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;
void Print(int _x) { printf("%d\n", _x); }
void Print(long long _x) { printf("%lld\n", _x); }
template <typename T>
void Print(T _a[], int _s, int _t) {
  for (int i = _s; i <= _t; i++) cout << _a[i] << " ";
  cout << endl;
}
struct InputReader {
  static const int bs = 100000;
  char buf[bs];
  int p;
  inline InputReader() : p(bs) {}
  inline void Flush() {
    p = 0;
    fread(buf, 1, bs, stdin);
  }
  inline char C() {
    if (p >= bs) Flush();
    return buf[p++];
  }
  int Getnum() {
    char ch = C();
    while (ch < '0' || ch > '9') ch = C();
    return (int)(ch - '0');
  }
  inline int operator()() {
    int ans = 0;
    char ch = C();
    int fu = 1;
    while (ch < '0' || ch > '9') {
      if (ch == '-') fu = -1;
      ch = C();
    }
    while (ch >= '0' && ch <= '9') {
      ans *= 10;
      ans += ch - '0';
      ch = C();
    }
    return ans * fu;
  }
  long long Readll() {
    long long ans = 0LL;
    long long fu = 1;
    char ch = C();
    while (ch < '0' || ch > '9') {
      if (ch == '-') fu = -1;
      ch = C();
    }
    while (ch >= '0' && ch <= '9') {
      ans *= 10LL;
      ans += ch - '0';
      ch = C();
    }
    return ans * fu;
  }
  inline void Readstring(string &x) {
    x.clear();
    char ch = C();
    while (!isdigit(ch) && !isalpha(ch)) ch = C();
    while (isdigit(ch) || isalpha(ch)) {
      x += ch;
      ch = C();
    }
  }
  inline void Readalpha(char &c) {
    c = C();
    while (!isalpha(c)) c = C();
  }
  inline void Specialread(char &c) {
    c = C();
    while (c != 'o' && c != 'x' && c != '?') c = C();
  }
} In;
inline void Read(int &x) { x = In(); }
inline void Read(int &x, int &y) {
  x = In();
  y = In();
}
inline void Read(int &x1, int &x2, int &x3) {
  x1 = In();
  x2 = In();
  x3 = In();
}
inline void Read(int &x1, int &x2, int &x3, int &x4) {
  x1 = In();
  x2 = In();
  x3 = In();
  x4 = In();
}
inline void Read(long long &x) { x = In.Readll(); }
inline void Read(long long &x, long long &y) {
  x = In.Readll();
  y = In.Readll();
}
inline void Read(long long &x1, long long &x2, long long &x3) {
  x1 = In.Readll();
  x2 = In.Readll();
  x3 = In.Readll();
}
inline void Read(long long &x1, long long &x2, long long &x3, long long &x4) {
  x1 = In.Readll();
  x2 = In.Readll();
  x3 = In.Readll();
  x4 = In.Readll();
}
clock_t _starttime, _endtime;
inline void FILEIO() {}
inline void END() { exit(0); }
int dx[] = {0, -1, 0, 1};
int dy[] = {1, 0, -1, 0};
int n, k;
long long dp[510][510][2];
long long Qpow(long long a, int b) {
  long long ret = 1;
  while (b) {
    if (b & 1) (ret *= a) %= 998244353LL;
    (a *= a) %= 998244353LL;
    b >>= 1;
  }
  return ret;
}
void Init() {
  memset(dp, 0, sizeof(dp));
  for (register int(l) = 1; (l) <= (n); (l)++) {
    for (register int(i) = 1; (i) <= (l); (i)++) {
      dp[l][i][0] = dp[l][i][1] = Qpow(2, i - 1);
    }
    for (register int(i) = (l + 1); (i) <= (n); (i)++) {
      for (register int(j) = (i - l); (j) <= (i - 1); (j)++) {
        (dp[l][i][0] += dp[l][j][1]) %= 998244353LL;
        (dp[l][i][1] += dp[l][j][0]) %= 998244353LL;
      }
    }
  }
}
long long ans = 0;
void Solve() {
  for (register int(i) = 1; (i) <= (n); (i)++)
    for (register int(j) = 1; (j) <= (n); (j)++) {
      if (i * j >= k) break;
      long long tans =
          (dp[i][n][0] - dp[i - 1][n][0]) * (dp[j][n][0] - dp[j - 1][n][0]);
      tans += (dp[i][n][1] - dp[i - 1][n][1]) * (dp[j][n][1] - dp[j - 1][n][1]);
      tans = (tans % 998244353LL + 998244353LL + 998244353LL) % 998244353LL;
      ans += tans;
      if (ans >= 998244353LL) ans -= 998244353LL;
      ;
    }
  cout << ans << endl;
}
int main() {
  FILEIO();
  Read(n, k);
  Init();
  Solve();
  END();
}
