#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; }