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>
using namespace std;
 
#define MOD 998244353
template<typename ty1,typename ty2>
inline int add(ty1 x, ty2 y) {
	if(y>=MOD)y%=MOD;
	if(x>=MOD)x%=MOD;
    x += y; return x < MOD ? x : x - MOD;
}
template<typename ty1,typename ty2>
inline void addto(ty1 &x, ty2 y) {
		  if(y>=MOD)y%=MOD;
		  if(x>=MOD)x%=MOD;
	      x += y; if (x >= MOD) x -= MOD;
}
template<typename ty1,typename ty2>
inline int sub(ty1 x, ty2 y) {
		if(y>=MOD)y%=MOD;
		 if(x>=MOD)x%=MOD;
        x -= y; return x < 0 ? x + MOD : x;
}
template<typename ty1,typename ty2>
inline void subto(ty1 &x, ty2 y) {
	if(y>=MOD)y%=MOD;
	if(x>=MOD)x%=MOD;
    x -= y; if (x < 0) x += MOD;
}
template<typename ty1,typename ty2>
inline int mul(ty1 x, ty2 y) {
        if(y>=MOD)y%=MOD;
		if(x>=MOD)x%=MOD;
        return 1ll * x * y % MOD;
}
template<typename ty1,typename ty2>
void multo(ty1 &x, ty2 y) {
        if(y>=MOD)y%=MOD;
		if(x>=MOD)x%=MOD;
        x=1ll * x * y % MOD;
}
 
 
long long int ppow(long long int i, long long int j){
		long long int res = 1LL;
		while (j){
			if ((j & 1LL)){
				res *= i;
				if (res >= MOD){
					res %= MOD;
				}
			}
			j >>= 1;
			i *= i;
			if (i >= MOD){
				i %= MOD;
			}
		}
		return res;
	}
class Combination{
	
public:
	vector<long long int> k;
	vector<long long int> r;
	void resize(int N){
		k.resize(N + 2);
		r.resize(N + 2);
		k[0] = 1;
		for (int i = 1; i < N+2; i++){
			k[i] = k[i - 1];
			k[i] *= i;
			if (k[i] >= MOD)k[i] %= MOD;
		}
		long long int al = k[k.size() - 1];
		long long int iv = ppow(k[k.size() - 1],MOD-2);
		r[k.size() - 1] = iv;
		for (int i = (int)(r.size()) - 2; i >= 0; i--){
			r[i] = r[i + 1] * (i + 1);
			if (r[i] >= MOD){
				r[i] %= MOD;
			}
		}
	}
	long long int C(int a, int b){
		if (a < b)return 0;
		long long int up = k[a];
		long long int dw = r[b] * r[a - b];
		dw %= MOD;
		up *= dw;
		up %= MOD;
		return up;
	}
	long long int H(int a, int b){
		return C(a + b - 1, b);
	}
	long long int catalan_number(int n){
		return (C(2 * n, n) + MOD - C(2 * n, n - 1)) % MOD;
	}
};
Combination C;

#define MAX 3002
#define MAX_X 6002

int n;
int x;

int dp[MAX_X][MAX_X];
int dp2[MAX_X][MAX_X];

int main(){
	cin>>n>>x;
	if(x==1){
		printf("%d\n",(int)ppow(2,n));
		return 0;
	}
	C.resize(100000);
	//x+1を超える 必ず1が含まれる場合
	long long int ans=0;
	for(int i=1;i<=n;i++){  //連続する2の個数
		int sum=2*i;
		int ret=x+1-sum-2;
		if(ret<0)break;
		for(int j=0;j<=ret;j++){
			long long int tmp=ans;
			if(ret==0&&j)continue;
			if(ret&&j==0)continue;
			if(ret==0&&false){
				int gap=i*2+1;
				int want=n-i*2;
				if(want<0)continue;
				addto(ans,C.C(gap+want-1,want));
				long long int tt=ans;
				//cerr<<i<<" "<<j<<" "<<tt-tmp<<endl;
				continue;
			}
			int w=ret-j;
			if(w>j)continue;
			int len=j+i*2;
			int gap=len+1;
			len=n-len;
			if(len<0)continue;
			if(j==w)continue;
			addto(ans,mul(C.C(j,w),C.C(len+gap-1,len)));
			long long int tt=ans;
			//cerr<<i<<" "<<j<<" "<<tt-tmp<<endl;
		}
	}
	if(x%2){  //0と2で構成される
		dp2[0][0]=1;
		for(int i=0;i<n;i++){
			for(int j=0;j<MAX_X;j++){
				if(dp2[i][j]==0)continue;
				addto(dp2[i+1][j+2],dp2[i][j]);
				addto(dp2[i+1][j],dp2[i][j]);
			}
		}
		for(int i=MAX_X-1;i>=x;i--){
			addto(ans,dp2[n][i]);
		}
	}
	dp[0][0]=1;
	for(int i=0;i<n;i++){
		for(int j=0;j<MAX_X;j++){
			if(dp[i][j]==0)continue;
			addto(dp[i+1][j+1],dp[i][j]);
			addto(dp[i+1][j+2],dp[i][j]);
			addto(dp[i+1][j],dp[i][j]);
		}
	}
	for(int i=x-1;i>=0;i--){
		addto(ans,dp[n][i]);
	}
	printf("%lld\n",ans);
	return 0;
}
