#include <bits/stdc++.h>
using namespace std;
template <typename t>
t in(t q) {
  cin >> q;
  return q;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < ((int)(v).size()); ++i) {
    os << v[i];
    if (i != ((int)(v).size()) - 1) os << ",";
  }
  os << "]";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  for (auto it : v) os << "(" << it.first << ":" << it.second << ")";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(" << v.first << "," << v.second << ")";
  return os;
}
const long double PI = acosl(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
inline int rand(int l, int r) {
  return uniform_int_distribution<int>(l, r)(rng);
}
inline long long rand(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
int ar[(505)][(505)][2];
int ar2[(505)][(505)];
int dp1(int left, int high, bool done) {
  if (left < 0) return 0;
  if (left == 0) return done;
  if (ar[left][high][done] != -1) {
    return ar[left][high][done];
  }
  int ans = 0;
  for (auto i = 0; i < high - 1; i++) {
    ans += dp1(left - i - 1, high, done);
    ans %= (998244353);
  }
  ans += dp1(left - high, high, 1);
  ar[left][high][done] = ans % (998244353);
  return ar[left][high][done];
}
int dp2(int left, int high) {
  if (left < 0) return 0;
  if (left == 0) return 1;
  if (ar2[left][high] != -1) {
    return ar2[left][high];
  }
  int ans = 0;
  for (auto i = 0; i < high; i++) {
    ans += dp2(left - i - 1, high);
    ans %= (998244353);
  }
  ar2[left][high] = ans;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(10);
  memset(ar, -1, sizeof ar);
  memset(ar2, -1, sizeof ar2);
  int n, k;
  cin >> n >> k;
  int ans = 0;
  for (auto i = 1; i < min(k, n) + 1; i++) {
    int sec = (k + i - 1) / i;
    ans += 2ll * dp1(n, i, 0) % (998244353) * dp2(n, min(n, sec - 1)) %
           (998244353);
    ans %= (998244353);
  }
  cout << ans;
  return 0;
}
