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;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define Ans(f, y, n) if(f) cout << y << endl; else cout << n << endl;
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i, v.size()){ cout << v[i] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i, m.size()){ REP(j, m[i].size()){ cout << m[i][j] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

template<typename T> void readv(vector<T> &a){ REP(i, a.size()) cin >> a[i]; }
void readv_m1(vector<int> &a){ REP(i, a.size()){cin >> a[i]; a[i]--;} }

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};

//int mod = 1000000007;
int mod = 998244353;
#define Add(x, y) x = (x + (y)) % mod
#define Mult(x, y) x = (x * (y)) % mod

int modpow(int x, int n, int m){
    int a = 1;
    IREP(i, 64){
        a = (a * a) % m;
        if(((n >> i) & 1) == 1) a = (a * x) % m;
    }
    return a;
}

class Combination
{
public:

    vec fact, invfact;
    int MAX_N, mod;

    Combination(int MAX_N, int mod): MAX_N(MAX_N), mod(mod) {
        initialize();
    }

    void initialize(){
        fact = vec(MAX_N + 1);
        invfact = vec(MAX_N + 1);
        fact[0] = 1;
        FOR(i, 1, MAX_N + 1){
            fact[i] = (fact[i - 1] * i) % mod;
        }
        invfact[MAX_N] = modpow(fact[MAX_N], mod - 2, mod);
        IREP(i, MAX_N){
            invfact[i] = (invfact[i + 1] * (i + 1)) % mod;
        }
    }

    int nCr(int n, int r){
        if(r > n || r < 0 || n < 0) return 0;
        if(n > MAX_N){
            MAX_N = n;
            initialize();
        }
        int a = fact[n];
        a = (a * invfact[r]) % mod;
        a = (a * invfact[n - r]) % mod;
        return a;
    }

    int nPr(int n, int r){
        if(r > n || r < 0 || n < 0) return 0;
        if(n > MAX_N){
            MAX_N = n;
            initialize();
        }
        int a = fact[n];
        a = (a * invfact[n - r]) % mod;
        return a;
    }
};

signed main(){

    int N, X; cin >> N >> X;
    
    mat dp(N + 1, vec(2 * N + 1, 0));
    dp[0][0] = 1;
    REP(i, N){
        REP(j, 2 * i + 1){
            Add(dp[i + 1][j + 1], dp[i][j]);
            Add(dp[i + 1][j + 2], dp[i][j]);
        }
    }

    Combination C(N, mod);

    vec a(N + 1, 0);
    REP(i, N + 1){
        REP(j, X) a[i] += dp[i][j];
        a[i] %= mod;
    }
    
    //sum=X-1+2k
    for(int k = 1; X - 1 + 2 * k <= 2 * N; k++){
        REP(i, N + 1){
            int i0 = i - 2 * k;
            int X0 = X - 1 - 2 * k;
            if(i0 >= 0 && X0 >= 0) Add(a[i], dp[i0][X0]);
            else if(X - 1 + 2 * k == 2 * i) Add(a[i], 1);
        }
    }


    int ans = 0;
    REP(i, N + 1) Add(ans, a[i] * C.nCr(N, i));
    cout << ans << endl;

    return 0;
}
