题意：有一个集合，求有多少形态不同的二叉树满足每个点的权值都属于这个集合并且总点权等于i
题解：先用生成函数搞出来(f(x)=f(x)^2*c(x)+1)
然后转化一下变成(f(x)=frac{2}{1+sqrt{1-4*c(x)}})
然后多项式开根和多项式求逆即可(先对下面的项开根，然后再求逆)
多项式开根：
(B(x)^2=A(x) mod x^{ lfloor frac{n}{2}
floor})
(B'(x)^2=A(x) mod x^{ lfloor frac{n}{2}
floor})
(B(x)^2-B'(x)^2equiv 0),((B(x)+B'(x))*(B(x)-B'(x))equiv 0)，取(B(x)=B'(x))
(B(x)^2-2*B(x)*B'(x)+B'(x)^2equiv0),
(B(x)equiv frac{A(x)+B'(x)^2}{2*B'(x)})
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f;
ll a[N<<3],b[N<<3],c[N<<3],d[N<<3],tmp[N<<3],inv2=qp(2,mod-2);
int rev[N<<3];
void getrev(int bit)
{
for(int i=0;i<(1<<bit);i++)
rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void ntt(ll *a,int n,int dft)
{
for(int i=0;i<n;i++)
if(i<rev[i])
swap(a[i],a[rev[i]]);
for(int step=1;step<n;step<<=1)
{
ll wn=qp(3,(mod-1)/(step*2));
if(dft==-1)wn=qp(wn,mod-2);
for(int j=0;j<n;j+=step<<1)
{
ll wnk=1;
for(int k=j;k<j+step;k++)
{
ll x=a[k];
ll y=wnk*a[k+step]%mod;
a[k]=(x+y)%mod;a[k+step]=(x-y+mod)%mod;
wnk=wnk*wn%mod;
}
}
}
if(dft==-1)
{
ll inv=qp(n,mod-2);
for(int i=0;i<n;i++)a[i]=a[i]*inv%mod;
}
}
void pol_inv(int deg,ll *a,ll *b)
{
if(deg==1){b[0]=qp(a[0],mod-2);return ;}
pol_inv((deg+1)>>1,a,b);
int sz=0;while((1<<sz)<=deg)sz++;
getrev(sz);int len=1<<sz;
for(int i=0;i<deg;i++)tmp[i]=a[i];
for(int i=deg;i<len;i++)tmp[i]=0;
ntt(tmp,len,1),ntt(b,len,1);
for(int i=0;i<len;i++)
b[i]=(2ll-tmp[i]*b[i]%mod+mod)%mod*b[i]%mod;
ntt(b,len,-1);
for(int i=deg;i<len;i++)b[i]=0;
}
void pol_sqrt(int deg,ll *a,ll *b)
{
if(deg==1){b[0]=1;return ;}
pol_sqrt((deg+1)>>1,a,b);
int sz=0;while((1<<sz)<=deg)sz++;
getrev(sz);int len=1<<sz;
for(int i=0;i<len;i++)d[i]=0;
pol_inv(deg,b,d);
for(int i=0;i<deg;i++)tmp[i]=a[i];
for(int i=deg;i<len;i++)tmp[i]=0;
ntt(tmp,len,1),ntt(b,len,1),ntt(d,len,1);
for(int i=0;i<len;i++)
b[i]=(tmp[i]*d[i]%mod+b[i])%mod*inv2%mod;
ntt(b,len,-1);
for(int i=deg;i<len;i++)b[i]=0;
}
int main()
{
int n,m;scanf("%d%d",&n,&m);
for(int i=0,x;i<n;i++)
{
scanf("%d",&x);
a[x]=mod-4;
}
a[0]=1;
int len=1;while(len<=m)len<<=1;
pol_sqrt(len,a,b);
++b[0];
pol_inv(len,b,c);
for(int i=1;i<=m;i++)printf("%lld
",c[i]*2%mod);
return 0;
}
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