#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
#define rep(i, n) for(int i=0;i<n;i++)
#define all(i, v) for(auto& i : v)
typedef long long ll;
using namespace std;
const ll mod=998244353, INF=(1LL<<60);
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;



///////modint
struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%mod+mod)%mod){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(ll t) const {
        if(t < 0) return mint(1) / pow(-t);
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false); 
    ll n, m;
    cin >> n >> m;
    ll d = 0, mask = 1;
    while((mask * 2) <= n) {
        mask *= 2;
        d++;
    }
    vector<mint> p(1000001, mint(1));
    rep(i, 1000000) p[i + 1] = p[i] * mint(2);
    mint ans = mint(0);
    if(m <= 10000) {
        vector<ll> s(1000000, 0);
        for(int i = 0; i <= d; i ++) {
            if(n >> i & 1) {
                for(int j = 0; j < m; j++) {
                    s[i+j] ^= 1;
                }
            }
        }
        rep(i, 1000000) if(s[i]) ans += p[i];
        cout << ans.x << endl;
        return 0;
    }
    vector<ll> s(100, 0), ss(100, 0);
    for(int i = 0; i < d; i++) if(n >> i & 1) {
        for(int j = 0; j + i <= d - 1; j ++) {
            s[i+j] ^= 1;
        }
    }
    for(int i = 0; i <= d; i++) if(n >> i & 1) {
        for(int j = m-d; j < m; j ++) {
            if(i + j < m) continue;
            ss[i + j - m] ^= 1;
        }
    }
    rep(i, 100) {
        if(s[i]) ans += p[i];
        if(ss[i]) ans += p[m] * p[i];
    }
    if(__builtin_popcountll(n) & 1) {
        ans += p[d] * (p[m - d] - mint(1));
    }
    cout << ans.x << endl;


    return 0;
}