Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.

Constraints

* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.

Input

Input is given from Standard Input in the following format:


N X


Output

Print the number, modulo 998244353, of sequences that satisfy the condition.

Examples

Input

3 3


Output

14


Input

8 6


Output

1179


Input

10 1


Output

1024


Input

9 13


Output

18402


Input

314 159


Output

459765451
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 3333
	#define mod 998244353ll
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char __sr[1<<21],__z[20];int __C=-1,__zz=0;
	inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
	inline void print(register int x)
	{
		if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
		while(__z[++__zz]=x%10+48,x/=10);
		while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
	}
	void file()
	{
		#ifdef NTFOrz
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifdef NTFOrz
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int n,X;
ll C[sz][sz];
ll f[sz][sz<<1];

int main()
{
	file();
	read(n,X);
	f[0][0]=1;
	rep(i,0,sz-5) rep(j,0,sz*2-5) (f[i+1][j]+=f[i][j])%=mod,(f[i+1][j+1]+=f[i][j])%=mod,(f[i+1][j+2]+=f[i][j])%=mod;
	rep(i,0,sz-1) C[i][0]=1;
	rep(i,1,sz-1) rep(j,1,i) (C[i][j]=C[i-1][j]+C[i-1][j-1])%=mod;
	ll ans=0;
	rep(i,0,X-1) (ans+=f[n][i])%=mod;
	rep(j,0,n-1) for (int k=1;X-2*k-1>=0&&k*2<=n-j;k++) (ans+=C[n-j-1][k*2-1]*f[j][X-2*k-1]%mod)%=mod;
	if (X&1) rep(i,X/2+X/2+1,n) (ans+=C[n][i])%=mod;
	cout<<ans;
	return 0;
}