#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc=998244353, MAX=1000000;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(modc-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{
        return x == a.x;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint &m) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){
        return x;
    }
};

vector<mint> f, finv;

mint inv(mint x){
    mint ans = 1;
    ll e = modc-2;

    while (e > 0){
        if ((e & 1LL)) ans *= x;
        e = e >> 1LL;
        x *= x;
    }

    return ans;
}

void init(){
    f.resize(MAX+1); finv.resize(MAX+1);
    f[0] = 1;
    for (int i=1; i<=MAX; i++) f[i] = f[i-1]*i;
    finv[MAX] = inv(f[MAX]);
    for (int i=MAX-1; i>=0; i--) finv[i] = finv[i+1] * (i+1);
}

mint C(ll n, ll k){ 
    if (n < k || k < 0) return 0;

    return f[n] * finv[k] * finv[n-k] ;
}

mint P(ll n, ll k){
    if (n < k || k < 0) return 0;

    return f[n] * finv[n-k];
}

mint H(ll n, ll k){
    if (n == 0 && k == 0) return 1;
    return C(n+k-1, k);
}

int main(){

    init();
    int N, K;
    mint ans;
    cin >> N >> K;
    if (N == 1){
        if (K == 1 || K == 5 || K == 6) cout << 1 << endl;
        else cout << 3 << endl; 
        return 0;
    }

    if (N % 2 == 0){
        ans = C(2*N+4, K) + (K % 2 == 0 ? C(N+2, K/2) * 3 : 0);
        ans /= 4;
        cout << ans << endl;
    }
    else{
        ans = C(2*N+4, K) + (K % 2 == 0 ? C(N+2, K/2) * 2 : 0);
        ans += (K % 2 == 0 ? C(N+1, K/2) + C(N+1, K/2-1) : C(N+1, (K-1)/2) * 2);
        ans /= 4;
        cout << ans << endl;
    }

    return 0;
}