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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
using namespace std;
const long long mod=998244353;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const long double EPS=1e-9;
const long double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
long double ABS(long double a){return max(a,-a);}
long long fact[110000];
long long finv[110000];
long long inv[110000];
long long dp[3100][6100];
long long C(int a,int b){
	if(b<0||a<b)return 0;
	return fact[a]*finv[b]%mod*finv[a-b]%mod;
}
int main(){
	fact[0]=finv[0]=inv[1]=1;
	for(int i=2;i<110000;i++){
		inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
	}
	for(int i=1;i<110000;i++){
		fact[i]=fact[i-1]*i%mod;
		finv[i]=finv[i-1]*inv[i]%mod;
	}
	int a,b;scanf("%d%d",&a,&b);
	dp[0][0]=1;
	for(int i=0;i<a;i++){
		for(int j=0;j<b;j++){
			if(!dp[i][j])continue;
			dp[i+1][j]=(dp[i+1][j]+dp[i][j])%mod;
			dp[i+1][j+1]=(dp[i+1][j+1]+dp[i][j])%mod;
			dp[i+1][j+2]=(dp[i+1][j+2]+dp[i][j])%mod;
		}
	}
	long long ret=0;
	for(int i=0;i<b;i++)ret=(ret+dp[a][i])%mod;
	for(int i=1;i<=a;i++){
		long long ks=C(a,i);
		for(int j=1;j*2<=a;j++){
			if(i+1>b||i*2-j*2+1<=b)continue;
		//	printf("%d %d: %lld\n",i,j,C(i-j*2,b-i-1));
			ret=(ret+C(i-j*2,b-i-1)*ks)%mod;
		}
		if(i*2>b&&b%2==1){
			ret=(ret+ks)%mod;
		}
	}
	printf("%lld\n",ret);
}
