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<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_set>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
#include<tuple>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=998244353;

class comb{
private:
    ll mod;
    ll mx;
    vector<ll> F;
    vector<ll> FR;
    
public:
    comb(ll mod=1000000007,ll mx=100000):mod(mod),mx(mx),F(mx+1,1),FR(mx+1,1){
        mk_F();
    }
    
    ll mod_pow(ll a,ll x){
        a%=mod;
        ll ans=1;
        while(x>0){
            if(x&1){ans*=a; ans%=mod;}
            a*=a;
            a%=mod;
            x>>=1;
        }
        return ans;
    }
    
    pll Ex_gcd(ll a,ll b){
        if(b==0){return {1,0};}
        pll ret=Ex_gcd(b,a%b);
        ret.F-=a/b*ret.S;
        return {ret.S,ret.F};
    }
    
    ll prime_R(ll a){
        return mod_pow(a,mod-2);
    }
    
    ll R(ll a){
        ll ret=Ex_gcd(a,mod).F;
        ret%=mod;
        if(ret<0){ret+=mod;}
        return ret;
    }
    
    void mk_F(){
        for(ll i=1;i<=mx;i++){F[i]=F[i-1]*i%mod; FR[i]=R(F[i]);}
    }
    
    ll c(ll n,ll k){
        if(n<0 || k<0){return 0;}
        if(n<k){return 0;}
        if(n==k || k==0){return 1;}
        return F[n]*FR[n-k]%mod*FR[k]%mod;
    }
};

comb C(MOD,1000000);


int main(){
    ll n,x;
    cin>>n>>x;
    ll ans=0;
    for(ll i=0;i<=n;i++){
        ll cnt=0;
        for(ll t=i;t<x && t<=2*i;t++){
            cnt+=C.c(i,t-i);
        }
        cnt%=MOD;
        for(ll t=1;t*2<i && t*2<x-1;t++){
            cnt+=C.c(i-2*t,x-1-i);
            if(i-2*t==x-1-i){cnt--;}
        }
        if(2*i>=x && (x&1)){cnt++;}
        cnt%=MOD;
        ans+=cnt*C.c(n,i)%MOD;
    }
    ans%=MOD;
    ans+=MOD; ans%=MOD;
    cout<<ans<<endl;
    
    
    return 0;
}
