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>
#define mems(a,x) memset(a,x,sizeof(a))
#define first fi
#define second se
using namespace std;
typedef long long ll;
const int mod=998244353,MS=105,N=3005;
ll inv(ll x){return x==1?x:(mod-mod/x)*inv(mod%x)%mod;}
ll inv(ll x,ll mod){return x==1?x:(mod-mod/x)*inv(mod%x,mod)%mod;}
ll qpow(ll a,ll n){ll ans=1;while(n){if(n&1) ans=ans*a%mod;a=a*a%mod;n>>=1;}return ans;}
ll mul(ll a,ll b){ll ans=0;while(b){if(b&1) ans=(ans+a)%mod;a=(a+a)%mod;b>>=1;};return ans;}
ll qpow(ll a,ll n,ll mod){ll ans=1;while(n){if(n&1) ans=ans*a%mod;a=a*a%mod;n>>=1;}return ans;}
ll mul(ll a,ll b,ll mod){ll ans=0;while(b){if(b&1) ans=(ans+a)%mod;a=(a+a)%mod;b>>=1;};return ans;}
struct vec
{
    double x,y;
    vec(double x=0,double y=0):x(x),y(y){}
    vec operator+(const vec&o)const{ return vec(x+o.x,y+o.y);}
    vec operator-(const vec&o)const{ return vec(x-o.x,y-o.y);}
    double operator*(const vec&o)const{ return x*o.x+y*o.y;}
    double operator^(const vec&o)const{ return x*o.y-y*o.x;}
    vec operator/(const double&o)const{ return vec(x/o,y/o);}
    vec operator*(const double&o)const{ return vec(x*o,y*o);}
    void sc(){scanf("%lf%lf",&x,&y);}
    double len(){return sqrt(x*x+y*y);}
};
struct Mat
{
    ll a[MS][MS];
    int n,m;
    Mat(int n=0,int m=0):n(n),m(m) { mems(a,0);}
    Mat operator*(const Mat&B)const
    {
        Mat C(n,B.m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=B.m;j++)
                for(int k=1;k<=m;k++)
                C.a[i][j]=(C.a[i][j]+a[i][k]*B.a[k][j])%mod;
        return C;
    }
};
int n,x,c[N][N];
ll dp[N][N<<1];
ll C(ll n,ll m)
{
    if(m==0||n==m) return 1;
    if(m==1) return n;
    if(c[n][m]) return c[n][m];
    return c[n][m]=(C(n-1,m-1)+C(n-1,m))%mod;
}
ll solve(int m)
{
    ll ans=0;
    for(int i=0;i<x;i++) (ans+=dp[m][i])%=mod;
    if((x&1)&&2*m>x) ans++;
    for(int i=x+1;i<2*m;i+=2)
    {
        int t=m-(i-(x-1)),s=i-(i-(x-1))*2;
        if(t>=0&&s>=0)
            (ans+=dp[t][s])%=mod;
    }
    return ans;
}
int main()
{
    scanf("%d%d",&n,&x);
    dp[0][0]=1;
    for(int i=0;i<n;i++)
        for(int j=i;j<=i*2;j++)
    {
        (dp[i+1][j+1]+=dp[i][j])%=mod;
        (dp[i+1][j+2]+=dp[i][j])%=mod;
    }
    ll ans=0;
    for(int i=0;i<=n;i++)
        ans=(ans+C(n,i)*solve(i))%mod;
    printf("%lld\n",ans);
}
