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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=998244353 ;

vector<ll> inv,fact,invfact;
void mod_build(int n=101010){
    fact.resize(n+1);
    inv.resize(n+1);
    invfact.resize(n+1);
    fact[0]=inv[0]=invfact[0]=1;
    inv[1]=1;
    rep(i,n){
        fact[i+1]=fact[i]*(i+1)%mod;
        if(i>0)inv[i+1]=mod-inv[mod%(i+1)]*(mod/(i+1))%mod;
        invfact[i+1]=invfact[i]*inv[i+1]%mod;
    }
}
ll perm(int n,int k){
    if(n<0||k<0||k>n)return 0;
    return fact[n]*invfact[n-k]%mod;
}
ll comb(int n,int k){
    if(n<0||k<0||k>n)return 0;
    return (fact[n]*invfact[n-k]%mod)*invfact[k]%mod;
}
ll powmod(ll n,ll k){
    k%=mod-1;
    if(k<0)k+=mod-1;
    ll ret=1;
    while(k){
        if(k&1)ret=ret*n%mod;
        n=n*n%mod;
        k>>=1;
    }
    return ret;
}


ll dp[3030][6060],sum[3030][6060];
int main(){
    mod_build();
    int n,x;
    cin>>n>>x;
    ll ans=0;
    rep(i,n+1)rep(j,2*n+1){
        dp[i][j]=comb(i,j-i);
        sum[i][j]=comb(i,j-i);
    }
    rep(i,n)rep(j,2*n+1){
        (dp[i+1][j]+=dp[i][j])%=mod;
    }
    rep(i,n+1)rep(j,2*n){
        (sum[i][j+1]+=sum[i][j])%=mod;
    }
    rep(k,n+1){
        ll res=0;
        rep(i,k){
            if(x<2*i+2)break;
            int s=x-2*i-2;
            res+=sum[k-i-1][x-2*i-3];
            res+=dp[k-i-1][x-2*i-2]-(k-2*i-2>=0?dp[k-2*i-2][x-2*i-2] : 0);
            res=(res+mod)%mod;
        }
        if(x%2==1||2*k<x)++res;
        ans+=res*comb(n,k)%mod;
    }
    cout<<ans%mod<<endl;
    return 0;
}