#include<bits/stdc++.h>
//#pragma GCC optimize(3)
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define int long long
#define ll long long
#define ull unsigned ll
#define db double
#define ldb long db
#define mp make_pair
#define fi first
#define se second
#define pii pair<int,int>
#define pbI push_back
#define inf 1e9
#define mdI int mid=(l+r)>>1
#define lowbit(x) (x&-x)
#define rep(a,b,c) for(int a=b;a<=c;a++)
#define MAXN 520
#define gint __int128
#define MOD 998244353
int qread(){
	char o=getchar();int f=1,x=0;
	for(;!isdigit(o);o=getchar())if(o=='-')f*=-1;
	for(;isdigit(o);o=getchar())x=x*10+o-'0';
	return f*x;
}
void chmx(int &x,int y){if(y>x)x=y;}
void chmi(int &x,int y){if(y<x)x=y;}
bool FLA;
namespace MathPart{
	#define rep(a,b,c) for(int a=b;a<=c;a++)
	#define vep(a,b,c) for(int a=b;a>=c;a--)
	void exgcd(ll a,ll b,ll &x,ll &y){if(b==0){x=1,y=0;return;}exgcd(b,a%b,y,x);y-=(a/b)*x;}
	ll inv(ll a,ll p=MOD){ll x,y;exgcd(a,p,x,y);return (x+p)%p;}
	gint qp(ll a,ll b,ll p=MOD){gint rs=1,bs=a;while(b){if(b&1){rs=rs*bs%p;}bs=bs*bs%p;b>>=1;}return rs;}
	ll fac[MAXN],ifac[MAXN];
	void initF(ll n,ll p=MOD){
		fac[0]=1;rep(i,1,n){fac[i]=fac[i-1]*i%p;}
		ifac[n]=inv(fac[n],p);vep(i,n-1,0){ifac[i]=ifac[i+1]*(i+1)%p;}
	}
	ll C(ll n,ll m){if(n<m){return 0;}return fac[n]*ifac[m]%MOD*ifac[n-m]%MOD;}
	ll CC(ll n,ll m,ll p=MOD){ll rs=1;rep(i,0,m-1){rs=rs*(n-i)%p*inv(i+1,MOD)%p;}return rs;}
	ll S2OL(ll n,ll m){
		ll rs=0,d=1,f=1;
		for(ll k=0;k<=m;k++){
			rs=(rs+f*C(m,k)*qp(m-k,n,MOD)%MOD+MOD)%MOD;
			d=d*max(k,1ll)%MOD;f*=-1;
		}return rs*inv(d,MOD)%MOD;
	}
	ll phi(ll a){ll rs=a,m=sqrt(a);
		rep(i,2,m){if(a%i==0){rs=rs/i*(i-1);while(a%i==0){a/=i;}}}
		if(a>1){rs=rs/a*(a-1);}return rs;
	}
	bool MR(ll pr){if(pr<2)return 0;if(pr==3)return 1;if(pr==2)return 1;
		ll a=pr-1,b=0;while(!(a&1))a>>=1,b++;
		for(int i=1;i<=8;++i){
		    ll x=rand()%(pr-2)+2,v=qp(x,a,pr),j;if(v==1||v==pr-1)continue;
		    for(j=0;j<b-1;j++){v=(gint)v*v%pr;if(v==pr-1)break;}
			if(v!=pr-1)return 0;
		}return 1;
	}
}
//using namespace MathPart;
int n,m,Pre;
int get(int m){
    if(m==1){return Pre;}
    if(m==0){return 1;}
    if(m&1){return (get(1)*get(m/2))%MOD;}
    else{return (get(m/2-1)+get(m/2))%MOD;}
}
struct Mat{
    int a[4][4];
    Mat(){memset(a,0,sizeof(a));}
    void one(){rep(i,0,3)a[i][i]=1;}
    Mat operator*(const Mat&b)const{
        Mat c;
        rep(k,0,3){rep(i,0,3){rep(j,0,3){
            c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j]%MOD)%MOD;
        }}}
        return c;
    }
    Mat operator=(Mat b){
        rep(i,0,3){rep(j,0,3){a[i][j]=b.a[i][j];}}
        return *this;
    }
}Pri,rs,bs;
bool FLB;
signed main(){
	cerr<<((&FLB)-(&FLA))/1024.0/1024<<endl;
	cin>>n>>m;
	Pri.a[1][2]=Pri.a[1][3]=Pri.a[3][2]=Pri.a[2][1]=1;
	int b=n;
	rs.one();bs=Pri;
	while(b){
        if(b&1){rs=rs*bs;}
        bs=bs*bs;b>>=1;
	}
	rep(i,0,3){Pre=(Pre+rs.a[i][i])%MOD;}
	cout<<get(m);
}
/*
*从小m考虑 m=1时01->10/11  11->10 10->01
m从低位开始，m奇->最低位不应有进位，最低一位独立，
m偶,如果偏要产生进位，应有三个1，但是可以由三个1扩展得知应当每一个都是1,实际上如果出现一个1，都应该全部为1
选择同奇,可以整体减一(化解进位问题)。

把所有的性质放在脑子里，随机应变

分段调试

不要执着于考虑环能带来什么性质，照常分析即可.(自信最重要)
*/