#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int mod=998244353;
inline int inc(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline int dec(int x,int y){return x-y<0?x-y+mod:x-y;}
inline int mul(int x,int y){return (ll)x*y%mod;}
inline int qpow(int x,int y){
	int res=1;
	for(;y;y>>=1)res=y&1?mul(res,x):res,x=mul(x,x);
	return res;
}
inline int inv(int x){return qpow(x,mod-2);}

const int N=1<<20;
namespace NTT{
int re[N],w[2][N];
inline int getre(int n){
	int len=1,bit=0;
	while(len<n)len<<=1,++bit;
	for(int i=1;i<len;++i)re[i]=(re[i>>1]>>1)|((i&1)<<(bit-1));
	w[0][0]=w[1][0]=1,w[0][1]=qpow(3,(mod-1)/len),w[1][1]=inv(w[0][1]);
	for(int o=0;o<2;++o)for(int i=2;i<len;++i)
		w[o][i]=mul(w[o][i-1],w[o][1]);
	return len;
}
inline void NTT(int* a,int n,int o=0){
	for(int i=1;i<n;++i)if(i<re[i])swap(a[i],a[re[i]]);
	int R;
	for(int k=1;k<n;k<<=1)
		for(int i=0,t=k<<1,st=n/t;i<n;i+=t)
			for(int j=0,nw=0;j<k;++j,nw+=st)
				R=mul(a[i+j+k],w[o][nw]),a[i+j+k]=dec(a[i+j],R),a[i+j]=inc(a[i+j],R);
	if(o){
		R=inv(n);
		for(int i=0;i<n;++i)a[i]=mul(a[i],R);
	}
}

int t0[N],t1[N],t2[N];
inline void multi(const int* a,const int* b,int* c,int n,int m){
	if(n+m<=32){
		memset(c,0,sizeof(int)*(n+m+1));
		for(int i=0;i<=n;++i)
			for(int j=0;j<=m;++j)
				c[i+j]=inc(c[i+j],mul(a[i],b[j]));
		return;
	}
	int len=getre(n+m+1);
	memset(t0,0,sizeof(int)*len),memcpy(t0,a,sizeof(int)*(n+1));
	memset(t1,0,sizeof(int)*len),memcpy(t1,b,sizeof(int)*(m+1));
	NTT(t0,len),NTT(t1,len);
	for(int i=0;i<len;++i)t0[i]=mul(t0[i],t1[i]);
	NTT(t0,len,1);
	memcpy(c,t0,sizeof(int)*(n+m+1));
}
inline void inver(const int* a,int* b,int n){
	int len=1;
	while(len<=n)len<<=1;
	memset(t0,0,sizeof(int)*len),memcpy(t0,a,sizeof(int)*(n+1));
	memset(t1,0,sizeof(int)*(len<<1));
	memset(t2,0,sizeof(int)*(len<<1));
	t2[0]=inv(t0[0]);
	for(int k=1;k<=len;k<<=1){
		memcpy(t1,t0,sizeof(int)*k);
		getre(k<<1);
		NTT(t1,k<<1),NTT(t2,k<<1);
		for(int i=0;i<(k<<1);++i)t2[i]=mul(t2[i],dec(2,mul(t1[i],t2[i])));
		NTT(t2,k<<1,1);
		for(int i=k;i<(k<<1);++i)t2[i]=0;
	}
	memcpy(b,t2,sizeof(int)*(n+1));
}
} //namespace NTT

struct poly:public vector<int>{
	inline int time()const{return size()-1;}
	inline poly(int tim=0,int c=0){
		resize(tim+1);
		if(tim>=0)at(0)=c;
	}
    inline poly(int *l,int *r) {
        resize((r-l)+1);
        for(int i = 0;i < size();i++) at(i) = *(l + i);
    }
	inline poly operator%(const int& n)const{
		poly r(*this);
		r.resize(n);
		return r;
	}
	inline poly operator%=(const int& n){
		resize(n);
		return *this;
	}
	inline poly operator+(const poly& p)const{
		int n=time(),m=p.time();
		poly r(*this);
		if(n<m)r.resize(m+1);
		for(int i=0;i<=m;++i)r[i]=inc(r[i],p[i]);
		return r;
	}
	inline poly operator-(const poly& p)const{
		int n=time(),m=p.time();
		poly r(*this);
		if(n<m)r.resize(m+1);
		for(int i=0;i<=m;++i)r[i]=dec(r[i],p[i]);
		return r;
	}
	inline poly operator*(const poly& p)const{
		poly r(time()+p.time());
		NTT::multi(&((*this)[0]),&p[0],&r[0],time(),p.time());
		return r;
	}
	inline poly operator*(const int& k)const{
		poly r(*this);
		int n=time();
		for(int i=0;i<=n;++i)r[i]=mul(r[i],k);
		return r;
	}
	inline poly operator~()const{
		poly r(*this);
		reverse(r.begin(),r.end());
		return r;
	}
};

int f[1000005];
int n , k;
void solve(int l,int r) {
    if(l == r) {
        for(int i = (l + 1) / 2 ; i <= (l - i) + k && i < l; i++) {
            f[l] = (f[l] + 1LL * f[i] * f[l - i]) % mod;
        }
        // printf("%d %d\n",l,f[l]) ;
        return ;
    }
    int md = (l + r >> 1);
    solve(l , md);
    if(l != 1) {
        poly a(f + l , f + md) ;
        poly b(f + 1 , f + r - l) ;
        a = a * b;
        for(int i = md + 1 ; i <= r;i++) {
            f[i] = (f[i] + a[i - (l + 1)]) % mod;
        }
    }
    solve(md + 1 , r) ;
    poly a(f + l , f+md);
    poly b(f + md + 1 , f + r) ;
    a = a * b;
    for(int i = 0;i < a.size() && i + md + 1 <= n;i++) {
        // printf("on %d %d : %d %d: contri %d\n",l,md,md+1,r,i+l+md+1);
        if(i + l + md + 1 > r)
            f[i + l + md + 1] = (f[i + l + md + 1] + a[i]) % mod;
    }
    return ;
}
int main() {
     cin >> n >> k;
     f[1] = 1;
    solve(1 , n );
    cout << f[n] << '\n';
}