#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
const int INF = 1001001;
const int mod = 998244353;

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) 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;}
};

istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

int modPow(long long a, long long n, long long p) {
  if (n == 0) return 1; // 0乗にも対応する場合
  if (n == 1) return a % p;
  if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;
  long long t = modPow(a, n / 2, p);
  return (t * t) % p;
}

int main(void) {
  ll n, l;
  cin >> n >> l;
  n = (n+l-1) / l;

  mint ans = modPow(2, n, mod) - 1;

  cout<<ans<<endl;
  return 0;
}