#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <cmath>
#include <numeric>
#include <functional>
#include <cassert>
#include <atcoder/modint>

#define debug_value(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << #x << "=" << x << endl;
#define debug(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << x << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using namespace std;
typedef long long ll;

template<typename T>
vector<vector<T>> vec2d(int n, int m, T v){
    return vector<vector<T>>(n, vector<T>(m, v));
}

template<typename T>
vector<vector<vector<T>>> vec3d(int n, int m, int k, T v){
    return vector<vector<vector<T>>>(n, vector<vector<T>>(m, vector<T>(k, v)));
}

template<typename T>
void print_vector(vector<T> v, char delimiter=' '){
    if(v.empty()) {
        cout << endl;
        return;
    }
    for(int i = 0; i+1 < v.size(); i++) cout << v[i] << delimiter;
    cout << v.back() << endl;
}

using mint = atcoder::modint998244353;

ostream& operator<<(ostream& os, const mint& m){
    os << m.val();
    return os;
}

#define N_MAX 1000002
mint inv[N_MAX],fac[N_MAX],finv[N_MAX];

void init(){
    const ll MOD = mint::mod();
    fac[0]=1;fac[1]=1;
    finv[0]=1;finv[1]=1;
    inv[1]=1;
    for(int i=2;i<N_MAX;i++){
        inv[i]=mint(MOD)-inv[MOD%i]*(MOD/i);
        fac[i]=fac[i-1]*(ll) i;
        finv[i]=finv[i-1]*inv[i];
    }
}

mint comb(ll n, ll r){
    if(n < r){
        return mint(0);
    }else{
        return fac[n]*finv[r]*finv[n-r];
    }
}


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    init();
    int n, k; cin >> n >> k;
    if(n == 1){
        if(k == 6) cout << 1 << endl;
        if(k == 1 || k == 5) cout << 1 << endl;
        if(k == 2 || k == 4) cout << 3 << endl;
        if(k == 3) cout << 3 << endl;
        return 0;
    }
    mint ans = comb(2*n+4, k);
    if(k%2 == 0){
        ans += comb(n+2, k/2)*2;
    }else{

    }
    if(n%2 == 0){
        if(k%2 == 0){
            ans += comb(n+2, k/2);
        }else{

        }
    }else{
        for(int x = 0; x <= 2; x++){
            if(x%2 == k%2 && k >= x){
                ans += comb(2, x)*comb(n+1, (k-x)/2);
            }
        }
    }
    ans /= 4;
    cout << ans << endl;
}