1ll*a[i]*ni%mod;
    }
}
void get_inv(int *a,int *b,int n){
    if(n==1){b[0]=pow(a[0],mod-2);return;}
    get_inv(a,b,n>>1);
    memcpy(tmp,a,sizeof(int)*n);memset(tmp+n,0,sizeof(int)*n);
    NTT(tmp,n<<1,1),NTT(b,n<<1,1);
    for(int i=0;i<n<<1;i++)tmp[i]=((1ll*b[i]*(2-1ll*tmp[i]*b[i]%mod))%mod+mod)%mod;
    NTT(tmp,n<<1,-1);
    memcpy(b,tmp,sizeof(int)*n);memset(b+n,0,sizeof(int)*n);
}
signed main(){
    int n,m,i;
    scanf("%d",&n);for(m=1;m<=n;m<<=1);
    for(fac[0]=1,i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
    for(i=0;i<=n;i++)B[i]=1ll*pow(2,(1ll*i*(i-1)/2)%(mod-1))*pow(fac[i],mod-2)%mod;
    for(i=0;i<=n;i++)C[i]=1ll*pow(2,(1ll*i*(i-1)/2)%(mod-1))*pow(fac[i-1],mod-2)%mod;
    get_inv(B,niB,m),NTT(niB,m,1);NTT(C,m,1);
    for(int i=0;i<m;i++)A[i]=1ll*niB[i]*C[i]%mod;
    NTT(A,m,-1);
    printf("%lld\n",1ll*A[n]*fac[n-1]%mod);
}

BZOJ 3456 NTT圖的計數 容斥