You are given a square board, consisting of n rows and n columns. Each tile in it should be colored either white or black.

Let's call some coloring beautiful if each pair of adjacent rows are either the same or different in every position. The same condition should be held for the columns as well.

Let's call some coloring suitable if it is beautiful and there is no rectangle of the single color, consisting of at least k tiles.

Your task is to count the number of suitable colorings of the board of the given size.

Since the answer can be very large, print it modulo 998244353.

Input

A single line contains two integers n and k (1 ≤ n ≤ 500, 1 ≤ k ≤ n^2) — the number of rows and columns of the board and the maximum number of tiles inside the rectangle of the single color, respectively.

Output

Print a single integer — the number of suitable colorings of the board of the given size modulo 998244353.

Examples

Input

1 1


Output

0


Input

2 3


Output

6


Input

49 1808


Output

359087121

Note

Board of size 1 × 1 is either a single black tile or a single white tile. Both of them include a rectangle of a single color, consisting of 1 tile.

Here are the beautiful colorings of a board of size 2 × 2 that don't include rectangles of a single color, consisting of at least 3 tiles:

<image>

The rest of beautiful colorings of a board of size 2 × 2 are the following:

<image>
#include <bits/stdc++.h>
using namespace std;
double const EPS = 1.0E-9;
int const MOD = (int)1e9 + 7;
inline int read(char* s) { return scanf("%s", s); }
inline int read(int& x) { return scanf("%d", &(x)); }
inline int read(long long& x) { return scanf("%lld", &x); }
inline int read(double& x) { return scanf("%lf", &x); }
inline int write(int x) { return printf("%d", x); }
inline int write(long long x) { return printf("%lld", x); }
inline int write(double x) { return printf("%.12f", x); }
inline int write(char c) { putchar(c); }
inline int write() { return 0; }
inline int write(const char* s) { return printf("%s", s); }
inline int write(std::string const& s) { return write(s.c_str()); }
template <typename T>
int write(std::vector<T> const& v) {
  int z = v.size();
  for (int i = (0), vzq_b = (z - 1); i < vzq_b; ++i) {
    write(v[i]);
    write(' ');
  }
  if (z > 0) write(v[z - 1]);
  return z;
}
template <typename... T>
int read(T&... args) {
  using swallow = int[];
  swallow w = {0, (read(args))...};
  return sizeof(w) / sizeof(w[0]) - 1;
}
template <typename T>
int read(std::vector<T>& v) {
  for (T& t : v) read(t);
  return v.size();
}
template <typename... T>
int write(T const&... args) {
  using swallow = int[];
  swallow w = {0, (write(args), write(' '))...};
  return sizeof(w) / sizeof(w[0]) - 1;
}
template <typename... T>
int writeln(T const&... args) {
  int r = write(args...);
  printf("\n");
  return r + 1;
}
inline long long gcd(long long x, long long y) {
  return y == 0 ? x : gcd(y, x % y);
}
inline long long lcm(long long x, long long y) {
  return x == 0 && y == 0 ? 0 : x / gcd(x, y) * y;
}
inline long long powmod(long long x, long long n, long long m = MOD) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * x) % m;
    x = (x * x) % m;
    n /= 2;
  }
  return r;
}
static int leap(int y) { return y % 4 == 0 && y % 100 != 0 || y % 400 == 0; }
static int const month[2][12] = {
    {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
    {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
int solve();
int main(int argc, char* argv[]) {
  solve();
  return 0;
}
static const int maxn = 505;
static const int maxm = 1 << 22;
static const int P = 998244353;
static const long long LL_INF = 1234567890123456789;
static int add(int a, int b) {
  a += b;
  if (a >= P) a -= P;
  return a;
}
static int mul(int a, int b) {
  long long r = 1ll * a * b;
  if (r >= P) r %= P;
  return r;
}
static int query(int l, int r, int arr[]) {
  if (l == 0) return arr[r];
  return (arr[r] - arr[l - 1] + P) % P;
}
int dp[maxn], dp2[maxn], pre[maxn];
int solve() {
  int n, k;
  read(n, k);
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      dp[j] = add(query(max(0, j - i), j - 1, dp), dp[j - 1]);
    }
    dp2[i] = query(n, n, dp);
  }
  for (int i = n; i > 0; --i) {
    dp2[i] = add(dp2[i], P - dp2[i - 1]);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (i * j < k) {
        ans = add(ans, mul(dp2[i], dp2[j]));
      }
    }
  }
  ans = add(ans, ans);
  writeln(ans);
  return 0;
}
