#include<bits/stdc++.h>
using namespace std;
#define IOS {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);}
#define rep(i,l,r) for(int i=(l); i<=(r); i++)
#define per(i,r,l) for(int i=(r); i>=(l); i--)
#define P pair<int,int>
#define ll long long
#define vi vector<int>
const int N =5e5+5, mod =998244353, G = 3;
ll qp(ll x, ll y) {
    ll res = 1;
    while (y) {
        if (y & 1)res = res * x % mod;
        x = x * x % mod; y >>= 1;
    }return res;
}
const int Gi = qp(G, mod - 2);
namespace Poly {
    typedef vi poly;
    int limit = 1, L = 0; int r[N * 4];
    void NTT(poly& A, int type) { //下标在[0,limit)范围内,数组开四倍即可
        A.resize(limit);
        for (int i = 0; i < limit; i++)
            if (i < r[i]) swap(A[i], A[r[i]]);
        for (int mid = 1; mid < limit; mid <<= 1) {
            ll Wn = qp(type == 1 ? G : Gi, (mod - 1) / (mid << 1)); //G是模数的原根,Gi是逆元
            for (int j = 0; j < limit; j += (mid << 1)) {
                ll w = 1; //ll不一定够
                for (int k = 0; k < mid; k++, w = (w * Wn) % mod) {
                    int x = A[j + k], y = w * A[j + k + mid] % mod; //int不一定够
                    A[j + k] = (x + y) % mod, A[j + k + mid] = (x - y + mod) % mod;
                }
            }
        }
    }
    poly operator + (poly a, poly b) {
        int n = max(a.size(), b.size());
        a.resize(n); b.resize(n);
        rep(i, 0, n - 1)a[i] = (a[i] + b[i]) % mod;
        return a;
    }
    poly operator - (poly a, poly b) {
        int n = max(a.size(), b.size());
        a.resize(n); b.resize(n);
        rep(i, 0, n - 1)a[i] = (a[i] - b[i] + mod) % mod;
        return a;
    }
    void poly_mul_init(poly& a, poly& b) {
        limit = 1; L = 0;
        int N = a.size() - 1, M = b.size() - 1;
        while (limit <= N + M) limit <<= 1, L++;
        rep(i, 0, limit - 1)r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
    }
    poly poly_mul(poly a, poly b) { //原先的a,b不需要维持原状的话,可以加&
        int n = a.size() + b.size() - 1;
        poly_mul_init(a, b);
        NTT(a, 1); NTT(b, 1);
        rep(i, 0, limit - 1)a[i] = 1ll * a[i] * b[i] % mod; //a[i]为ll不一定够
        NTT(a, -1);
        ll INV = qp(limit, mod - 2);
        rep(i, 0, limit - 1)a[i] = a[i] * INV % mod;
        a.resize(n); return a;
    }
}
signed main(){
    IOS
    const int M=2e6;
    vector<ll>fac(M+1),inv(M+1);
    fac[0]=1; rep(i,1,M)fac[i]=i*fac[i-1]%mod;
    inv[M]=qp(fac[M],mod-2); per(i,M,1)inv[i-1]=inv[i]*i%mod;
    int n,m; cin>>n>>m;
    vi fc(m+1),fd(m+1);
    auto p2=[&](ll x){
        return x*x%mod;
    };
    rep(c,1,m){
        if(n+1-2*c<0)break;
        fc[c]=inv[c]*inv[c-1]%mod*inv[n+1-2*c]%mod;
    }
    rep(d,0,m)fd[d]=fac[n+1+2*d]*inv[d]%mod*inv[d+1]%mod;
    auto res=Poly::poly_mul(fc,fd);
    ll ans=1;
    // auto C=[&](int i,int j){
    //     if(i<j)return 0ll;
    //     return fac[i]*inv[j]%mod*inv[i-j]%mod;
    // };
    rep(k,1,m){
        ans=(ans + fac[k-1]*fac[k]%mod*inv[2*k]%mod*res[k])%mod;
        // rep(c,1,k) {
        //     ans=(ans+ C(n+1-2*c+2*k,2*k)*p2(C(k-1,c-1))%mod*qp(1,mod-2) )%mod;
        //     // cout<<c<<' '<<k<<" "<<p2(C(k-1,c-1))%mod*qp(1,mod-2)%mod<<endl;
        // }
    }
    cout<<ans<<'\n';
}